Commutative algebra studies the structure of commutative rings and their modules, laying the algebraic foundation for algebraic geometry and number theory. This course builds systematically from basic definitions through to dimension theory, equipping you with both foundational concepts and sophisticated tools for understanding polynomial rings, integral extensions, and the geometry hidden within algebraic structures. The chapters progress from concrete ring-theoretic properties toward increasingly abstract structural results, with a clear arc from local phenomena (localization, primary decomposition) to global invariants (dimension, integral closure). The course begins with essentials: rings, ideals, and modules as the central objects of study, then introduces finiteness conditions through chain conditions that force rings and modules to behave well. Tensor products appear early as a fundamental operation, while localization emerges as a technique to isolate local behavior — a theme that recurs throughout. Nakayama's lemma and integral extensions anchor the middle chapters, providing tools to relate rings to their integral closures and to understand when a ring extension is finite. The Cohen-Seidenberg theorems capture how dimension and prime ideals behave under integral extensions, while primary decompositions generalize unique factorization beyond factorial rings. Later chapters shift focus to homological and invariant structures: direct and inverse limits formalize taking colimits and their duals, completions produce new rings from old ones, and filtrations and gradings organize rings by degree or other decorations. Dimension theory synthesizes much of what comes before, quantifying the "size" of a ring through the lengths of chains of prime ideals. # 1. Rings and Modules This chapter sets up the two central objects of commutative algebra: modules over a commutative ring, and the finiteness conditions — the noetherian and artinian properties — that make them tractable. Modules generalise both vector spaces and abelian groups within a single framework, and the noetherian condition is the precise algebraic substitute for compactness. The chapter concludes with Hilbert's basis theorem, which is the engine behind the finiteness properties of polynomial rings and algebraic varieties. Throughout, a **ring** means a commutative ring $R$ with $1 \in R$, unless stated otherwise. The one non-commutative exception that appears is $\operatorname{End}(M)$: for an abelian group $(M, +)$, the set $\operatorname{End}(M) = \{f: M \to M \mid f \text{ is a group homomorphism}\}$ forms a ring under composition as multiplication. ## Modules and the Structure Homomorphism The idea behind an $R$-module is to have a set on which $R$ acts by scalar multiplication, generalising the notion of a vector space over a field. The key insight is to package this action as a single ring homomorphism. [definition: R-Module] Let $R$ be a commutative ring with $1$. An **$R$-module** is an abelian group $M$ together with a fixed ring homomorphism $\rho: R \to \operatorname{End}(M)$, called the **structure homomorphism**. The scalar multiplication is defined by \begin{align*} r \cdot m := \rho(r)(m) \end{align*} for $r \in R$ and $m \in M$. [/definition] The structure homomorphism viewpoint is conceptually clean: it packages all of the scalar multiplication axioms into the single requirement that $\rho$ be a ring homomorphism. Unpacking this, one obtains the familiar identities. Since $\rho(r) \in \operatorname{End}(M)$ is a group homomorphism, we have $r(m + m') = rm + rm'$. Since $\rho$ is a ring homomorphism, $\rho(r_1 + r_2) = \rho(r_1) + \rho(r_2)$ in $\operatorname{End}(M)$, giving $(r_1 + r_2)m = r_1 m + r_2 m$. Multiplicativity of $\rho$ gives $(r_1 r_2)m = r_1(r_2 m)$, and $\rho(1_R) = \operatorname{id}_M$ gives $1 \cdot m = m$. [example: Standard Examples of Modules] The following illustrate the range of the module concept. **Vector spaces.** A $k$-module is exactly a $k$-vector space, for $k$ a field. The structure homomorphism is the natural scalar multiplication. **Abelian groups as $\mathbb{Z}$-modules.** Every abelian group $M$ is a $\mathbb{Z}$-module in a unique way. A ring homomorphism $\rho: \mathbb{Z} \to \operatorname{End}(M)$ is completely determined by $\rho(1_{\mathbb{Z}}) = \operatorname{id}_M$, so there is no choice. Conversely, any $\mathbb{Z}$-module is an abelian group. Thus: abelian group $\iff$ $\mathbb{Z}$-module. **The ring itself.** Every ring $R$ is an $R$-module via the structure homomorphism \begin{align*} R &\to \operatorname{End}(R) \\ r_0 &\mapsto (r \mapsto r_0 r). \end{align*} **Direct sums and products.** The direct sum $R^{\oplus \mathbb{N}}$ (elements with only finitely many non-zero coordinates) and the direct product $R^{\mathbb{N}}$ (all $\mathbb{N}$-indexed tuples) are both $R$-modules, with componentwise scalar multiplication. **Ideals.** An ideal of the ring $R$ is exactly the same thing as an $R$-submodule of $R$ (as an $R$-module). This unifies the theory of ideals with the theory of submodules. [/example] [definition: Submodule] An **$R$-submodule** of an $R$-module $M$ is an additive subgroup $N \subseteq M$ such that $rN \subseteq N$ for all $r \in R$. [/definition] For a subset $S$ of an $R$-module $M$, the $R$-submodule **generated by $S$** is the intersection of all $R$-submodules of $M$ containing $S$. Equivalently, it is the set of all $R$-linear combinations \begin{align*} \left\{ \sum_{i=1}^{n} r_i x_i : n \geq 0,\, r_i \in R,\, x_i \in S \right\}. \end{align*} Note that each sum is finite. A module is **finitely generated** if it is generated by a finite set $S = \{x_1, \dots, x_k\}$. ## Quotient Modules and Ideals When can we collapse part of a module to zero without losing the $R$-module structure? The obstacle is that a bare subgroup quotient $M/N$ need not respect scalar multiplication: if $N$ is not closed under the $R$-action, the coset $r \cdot (m + N)$ is not well-defined. Requiring $rN \subseteq N$ for all $r \in R$ is precisely what fixes this — it is the definition of a submodule. For the special case where $M = R$ and $N = I$ is an ideal, we recover the familiar ring-quotient construction. Given an ideal $I \trianglelefteq R$, the quotient $R/I$ inherits a ring structure. As an abelian group, $R/I$ is the usual group quotient. The multiplication is $(r_1 + I)(r_2 + I) = r_1 r_2 + I$. One should check this is well-defined: the product set $\{(r_1 + x_1)(r_2 + x_2) : x_1, x_2 \in I\}$ is contained in $r_1 r_2 + I$ (since $r_1 x_2 + r_2 x_1 + x_1 x_2 \in I$), so the product of two cosets lands in a single coset. Note that this containment may be proper, unlike in the group case where $g_1 N \cdot g_2 N = g_1 g_2 N$ exactly. [remark: Characterisation of Ideals via Kernels] A subset $I \subseteq R$ is an ideal if and only if there exists a ring $S$ and a ring homomorphism $\varphi: R \to S$ whose kernel is $I$. The forward direction is the first isomorphism theorem: $\ker \varphi \trianglelefteq R$. The backward direction: take $S = R/I$ and $\varphi$ the quotient map. [/remark] ## Free Modules and Homomorphisms from $R^{\oplus n}$ How do we produce $R$-module homomorphisms in practice? For vector spaces over a field, a linear map is determined entirely by where it sends a basis. Is there an analogue for modules? The answer leads to the notion of a free module: those for which every element decomposes uniquely as an $R$-linear combination of basis vectors. Unlike vector spaces, not every module is free — the $\mathbb{Z}$-module $\mathbb{Z}/2\mathbb{Z}$ has no basis, since $1 \cdot \bar{1} = \bar{1}$ but also $2 \cdot \bar{1} = \bar{0}$, so no generating set can be linearly independent. Free modules are, however, the workhorses for building arbitrary finitely generated modules: every such module is a quotient of a free module, a fact that turns finiteness questions into linear algebra. For an $R$-module homomorphism $\varphi: R \to M$, one has $\varphi(r) = \varphi(r \cdot 1) = r\varphi(1)$, so $\varphi$ is completely determined by the image of $1$. Conversely, for any $m \in M$, the map $r \mapsto rm$ is an $R$-module homomorphism $R \to M$. Thus $\operatorname{Hom}_R(R, M) \cong M$ as sets (in fact as $R$-modules). The **universal property of the direct sum** extends this: for $R$-modules $(M_t)_{t \in T}$, the direct sum $\bigoplus_{t \in T} M_t$ with natural embeddings $\rho_t: M_t \to \bigoplus_{t \in T} M_t$ satisfies: for any $R$-module $N$ and homomorphisms $\varphi_t: M_t \to N$, there is a unique homomorphism $\varphi: \bigoplus_{t \in T} M_t \to N$ with $\varphi \circ \rho_t = \varphi_t$ for all $t$. Combining these observations: an $R$-module homomorphism $\varphi: R^{\oplus n} \to M$ is exactly a map of the form $(r_1, \dots, r_n) \mapsto \sum_{i=1}^n r_i m_i$ for fixed elements $m_1, \dots, m_n \in M$. Such a $\varphi$ is surjective precisely when $M$ is generated by $m_1, \dots, m_n$. Therefore: a finitely generated $R$-module is the same as a quotient of $R^{\oplus n}$ for some $n \geq 1$. [definition: Free Module] An $R$-module $M$ is **free** if it is isomorphic to $R^{\oplus S}$ for some set $S$ (the direct sum of $|S|$ copies of $R$). Equivalently, $M$ is free if it has a **free basis**: a subset $B \subseteq M$ such that every $x \in M$ can be written uniquely as $x = \sum_{i=1}^{\ell} r_i x_i$ with $r_i \in R$, $x_i \in B$, $r_i \neq 0$, and $\ell \geq 0$. [/definition] For a field $k$, the statement "every $k$-vector space has a basis" is equivalent to "every $k$-module is free". In particular, even $k^{\mathbb{N}}$ — the product of countably many copies of $k$ — is isomorphic to $k^{\oplus S}$ for some set $S$ (the proof requires the axiom of choice, and the resulting $S$ is uncountable even though $k^{\mathbb{N}}$ looks countably infinite). ## Chain Conditions: Noetherian and Artinian Modules One of the central themes in commutative algebra is the question of when processes terminate — whether ascending chains of submodules stabilise, whether ideals are finitely generated. The noetherian condition formalises exactly this. [definition: Noetherian Module] An $R$-module $M$ is **noetherian** if either of the following equivalent conditions holds: 1. (**Ascending chain condition, ACC**) Every ascending chain of submodules $M_0 \subseteq M_1 \subseteq M_2 \subseteq \cdots$ stabilises: there exists $k$ such that $M_{k+j} = M_k$ for all $j \geq 0$. 2. (**Maximal condition**) Every non-empty set $\Sigma$ of submodules of $M$ has a maximal element. [/definition] The equivalence of these two conditions relies on the Axiom of Choice (specifically Zorn's lemma): given an ascending chain, its set of terms has a maximal element by (2), which gives stability; conversely, given a non-empty set without a maximal element one can construct a non-stabilising chain. [definition: Artinian Module] An $R$-module $M$ is **artinian** if the analogous conditions hold for descending chains: every descending chain of submodules stabilises (**descending chain condition, DCC**), or equivalently, every non-empty set of submodules has a minimal element. [/definition] The link between noetherian modules and finite generation is fundamental: [quotetheorem:2810] [citeproof:2810] [remark: Finitely Generated Does Not Imply Noetherian] The theorem implies every noetherian module is finitely generated (take $N = M$). The converse fails. Every ring $R$ is finitely generated as an $R$-module (by $1_R$). In particular, the polynomial ring $R = \mathbb{Z}[T_1, T_2, \dots]$ in countably many variables is finitely generated as an $R$-module. But the submodule $M = (T_1, T_2, \dots) \subseteq R$ (polynomials with constant term zero) is not finitely generated: any finite subset $S \subseteq M$ involves only finitely many variables, so $S \subseteq R \cdot T_1 + \cdots + R \cdot T_\ell$ for some $\ell$, and this submodule does not contain $T_{\ell+1}$. [/remark] [definition: Noetherian and Artinian Rings] A ring $R$ is **noetherian** (resp. **artinian**) if $R$, viewed as an $R$-module, is noetherian (resp. artinian). [/definition] [example: Noetherian and Artinian Examples] **$\mathbb{Z}$-modules.** - $\mathbb{Z}$ is a noetherian $\mathbb{Z}$-module (it is a PID, so all submodules are principal, hence finitely generated). It is not artinian: the chain $(2) \supset (4) \supset (8) \supset \cdots$ does not stabilise, since $(2^{n+1}) \subsetneq (2^n)$ for all $n$. - The $\mathbb{Z}$-module $\mathbb{Z}[\frac{1}{2}]/\mathbb{Z}$, where $\mathbb{Z}[\frac{1}{2}] = \{\frac{a}{2^m} : a, m \in \mathbb{Z}\}$, is artinian but not noetherian. To see it is not noetherian: let $N_k$ be the submodule generated by $\frac{1}{2^k} + \mathbb{Z}$. Since $\frac{1}{2^{k+1}}$ is not an integer multiple of $\frac{1}{2^k}$ modulo $\mathbb{Z}$, we have $N_k \subsetneq N_{k+1}$, giving a non-stabilising ascending chain. To see it is artinian: every non-zero submodule is one of the $N_k$. If $P$ is a non-zero submodule, pick any $\frac{a}{2^j} + \mathbb{Z} \in P$ with $\gcd(a,2) = 1$ (i.e., the element has exact $2$-power denominator $2^j$); then $P$ contains $\frac{1}{2^j} + \mathbb{Z}$, so $N_j \subseteq P$. Since $P \subseteq \mathbb{Z}[\frac{1}{2}]/\mathbb{Z}$ and every element of $\mathbb{Z}[\frac{1}{2}]/\mathbb{Z}$ has the form $\frac{a}{2^m} + \mathbb{Z}$, the submodule $P$ is contained in $N_m$ for the largest $m$ appearing, giving $P = N_j$ for some $j$. The descending chain $N_1 \supset N_2 \supset \cdots$ is thus the unique infinite strictly descending chain, and every descending chain of non-zero submodules eventually enters this chain and stabilises. **Rings.** - $\mathbb{Z}$ is a noetherian ring, but not an artinian ring (the same descending chain of ideals shows this). - Every artinian ring is noetherian. In fact, a ring $R$ is artinian if and only if $R$ is noetherian and has Krull dimension zero (all prime ideals are maximal). [/example] ## Exact Sequences and Noetherian Properties If $M$ is noetherian, must every submodule and quotient be noetherian? Must every extension of noetherian modules be noetherian? These questions cannot even be posed cleanly without a language for expressing how modules fit together. A bare isomorphism $M \cong N \oplus L$ tells us everything, but in practice modules arise as extensions where we only have a submodule $N \hookrightarrow M$ and a quotient $M \twoheadrightarrow L$, without a splitting. The language of exact sequences records exactly this data: how an injection and a surjection compose to zero, encoding the relationship between a module, one of its submodules, and the resulting quotient. Without it, statements about "$M$ has a submodule isomorphic to $N$ with quotient $L$" become unwieldy. To study how the noetherian property behaves with respect to submodules and quotients, the language of exact sequences is essential. [definition: Exact Sequence] A sequence of $R$-modules and $R$-module homomorphisms \begin{align*} \cdots \to M_{i-1} \xrightarrow{f_i} M_i \xrightarrow{f_{i+1}} M_{i+1} \to \cdots \end{align*} is **exact** if $\operatorname{im}(f_i) = \ker(f_{i+1})$ for all $i$. [/definition] [definition: Short Exact Sequence] A **short exact sequence** is an exact sequence of the form \begin{align*} 0 \to M' \xrightarrow{f} M \xrightarrow{g} M'' \to 0. \end{align*} [/definition] In a short exact sequence, exactness at $M'$ forces $f$ to be injective; exactness at $M''$ forces $g$ to be surjective; and exactness at $M$ says $\ker g = \operatorname{im} f$, so $M'' \cong M/f(M')$. [quotetheorem:2901] [citeproof:2901] This result is sometimes called the "three-for-two" property of the noetherian condition. Notice what the theorem does not say: it does not claim that if two of $N$, $M$, $L$ are noetherian then the third is automatically noetherian — the third hypothesis is always also required. The theorem also does not apply to longer exact sequences without decomposing them into short ones. The artinian analogue holds by the same proof with all chain directions reversed. [quotetheorem:2812] [citeproof:2812] This result, combined with the theorem on noetherian rings defined below, gives the fundamental structural fact of this course: over a noetherian ring, finitely generated modules are extremely well-behaved. It is worth pausing on the finiteness hypothesis: the result fails for infinite direct sums. The $\mathbb{Z}$-module $\bigoplus_{n \geq 1} \mathbb{Z}/n\mathbb{Z}$ is an infinite direct sum of noetherian (artinian, even) modules, but it is not finitely generated over $\mathbb{Z}$, and the submodule generated by all the distinct torsion elements is not finitely generated — so the direct sum theorem must be read as a strictly finite statement. ## Algebras over a Ring Why do we need algebras at all — is the module formalism not enough? The module concept captures the idea of $R$ acting on an abelian group $M$ by scalars, but it says nothing about how elements of $M$ multiply together. When $M$ itself has a ring structure (think of $\mathbb{C}$ as an $\mathbb{R}$-module, or $k[T]$ as a $k$-module), the module language loses track of this extra structure. An algebra packages both: $R$ acts on $A$, and $A$ has its own multiplication compatible with this action. The standard example to keep in mind is the polynomial ring $k[T]$: it is a $k$-module (a vector space), but it also has multiplication of polynomials, making it a $k$-algebra. Forgetting the multiplication and treating $k[T]$ as merely a $k$-module loses the central content. Having established the basics of modules, we introduce a closely related notion: algebras. An algebra combines a ring structure on $A$ with the $R$-module structure. [definition: R-Algebra] An **$R$-algebra** is a ring $A$ together with a fixed ring homomorphism $\rho: R \to A$, called the **structure homomorphism**. One writes $ra = \rho(r) \cdot a$ for the action of $R$ on $A$. [/definition] Specifying an $R$-algebra is superficially similar to specifying an $R$-module: one starts with a ring $A$ (rather than merely an abelian group $M$), and one fixes a ring homomorphism $R \to A$ (rather than $R \to \operatorname{End}(M)$). The key difference is that $A$ already has its own multiplication, and $R$ acts by multiplying via $\rho$. Every $R$-algebra is also an $R$-module: given the structure homomorphism $\rho: R \to A$, the map $r \mapsto (x \mapsto \rho(r) x)$ defines a ring homomorphism $R \to \operatorname{End}(A, +)$, making $A$ into an $R$-module. Every ring $A$ is a $\mathbb{Z}$-algebra in exactly one way, since there is exactly one ring homomorphism $\mathbb{Z} \to A$ (it must send $1 \mapsto 1$, and this determines everything). [remark: Notation Simplification] Notice that $\rho(r) = \rho(r) \cdot 1_A = r \cdot 1_A$ in the algebra notation. This means the image of $R$ in $A$ consists precisely of the elements $\{r \cdot 1_A : r \in R\}$. One can therefore often avoid naming the structure homomorphism $\rho$ explicitly. [/remark] [remark: Non-zero Algebras over a Field] Let $k$ be a field and $A \neq \{0\}$ a $k$-algebra. The zero ring $\{0\}$ is the only ring where $1 = 0$. Since $\rho: k \to A$ must send $1 \mapsto 1$ and $1 \neq 0$ in $A$, we have $1 \notin \ker \rho$. Since the only ideals of $k$ are $\{0\}$ and $k$, we get $\ker \rho = \{0\}$, so $\rho$ is injective. Every non-zero $k$-algebra contains an embedded copy of $k$. [/remark] A **subalgebra** of an $R$-algebra $A$ is a subring $B \subseteq A$ such that $rx \in B$ for all $r \in R$, $x \in B$. This forces $B$ to contain the image of $R$ in $A$. For a subset $S \subseteq A$, the subalgebra generated by $S$ consists of all elements of the form $p(x_1, \dots, x_m)$ where $p \in R[T_1, \dots, T_m]$ is a polynomial with coefficients in $R$ and $x_1, \dots, x_m \in S$. One should contrast this with the $R$-submodule generated by $S$, which requires $p$ to have degree exactly $1$ with no constant term — only $R$-linear combinations, not arbitrary polynomials. [definition: R-Algebra Homomorphism] For $R$-algebras $A$ and $B$ with structure homomorphisms $\rho_A: R \to A$ and $\rho_B: R \to B$, an **$R$-algebra homomorphism** is a ring homomorphism $\varphi: A \to B$ such that $\varphi \circ \rho_A = \rho_B$, i.e., $\varphi(r \cdot 1_A) = r \cdot 1_B$ for all $r \in R$. [/definition] Equivalently, $\varphi$ is an $R$-algebra homomorphism if and only if it is an $R$-module homomorphism satisfying $\varphi(1_A) = 1_B$ and $\varphi(a_1 a_2) = \varphi(a_1)\varphi(a_2)$. [example: Polynomial Algebras] Let $k$ be a field and consider $A = k[T_1, \dots, T_m]$. Then $A$ is both a $k$-algebra and a $k$-module. It is a finitely generated $k$-algebra (generated by $T_1, \dots, T_m$), but it is not a finitely generated $k$-module: any finite set of polynomials generates only a finite-dimensional $k$-vector subspace of $A$, whereas $A$ is infinite-dimensional over $k$. [/example] **The universal property of polynomial algebras.** For an $R$-algebra $A$ and elements $a_1, \dots, a_n \in A$, there is a unique $R$-algebra homomorphism $R[T_1, \dots, T_n] \to A$ sending $T_i \mapsto a_i$ for $1 \leq i \leq n$. This works even for infinitely many variables, since each polynomial involves only finitely many. This makes it easy to produce homomorphisms out of polynomial algebras — analogous to the way free modules make it easy to produce homomorphisms out of $R^{\oplus n}$. [quotetheorem:2902] [citeproof:2902] This theorem is the algebra analogue of the earlier fact that a finitely generated $R$-module is a quotient of $R^{\oplus n}$: just as $R^{\oplus n}$ is the free $R$-module on $n$ generators, $R[T_1, \dots, T_n]$ is the free $R$-algebra on $n$ generators, and surjections from it classify finitely generated $R$-algebras. Notice what the theorem does not claim: a surjection $R[T_1, \dots, T_n] \twoheadrightarrow A$ does not imply $A$ is a finitely generated $R$-module — merely a finitely generated $R$-algebra. The polynomial algebra $k[T]$ is generated by one element as a $k$-algebra, but requires infinitely many generators as a $k$-module ($k$-vector space), since monomials $1, T, T^2, \dots$ are linearly independent over $k$. ## Finitely Generated Modules over Noetherian Rings At this point we have the noetherian condition for rings and the noetherian condition for modules as separate definitions. The key question is: when does noetherianness of the ring force noetherianness of its modules? The answer — that finitely generated modules over noetherian rings are noetherian — is the fact that makes the whole theory tractable. Without it, we would know that $\mathbb{Z}$ is noetherian as a ring, but could say nothing structural about $\mathbb{Z}$-modules beyond what we already know about abelian groups. With it, the theory of finitely generated abelian groups and, more generally, modules over PIDs becomes a consequence of the noetherian framework. We can now connect the noetherian property of a ring to that of its modules. [quotetheorem:2903] [citeproof:2903] The finite generation hypothesis is indispensable. Over a noetherian ring, an infinitely generated module need not be noetherian: the $\mathbb{Z}$-module $\mathbb{Q}$ is not noetherian — the ascending chain $\mathbb{Z} \cdot \frac{1}{2} \subset \mathbb{Z} \cdot \frac{1}{4} \subset \mathbb{Z} \cdot \frac{1}{8} \subset \cdots$ does not stabilise — even though $\mathbb{Z}$ is noetherian. The theorem also feeds directly into Hilbert's basis theorem: once we know finitely generated algebras over noetherian rings are noetherian as rings, we immediately know that all their finitely generated modules are noetherian as well. ## Hilbert's Basis Theorem The most important consequence of the module-theoretic framework developed so far is Hilbert's basis theorem, which asserts that passing to polynomial rings preserves the noetherian condition. [quotetheorem:2904] [citeproof:2904] [example: Polynomial Rings over Fields] If $k$ is a field, then $k[T_1, \dots, T_n]$ is noetherian for all $n \geq 1$. Since $k$ is a field, every non-zero ideal is the whole ring, so $k$ is trivially noetherian. Hilbert's basis theorem then gives noetherianness for all polynomial extensions. [/example] The geometric content of Hilbert's basis theorem is striking. For a field $k$ and a subset $S \subseteq k[T_1, \dots, T_n]$, the variety $V(S) = \{x \in k^n : f(x) = 0 \text{ for all } f \in S\}$ depends only on the ideal $I = (S)$ generated by $S$. Since $I$ is finitely generated by the basis theorem — say $I = (f_1, \dots, f_\ell)$ — the variety $V(S) = V(f_1, \dots, f_\ell)$ is defined by finitely many equations. Consequently, every infinite system of polynomial equations over a field admits an equivalent finite subsystem. [remark: No Artinian Analogue] There is no artinian analogue of Hilbert's basis theorem. If $k$ is a field (which is artinian), the polynomial ring $k[X]$ is not artinian: the chain of ideals $\langle X \rangle \supset \langle X^2 \rangle \supset \langle X^3 \rangle \supset \cdots$ does not stabilise. [/remark] [remark: Gröbner Bases (Non-Examinable)] A more computational approach to Hilbert's basis theorem passes through the theory of Gröbner bases. Fix a monomial order $\succ$ on $k[T_1, \dots, T_n]$ (for example, degree-lexicographic order: $T_1^{\alpha_1} \cdots T_n^{\alpha_n} \succ T_1^{\beta_1} \cdots T_n^{\beta_n}$ if $\sum \alpha_i > \sum \beta_i$, or if the totals are equal and $(\alpha_1, \dots, \alpha_n)$ exceeds $(\beta_1, \dots, \beta_n)$ lexicographically). For a generating set $\{f_1, \dots, f_\ell\}$ of an ideal $I \trianglelefteq k[T_1, \dots, T_n]$, let $A = \operatorname{LM}(I)$ be the set of leading monomials of all elements of $I$, and let $B$ be the set of monomials divisible by the leading monomial of at least one $f_j$. One always has $B \subseteq A$; when $B = A$, the set $\{f_1, \dots, f_\ell\}$ is a **Gröbner basis** for $I$. Every ideal has a finite Gröbner basis, and Buchberger's algorithm computes it. Gröbner bases are powerful computational tools: they solve membership problems ($h \in I$?), compute ideal intersections $I \cap J$, and address many other problems in computational commutative algebra. [/remark] Gröbner bases and membership problems show how module theory enables computational answers. Chain conditions formalize the finiteness that makes these algorithms terminate. # 2. Chain Conditions Chapter 1 established the foundational language of rings and modules. This chapter introduces two finiteness conditions — the noetherian and artinian conditions — that tame the potentially wild behaviour of infinite ascending or descending chains of submodules. These conditions are among the most useful structural hypotheses in commutative algebra: they appear in almost every theorem that gives a definitive classification or makes an existence argument tractable. The chapter culminates in Hilbert's basis theorem, which shows that finitely generated algebras over noetherian rings inherit noetherianness — a result with immediate applications to algebraic geometry. ## Noetherian and Artinian Modules A module can fail to be "finite" in many senses. The most practically important finiteness conditions are captured by the behaviour of chains of submodules: can one form an infinite strictly ascending chain $M_0 \subsetneq M_1 \subsetneq M_2 \subsetneq \cdots$, or an infinite strictly descending one? [definition: Noetherian Module] An $R$-module $M$ is **noetherian** if either of the following equivalent conditions holds: 1. **(Ascending chain condition, ACC)** Every ascending chain of submodules $M_0 \subseteq M_1 \subseteq M_2 \subseteq \cdots$ of $M$ stabilizes: there exists $k \geq 0$ such that $M_{k+j} = M_k$ for all $j \geq 0$. 2. **(Maximal condition)** Every nonempty set $\Sigma$ of submodules of $M$ has a maximal element (a submodule $N \in \Sigma$ such that no $N' \in \Sigma$ properly contains $N$). [/definition] The equivalence of these two conditions relies on the Axiom of Choice via Zorn's lemma: the maximal condition is precisely the conclusion of Zorn applied to $\Sigma$ once one has that every ascending chain in $\Sigma$ is bounded above (by its union), and conversely the ACC follows from the maximal condition applied to the set of terms in any ascending chain. [definition: Artinian Module] An $R$-module $M$ is **artinian** if the descending analogue holds: every descending chain $M_0 \supseteq M_1 \supseteq M_2 \supseteq \cdots$ of submodules stabilizes, or equivalently, every nonempty set $\Sigma$ of submodules has a minimal element. [/definition] These two definitions are parallel in statement but very different in character. Noetherian modules are far more common in practice — the ring $\mathbb{Z}$ is noetherian but not artinian — while artinian modules are a richer and more constrained class. [quotetheorem:2905] [citeproof:2905] [remark: Finitely Generated Does Not Imply Noetherian] Every noetherian module is finitely generated (take $M$ itself as the submodule — it must be finitely generated). The converse fails. Consider the polynomial ring $R = \mathbb{Z}[T_1, T_2, \ldots]$ in countably many variables. As an $R$-module, $R$ is generated by the single element $1_R$, so it is finitely generated. But the submodule $M$ of $R$ generated by $\{T_1, T_2, \ldots\}$ (all polynomials with zero constant term) is not finitely generated: any finite subset $S \subset M$ is contained in $R \cdot T_1 + \cdots + R \cdot T_\ell$ for some $\ell$, and this submodule does not contain $T_{\ell+1}$, so $S$ cannot generate $M$. [/remark] [definition: Noetherian and Artinian Rings] A ring $R$ is **noetherian** (respectively **artinian**) if $R$, viewed as an $R$-module over itself, is noetherian (respectively artinian). Equivalently, $R$ is noetherian if every ideal of $R$ is finitely generated. [/definition] [example: Noetherian and Artinian Examples] Consider first the $\mathbb{Z}$-modules. The ring $\mathbb{Z}$ is a PID, so every ideal is principal and in particular finitely generated. Thus $\mathbb{Z}$ is noetherian as a module and as a ring. It is not artinian: the chain of ideals $(2) \supset (4) \supset (8) \supset \cdots$ is strictly descending and does not stabilize. The $\mathbb{Z}$-module $\mathbb{Z}[\frac{1}{2}]/\mathbb{Z}$, where $\mathbb{Z}[\frac{1}{2}] = \{\frac{a}{2^m} : a, m \in \mathbb{Z}\}$, is artinian but not noetherian. To see it is artinian, note that every submodule is of the form $\frac{1}{2^k}\mathbb{Z}/\mathbb{Z}$ for some $k \geq 0$, and these form a descending chain that terminates. To see it is not noetherian, the ascending chain $\frac{1}{2}\mathbb{Z}/\mathbb{Z} \subsetneq \frac{1}{4}\mathbb{Z}/\mathbb{Z} \subsetneq \cdots$ does not stabilize. For rings, $\mathbb{Z}$ is noetherian but not artinian, as the same descending chain of ideals shows. The relationship between artinian and noetherian rings is more subtle than for modules: it turns out that every artinian ring is noetherian, but not conversely. A precise characterisation is proved later in the course: a ring $R$ is artinian if and only if $R$ is noetherian and has Krull dimension zero — that is, every prime ideal of $R$ is maximal. This will follow from the structure theorem for artinian rings. [/example] ## Exact Sequences and Inheritance of Chain Conditions We now face a basic structural question: if $M$ has a submodule $N$ and quotient $M/N$, and both $N$ and $M/N$ are noetherian, is $M$? Conversely, does noetherianness pass to submodules and quotients? These questions cannot be answered module by module — one needs a framework that makes the relationship between $M$, $N$, and $M/N$ explicit. That framework is exact sequences. To understand how the noetherian and artinian properties interact with submodules and quotients, one needs the language of exact sequences. [definition: Exact Sequence] A sequence of $R$-modules and $R$-module homomorphisms \begin{align*} \cdots \to M_{i-1} \xrightarrow{f_i} M_i \xrightarrow{f_{i+1}} M_{i+1} \to \cdots \end{align*} is **exact** if $\operatorname{im}(f_i) = \ker(f_{i+1})$ for every $i$. [/definition] Exactness is a precise way of saying that consecutive maps compose to zero in the sharpest possible way: the image of each map is exactly the kernel of the next, with no excess and no gap. The key special case is the short exact sequence, which encodes the relationship between a module, a submodule, and the corresponding quotient. [definition: Short Exact Sequence] A **short exact sequence** (SES) is an exact sequence of the form \begin{align*} 0 \to M' \xrightarrow{f} M \xrightarrow{g} M'' \to 0. \end{align*} [/definition] Exactness at $M'$ means $f$ is injective; exactness at $M''$ means $g$ is surjective; and exactness at $M$ means $\ker(g) = \operatorname{im}(f)$. Thus $M'' \cong M/f(M')$: a short exact sequence expresses $M$ as an extension of $M''$ by $M'$. The following lemma is the key tool for propagating chain conditions through constructions. [quotetheorem:2906] [citeproof:2906] This theorem is the engine behind almost every result in this chapter. It says noetherianness is an extension property: to check that $M$ is noetherian, it suffices to check the sub and the quotient separately. In concrete terms, if $N \leq M$ and both $N$ and $M/N$ satisfy the ACC, then so does $M$ — even though a chain in $M$ can involve elements drawn from both $N$ and its complement. The proof works because the quotient map $\pi: M \to M/N$ and the inclusion $N \hookrightarrow M$ together triangulate any ascending chain in $M$. [quotetheorem:2818] [citeproof:2818] The finiteness hypothesis here is essential. An infinite direct sum of noetherian modules need not be noetherian: the countable direct sum $\bigoplus_{n \geq 1} \mathbb{Z}$ is a $\mathbb{Z}$-module, and it is not noetherian because the submodule generated by $e_1, e_2, \ldots$ (the standard basis vectors) has the strictly ascending chain $(e_1) \subsetneq (e_1, e_2) \subsetneq (e_1, e_2, e_3) \subsetneq \cdots$ that never stabilises. This is why the theory of noetherian modules will ultimately focus on finitely generated modules. ## Free Modules, Finitely Generated Modules, and the Homomorphism Perspective If $R$ is a noetherian ring, which $R$-modules inherit this property? The whole category of $R$-modules is too broad: the direct sum $\bigoplus_{n \geq 1} R$ is an $R$-module but fails the ACC, as the previous remark showed. The right class turns out to be finitely generated modules — but to prove this one first needs to understand the homomorphism perspective on finitely generated modules, and in particular the role of free modules as building blocks. For any $R$-module homomorphism $\varphi: R \to M$, the value of $\varphi$ is determined by $\varphi(1_R)$: one has $\varphi(r) = \varphi(r \cdot 1_R) = r \varphi(1_R)$. Conversely, for any $m \in M$, the map $r \mapsto rm$ is an $R$-module homomorphism $R \to M$. Thus $R$-module homomorphisms $R \to M$ correspond exactly to elements of $M$. The universal property of the direct sum says: for $R$-modules $(M_t)_{t \in T}$ with embeddings $\iota_t: M_t \hookrightarrow \bigoplus_{t \in T} M_t$, any collection of homomorphisms $\varphi_t: M_t \to N$ extends uniquely to a homomorphism $\varphi: \bigoplus_{t \in T} M_t \to N$ with $\varphi \circ \iota_t = \varphi_t$. Combining: $R$-module homomorphisms $\varphi: R^{\oplus n} \to M$ are exactly the maps $(r_1, \ldots, r_n) \mapsto \sum_{i=1}^n r_i m_i$ for fixed $m_1, \ldots, m_n \in M$. A module $M$ is finitely generated precisely when there exists a surjective homomorphism $R^{\oplus n} \to M$ for some $n \geq 1$ — equivalently, $M$ is a quotient of $R^{\oplus n}$. [definition: Free Module] An $R$-module $M$ is **free** if it is isomorphic to $R^{\oplus S}$ for some set $S$ — that is, a direct sum of copies of $R$ indexed by $S$. Equivalently, $M$ is free if it has a **free basis**: a subset $B \subset M$ such that every $x \in M$ has a unique expression $x = \sum_{i=1}^\ell r_i x_i$ with $\ell \geq 0$, $0 \neq r_i \in R$, $x_i \in B$ distinct. [/definition] Free modules behave like vector spaces in the sense that homomorphisms from them are easy to specify: one chooses the image of each basis element independently. When the base ring is a field $k$, every $k$-module (i.e., every $k$-vector space) is free. In general, this fails: $\mathbb{Z}/2\mathbb{Z}$ is a $\mathbb{Z}$-module that is not free (it has no free basis, since $2 \cdot \bar{1} = 0$ in $\mathbb{Z}/2\mathbb{Z}$). [quotetheorem:2819] [citeproof:2819] The finitely generated hypothesis cannot be dropped. Take $R = \mathbb{Z}$ — a noetherian ring — and consider $M = \bigoplus_{n \geq 1} \mathbb{Z}/n\mathbb{Z}$. This is a $\mathbb{Z}$-module that is not finitely generated, and it is not noetherian: the submodule generated by the first $k$ components has a strictly ascending chain as $k$ increases. For artinian modules the story is similar but sharper: if $R$ is artinian and $M$ is finitely generated, then $M$ is artinian. This mirrors the noetherian case, and the same SES argument applies. ## Algebras and Their Module Structure The module framework captures linear structure over a ring $R$, but it does not see ring multiplication on the module itself. For example, the polynomial ring $R[T]$ is an $R$-module, but the product of two polynomials is invisible to the module structure — the module axioms cannot express that $T \cdot T = T^2$. To study rings built from $R$ by adjoining elements (like $T$), one needs a richer structure that sees both the $R$-module structure and the ring structure simultaneously. This is the notion of an $R$-algebra. [definition: R-Algebra] An **$R$-algebra** is a ring $A$ together with a fixed ring homomorphism $\rho: R \to A$. One writes $r \cdot a = \rho(r) \cdot a$ for $r \in R$, $a \in A$. [/definition] Specifying an $R$-algebra is superficially similar to specifying an $R$-module, with two key differences: $A$ is itself a ring (not merely an abelian group), and the structural map is a ring homomorphism $R \to A$ (rather than $R \to \operatorname{End}(M)$). Every $R$-algebra $A$ is also an $R$-module, via the composition $R \to A \hookrightarrow \operatorname{End}(A, +)$ given by $r \mapsto (x \mapsto \rho(r) x)$. A convenient simplification: since $\rho(r) = \rho(r) \cdot 1_A = r \cdot 1_A$, the notation $\rho$ is often suppressed. One simply says "$A$ is an $R$-algebra" and writes $r \cdot a$ for $\rho(r) a$. Every ring $A$ is a $\mathbb{Z}$-algebra in exactly one way, because there is exactly one ring homomorphism $\mathbb{Z} \to A$ (sending $n$ to $n \cdot 1_A$). This makes the theory of $\mathbb{Z}$-algebras the same as the theory of (commutative, unital) rings. [definition: R-Algebra Homomorphism] For $R$-algebras $A$ and $B$ with structural homomorphisms $\rho_A: R \to A$ and $\rho_B: R \to B$, an **$R$-algebra homomorphism** is a ring homomorphism $\varphi: A \to B$ such that $\varphi \circ \rho_A = \rho_B$. Equivalently, $\varphi$ is an $R$-module homomorphism that satisfies $\varphi(1_A) = 1_B$ and $\varphi(a_1 a_2) = \varphi(a_1)\varphi(a_2)$ for all $a_1, a_2 \in A$. [/definition] There is a crucial distinction in how "finitely generated" works for algebras versus modules: [example: Finitely Generated Algebra vs Module] Let $k$ be a field and consider the polynomial ring $A = k[T_1, \ldots, T_m]$. This is a $k$-algebra (with $\rho: k \hookrightarrow A$ the natural inclusion) and also a $k$-module. As a $k$-algebra, $A$ is finitely generated: the elements $T_1, \ldots, T_m$ generate $A$ (every polynomial is a $k$-linear combination of monomials in $T_1, \ldots, T_m$). But as a $k$-module, $A$ is not finitely generated: the monomials $T_1^{a_1} \cdots T_m^{a_m}$ form an infinite basis, and no finite set of polynomials spans all of $A$ over $k$. [/example] [remark: Nonzero k-Algebra Contains k] For a nonzero $k$-algebra $A$ over a field $k$, the structural homomorphism $\rho: k \to A$ is necessarily injective. Indeed, $\ker \rho$ is an ideal of $k$; since $A \neq \{0\}$ and $\rho(1_k) = 1_A \neq 0$, the kernel does not contain $1_k$. But the only ideals of $k$ are $\{0\}$ and $k$ itself, so $\ker \rho = \{0\}$. Thus $A$ contains a copy of $k$ embedded via $\rho$. [/remark] The polynomial algebra $R[T_1, \ldots, T_n]$ plays the role of a free object in the category of $R$-algebras: [quotetheorem:2820] This means specifying a homomorphism out of $R[T_1, \ldots, T_n]$ is the same as choosing $n$ elements of $A$ — a direct parallel with the characterisation of free module homomorphisms. Consequently, an $R$-algebra $A$ is finitely generated if and only if there exists $n \geq 0$ and a surjective $R$-algebra homomorphism $R[T_1, \ldots, T_n] \to A$, i.e., $A \cong R[T_1, \ldots, T_n]/\mathfrak{a}$ for some ideal $\mathfrak{a}$. ## Hilbert's Basis Theorem We have shown that noetherian rings behave well under finitely generated module quotients. A natural question is whether the same holds for polynomial extensions: if $R$ is noetherian, is $R[T]$ noetherian? This is not immediate — one would need every ideal of $R[T]$ to be finitely generated, but ideals of $R[T]$ can involve polynomials of arbitrarily large degree. The artinian analogue fails entirely: a field $k$ is artinian, but $k[T]$ is not artinian (the chain $\langle T \rangle \supset \langle T^2 \rangle \supset \cdots$ does not stabilise). For the noetherian condition, however, the answer is yes — and this is Hilbert's basis theorem. [quotetheorem:2907] [citeproof:2907] [example: Polynomial Rings over Fields] If $k$ is a field, then $k$ is noetherian (it has only two ideals: $\{0\}$ and $k$). Hilbert's basis theorem therefore implies that $k[T_1, \ldots, T_n]$ is noetherian for every $n \geq 1$. More generally, $\mathbb{Z}[T_1, \ldots, T_n]$ is noetherian. [/example] Hilbert's theorem has a striking application to systems of polynomial equations. Given a set $S$ of polynomials in $k[T_1, \ldots, T_n]$, the variety $V(S)$ (the set of simultaneous zeros in $k^n$) equals $V(I)$ where $I = (S)$ is the ideal generated by $S$. Since $k[T_1, \ldots, T_n]$ is noetherian, $I$ is finitely generated, say $I = (g_1, \ldots, g_r)$. Thus $V(S) = V(g_1, \ldots, g_r)$: every system of polynomial equations over $k$ is equivalent to a finite subsystem. In particular, as one adds polynomials to $S$ one by one, the variety $V(S)$ can only shrink, and this shrinking process must stabilize after finitely many steps. [remark: No Artinian Analogue] There is no analogue of Hilbert's basis theorem for artinian rings. A field $k$ is artinian, but the polynomial ring $k[X]$ is not artinian: the chain $\langle X \rangle \supset \langle X^2 \rangle \supset \langle X^3 \rangle \supset \cdots$ is strictly descending and does not stabilize. [/remark] This failure is not a coincidence: artinian rings have Krull dimension zero, and polynomial rings have positive Krull dimension. Hilbert's basis theorem succeeds precisely because the noetherian condition is about ascending chains of ideals, and the degree argument in the proof converts an ascending chain problem in $R[T]$ into a stabilisation problem for ideals in $R$. [remark: Gröbner Bases (Non-Examinable)] Let $k$ be a field. On the polynomial algebra $k[T_1, \ldots, T_n]$, one can order monomials by first comparing total degree $\alpha_1 + \cdots + \alpha_n$ and then breaking ties lexicographically; this is the **graded lexicographic order** $\succ$. For a polynomial $f$, let $\operatorname{LM}(f)$ denote its leading monomial under $\succ$. For an ideal $I = (f_1, \ldots, f_\ell)$, let $A = \operatorname{LM}(I)$ denote the set of leading monomials of all elements of $I$, and let $B$ be the set of monomials divisible by $\operatorname{LM}(f_i)$ for some $i$. Since $B \subseteq A$, one says $(f_1, \ldots, f_\ell)$ is a **Gröbner basis** of $I$ if $A = B$. It is a theorem that every ideal has a finite Gröbner basis, and every Gröbner basis generates the ideal. Gröbner bases provide an algorithmic approach to ideal membership ($h \in I$?), ideal intersection, and dimension computations in polynomial rings. [/remark] Noetherian rings guarantee that ascending chains stabilize, allowing algorithms to give constructive proofs. Tensor products build modules horizontally, extending scalars and combining structures across different algebras. # 3. Tensor Products ## Tensor Products of Modules The previous chapter established that noetherian and artinian conditions control how modules behave "in the vertical direction" — ascending and descending chains of submodules. Tensor products, by contrast, give a way to build new modules from old ones "horizontally," combining an $R$-module $M$ with an $R$-module $N$ into a single object $M \otimes_R N$ that encodes how $M$ and $N$ interact over $R$. This construction is indispensable throughout commutative algebra: it appears in base change, in the definition of algebras, and later in the study of exactness and flatness. [motivation] ### Why bilinear maps are not enough Given two $R$-modules $M$ and $N$, the most natural thing to consider is a bilinear map $f: M \times N \to L$ into some target module $L$. But $M \times N$ itself does not carry a canonical module structure that makes bilinear maps into module homomorphisms. The direct sum $M \oplus N$ linearises the variables separately — a homomorphism $M \oplus N \to L$ amounts to specifying a pair of linear maps, one from $M$ and one from $N$ independently. A bilinear map couples the two variables: scaling in the first entry should have the same effect as scaling in the second. No subquotient of $M \oplus N$ captures this coupling automatically. The tensor product $M \otimes_R N$ is precisely the $R$-module built to universally linearise bilinear maps: every bilinear map $M \times N \to L$ factors uniquely through a linear map $M \otimes_R N \to L$. This universal property both characterises $M \otimes_R N$ up to unique isomorphism and provides the main tool for constructing homomorphisms out of it. [/motivation] ### The construction Before giving the formal definition, it is helpful to describe $M \otimes_R N$ informally. One introduces a formal symbol $m \otimes n$ for each pair $(m, n) \in M \times N$, takes all finite $R$-linear combinations, and then imposes the bilinearity relations: \begin{align*} (m_1 + m_2) \otimes n &= m_1 \otimes n + m_2 \otimes n, \\ m \otimes (n_1 + n_2) &= m \otimes n_1 + m \otimes n_2, \\ (rm) \otimes n &= r(m \otimes n) = m \otimes (rn). \end{align*} From the third relation, $0 \otimes n = 0 \cdot (1 \otimes n) = 0$ and $m \otimes 0 = 0$ for all $m \in M$, $n \in N$. The formal definition makes this precise via a quotient construction. [definition: R-Bilinear Map] Let $M$, $N$, $L$ be $R$-modules. A map $f: M \times N \to L$ is **$R$-bilinear** if for every fixed $m_0 \in M$ the map $n \mapsto f(m_0, n)$ is $R$-linear, and for every fixed $n_0 \in N$ the map $m \mapsto f(m, n_0)$ is $R$-linear. [/definition] [definition: Tensor Product of Modules] Let $M$ and $N$ be $R$-modules. Let $\mathcal{F} = R^{\oplus(M \times N)}$ be the free $R$-module with one basis element $e_{(m,n)}$ for each pair $(m, n) \in M \times N$. Let $K \subseteq \mathcal{F}$ be the submodule generated by the elements: \begin{align*} &e_{(m, n_1)} + e_{(m, n_2)} - e_{(m, n_1 + n_2)}, \\ &e_{(m_1, n)} + e_{(m_2, n)} - e_{(m_1 + m_2, n)}, \\ &r \cdot e_{(m,n)} - e_{(rm, n)}, \\ &r \cdot e_{(m,n)} - e_{(m, rn)}, \end{align*} for all $m, m_1, m_2 \in M$, $n, n_1, n_2 \in N$, $r \in R$. The **tensor product** of $M$ and $N$ over $R$ is the quotient \begin{align*} M \otimes_R N := \mathcal{F} / K. \end{align*} The image of $e_{(m,n)}$ in this quotient is written $m \otimes n$, and the canonical map $i_{M \otimes N}: M \times N \to M \otimes_R N$ sending $(m, n) \mapsto m \otimes n$ is $R$-bilinear by construction. [/definition] An element of the form $m \otimes n$ with $m \in M$, $n \in N$ is called a **pure tensor**. The pure tensors generate $M \otimes_R N$ as an $R$-module: every element of $M \otimes_R N$ can be written as a finite sum $\sum_{i=1}^\ell m_i \otimes n_i$. However, it is not the case in general that every element is pure — distinguishing pure from non-pure tensors is a recurring theme in the subject. [example: Vanishing and Free Tensors] Consider $\mathbb{Z}/2 \otimes_\mathbb{Z} \mathbb{Z}/3$. For any $x \in \mathbb{Z}/2$ and $y \in \mathbb{Z}/3$, one can write $x = 3x$ in $\mathbb{Z}/2$ (since $3 \equiv 1 \pmod 2$), so \begin{align*} x \otimes y = (3x) \otimes y = x \otimes (3y) = x \otimes 0 = 0. \end{align*} Since every pure tensor is zero, and pure tensors generate, $\mathbb{Z}/2 \otimes_\mathbb{Z} \mathbb{Z}/3 = \{0\}$. On the other hand, for any ring $R$ one has $R^m \otimes_R R^n \cong R^{mn}$. Over a field $k$, this follows because $\{e_i \otimes f_j\}_{i,j}$ is a $k$-basis of $k^m \otimes_k k^n$ (proved in full below), giving $k^m \otimes_k k^n \cong k^{mn}$. [/example] ### The universal property The central fact about the tensor product is that it linearises bilinear maps. This is the content of the following proposition, which also characterises $M \otimes_R N$ up to unique isomorphism. [quotetheorem:2908] [citeproof:2908] [remark: Bilinear-Hom adjunction] The universal property can be rephrased as a natural bijection of sets \begin{align*} \operatorname{Bilin}_R(M \times N, L) \xrightarrow{\;\sim\;} \operatorname{Hom}_R(M \otimes_R N, L), \end{align*} for every $R$-module $L$, where the left side denotes the set of $R$-bilinear maps $M \times N \to L$ and the right side is the set of $R$-module homomorphisms. The bijection sends $f$ to the unique $h$ with $h \circ i_{M \otimes N} = f$. [/remark] The universal property immediately implies that any two pairs $(T, j)$ satisfying the same universal property are uniquely isomorphic. [quotetheorem:2909] [citeproof:2909] ### When does a sum of pure tensors vanish? Working concretely in a tensor product requires a criterion for when a finite sum $\sum_i m_i \otimes n_i$ equals zero. Manipulating the bilinearity relations is one approach; the following characterisation makes this systematic. [quotetheorem:2824] [citeproof:2824] This criterion is useful both in computations and in proofs. One important consequence: if $(m_i) \subset M$ and $(n_i) \subset N$ are finite lists, whether $\sum_i m_i \otimes n_i = 0$ depends only on a finitely generated piece of the problem. [quotetheorem:2910] [citeproof:2910] [remark: Submodule direction] The finite generation proposition above goes one way: vanishing in $M' \otimes_R N'$ need not imply anything about the ambient tensor product, and — as the example below shows — the tensor product of submodules cannot be naively embedded in the tensor product of the ambient modules. However, the other direction always works: if $M' \subseteq M$ and $N' \subseteq N$ are submodules and $\sum_i m_i \otimes n_i = 0$ in $M \otimes_R N$, then for any $R$-bilinear map $f: M' \times N' \to L$ one can extend to $M \times N$ (by setting $f = 0$ outside $M' \times N'$ is not an extension in general, but the zero criterion applies directly to $i_{M \otimes N}$) — more precisely, the universal property applied to the composition with the inclusion shows $\sum_i m_i \otimes n_i = 0$ in $M' \otimes_R N'$ whenever it holds in $M \otimes_R N$. [/remark] [example: Tensor product depends on the ambient modules] Consider the $\mathbb{Z}$-modules $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$, and the submodule $2\mathbb{Z} \subset \mathbb{Z}$. In $\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$: \begin{align*} 2 \otimes (1 + 2\mathbb{Z}) = 2(1 \otimes (1 + 2\mathbb{Z})) = 1 \otimes (2 + 2\mathbb{Z}) = 1 \otimes 0 = 0. \end{align*} However, the same computation is invalid in $(2\mathbb{Z}) \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$, where $2 \otimes (1 + 2\mathbb{Z}) \neq 0$. To verify this, define a $\mathbb{Z}$-bilinear map $g: 2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ by $g(2x, y + 2\mathbb{Z}) = xy + 2\mathbb{Z}$. This is well defined and bilinear, and $g(2, 1 + 2\mathbb{Z}) = 1 + 2\mathbb{Z} \neq 0$. By the zero criterion, $2 \otimes (1 + 2\mathbb{Z}) \neq 0$ in $(2\mathbb{Z}) \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$. The moral: the element $2 \otimes (1 + 2\mathbb{Z})$ looks the same in both tensor products, but the tensor product is highly sensitive to the ambient modules. It is wrong to view $(2\mathbb{Z}) \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$ as a submodule of $\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$. [/example] As an application of the finite generation result, one obtains a structural property of tensor products of torsion-free abelian groups. [quotetheorem:2911] [citeproof:2911] ### Basic structural isomorphisms The following proposition collects the main formal properties of the tensor product. [quotetheorem:2827] [citeproof:2827] [remark: Tensor products of free modules] The distributivity isomorphism immediately recovers $R^m \otimes_R R^n \cong R^{mn}$ for any ring $R$: \begin{align*} R^m \otimes_R R^n = \left(\bigoplus_{1 \leq i \leq m} R\right) \otimes_R \left(\bigoplus_{1 \leq j \leq n} R\right) \cong \bigoplus_{i, j} (R \otimes_R R) \cong \bigoplus_{i, j} R \cong R^{mn}, \end{align*} where the unit isomorphism $R \otimes_R R \cong R$ is used. In particular $k^m \otimes_k k^n \cong k^{mn}$ for any field $k$. [/remark] [remark: Quotients and sum of ideals] From the quotient isomorphism, for ideals $I$ and $J$ of $R$ one has \begin{align*} R/I \otimes_R R/J \cong R/(I + J), \end{align*} via $(r_1 + I) \otimes (r_2 + J) \mapsto r_1 r_2 + (I + J)$. This means the kernel of $R \to R/I \otimes_R R/J$ is $I + J$, in contrast to the kernel of $R \to R/I \times R/J$, which is $I \cap J$. The product ideal $IJ$ does not appear as the kernel of a natural map of this form. [/remark] ### Tensor products of homomorphisms The tensor product construction extends to morphisms: given $R$-module homomorphisms $f: M \to M'$ and $g: N \to N'$, one can form a tensor product map. [quotetheorem:2912] [citeproof:2912] This construction is functorial: for composable chains $M_1 \xrightarrow{h} M_2 \xrightarrow{f} M_3$ and $N_1 \xrightarrow{k} N_2 \xrightarrow{g} N_3$, one has $(f \otimes g) \circ (h \otimes k) = (f \circ h) \otimes (g \circ k)$. This follows by evaluating on pure tensors and invoking uniqueness. The tensor product preserves isomorphisms and surjections, but notably does not preserve injections in general. [quotetheorem:2913] [citeproof:2913] The failure of tensor products to preserve injections is one of the most important phenomena in commutative algebra, and it will be treated systematically in Section 4 on exactness. For now, the following example illustrates it. [example: Tensor product need not preserve injections] Let $f: \mathbb{Z} \to \mathbb{Z}$ be multiplication by a prime $p$, so $f(x) = px$, and let $\operatorname{id}: \mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$ be the identity. Both $f$ and $\operatorname{id}$ are injective $\mathbb{Z}$-module homomorphisms. But for any $a \in \mathbb{Z}$ and $b \in \mathbb{Z}/p\mathbb{Z}$: \begin{align*} (f \otimes \operatorname{id})(a \otimes b) = (pa) \otimes b = a \otimes (pb) = a \otimes 0 = 0, \end{align*} since $pb = 0$ in $\mathbb{Z}/p\mathbb{Z}$. Thus $f \otimes \operatorname{id}$ is the zero map, even though both $f$ and $\operatorname{id}$ are injective. [/example] ### The base ring matters: an illustration The base ring over which one tensors has a substantial effect on the result. [example: Base ring and dimension] Compare $\mathbb{C}^2 \otimes_\mathbb{C} \mathbb{C}^3$ and $\mathbb{C}^2 \otimes_\mathbb{R} \mathbb{C}^3$, where $\mathbb{C}^n$ is viewed as an $\mathbb{R}$-module in the natural way (identifying $\mathbb{C}^n \cong \mathbb{R}^{2n}$ as $\mathbb{R}$-modules). Over $\mathbb{C}$: $\mathbb{C}^2 \otimes_\mathbb{C} \mathbb{C}^3 \cong \mathbb{C}^6$, which is $12$-dimensional over $\mathbb{R}$. Over $\mathbb{R}$: $\mathbb{C}^2 \otimes_\mathbb{R} \mathbb{C}^3 \cong \mathbb{R}^4 \otimes_\mathbb{R} \mathbb{R}^6 \cong \mathbb{R}^{24}$, which is $24$-dimensional over $\mathbb{R}$. The discrepancy arises because tensoring over $\mathbb{R}$ allows fewer identifications than tensoring over $\mathbb{C}$. In $\mathbb{C}^2 \otimes_\mathbb{C} \mathbb{C}^3$ one has the relation $(iv) \otimes w = v \otimes (iw)$ for all $v \in \mathbb{C}^2$, $w \in \mathbb{C}^3$, since $i$ is a $\mathbb{C}$-scalar. In $\mathbb{C}^2 \otimes_\mathbb{R} \mathbb{C}^3$ this identification does not hold — only real scalars can be moved across the tensor symbol. [/example] ### Pure versus non-pure tensors: a detailed example It is worth working out in detail when an element of $k^m \otimes_k k^n$ is pure, since this illustrates both the basis for the tensor product and the subtlety of pure tensors. [example: Pure tensors in k^m tensored with k^n] Let $k$ be a field and $m, n \geq 0$. Write $\{e_1, \dots, e_m\}$ for the standard basis of $k^m$ and $\{f_1, \dots, f_n\}$ for the standard basis of $k^n$. Since $k^m$ is generated by $\{e_i\}$ and $k^n$ by $\{f_j\}$, the set $\mathcal{B} = \{e_i \otimes f_j\}_{1 \leq i \leq m,\, 1 \leq j \leq n}$ spans $k^m \otimes_k k^n$. To show $\mathcal{B}$ is linearly independent, suppose $\sum_{i,j} \alpha_{ij} (e_i \otimes f_j) = 0$ with $\alpha_{ij} \in k$. For fixed $a, b$, define a $k$-bilinear map $T: k^m \times k^n \to k$ by $T(v, w) = \pi_a(v) \cdot \pi_b(w)$, where $\pi_a$ and $\pi_b$ denote projection onto the $a$-th and $b$-th coordinates. By the zero criterion, \begin{align*} \sum_{i,j} T(\alpha_{ij} e_i, f_j) = 0, \end{align*} but the left side equals $\alpha_{ab}$, since $\pi_a(e_i) = \delta_{ai}$ and $\pi_b(f_j) = \delta_{bj}$. Hence $\alpha_{ab} = 0$ for all $a, b$, so $\mathcal{B}$ is a $k$-basis and $k^m \otimes_k k^n \cong k^{mn}$. Now consider $\mathbb{R}^2 \otimes_\mathbb{R} \mathbb{R}^2$, which has basis $\{e_1 \otimes e_1,\, e_1 \otimes e_2,\, e_2 \otimes e_1,\, e_2 \otimes e_2\}$. A pure tensor has the form \begin{align*} (\alpha e_1 + \beta e_2) \otimes (\gamma e_1 + \delta e_2) = (\alpha\gamma)(e_1 \otimes e_1) + (\alpha\delta)(e_1 \otimes e_2) + (\beta\gamma)(e_2 \otimes e_1) + (\beta\delta)(e_2 \otimes e_2) \end{align*} for some $\alpha, \beta, \gamma, \delta \in \mathbb{R}$. The coefficient vectors $(\alpha\gamma, \alpha\delta)$ and $(\beta\gamma, \beta\delta)$ of the two "rows" are proportional (both are multiples of $(\gamma, \delta)$), so the $2 \times 2$ matrix of coefficients has rank at most $1$. For instance, $3(e_1 \otimes e_1) + 4(e_1 \otimes e_2) + 6(e_2 \otimes e_1) + 8(e_2 \otimes e_2)$ is pure: the coefficient matrix is $\begin{pmatrix} 3 & 4 \\ 6 & 8 \end{pmatrix}$, which has rank $1$, and indeed the expression equals $(e_1 + 2e_2) \otimes (3e_1 + 4e_2)$. On the other hand, $1 \cdot (e_1 \otimes e_1) + 2 \cdot (e_1 \otimes e_2) + 3 \cdot (e_2 \otimes e_1) + 4 \cdot (e_2 \otimes e_2)$ corresponds to the matrix $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, which has rank $2$ and determinant $-2 \neq 0$. Therefore this element is not a pure tensor in $\mathbb{R}^2 \otimes_\mathbb{R} \mathbb{R}^2$. [/example] ### The Kronecker product Tensor products of linear maps between finite-dimensional spaces have a concrete matrix description known as the Kronecker product. [example: Kronecker Product] Let $k$ be a field, and let $T: k^a \to k^b$ and $S: k^c \to k^d$ be $k$-linear maps. Write $[T] \in M_{b \times a}(k)$ and $[S] \in M_{d \times c}(k)$ for their matrices with respect to standard bases. The tensor product $T \otimes S: k^a \otimes_k k^c \to k^b \otimes_k k^d$ satisfies: \begin{align*} (T \otimes S)(e_i \otimes e_j) = (Te_i) \otimes (Se_j) = \left(\sum_{\ell=1}^b [T]_{\ell i} f_\ell\right) \otimes \left(\sum_{t=1}^d [S]_{tj} f_t\right) = \sum_{\ell, t} [T]_{\ell i} [S]_{tj} \cdot f_\ell \otimes f_t. \end{align*} Ordering the basis $\{e_i \otimes e_j\}$ of $k^a \otimes_k k^c$ as $e_1 \otimes e_1, \dots, e_1 \otimes e_c, e_2 \otimes e_1, \dots, e_a \otimes e_c$, and similarly for $\{f_i \otimes f_j\}$, the matrix representing $T \otimes S$ is the **Kronecker product**: \begin{align*} [T \otimes S] = \begin{pmatrix} [T]_{11}[S] & \cdots & [T]_{1a}[S] \\ \vdots & \ddots & \vdots \\ [T]_{b1}[S] & \cdots & [T]_{ba}[S] \end{pmatrix} \in M_{bd \times ac}(k). \end{align*} This is a block matrix where each entry of $[T]$ is replaced by the scalar multiple $[T]_{\ell i} \cdot [S]$. [/example] ## Tensor Products of Algebras The tensor product $M \otimes_R N$ of two $R$-modules is an $R$-module. When the modules in question carry additional ring structure — that is, when $B$ and $C$ are $R$-algebras — the tensor product $B \otimes_R C$ inherits a natural ring structure as well. This section constructs that structure, identifies $B \otimes_R C$ as an $R$-algebra, and establishes its universal property, which characterises it as the coproduct in the category of $R$-algebras. ### The ring structure on $B \otimes_R C$ Let $B$ and $C$ be $R$-algebras. Regarded as $R$-modules, one can form the tensor product $B \otimes_R C$. The goal is to promote this to a ring by defining a multiplication. The natural candidate is \begin{align*} (b \otimes c)(b' \otimes c') = (bb') \otimes (cc') \end{align*} extended $R$-linearly to all of $B \otimes_R C$. The key point is that this must be checked to be well-defined: since elements of $B \otimes_R C$ are not in general single pure tensors, one must verify that the formula is independent of how an element is represented as a sum of pure tensors. [definition: Algebra Tensor Product] Let $B$ and $C$ be $R$-algebras. The **tensor product of algebras** $B \otimes_R C$ is the $R$-module $B \otimes_R C$ equipped with the ring multiplication defined on pure tensors by \begin{align*} (b \otimes c)(b' \otimes c') = (bb') \otimes (cc'), \end{align*} extended $R$-linearly to all elements of $B \otimes_R C$, and with multiplicative identity $1_B \otimes 1_C$. [/definition] [quotetheorem:2914] [citeproof:2914] [remark: Alternative Well-Definedness Argument] The source also gives a direct argument: for fixed $(b, c) \in B \times C$, the map $(b', c') \mapsto (bb') \otimes (cc')$ is $R$-bilinear, so it induces an $R$-linear map $\mu_{b,c}: B \otimes_R C \to B \otimes_R C$ with $\mu_{b,c}(b' \otimes c') = (bb') \otimes (cc')$. This shows that multiplication on the left by a pure tensor is well-defined. For general elements, if $\sum_i b_i \otimes c_i = \sum_j b'_j \otimes c'_j$ in $B \otimes_R C$, then using the $R$-linearity of each $\mu_{b,c}$ one can check that multiplying either representative by any element gives the same result. The key computation uses the fact that $\mu_{b,c}$ depends on $(b, c)$ linearly. [/remark] ### $B \otimes_R C$ as an $R$-algebra Having a ring structure on $B \otimes_R C$, one makes it into an $R$-algebra by specifying a ring homomorphism $R \to B \otimes_R C$. There are two natural candidates: the map $\varphi: r \mapsto \rho_B(r) \otimes 1_C$ arising from the structure homomorphism $\rho_B: R \to B$, and the map $\psi: r \mapsto 1_B \otimes \rho_C(r)$ arising from $\rho_C: R \to C$. These two maps coincide: since $\rho_C(r) = r \cdot 1_C$ and $r \cdot 1_C \otimes 1_B = \rho_B(r) \otimes 1_B$ by the $R$-bilinearity of $\otimes$, one computes \begin{align*} 1_B \otimes \rho_C(r) = 1_B \otimes (r \cdot 1_C) = (r \cdot 1_B) \otimes 1_C = \rho_B(r) \otimes 1_C. \end{align*} So $\varphi = \psi$, and there is a single well-defined ring homomorphism $R \to B \otimes_R C$ making $B \otimes_R C$ into an $R$-algebra. Accompanying this $R$-algebra structure are two natural $R$-algebra homomorphisms \begin{align*} i_B: B \to B \otimes_R C, \quad b \mapsto b \otimes 1_C, \end{align*} \begin{align*} i_C: C \to B \otimes_R C, \quad c \mapsto 1_B \otimes c. \end{align*} These are ring homomorphisms: $i_B(bb') = bb' \otimes 1 = (b \otimes 1)(b' \otimes 1) = i_B(b) i_B(b')$, and similarly for $i_C$. They are also compatible with the $R$-algebra structure by construction. The images of $i_B$ and $i_C$ commute with each other inside $B \otimes_R C$: $(b \otimes 1)(1 \otimes c) = b \otimes c = (1 \otimes c)(b \otimes 1)$. ### The universal property The pair $(B \otimes_R C, i_B, i_C)$ is characterised by a universal property that mirrors the universal property of the direct sum for modules — except now in the category of $R$-algebras and with the roles of "initial" and "terminal" reversed compared to the module case. [quotetheorem:2915] [citeproof:2915] The universal property says precisely that $A \otimes_R B$ is the **coproduct** of $A$ and $B$ in the category of $R$-algebras. Part (1) is the defining universal property of the coproduct, and part (2) says that any two coproducts are uniquely isomorphic. [remark: Coproducts in Other Categories] In the category of sets, the coproduct is the disjoint union. In the category of groups, it is the free product $A * B$. In the category of $R$-modules, it is the direct sum $A \oplus B$. In each case, a map out of the coproduct is equivalent to specifying maps out of each factor individually. The tensor product plays this role for $R$-algebras. [/remark] ### Examples: tensor products of polynomial algebras The most important class of examples comes from polynomial algebras. These make the abstract structure completely explicit and connect to algebraic geometry. [example: Tensor Product of Polynomial Algebras] Let $R$ be a commutative ring. There is an $R$-algebra isomorphism \begin{align*} \varphi: R[X_1, \dots, X_n] \otimes_R R[T_1, \dots, T_r] \xrightarrow{\;\sim\;} R[X_1, \dots, X_n, T_1, \dots, T_r]. \end{align*} There are two ways to see this. **Direct construction.** Both sides are free $R$-modules with a basis indexed by monomials. On the left, the monomials $X_1^{a_1} \cdots X_n^{a_n} \otimes T_1^{b_1} \cdots T_r^{b_r}$ form an $R$-basis. On the right, the monomials $X_1^{a_1} \cdots X_n^{a_n} T_1^{b_1} \cdots T_r^{b_r}$ form an $R$-basis. The map $\varphi(f \otimes g) = fg$ sends one basis to the other, giving an $R$-module isomorphism. To check it is a ring homomorphism, one computes on general elements: for $\sum_i p_i \otimes q_i$ and $\sum_j h_j \otimes g_j$, \begin{align*} \varphi\!\left(\left(\sum_{i} p_i \otimes q_i\right)\left(\sum_{j} h_j \otimes g_j\right)\right) &= \varphi\!\left(\sum_{i,j} (p_i h_j) \otimes (q_i g_j)\right) \\ &= \sum_{i,j} p_i h_j q_i g_j \\ &= \left(\sum_i p_i q_i\right)\!\left(\sum_j h_j g_j\right) \\ &= \varphi\!\left(\sum_i p_i \otimes q_i\right) \varphi\!\left(\sum_j h_j \otimes g_j\right), \end{align*} where the third equality uses commutativity of $R[X_1, \dots, X_n, T_1, \dots, T_r]$. Since $\varphi(1 \otimes 1) = 1$, $\varphi$ is an $R$-algebra isomorphism. **Via the universal property.** Define $j_A: R[X_1, \dots, X_n] \to R[X_1, \dots, X_n, T_1, \dots, T_r]$ and $j_B: R[T_1, \dots, T_r] \to R[X_1, \dots, X_n, T_1, \dots, T_r]$ as the natural inclusions. To check that the RHS is the coproduct, take any $R$-algebra $C$ and $R$-algebra homomorphisms $f_1: R[X_1, \dots, X_n] \to C$ and $f_2: R[T_1, \dots, T_r] \to C$. A map $h: R[X_1, \dots, X_n, T_1, \dots, T_r] \to C$ with $h \circ j_A = f_1$ and $h \circ j_B = f_2$ must satisfy $h(X_i) = f_1(X_i)$ and $h(T_j) = f_2(T_j)$. The universal property of the polynomial ring provides a unique such $h$. By the uniqueness clause in the universal property of the tensor product (part (2)), this yields the desired isomorphism. This approach also recovers the formula $\varphi(p \otimes q) = pq$ on pure tensors from the universal property itself, without a separate computation. [/example] ### Quotients: tensor products of finitely presented algebras The polynomial algebra isomorphism has a useful extension to quotients. If $I \trianglelefteq R[X_1, \dots, X_n]$ and $J \trianglelefteq R[T_1, \dots, T_r]$ are ideals, one can compute the tensor product $R[X_1, \dots, X_n]/I \otimes_R R[T_1, \dots, T_r]/J$ explicitly. Using the quotient isomorphism for tensor products (which says that for an $R$-module $M$ and an ideal $I$, one has $M/IM \cong M \otimes_R R/I$, applied twice), one obtains an $R$-module isomorphism \begin{align*} R[X_1, \dots, X_n]/I \otimes_R R[T_1, \dots, T_r]/J \;\cong\; R[X_1, \dots, X_n, T_1, \dots, T_r]/(I^e + J^e), \end{align*} where $I^e$ denotes the extension of $I$ to $R[X_1, \dots, X_n, T_1, \dots, T_r]$ (the ideal generated by $I$ in the larger ring), and similarly for $J^e$. More precisely, under the isomorphism $\varphi$ from the polynomial algebra example, the submodule of $R[X_1, \dots, X_n] \otimes_R R[T_1, \dots, T_r]$ generated by \begin{align*} \{p \otimes q : p \in I,\, q \in R[T_1, \dots, T_r]\} \cup \{f \otimes h : f \in R[X_1, \dots, X_n],\, h \in J\} \end{align*} maps onto $I^e + J^e$. This is an isomorphism of $R$-algebras, not just $R$-modules. [example: Explicit Computation] As a concrete instance, one has an isomorphism of $\mathbb{C}$-algebras \begin{align*} \mathbb{C}[X, Y, Z]/(f, g) \otimes_{\mathbb{C}} \mathbb{C}[W, U]/(h) \;\cong\; \mathbb{C}[X, Y, Z, W, U]/(f, g, h), \end{align*} given by $\bigl(p + (f, g)\bigr) \otimes \bigl(q + (h)\bigr) \mapsto pq + (f, g, h)$. One checks this is well-defined: if $p' \equiv p \pmod{(f,g)}$ and $q' \equiv q \pmod{(h)}$, then $p'q' - pq = (p' - p)q' + p(q' - q)$, which lies in $(f,g,h)$ since $p' - p \in (f,g)$ and $q' - q \in (h)$. [/example] [remark: Connection to Algebraic Geometry] The calculation above is non-examinable but illuminating. In algebraic geometry over $\mathbb{C}$, the affine variety $V(f, g) \subset \mathbb{A}^3_\mathbb{C}$ has coordinate ring $\mathbb{C}[X,Y,Z]/(f,g)$, and $V(h) \subset \mathbb{A}^2_\mathbb{C}$ has coordinate ring $\mathbb{C}[W,U]/(h)$. The isomorphism $\mathbb{C}[X,Y,Z]/(f,g) \otimes_\mathbb{C} \mathbb{C}[W,U]/(h) \cong \mathbb{C}[X,Y,Z,W,U]/(f,g,h)$ says that the product variety $V(f,g) \times V(h) \subset \mathbb{A}^5_\mathbb{C}$ has coordinate ring equal to this tensor product. More generally, the tensor product of algebras is the algebraic mechanism that defines products of affine schemes over $\operatorname{Spec}(R)$. [/remark] ### Functoriality and further algebra isomorphisms The tensor product of algebras is functorial: if $f: A \to A'$ and $g: B \to B'$ are $R$-algebra homomorphisms, then $f \otimes g: A \otimes_R B \to A' \otimes_R B'$ is also an $R$-algebra homomorphism. Many of the $R$-module isomorphisms from Section 1 are in fact $R$-algebra isomorphisms. To verify this, it suffices in each case to check that $1 \mapsto 1$ (which the module isomorphism already gives, since it sends the identity tensor to the identity) and that the map respects multiplication — and this one checks on a set of $R$-module generators (i.e., on pure tensors). The following hold as $R$-algebra isomorphisms: - **Quotient ring tensor product:** $R/I \otimes_R R/J \cong R/(I + J)$, where the module isomorphism $r_1 + I) \otimes (r_2 + J) \mapsto r_1 r_2 + (I+J)$ respects multiplication since $(r_1 r_2)(r_3 r_4) = (r_1 r_3)(r_2 r_4)$. - **Commutativity:** $A \otimes_R B \cong B \otimes_R A$, via $a \otimes b \mapsto b \otimes a$. - **Distributivity over finite products:** $A \otimes_R (B \times C) \cong (A \otimes_R B) \times (A \otimes_R C)$, and in particular $A \otimes_R B^n \cong (A \otimes_R B)^n$. - **Associativity:** $(A \otimes_R B) \otimes_R C \cong A \otimes_R (B \otimes_R C)$. Each of these is an $R$-algebra isomorphism, not merely an $R$-module isomorphism. In the next section, these isomorphisms will be applied in the context of restriction and extension of scalars. ## Restriction and Extension of Scalars A ring homomorphism $f: R \to S$ creates a bridge between the module categories of $R$ and $S$. One can travel in either direction: from $S$-modules to $R$-modules (restriction of scalars), or from $R$-modules to $S$-modules (extension of scalars). The tensor product is the key tool for the latter direction, and this section makes the construction precise and explores its properties. ### Restriction of Scalars Given a ring homomorphism $f: R \to S$, any $S$-module automatically carries an $R$-module structure. The idea is straightforward: if $\rho: S \to \operatorname{End}(M)$ is the structural homomorphism of $M$ as an $S$-module, then the composite $\rho \circ f: R \to \operatorname{End}(M)$ makes $M$ an $R$-module. Concretely, the action is $r \cdot m := f(r)m$ for all $r \in R$, $m \in M$. [definition: Restriction of Scalars] Let $f: R \to S$ be a ring homomorphism and let $M$ be an $S$-module. The **restriction of scalars** of $M$ along $f$ is the $R$-module whose underlying abelian group is $M$ and whose $R$-action is \begin{align*} r \cdot m := f(r)m \quad \text{for all } r \in R,\; m \in M. \end{align*} [/definition] The canonical example illustrates the dimension-doubling phenomenon. For the inclusion $f: \mathbb{R} \hookrightarrow \mathbb{C}$ and the $\mathbb{C}$-module $\mathbb{C}^n$, restricting scalars to $\mathbb{R}$ yields $\mathbb{R}^{2n}$: each copy of $\mathbb{C}$ becomes a two-dimensional real vector space spanned by $1$ and $i$, so $n$ complex dimensions become $2n$ real dimensions. ### Extension of Scalars Restriction goes from $S$-modules to $R$-modules by forgetting structure. Extension of scalars goes the other way, using the tensor product to construct an $S$-module from an $R$-module. The idea is that $S$ itself is an $R$-module via restriction of scalars, and tensoring an $R$-module $N$ with $S$ over $R$ produces a new module that $S$ can act on. [definition: Extension of Scalars] Let $f: R \to S$ be a ring homomorphism and let $N$ be an $R$-module. The **extension of scalars** of $N$ along $f$ is the tensor product $S \otimes_R N$, equipped with the $S$-module structure \begin{align*} s' \cdot (s \otimes n) := (s's) \otimes n \quad \text{for all } s, s' \in S,\; n \in N. \end{align*} [/definition] More generally, if $M$ is any $S$-module (not just $S$ itself), one can form the tensor product $M \otimes_R N$ and give it an $S$-module structure. The action on pure tensors is \begin{align*} s(m \otimes n) := (sm) \otimes n \quad \text{for all } s \in S,\; m \in M,\; n \in N. \end{align*} To verify this is well-defined, fix $s \in S$ and consider the map $M \times N \to M \otimes_R N$ given by $(m, n) \mapsto (sm) \otimes n$. This is $R$-bilinear (the $R$-linearity in $m$ uses the fact that $s$ acts $R$-linearly on $M$, since $M$ is an $S$-module and $R$ acts via $f$). By the universal property of $M \otimes_R N$, it induces an $R$-module endomorphism $h_s: M \otimes_R N \to M \otimes_R N$ satisfying $h_s(m \otimes n) = (sm) \otimes n$. The assignment $s \mapsto h_s$ defines a map $\varphi: S \to \operatorname{End}(M \otimes_R N)$; checking on pure tensors shows $\varphi$ is a ring homomorphism, so $M \otimes_R N$ becomes an $S$-module. [example: Basic Extension of Scalars Computations] Fix the inclusion $\mathbb{R} \hookrightarrow \mathbb{C}$. **$S \otimes_R R \cong S$.** The $R$-module isomorphism $S \otimes_R R \xrightarrow{\sim} S$ given by $s \otimes r \mapsto sr$ is also an $S$-module isomorphism. To check: the $S$-action on the left sends $s'(s \otimes r) = (s's) \otimes r$, which maps to $(s's)r = s'(sr)$, matching the $S$-action on the right. In particular, $\mathbb{C} \otimes_\mathbb{R} \mathbb{R} \cong \mathbb{C}$ as $\mathbb{C}$-modules. **Extension commutes with direct sums.** For an $S$-module $M$ and a family of $R$-modules $(N_i)_{i \in I}$, the $R$-module isomorphism \begin{align*} M \otimes_R \bigoplus_{i \in I} N_i \xrightarrow{\sim} \bigoplus_{i \in I} (M \otimes_R N_i) \end{align*} is also an $S$-module isomorphism. In particular, $\mathbb{C} \otimes_\mathbb{R} \mathbb{R}^n \cong \mathbb{C}^n$ as $\mathbb{C}$-modules. **Restricting and then extending.** Take the $\mathbb{C}$-module $\mathbb{C}^n$. Restrict scalars to $\mathbb{R}$ to obtain $\mathbb{R}^{2n}$. Extend scalars back to $\mathbb{C}$ to obtain $\mathbb{C} \otimes_\mathbb{R} \mathbb{R}^{2n} \cong \mathbb{C}^{2n}$. The round trip doubles the dimension. **Extending and then restricting.** Take the $\mathbb{R}$-module $\mathbb{R}^n$. Extend scalars to $\mathbb{C}$ to obtain $\mathbb{C} \otimes_\mathbb{R} \mathbb{R}^n \cong \mathbb{C}^n$. Restrict scalars back to $\mathbb{R}$ to obtain $\mathbb{R}^{2n}$. Again the dimension doubles. **Extension over a quotient ring.** Take the $\mathbb{Z}$-module $\mathbb{Z}^m$ and extend scalars along $\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ to obtain $(\mathbb{Z}/n\mathbb{Z}) \otimes_\mathbb{Z} \mathbb{Z}^m \cong (\mathbb{Z}/n\mathbb{Z})^m$. [/example] ### The Key Proposition: Replacing the Base Ring The examples above all use $M = S$. When $M$ is a general $S$-module, the computation is less direct. A fundamental proposition reduces such computations to the $M = S$ case: to tensor an $R$-module $N$ against an $S$-module $M$ over $R$, one can first extend $N$ to an $S$-module via $S \otimes_R N$, and then tensor over $S$. To see why this is useful concretely, consider $\mathbb{C}^n \otimes_\mathbb{R} \mathbb{R}^\ell$. As an $\mathbb{R}$-module this is isomorphic to $\mathbb{C}^{n\ell}$, but the $\mathbb{C}$-module structure is not immediately transparent. The proposition says: \begin{align*} \mathbb{C}^n \otimes_\mathbb{R} \mathbb{R}^\ell \cong \mathbb{C}^n \otimes_\mathbb{C} (\mathbb{C} \otimes_\mathbb{R} \mathbb{R}^\ell) \cong \mathbb{C}^n \otimes_\mathbb{C} \mathbb{C}^\ell \cong \mathbb{C}^{n\ell} \end{align*} as $\mathbb{C}$-modules, where the last step follows because $\mathbb{C}^n \otimes_\mathbb{C} \mathbb{C}^\ell$ is a free $\mathbb{C}$-module of rank $n\ell$. [quotetheorem:2916] [citeproof:2916] ### Further Isomorphisms for Extension of Scalars With the base-change proposition in hand, one can derive a collection of useful $S$-module isomorphisms. The next theorem records the most important ones. [quotetheorem:2917] [citeproof:2917] [remark: Note on Isomorphism (2)] In isomorphism (2), the left-hand side $(M \otimes_R N) \otimes_R N'$ is the result of two separate extensions of scalars (first $N$ extends to an $S$-module via $M$, then $N'$ extends via the result), while the right-hand side $M \otimes_R (N \otimes_R N')$ involves only one extension (the $R$-module $N \otimes_R N'$ is extended by $M$). [/remark] A particularly clean consequence describes how extending scalars interacts with tensor products of $R$-modules. When $N$ and $N'$ are both $R$-modules, extending $N \otimes_R N'$ to an $S$-module is the same as first extending each factor separately and then tensoring over $S$. [quotetheorem:2918] [citeproof:2918] [example: Complexification of a Real Tensor Product] As $\mathbb{C}$-modules: \begin{align*} \mathbb{C} \otimes_\mathbb{R} (\mathbb{R}^n \otimes_\mathbb{R} \mathbb{R}^\ell) &\cong (\mathbb{C} \otimes_\mathbb{R} \mathbb{R}^n) \otimes_\mathbb{C} (\mathbb{C} \otimes_\mathbb{R} \mathbb{R}^\ell) \\ &\cong \mathbb{C}^n \otimes_\mathbb{C} \mathbb{C}^\ell \\ &\cong \mathbb{C}^{n\ell}. \end{align*} This is consistent with the direct calculation: $\mathbb{R}^n \otimes_\mathbb{R} \mathbb{R}^\ell \cong \mathbb{R}^{n\ell}$ as $\mathbb{R}$-modules, so $\mathbb{C} \otimes_\mathbb{R} \mathbb{R}^{n\ell} \cong \mathbb{C}^{n\ell}$ as $\mathbb{C}$-modules. [/example] [remark: The Corollary Does Not Hold for General $S$-modules] The isomorphism $S \otimes_R (N \otimes_R N') \cong (S \otimes_R N) \otimes_S (S \otimes_R N')$ is specific to the case where the $S$-module on the left is $S$ itself. For a general $S$-module $M$ in place of $S$, no analogous isomorphism holds. [/remark] Applying the corollary repeatedly by induction gives a more general statement: for $R$-modules $N_1, \ldots, N_\ell$, there is an $S$-module isomorphism \begin{align*} S \otimes_R (N_1 \otimes_R \cdots \otimes_R N_\ell) \cong (S \otimes_R N_1) \otimes_S \cdots \otimes_S (S \otimes_R N_\ell). \end{align*} ### Extension of Scalars Acts on Morphisms Extension of scalars is not merely an operation on modules — it is functorial. Given an $R$-module homomorphism $f: N \to N'$ and an $S$-module $M$, the map $\operatorname{id}_M \otimes f: M \otimes_R N \to M \otimes_R N'$ is an $S$-module homomorphism (not just an $R$-module homomorphism). This follows from the $S$-linearity calculation on pure tensors: \begin{align*} (\operatorname{id}_M \otimes f)(s(m \otimes n)) = (\operatorname{id}_M \otimes f)((sm) \otimes n) = (sm) \otimes f(n) = s(m \otimes f(n)) = s(\operatorname{id}_M \otimes f)(m \otimes n). \end{align*} [example: Complexification of a Real Linear Map] Let $T: \mathbb{R}^n \to \mathbb{R}^m$ be a linear map. Tensor with $\operatorname{id}_\mathbb{C}: \mathbb{C} \to \mathbb{C}$ to obtain $\operatorname{id}_\mathbb{C} \otimes T: \mathbb{C} \otimes_\mathbb{R} \mathbb{R}^n \to \mathbb{C} \otimes_\mathbb{R} \mathbb{R}^m$, which is a $\mathbb{C}$-module homomorphism. Under the identifications $\mathbb{C} \otimes_\mathbb{R} \mathbb{R}^n \cong \mathbb{C}^n$ and $\mathbb{C} \otimes_\mathbb{R} \mathbb{R}^m \cong \mathbb{C}^m$, the $\mathbb{C}$-bases are $1 \otimes e_1, \ldots, 1 \otimes e_n$ and $1 \otimes f_1, \ldots, 1 \otimes f_m$ where $e_1, \ldots, e_n$ and $f_1, \ldots, f_m$ are the standard $\mathbb{R}$-bases. Writing $[T]$ for the matrix of $T$ in these bases, we compute: \begin{align*} (\operatorname{id}_\mathbb{C} \otimes T)(1 \otimes e_i) &= 1 \otimes Te_i = 1 \otimes \sum_{\ell=1}^m [T]_{\ell i} f_\ell = \sum_{\ell=1}^m [T]_{\ell i} (1 \otimes f_\ell). \end{align*} The matrix of $\operatorname{id}_\mathbb{C} \otimes T$ in the $\mathbb{C}$-bases is exactly $[T]$, now viewed as a matrix with complex entries (all of which happen to be real). Complexification of a real linear map does not change the matrix representation — it only reinterprets the scalars. [/example] ### Extension of Scalars for Algebras When $R$, $S$, $A$, $B$ are rings and algebra structures are present, extension of scalars upgrades from $S$-module isomorphisms to $S$-algebra isomorphisms. Recall from the previous section that $A \otimes_R B$ is simultaneously an $A$-algebra (via $a \mapsto a \otimes 1$) and a $B$-algebra (via $b \mapsto 1 \otimes b$), and its $R$-algebra structure is obtained by restricting scalars along either of these algebra maps. The most important example is the complexification of polynomial algebras. [quotetheorem:2835] [citeproof:2835] The same argument gives $S \otimes_R R[T_1, \ldots, T_n] \cong S[T_1, \ldots, T_n]$ as $S$-algebras for any ring homomorphism $f: R \to S$. More generally, if $I$ is an ideal of $R[T_1, \ldots, T_n]$, then there is an $S$-algebra isomorphism \begin{align*} S \otimes_R (R[T_1, \ldots, T_n]/I) \cong S[T_1, \ldots, T_n]/I^e, \end{align*} where $I^e$ is the ideal of $S[T_1, \ldots, T_n]$ generated by the image of $I$ under the map $R[T_1, \ldots, T_n] \to S[T_1, \ldots, T_n]$ induced by $f$. The module-level results from earlier in this section have algebra-level analogues. For an $R$-algebra $A$ and an $S$-algebra $B$ (both connected to $R$ via $f: R \to S$), the tensor product $A \otimes_R B$ acquires an $S$-algebra structure via $B$, and there is an $S$-algebra isomorphism \begin{align*} A \otimes_R B \cong (A \otimes_R S) \otimes_S B. \end{align*} For $R$-algebras $A$ and $B$, there is also an $S$-algebra isomorphism \begin{align*} S \otimes_R (A \otimes_R B) \cong (S \otimes_R A) \otimes_S (S \otimes_R B), \end{align*} sending $s \otimes (a \otimes b) \mapsto s((1 \otimes a) \otimes (1 \otimes b))$. Both isomorphisms are verified by checking multiplicativity on pure tensors and using the corresponding module isomorphisms. ## Exactness Properties of the Tensor Product The tensor product construction is functorial: fixing an $R$-module $M$, the assignment $N \mapsto M \otimes_R N$ extends to a functor $T_M$ on the category of $R$-modules, sending a homomorphism $f: N \to N'$ to $\mathrm{id}_M \otimes f: M \otimes_R N \to M \otimes_R N'$. The central question of this section is: how does $T_M$ interact with exact sequences? The answer is asymmetric and consequential — $T_M$ preserves exactness on the right but not necessarily on the left. ### The Hom Functors and Left Exactness Before addressing $T_M$ directly, it is useful to understand the behaviour of the $\operatorname{Hom}$ functors, which will serve as a key technical tool. [definition: Module of Homomorphisms] For $R$-modules $Q$ and $P$, let $\operatorname{Hom}_R(Q, P)$ denote the set of $R$-module homomorphisms $Q \to P$. This set carries the structure of an $R$-module: addition is pointwise, and the $R$-action is defined by \begin{align*} (rf)(x) = r \cdot f(x) \qquad \forall\, r \in R,\ f \in \operatorname{Hom}_R(Q, P),\ x \in Q. \end{align*} [/definition] Fixing $Q$, the assignment $P \mapsto \operatorname{Hom}_R(Q, P)$ defines a covariant functor: a homomorphism $f: M \to N$ induces $\operatorname{Hom}_R(Q, f): \operatorname{Hom}_R(Q, M) \to \operatorname{Hom}_R(Q, N)$ by post-composition, $\varphi \mapsto f_*(\varphi) := f \circ \varphi$. Fixing $P$, the assignment $Q \mapsto \operatorname{Hom}_R(Q, P)$ defines a contravariant functor: the same homomorphism $f: M \to N$ induces $\operatorname{Hom}_R(f, P): \operatorname{Hom}_R(N, P) \to \operatorname{Hom}_R(M, P)$ by pre-composition, $\varphi \mapsto f^*(\varphi) := \varphi \circ f$. The reversal of direction is why this functor is called contravariant. [quotetheorem:2836] The theorem says that both Hom functors are left exact: they preserve a zero on the left but not, in general, a zero on the right (that would require injectivity or surjectivity conditions on the module $Q$ or $P$). A key lemma lets us detect exactness of a sequence by testing it against $\operatorname{Hom}(\cdot, P)$ for all $P$: [quotetheorem:2919] [citeproof:2919] ### Tensor-Hom Adjunction There is a fundamental relationship between $T_M$ and $\operatorname{Hom}_R(M, \cdot)$. Recall from the universal property of the tensor product that there is a natural bijection \begin{align*} \operatorname{Hom}_R(M \otimes_R N, L) \xrightarrow{\ \sim\ } \mathrm{Bilin}_R(M \times N, L). \end{align*} A second natural bijection sends bilinear maps to homomorphisms by currying: \begin{align*} \mathrm{Bilin}_R(M \times N, L) \xrightarrow{\ \sim\ } \operatorname{Hom}_R\!\left(N, \operatorname{Hom}_R(M, L)\right), \qquad b \mapsto \bigl(n \mapsto (m \mapsto b(m, n))\bigr). \end{align*} Composing these two bijections gives the **tensor-hom adjunction**: \begin{align*} \operatorname{Hom}_R(M \otimes_R N, L) \xrightarrow{\ \sim\ } \operatorname{Hom}_R\!\left(N, \operatorname{Hom}_R(M, L)\right). \end{align*} [remark: Categorical Interpretation] In categorical language, this natural bijection in both $N$ and $L$ says that $T_M(\cdot) = M \otimes_R (\cdot)$ and $\operatorname{Hom}_R(M, \cdot)$ form an adjoint pair. The tensor functor $T_M$ is the left adjoint and $\operatorname{Hom}_R(M, \cdot)$ is the right adjoint. A general categorical principle then accounts for the exactness properties we are about to prove: left adjoint functors are right exact, and right adjoint functors are left exact. [/remark] ### Right Exactness of the Tensor Product We can now prove the main exactness result. [quotetheorem:2838] [citeproof:2838] [remark: Left Adjoints Are Cocontinuous] A more general principle underlies this result: a left adjoint functor commutes with colimits (it is "cocontinuous"), and cokernels are colimits. Right exactness is therefore a formal consequence of being a left adjoint. An example of the same phenomenon: $M \otimes_R (A \oplus B) \cong (M \otimes_R A) \oplus (M \otimes_R B)$, since the direct sum is the coproduct (a colimit) in the category of $R$-modules. [/remark] ### Tensor Products Can Fail to Be Exact The right-exactness theorem requires that the sequence ends with $\to 0$. Without this condition, tensoring can destroy exactness. [example: Tensoring Kills Injectivity] Consider the exact sequence of $\mathbb{Z}$-modules \begin{align*} 0 \to \mathbb{Z} \xrightarrow{\ x \mapsto 2x\ } \mathbb{Z}. \end{align*} The map $x \mapsto 2x$ is injective, so this sequence is exact at $\mathbb{Z}$. Now tensor with $\mathbb{Z}/2\mathbb{Z}$. Using the isomorphism $\mathbb{Z}/2\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z}$ (sending $[a] \otimes n \mapsto [an]$), the tensored sequence becomes \begin{align*} 0 \to \mathbb{Z}/2\mathbb{Z} \xrightarrow{\ [x] \mapsto [2x]\ } \mathbb{Z}/2\mathbb{Z}. \end{align*} But $[2x] = [0]$ for every $[x] \in \mathbb{Z}/2\mathbb{Z}$, so the map $[x] \mapsto [2x]$ is the zero map. This is not injective, so the tensored sequence is not exact. [/example] This example is an instance of a morphism of exact sequences. The original sequence and the tensored sequence fit into a commutative diagram connected by vertical maps (the identifications $\mathbb{Z}/2\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z}$ and $\mathrm{id} \otimes (x \mapsto 2x)$ becoming $[x] \mapsto [2x]$). A **morphism of sequences** is a collection of module homomorphisms making every square commute; an **isomorphism of sequences** is a morphism where all vertical maps are isomorphisms. This notion of isomorphism is the correct one for identifying two sequences as essentially the same. ### Flat Modules The failure of left-exactness in the example above raises a natural question: for which modules $M$ does $T_M$ preserve all exact sequences? [definition: Flat Module] An $R$-module $M$ is **flat** if for every injective $R$-module homomorphism $f: N \to N'$, the induced map \begin{align*} \mathrm{id}_M \otimes f: M \otimes_R N \to M \otimes_R N' \end{align*} is also injective. [/definition] The definition says precisely that $T_M$ preserves injections — the only thing right-exactness does not already guarantee. [example: Flat and Non-Flat Modules] 1. **$\mathbb{Z}/2\mathbb{Z}$ is not a flat $\mathbb{Z}$-module.** The example above shows this directly: the injection $\mathbb{Z} \xrightarrow{x \mapsto 2x} \mathbb{Z}$ becomes the zero map after tensoring with $\mathbb{Z}/2\mathbb{Z}$. 2. **Free modules are flat.** If $f: M \to M'$ is an injective $R$-module homomorphism and $I$ is any set, then $\mathrm{id}_{R^{\oplus I}} \otimes f$ corresponds, under the isomorphism $R^{\oplus I} \otimes_R M \cong M^{\oplus I}$, to the map $(m_i)_{i \in I} \mapsto (f(m_i))_{i \in I}$. This is injective because $f$ is injective on each component. 3. **The base ring matters.** The module $\mathbb{Z}/2\mathbb{Z}$ is flat as a $\mathbb{Z}/2\mathbb{Z}$-module (since $\mathbb{Z}/2\mathbb{Z}$ is free over itself), but not as a $\mathbb{Z}$-module. Flatness is a property of the module relative to its base ring. 4. **Flat modules are torsion-free.** Recall that an $R$-module $M$ is **torsion-free** if $rm \neq 0$ whenever $r \in R$ is not a zero divisor and $m \neq 0$. If $M$ is not torsion-free, there exist $r_0 \in R$ (not a zero divisor) and $0 \neq m_0 \in M$ with $r_0 m_0 = 0$. The multiplication map $\mu_{r_0}: R \to R$, $r \mapsto r_0 r$, is injective because $r_0$ is not a zero divisor. However, $\mathrm{id}_M \otimes \mu_{r_0}: M \otimes_R R \to M \otimes_R R$ sends $m_0 \otimes 1 \mapsto m_0 \otimes r_0 = r_0 m_0 \otimes 1 = 0$. Under the isomorphism $M \otimes_R R \cong M$ (sending $m \otimes r \mapsto rm$), the element $m_0 \otimes 1$ maps to $m_0 \neq 0$, so $m_0 \otimes 1 \neq 0$ in $M \otimes_R R$. Thus $\mathrm{id}_M \otimes \mu_{r_0}$ is not injective, and $M$ is not flat. 5. **Proper quotients of integral domains are not flat.** If $R$ is an integral domain and $(0) \subsetneq I \subsetneq R$ is a proper nonzero ideal, then $R/I$ is not a flat $R$-module: pick any nonzero $r \in I$; then $r$ is not a zero divisor (since $R$ is a domain), yet $r \cdot \bar{1} = 0$ in $R/I$, so $R/I$ is not torsion-free. [/example] ### Characterization of Flat Modules [quotetheorem:2839] [citeproof:2839] [remark: Torsion-Free vs Flat] Flatness implies torsion-freeness but the converse fails in general. Over a principal ideal domain, however, the two notions coincide for finitely generated modules: a finitely generated module over a PID is flat if and only if it is free (and torsion-free modules over a PID are free by the structure theorem). [/remark] ### Extension of Scalars Preserves Flatness [quotetheorem:2920] [citeproof:2920] The non-left-exactness of $T_M$ for modules that are not flat is not merely a deficiency — it opens the door to homological algebra. The failure of $T_M$ to be exact is measured by the derived functors $\operatorname{Tor}_n^R(M, N)$, which form the starting point for a systematic study of resolutions and homological invariants of modules and rings. <!-- illustration-needed: a commutative diagram for the extension of scalars flatness proof — show (S ⊗_R M) ⊗_S N on the top left, M ⊗_R N on the bottom left, with the corresponding terms on the right, connected by horizontal maps (tensored with g) and vertical isomorphisms, to make the diagram-chasing argument visible --> ## Further Examples of Tensor Products The abstract machinery of tensor products — universal properties, extension of scalars, flatness — is best understood through concrete calculations. This final section collects a range of illuminating examples that test the theory and reveal subtleties not visible from the definitions alone. The examples range from torsion and divisible groups, through fraction fields and ideals in number rings, to the complex calculation of $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$. ### Divisible and Torsion Groups The first family of examples concerns the interaction between torsion and divisibility when tensoring over $\mathbb{Z}$. [example: Q Tensor Z/n is Zero] For any pure tensor $x \otimes y \in \mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}/n\mathbb{Z}$, write $x = n \cdot \frac{x}{n}$ to move the factor of $n$ across the tensor: \begin{align*} x \otimes y = \left(n \cdot \frac{x}{n}\right) \otimes y = \frac{x}{n} \otimes ny = \frac{x}{n} \otimes 0 = 0. \end{align*} Since every element of $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}/n\mathbb{Z}$ is a finite sum of such pure tensors, the entire module is zero: $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}/n\mathbb{Z} = 0$. The two properties of $\mathbb{Q}$ and $\mathbb{Z}/n\mathbb{Z}$ used here are: - $\mathbb{Q}$ is a **divisible** abelian group: for every $n \geq 1$ and every $a \in \mathbb{Q}$, there exists $a' \in \mathbb{Q}$ with $na' = a$. - $\mathbb{Z}/n\mathbb{Z}$ is a **torsion** abelian group: every element has finite order. The same argument shows $A \otimes_{\mathbb{Z}} B = 0$ whenever $A$ is a divisible abelian group and $B$ is an abelian torsion group. Since $\mathbb{Q}/\mathbb{Z}$ is both torsion and divisible (every element $[a/b]$ has finite order $b$, and for any $n$ and $[a/b] \in \mathbb{Q}/\mathbb{Z}$, the element $[a/(nb)]$ satisfies $n \cdot [a/(nb)] = [a/b]$), one obtains \begin{align*} (\mathbb{Q}/\mathbb{Z})^{\otimes 2} = (\mathbb{Q}/\mathbb{Z}) \otimes_{\mathbb{Z}} (\mathbb{Q}/\mathbb{Z}) = 0. \end{align*} [/example] This example illustrates a general principle: torsion phenomena in one factor can annihilate the entire tensor product. The following proposition, shows the opposite extreme: nonzero finitely generated modules cannot be killed by iterated tensoring over a commutative ring. [quotetheorem:2841] ### Multiple Tensor Products Before continuing with further examples, it is worth noting that tensor products generalise to arbitrary finite collections of modules. Given $R$-modules $M_1, \ldots, M_n$, one forms $M_1 \otimes_R M_2 \otimes_R \cdots \otimes_R M_n$ using the same construction as for $n = 2$. The natural isomorphisms \begin{align*} M_1 \otimes_R (M_2 \otimes_R M_3) \cong M_1 \otimes_R M_2 \otimes_R M_3 \cong (M_1 \otimes_R M_2) \otimes_R M_3 \end{align*} show that parenthesisation does not matter. The universal property extends: $M_1 \otimes_R \cdots \otimes_R M_n$ is the target of a universal $R$-multilinear map from $M_1 \times \cdots \times M_n$, in the sense that every $R$-multilinear map $M_1 \times \cdots \times M_n \to L$ factors uniquely through it. ### Extension of Scalars from $\mathbb{Z}$ to $\mathbb{Q}$ The next example shows that extending scalars from an integral domain to its fraction field recovers a module in a very clean sense. [example: Q Tensor V over Z vs Q] Let $V$ be a $\mathbb{Q}$-vector space. Working over $\mathbb{Q}$, the isomorphism $\mathbb{Q} \otimes_{\mathbb{Q}} V \cong V$ is clear: every pure tensor reduces as \begin{align*} \sum_i x_i \otimes v_i = \sum_i 1 \otimes x_i v_i = 1 \otimes \sum_i x_i v_i, \end{align*} so every element is a pure tensor of the form $1 \otimes v$, and the map $x \otimes v \mapsto xv$ is an isomorphism. Now consider $\mathbb{Q} \otimes_{\mathbb{Z}} V$, where $V$ is first restricted to a $\mathbb{Z}$-module (by forgetting the $\mathbb{Q}$-module structure) and then the scalars are extended back to $\mathbb{Q}$. Moving rational scalars across the tensor sign now requires only $\mathbb{Z}$-linearity, which means one can move integer multiples but must handle the denominators separately. Writing $a_i, b_i \in \mathbb{Z}$ with $b_i \neq 0$: \begin{align*} \sum_i \frac{a_i}{b_i} \otimes v_i &= \sum_i \frac{1}{b_i} \otimes a_i v_i \\ &= \sum_i \frac{1}{b_i} \otimes \frac{a_i b_i}{b_i} v_i \\ &= \sum_i 1 \otimes \frac{a_i}{b_i} v_i \\ &= 1 \otimes \sum_i \frac{a_i}{b_i} v_i. \end{align*} Here the third step uses the relation $\frac{1}{b_i} \otimes b_i w = 1 \otimes w$ in $\mathbb{Q} \otimes_{\mathbb{Z}} V$ (valid because $b_i \cdot \frac{1}{b_i} = 1$). This shows every element of $\mathbb{Q} \otimes_{\mathbb{Z}} V$ is a pure tensor $1 \otimes v$ for some $v \in V$. The $\mathbb{Z}$-bilinear map $\mathbb{Q} \times V \to V$, $(x, v) \mapsto xv$, induces a $\mathbb{Z}$-module homomorphism $\varphi: \mathbb{Q} \otimes_{\mathbb{Z}} V \to V$ with $\varphi(x \otimes v) = xv$. This map is surjective (since $v = \varphi(1 \otimes v)$). For injectivity: a nonzero pure tensor $x \otimes v$ with $x \neq 0$ and $v \neq 0$ maps to $xv \neq 0$. Since every element is pure, $\varphi$ is injective. Thus $\varphi$ is an isomorphism of $\mathbb{Z}$-modules (and also of $\mathbb{Q}$-modules). [/example] [remark: Injectivity on Pure Tensors Does Not Imply Injectivity] The example above uses the fact that every element of $\mathbb{Q} \otimes_{\mathbb{Z}} V$ is pure. In general, a $R$-module homomorphism $f: M \otimes_R N \to L$ that is injective on pure tensors need not be injective. Indeed, the argument in the example depends crucially on this special purity property of $\mathbb{Q} \otimes_{\mathbb{Z}} V$, and fails without it. [/remark] The argument generalises directly: if $R$ is an integral domain, $K = \operatorname{Frac}(R)$, and $V$ is a $K$-module (viewed as an $R$-module by restriction of scalars), the same calculation gives $K \otimes_R V \cong V$ as $K$-modules. A further generalisation is: [quotetheorem:2921] [citeproof:2921] ### Ideals in $\mathbb{Z}[\sqrt{-5}]$ The above theorem has a striking application to the ring $R = \mathbb{Z}[\sqrt{-5}]$, which provides a classic example of a Dedekind domain that is not a PID. [example: Nonprincipal Ideal in Z[sqrt(-5)] Still Tensors to Fraction Field] Let $R = \mathbb{Z}[\sqrt{-5}]$, so $K = \operatorname{Frac}(R) = \mathbb{Q}(\sqrt{-5})$. **Principal ideals.** If $I = (x)$ is a principal ideal of $R$ with $x \neq 0$, then the map $R \to I$, $r \mapsto rx$, is a surjective $R$-module homomorphism, and injective because $R$ is an integral domain. So $I \cong R$ as $R$-modules, and \begin{align*} I \otimes_R K \cong R \otimes_R K \cong K. \end{align*} **Nonprincipal ideals.** The ideal $I = (2, 1 + \sqrt{-5})$ is nonprincipal in $\mathbb{Z}[\sqrt{-5}]$: the ring $R$ is generated as an $R$-module by the single element $1_R$, while $I$ cannot be generated by a single element (one can verify this using the norm $N(a + b\sqrt{-5}) = a^2 + 5b^2$; any generator $g$ would need $N(g) \mid N(2) = 4$ and $N(g) \mid N(1 + \sqrt{-5}) = 6$, so $N(g) \mid \gcd(4,6) = 2$; but $a^2 + 5b^2 = 2$ has no integer solutions). Thus $R \not\cong I$ as $R$-modules. However, $I$ is a nonzero $R$-submodule of $K = \mathbb{Q}(\sqrt{-5})$, so the theorem applies directly: \begin{align*} (2, 1 + \sqrt{-5}) \otimes_{\mathbb{Z}[\sqrt{-5}]} \mathbb{Q}(\sqrt{-5}) \cong \mathbb{Q}(\sqrt{-5}). \end{align*} The nonprincipal ideal $I$ and the free module $R$ are not isomorphic as $R$-modules, but they become isomorphic after tensoring with the fraction field. [/example] This phenomenon — non-isomorphic modules becoming isomorphic after base change to the fraction field — reflects the fact that tensoring with $K$ "inverts" all nonzero elements of $R$ and in doing so collapses the distinction between $R$ and any nonzero fractional ideal. ### Tensor Products and Direct Products The tensor product distributes over direct sums: for any $R$-module $M$ and any family $(M_i)_{i \in I}$ of $R$-modules, there is a natural isomorphism \begin{align*} M \otimes_R \bigoplus_{i \in I} M_i \xrightarrow{\;\sim\;} \bigoplus_{i \in I} (M \otimes_R M_i), \quad m \otimes (m_i)_{i \in I} \mapsto (m \otimes m_i)_{i \in I}. \end{align*} The analogous map for direct products, $M \otimes_R \prod_{i \in I} M_i \to \prod_{i \in I}(M \otimes_R M_i)$, exists but is generally not an isomorphism. [example: Tensor Does Not Commute with Infinite Products] Consider $M = \mathbb{Q}$ and the family $\mathbb{Z}/p^n\mathbb{Z}$ for $n \geq 1$, tensored over $\mathbb{Z}$. For the direct product side: $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}/p^n\mathbb{Z} = 0$ for each $n$ (since $\mathbb{Z}/p^n\mathbb{Z}$ is torsion and $\mathbb{Q}$ is divisible), so \begin{align*} \prod_{n \geq 1} \left(\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}/p^n\mathbb{Z}\right) = \prod_{n \geq 1} 0 = 0. \end{align*} For the tensor side: $\mathbb{Q} \otimes_{\mathbb{Z}} \prod_{n \geq 1} \mathbb{Z}/p^n\mathbb{Z}$ is nonzero. To see why, let $x = (1, 1, 1, \ldots) \in \prod_{n \geq 1} \mathbb{Z}/p^n\mathbb{Z}$. The element $x$ has infinite order in this product, since no nonzero multiple of $x$ is zero in every coordinate simultaneously (the $n$-th coordinate of $kx$ is $k \bmod p^n$, which is zero only when $p^n \mid k$, so no single $k \neq 0$ works for all $n$). Let $\langle x \rangle$ be the cyclic subgroup generated by $x$; since $x$ has infinite order, $\langle x \rangle \cong \mathbb{Z}$ as abelian groups, so \begin{align*} \mathbb{Q} \otimes_{\mathbb{Z}} \langle x \rangle \cong \mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z} \cong \mathbb{Q} \neq 0. \end{align*} Since $\mathbb{Q}$ is a flat $\mathbb{Z}$-module, tensoring the inclusion $\langle x \rangle \hookrightarrow \prod_{n \geq 1} \mathbb{Z}/p^n\mathbb{Z}$ with $\mathbb{Q}$ yields an injection \begin{align*} \mathbb{Q} \otimes_{\mathbb{Z}} \langle x \rangle \hookrightarrow \mathbb{Q} \otimes_{\mathbb{Z}} \prod_{n \geq 1} \mathbb{Z}/p^n\mathbb{Z}. \end{align*} The left side is nonzero, so the right side is nonzero as well. The two sides of the (non-)isomorphism $\mathbb{Q} \otimes_{\mathbb{Z}} \prod_{n \geq 1} \mathbb{Z}/p^n\mathbb{Z} \not\cong \prod_{n \geq 1}\left(\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}/p^n\mathbb{Z}\right)$ are therefore $\mathbb{Q} \otimes_{\mathbb{Z}} \prod_{n \geq 1} \mathbb{Z}/p^n\mathbb{Z} \neq 0$ on the left and $0$ on the right. [/example] ### The Complex Tensor Product $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ The most intricate example of the section is the computation of $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$. It combines extension of scalars, the Chinese Remainder Theorem for algebras, and a careful analysis of how the $\mathbb{C}$-module structure depends on which copy of $\mathbb{C}$ is used to scale. [example: C Tensor R C] **Step 1: Dimension as a $\mathbb{C}$-module.** Viewing the right copy of $\mathbb{C}$ as an $\mathbb{R}$-vector space with basis $\{1, i\}$, the tensor product $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ is an extension of scalars of the right copy from $\mathbb{R}$ to $\mathbb{C}$ (via the left copy). It therefore has $\mathbb{C}$-basis $\{1 \otimes 1, 1 \otimes i\}$, making it a two-dimensional $\mathbb{C}$-vector space. **Step 2: Algebra isomorphism with $\mathbb{C} \times \mathbb{C}$.** Using the isomorphism $\mathbb{C} \cong \mathbb{R}[T]/(T^2+1)$, one computes: \begin{align*} \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} &\cong \mathbb{C} \otimes_{\mathbb{R}} \mathbb{R}[T]/(T^2+1) \\ &\cong \mathbb{C}[T]/\bigl((T-i)(T+i)\bigr) \\ &\cong \mathbb{C}[T]/(T-i) \times \mathbb{C}[T]/(T+i) \\ &\cong \mathbb{C} \times \mathbb{C}, \end{align*} where the third isomorphism is the Chinese Remainder Theorem (the ideals $(T-i)$ and $(T+i)$ are coprime since $T-i$ and $T+i$ differ by $2i$). The explicit isomorphism $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \xrightarrow{\;\sim\;} \mathbb{C} \times \mathbb{C}$ sends $x \otimes y \mapsto (xy, x\bar{y})$ where $\bar{y}$ denotes complex conjugation. To verify: writing $x = a+bi$ and $y = c+di$, the computation through the chain of isomorphisms gives \begin{align*} (a+bi) \otimes (c+di) \mapsto \bigl((ac-bd)+i(bc+ad),\; (ac+bd)+i(bc-ad)\bigr) = (xy,\, x\bar{y}). \end{align*} **Step 3: Identifying the idempotents.** Writing a general element in the $\mathbb{C}$-basis as $\alpha(1 \otimes 1) + \beta(1 \otimes i)$, the isomorphism sends \begin{align*} \alpha(1 \otimes 1) + \beta(1 \otimes i) \mapsto (\alpha + \beta i,\; \alpha - \beta i). \end{align*} The preimages of the standard idempotents $(1,0)$ and $(0,1)$ in $\mathbb{C} \times \mathbb{C}$ are obtained by solving $\alpha + \beta i = 1$, $\alpha - \beta i = 0$ and vice versa: \begin{align*} \tfrac{1}{2}(1 \otimes 1) - \tfrac{i}{2}(1 \otimes i) &\mapsto (1, 0), \\ \tfrac{1}{2}(1 \otimes 1) + \tfrac{i}{2}(1 \otimes i) &\mapsto (0, 1). \end{align*} These are the two orthogonal idempotents in $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ corresponding to the two factors of $\mathbb{C} \times \mathbb{C}$. **Step 4: The two $\mathbb{C}$-module structures.** Since $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ is a tensor product over $\mathbb{R}$, one is only permitted to move real scalars across $\otimes$. There are two natural ways to make $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ into a $\mathbb{C}$-module: let $\mathbb{C}$ act by scaling on the left copy, or on the right copy. These two structures are not the same: the identity map $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \to \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ is an $\mathbb{R}$-module isomorphism between them but is not a $\mathbb{C}$-module homomorphism. This is because moving the $\mathbb{C}$-action from left to right involves conjugation. The same issue arises for modules in general: making $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ into a $\mathbb{C}$-module via the left copy vs. the right copy gives different $\mathbb{C}$-module structures (though there is a $\mathbb{C}$-algebra isomorphism between them, implemented by complex conjugation on the right factor). [/example] [remark: Algebra Homomorphisms from $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ to $\mathbb{C}$] **(Non-examinable.)** One can classify all $\mathbb{C}$-algebra homomorphisms $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \to \mathbb{C}$. Any such homomorphism is $\mathbb{R}$-linear, hence corresponds to an $\mathbb{R}$-bilinear map $\mathbb{C} \times \mathbb{C} \to \mathbb{C}$ of the form $(x,y) \mapsto T(x)S(y)$ for $\mathbb{R}$-linear maps $T, S: \mathbb{C} \to \mathbb{C}$. For this to be a $\mathbb{C}$-algebra homomorphism, the condition $x \otimes 1 \mapsto x$ forces $T = \mathrm{id}_{\mathbb{C}}$. The remaining condition is that $S$ is a ring homomorphism $\mathbb{C} \to \mathbb{C}$ that is $\mathbb{R}$-linear; such a map must fix $\mathbb{R}$ pointwise and send $i$ to a root of $T^2 + 1$ in $\mathbb{C}$. The only possibilities are $S = \mathrm{id}$ and $S = $ complex conjugation. This yields exactly two $\mathbb{C}$-algebra homomorphisms: $x \otimes y \mapsto xy$ and $x \otimes y \mapsto x\bar{y}$. Packaging them together gives the isomorphism $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \xrightarrow{\;\sim\;} \mathbb{C} \times \mathbb{C}$, $x \otimes y \mapsto (xy, x\bar{y})$, confirming the computation above. The surjectivity of this map is not coincidental: for a finite-dimensional commutative $\mathbb{C}$-algebra $A$ with no nonzero nilpotents, the collection of all $\mathbb{C}$-algebra homomorphisms $A \to \mathbb{C}$ suffices to separate points, and the combined map $A \to \mathbb{C}^n$ is an isomorphism (this can be seen as an instance of the Artin–Wedderburn theorem). [/remark] Tensor products and localization both modify the ring, but the Artin-Wedderburn theorem shows when semisimple structure emerges. Localization inverts specific elements to isolate local phenomena. # 4. Localization Localization is the process of systematically making elements of a ring invertible. The idea is ancient — one constructs the rational numbers $\mathbb{Q}$ from the integers $\mathbb{Z}$ by formally adding fractions — but the algebraic abstraction allows far greater generality and is one of the central tools of commutative algebra. This chapter develops the theory of localization in full: we construct the localized ring $S^{-1}R$ for an arbitrary multiplicative subset $S \subset R$, establish its universal property, study how ideals and modules behave under the localization map, and examine the important special cases of localizing at a prime ideal or at a principal element. The preceding chapters established the basic language of modules (including chain conditions) and the tensor product; both will appear throughout this chapter, as localization interacts with all of these constructions. ## The Idea of Inverting Elements [motivation] ### From fractions to algebra The most familiar ring that is not a field is $\mathbb{Z}$. Its fraction field $\mathbb{Q}$ is constructed by formally allowing division by any nonzero integer: one declares that elements are fractions $a/b$ with $a \in \mathbb{Z}$ and $b \in \mathbb{Z} \setminus \{0\}$, subject to the identification $a/b = c/d$ whenever $ad = bc$. This works because $\mathbb{Z}$ is an integral domain — it has no zero divisors — so the set $S = \mathbb{Z} \setminus \{0\}$ is closed under multiplication and $a \cdot c = 0$ with $a \neq 0$ forces $c = 0$. But one often needs something more selective: not all of $\mathbb{Q}$, but only fractions whose denominators are powers of a fixed prime $p$. The ring $\mathbb{Z}_{(p)} = \{a/b \in \mathbb{Q} : p \nmid b\}$ of integers localized at $p$ inverts every integer coprime to $p$ while leaving multiples of $p$ non-invertible. This ring remembers the arithmetic at $p$ faithfully, while making everything else invertible. ### The passage from global to local A ring $R$ encodes "global" information — properties that hold everywhere. But many questions are fundamentally local: does a polynomial have a root in a neighborhood of a given point? Is a module free near a specific prime? Localization provides a way to zoom in: by inverting all elements of $R$ outside a prime ideal $\mathfrak{p}$, one obtains the local ring $R_\mathfrak{p}$, which contains only information relevant to $\mathfrak{p}$. This is the algebraic analogue of passing from global functions on a manifold to germs of functions at a point. The local ring $R_\mathfrak{p}$ has a unique maximal ideal $\mathfrak{p} R_\mathfrak{p}$, so questions about the structure of $R$ at $\mathfrak{p}$ reduce to questions about a local ring, which are often much more tractable. ### Why multiplicative subsets In any ring $A$, the product of two invertible elements is invertible: if $a_1 a_2 \in A$ and both $a_1^{-1}$ and $a_2^{-1}$ exist, then $(a_1 a_2)^{-1} = a_2^{-1} a_1^{-1}$. So if one wants to "add inverses" to a set $U \subset R$, the inverses of products of elements of $U$ are automatically forced to exist as well. This means the right datum to work with is not an arbitrary subset $U$ but its multiplicative closure — the smallest multiplicative subset $S \subset R$ containing $U$. Working with multiplicative subsets from the outset therefore loses no generality. [/motivation] The following two definitions set up the basic language. [definition: Multiplicative Subset] A **multiplicative subset** of a ring $R$ is a subset $S \subset R$ such that $1 \in S$ and $ab \in S$ whenever $a, b \in S$. [/definition] The condition $1 \in S$ ensures the canonical map $R \to S^{-1}R$ sends $r \mapsto r/1$, and closure under products ensures the product of two fractions with denominators in $S$ again has its denominator in $S$. [definition: Multiplicative Closure] The **multiplicative closure** of a subset $U \subset R$ is the intersection of all multiplicative subsets of $R$ that contain $U$. Equivalently, it is the set of all products $s_1 s_2 \cdots s_n$ with $s_i \in U$ and $n \geq 0$ (the empty product being $1$). [/definition] [example: Key examples of multiplicative subsets] Three families of multiplicative subsets appear throughout the subject. **Powers of a single element.** Fix $f \in R$ and let $S = \{1, f, f^2, f^3, \ldots\} = \{f^n : n \geq 0\}$. This is a multiplicative subset because $f^m \cdot f^n = f^{m+n} \in S$ and $f^0 = 1 \in S$. The localization $S^{-1}R$ is usually written $R_f$ or $R[f^{-1}]$. For $R = \mathbb{Z}$ and $f = 2$, this gives $\mathbb{Z}[1/2] = \{a/2^n : a \in \mathbb{Z},\, n \geq 0\}$, the ring of dyadic rationals. **Complement of a prime ideal.** If $\mathfrak{p}$ is a prime ideal of $R$, then $S = R \setminus \mathfrak{p}$ is a multiplicative subset. The definition of a prime ideal says: $ab \in \mathfrak{p}$ implies $a \in \mathfrak{p}$ or $b \in \mathfrak{p}$. Equivalently, $a \notin \mathfrak{p}$ and $b \notin \mathfrak{p}$ imply $ab \notin \mathfrak{p}$, which is exactly closure under multiplication. The localization $(R \setminus \mathfrak{p})^{-1} R$ is denoted $R_\mathfrak{p}$ and is called the **localization of $R$ at $\mathfrak{p}$**. **All nonzerodivisors.** The set of nonzerodivisors of $R$ — elements $s$ for which $sr = 0$ implies $r = 0$ — is a multiplicative subset: if $s_1$ and $s_2$ are nonzerodivisors and $s_1 s_2 r = 0$, then $s_1(s_2 r) = 0$, so $s_2 r = 0$ (since $s_1$ is a nonzerodivisor), and then $r = 0$ (since $s_2$ is a nonzerodivisor). The resulting localization is called the **total ring of fractions** of $R$, generalizing the fraction field to rings with zero divisors. [/example] ### The overview of the construction The source material for this section describes the central objects and their relationships in full generality. We record the guiding picture here as orientation for the construction that follows in the next section. Starting from a ring $R$ and a multiplicative subset $S \subset R$, one constructs a new ring $S^{-1}R$ — the **localization of $R$ at $S$** — and a canonical ring homomorphism \begin{align*} \iota: R \to S^{-1}R, \qquad r \mapsto \frac{r}{1}. \end{align*} The elements of $S^{-1}R$ are fractions $r/s$ with $r \in R$ and $s \in S$, subject to an appropriate equivalence relation. The map $\iota$ is not always injective: if $s \in S$ is a zerodivisor in $R$, then there exists $r \in R$ with $sr = 0$, and in $S^{-1}R$ one computes $r/1 = sr/(s \cdot 1) = 0/s = 0$, so $r \in \ker \iota$ even though $r \neq 0$ in $R$. The map $\iota$ is injective if and only if $S$ contains no zerodivisors of $R$. The construction extends to modules: for an $R$-module $M$, one forms the $S^{-1}R$-module \begin{align*} S^{-1}M = \left\{ \frac{m}{s} : m \in M,\, s \in S \right\} \end{align*} with the natural addition and scalar multiplication. Taking $M = R$ and including the ring multiplication recovers $S^{-1}R$ as a special case. Crucially, $S^{-1}M$ is not only an $R$-module (via restriction of scalars along $\iota$) but naturally an $S^{-1}R$-module: the structure map $R \to \operatorname{End}(S^{-1}M)$ factors through $\iota$, since $s \cdot (m/s) = m/1$ in $S^{-1}M$ for $s \in S$. The ring $S^{-1}R$ is characterized by a universal property: it is the initial ring over $R$ in which every element of $S$ is invertible. Precisely, for any ring homomorphism $\varphi: R \to A$ such that $\varphi(s) \in A^\times$ for every $s \in S$, there is a unique ring homomorphism $\tilde{\varphi}: S^{-1}R \to A$ making the triangle commute: \begin{align*} \tilde{\varphi}\!\left(\frac{r}{s}\right) = \varphi(r) \cdot \varphi(s)^{-1}. \end{align*} This universal property pins down $S^{-1}R$ up to unique isomorphism, and it is the right way to think about why the construction is natural — any ring that inverts $S$ must factor through $S^{-1}R$. Finally, since any ring homomorphism that inverts all elements of a set $U$ automatically inverts all elements of the multiplicative closure $S$ of $U$ (products of invertible elements are invertible), the universal property for $U$ and the universal property for $S$ coincide. Working with multiplicative subsets from the outset is therefore fully general. The formal construction of $S^{-1}R$ and $S^{-1}M$, together with the proof of the universal property, occupies the next section. ## The Construction and Universal Property How do we make fractions rigorous when $S$ contains zerodivisors? Over an integral domain, the naive relation $a/b = c/d \iff ad = bc$ works because cancellation holds. But if $s \in S$ satisfies $sr = 0$ for some nonzero $r$, then the fraction $r/1$ should equal $0/s = 0$ in $S^{-1}R$ — yet $s \cdot 1 \cdot 0 \neq s \cdot r \cdot 1$ in $R$. The fix is to build the effect of "further cancellation by elements of $S$" into the equivalence relation itself. With that correction in hand, we construct the localization explicitly as a set of equivalence classes of fractions, establish the ring and module structures, and pin down the construction via its universal property. ### Constructing $S^{-1}M$ as a Set of Fractions The first step is to make precise what it means for two fractions $m_1/s_1$ and $m_2/s_2$ to be equal. Over an integral domain this would simply be $s_2 m_1 = s_1 m_2$, but in general rings elements of $S$ might be zero divisors, so this naive relation can fail to be transitive. The fix is to build in the effect of further multiplication by elements of $S$. [definition: Localization of a Module] Let $S$ be a multiplicative subset of a commutative ring $R$, and let $M$ be an $R$-module. On the set of pairs $(m, s)$ with $m \in M$ and $s \in S$, define the relation \begin{align*} (m_1, s_1) \sim (m_2, s_2) \iff \exists\, u \in S \text{ such that } u(s_2 m_1 - s_1 m_2) = 0. \end{align*} This is an equivalence relation. Write $\tfrac{m}{s}$ for the equivalence class of $(m, s)$, and set \begin{align*} S^{-1}M = \left\{ \frac{m}{s} \;\middle|\; m \in M,\, s \in S \right\}. \end{align*} Give $S^{-1}M$ the structure of an abelian group by \begin{align*} \frac{m_1}{s_1} + \frac{m_2}{s_2} = \frac{s_2 m_1 + s_1 m_2}{s_1 s_2}, \end{align*} with zero element $\tfrac{0}{1}$. Give it the structure of an $R$-module by \begin{align*} r \cdot \frac{m}{s} = \frac{rm}{s}. \end{align*} Both operations are well-defined with respect to $\sim$. When $M = R$ (viewed as an $R$-module), the set $S^{-1}R$ becomes a commutative ring under \begin{align*} \frac{r_1}{s_1} \cdot \frac{r_2}{s_2} = \frac{r_1 r_2}{s_1 s_2}, \end{align*} with multiplicative identity $\tfrac{1}{1}$. For a general $R$-module $M$, the $R$-module $S^{-1}M$ is moreover an $S^{-1}R$-module via $\tfrac{r}{s} \cdot \tfrac{m}{t} = \tfrac{rm}{st}$. [/definition] [remark: Why the Witness $u$ Is Necessary] The role of $u$ in the equivalence relation is easy to misread as a technicality. It is in fact essential. In $S^{-1}R$, every element of $S$ becomes a unit. But if $x \in R$ is a unit and $xm = 0$ in some module, then multiplying both sides by $x^{-1}$ gives $m = 0$. So once $s \in S$ is made invertible, any equation $u(s_2 m_1 - s_1 m_2) = 0$ with $u \in S$ forces $s_2 m_1 - s_1 m_2 = 0$ in $S^{-1}M$. The surprise is not that $u$ is needed, but that incorporating $u$ into the definition of $\sim$ is sufficient to make everything work. [/remark] We verify that $\sim$ is an equivalence relation. Reflexivity and symmetry are immediate. For transitivity, suppose $(m_1, s_1) \sim (m_2, s_2)$ and $(m_2, s_2) \sim (m_3, s_3)$, witnessed by $u, v \in S$ respectively: \begin{align*} u(s_2 m_1 - s_1 m_2) = 0, \qquad v(s_3 m_2 - s_2 m_3) = 0. \end{align*} Multiply the first equation by $vs_3$ and the second by $us_1$, then add: \begin{align*} uvs_2(s_3 m_1 - s_1 m_3) = 0. \end{align*} Since $uvs_2 \in S$ (as $S$ is multiplicative), this shows $(m_1, s_1) \sim (m_3, s_3)$. The well-definedness of addition, scalar multiplication, and ring multiplication are straightforward verifications: each reduces to showing that if the inputs represent equal elements then the output does too, using the existence of a single witness $u \in S$. ### The Canonical Map and When It Is Injective There is a natural ring homomorphism from $R$ into its localization, obtained by placing elements over $1$. [definition: Canonical Map into the Localization] Let $S \subset R$ be a multiplicative subset. The **canonical ring homomorphism** is \begin{align*} \iota \colon R \to S^{-1}R, \qquad \iota(r) = \frac{r}{1}. \end{align*} By definition $\iota(s)$ is a unit of $S^{-1}R$ for every $s \in S$, with inverse $\tfrac{1}{s}$. Similarly, for an $R$-module $M$, the canonical $R$-linear map $\iota_M \colon M \to S^{-1}M$ is defined by $\iota_M(m) = \tfrac{m}{1}$. [/definition] The kernel of $\iota$ is easy to identify directly from the definition of $\sim$: \begin{align*} \ker \iota = \{ r \in R \mid \exists\, u \in S \text{ with } ur = 0 \}. \end{align*} In particular, $\iota$ is injective if and only if $S$ contains no zero divisors of $R$. Two further elementary observations: - $S^{-1}R = 0$ if and only if $0 \in S$, since $\tfrac{1}{1} = \tfrac{0}{1}$ requires a $u \in S$ with $u \cdot 1 = 0$. - An element $\tfrac{r}{s} \in S^{-1}R$ is zero if and only if there exists $u \in S$ with $ur = 0$. [remark: $\iota$ Is an Epimorphism of Rings] Although $\iota$ is usually not surjective, it is always an **epimorphism** in the category of rings: if $g, h \colon S^{-1}R \to B$ are ring homomorphisms with $g \circ \iota = h \circ \iota$, then $g = h$. This mirrors the classical fact that the inclusion $\mathbb{Z} \hookrightarrow \mathbb{Q}$ is an epimorphism — two ring homomorphisms out of $\mathbb{Q}$ that agree on $\mathbb{Z}$ must agree everywhere, because every element of $\mathbb{Q}$ is a ratio of integers. The same logic applies: any element $\tfrac{r}{s} \in S^{-1}R$ satisfies $g(\tfrac{r}{s}) = g(\iota(r)) g(\iota(s))^{-1}$, which is already determined by $g \circ \iota$. Contrast this with the situation for sets, groups, or topological spaces: in those categories, epimorphisms are precisely the surjective maps. [/remark] ### The Universal Property The universal property characterizes $S^{-1}R$ up to unique isomorphism: it is the initial ring receiving $R$ in which all elements of $S$ become units. Precisely: [quotetheorem:2922] [citeproof:2922] The universal property is what makes localization functorially well-behaved: any ring map out of $R$ that sends $S$ to units factors uniquely through $S^{-1}R$. It is worth pausing on what the universal property does and does not say. It characterises morphisms **from** $S^{-1}R$ to other rings, not the internal structure of $S^{-1}R$ itself. In particular, it does not tell us which ideals $S^{-1}R$ has or how its elements multiply — for that, the fraction description is indispensable. As a concrete application: to show that $\mathbb{Z}[1/6] \cong \mathbb{Z}[1/2][1/3]$, observe that both rings receive $\mathbb{Z}$ and make $2$ and $3$ invertible, hence make $6 = 2 \cdot 3$ invertible; the universal property of $\mathbb{Z}[1/6]$ then supplies unique isomorphisms in both directions. ### Standard Examples [example: Inverting a Single Element] Let $f \in R$. The set $S = \{f^n \mid n \geq 0\}$ is multiplicative. The localization $S^{-1}R$ is written $R_f$, "R with $f$ inverted". - Take $R = \mathbb{Z}$ and $f = 2$. Then $\mathbb{Z}_2 = \{ a/2^n \mid a \in \mathbb{Z},\, n \geq 0 \}$ is the ring of dyadic rationals. It is isomorphic to the subring $\mathbb{Z}[1/2]$ of $\mathbb{Q}$ generated by $\mathbb{Z}$ and $1/2$, i.e., the image of the unique $\mathbb{Z}$-algebra homomorphism $\mathbb{Z}[T] \to \mathbb{Q}$ sending $T \mapsto 1/2$. Note: when $n = p$ is prime, $\mathbb{Z}_p$ conventionally denotes the $p$-adic integers, not $\{p^n \mid n \geq 0\}^{-1}\mathbb{Z}$; to avoid ambiguity, write $\mathbb{Z}[1/p]$ for the latter. - $R_f = 0$ if and only if $f$ is nilpotent in $R$, since $0 = f^k \in S$ for some $k$ implies $S^{-1}R = 0$. [/example] [example: Localizing at a Prime Ideal] Let $\mathfrak{p}$ be a prime ideal of $R$. The complement $S = R \setminus \mathfrak{p}$ is a multiplicative subset (this is equivalent to $\mathfrak{p}$ being prime). The localization \begin{align*} R_{\mathfrak{p}} := (R \setminus \mathfrak{p})^{-1} R \end{align*} is called $R$ **localized at $\mathfrak{p}}$. It is a local ring with unique maximal ideal $\mathfrak{p} R_{\mathfrak{p}} = \{ r/s \mid r \in \mathfrak{p},\, s \notin \mathfrak{p} \}$. For instance, take $p$ a prime number and $\mathfrak{p} = (p) \subset \mathbb{Z}$. Then \begin{align*} \mathbb{Z}_{(p)} = \left\{ \frac{m}{n} \;\middle|\; m, n \in \mathbb{Z},\, p \nmid n \right\}, \end{align*} the ring of rational numbers whose denominators are coprime to $p$. This keeps $p$ as a non-unit while inverting everything else. [/example] ### The Module Universal Property An analogous universal property holds for $S^{-1}M$, characterizing it as the $S^{-1}R$-module generated by $M$ in the most efficient way. [quotetheorem:2844] [citeproof:2844] In categorical language, the bijection $\operatorname{Hom}_R(M, L) \xrightarrow{\sim} \operatorname{Hom}_{S^{-1}R}(S^{-1}M, L)$ says that $S^{-1}(\cdot)$ is **left adjoint** to the restriction-of-scalars functor along $\iota \colon R \to S^{-1}R$. This adjunction is the reason why localization interacts so cleanly with maps: to define an $S^{-1}R$-linear map out of $S^{-1}M$, it suffices to specify an $R$-linear map out of $M$ whose codomain is already an $S^{-1}R$-module. The theorem also foreshadows the tensor product identification: the natural isomorphism $S^{-1}M \cong S^{-1}R \otimes_R M$ that appears in the proof is not an accident, but the concrete manifestation of this adjunction. ### Localization as a Tensor Product The identification between $S^{-1}M$ and the tensor product $S^{-1}R \otimes_R M$ is more than a notational convenience — it makes the functorial properties of localization transparent. [quotetheorem:2923] [citeproof:2923] This identification makes $S^{-1}(\cdot)$ into a functor from $R$-modules to $S^{-1}R$-modules: for an $R$-linear map $f \colon N \to N'$, define \begin{align*} S^{-1}f \colon S^{-1}N \to S^{-1}N', \qquad \frac{n}{s} \longmapsto \frac{f(n)}{s}, \end{align*} which is the composition of the isomorphism $S^{-1}N \cong S^{-1}R \otimes_R N$, then $\mathrm{id} \otimes f$, then $S^{-1}R \otimes_R N' \cong S^{-1}N'$. [remark: Natural Isomorphism and Category Theory] In categorical language, the functors $S^{-1}R \otimes_R (\cdot)$ and $S^{-1}(\cdot)$ from $R$-modules to $S^{-1}R$-modules are **naturally isomorphic**: the isomorphisms $\varepsilon_M \colon S^{-1}R \otimes_R M \to S^{-1}M$ are compatible with all $R$-linear maps, in the sense that for every $f \colon N \to N'$ the diagram with $\varepsilon_N$, $\varepsilon_{N'}$, $\mathrm{id} \otimes f$, and $S^{-1}f$ commutes. A **natural transformation** is a family of morphisms $(\varepsilon_M)_M$ with this commutativity property; when each $\varepsilon_M$ is an isomorphism it is a natural isomorphism. [/remark] Since $S^{-1}(\cdot) \cong S^{-1}R \otimes_R (\cdot)$ as functors, the algebraic properties of the tensor product carry over to localization. In particular, the tensor product of two algebras over $R$ localizes as follows: for an $R$-algebra $A$, the natural map $S^{-1}R \otimes_R A \to S^{-1}A$ is an $S^{-1}R$-algebra isomorphism (not just a module isomorphism), and $S^{-1}(\cdot)$ sends $R$-algebra homomorphisms to $S^{-1}R$-algebra homomorphisms. ### $S^{-1}R$ Is Already Localized A useful consistency check: if $M$ is already an $S^{-1}R$-module (rather than merely an $R$-module), then localizing at $S$ does nothing new. [quotetheorem:2924] [citeproof:2924] Equivalently, $S^{-1}R \otimes_R M \cong M$ as $S^{-1}R$-modules via $\tfrac{r}{s} \otimes m \mapsto \tfrac{r}{s} \cdot m$. This says that $S^{-1}R$ is **idempotent** as an $R$-module in the sense that tensoring with it twice is the same as tensoring once. ### Exactness of Localization One of the most important properties of localization is that it preserves exact sequences — a property that will be used repeatedly when comparing ideals and modules before and after localizing. [quotetheorem:2847] [citeproof:2847] Flatness has an important consequence for submodules. If $N \subset M$ is an inclusion of $R$-modules, then $S^{-1}N \to S^{-1}M$ is injective, so $S^{-1}N$ embeds as a submodule of $S^{-1}M$ and the fraction notation $\tfrac{n}{s}$ for elements of $S^{-1}N$ is consistent with the same notation for elements of $S^{-1}M$. ### Submodule Operations Commute with Localization Let $N, P$ be submodules of an $R$-module $M$. [quotetheorem:2925] [citeproof:2925] Part (2) — that intersection commutes with localization — deserves emphasis. It relies crucially on the flatness of $S^{-1}R$ as an $R$-module. For a general change of rings $R \to A$ (even a surjection), the formula $A \otimes_R (N \cap P) = (A \otimes_R N) \cap (A \otimes_R P)$ can fail: the tensor product is right-exact but not left-exact in general, so it does not preserve intersections. Localization escapes this failure precisely because it is flat. ### Tensor Products Localize Componentwise Finally, localization commutes with tensor products in the following sense. [quotetheorem:2926] [citeproof:2926] This identity is the key tool for computing **stalks** in algebraic geometry. If $\mathcal{F}$ and $\mathcal{G}$ are quasi-coherent sheaves on $\operatorname{Spec}(R)$ corresponding to $R$-modules $M$ and $N$, then the stalk of $\mathcal{F} \otimes \mathcal{G}$ at a point $\mathfrak{p}$ is $(M \otimes_R N)_\mathfrak{p} \cong M_\mathfrak{p} \otimes_{R_\mathfrak{p}} N_\mathfrak{p}$, so tensor products of sheaves are computed fibrewise. The result depends on $S^{-1}R$ being flat over $R$: without flatness, the formula $S^{-1}(M \otimes_R N) \cong S^{-1}M \otimes_{S^{-1}R} S^{-1}N$ can fail for general base change $R \to A$. The canonical example of failure is $A = R/I$ for an ideal $I \not\subset \operatorname{Ann}(M)$: the tensor product $R/I \otimes_R M = M/IM$ does not commute with intersection in $M$ unless $I$ acts on $M$ in a controlled way. ## Ideals Under Localization: Extension and Contraction When we pass from a ring $R$ to its localization $S^{-1}R$ via the map $\iota: R \to S^{-1}R$, $r \mapsto \frac{r}{1}$, ideals of $R$ can be pushed forward to ideals of $S^{-1}R$ and ideals of $S^{-1}R$ can be pulled back to ideals of $R$. Understanding exactly which ideals survive this process, and how, is the key to the correspondence between $\operatorname{Spec}(R)$ and $\operatorname{Spec}(S^{-1}R)$. ### Extension and Contraction Recalled For any ring homomorphism $f: A \to B$, we have two operations on ideals. Given an ideal $\mathfrak{b}$ of $B$, its **contraction** is the ideal $\mathfrak{b}^c = f^{-1}(\mathfrak{b})$ of $A$. Given an ideal $\mathfrak{a}$ of $A$, its **extension** is the ideal $\mathfrak{a}^e = (f(\mathfrak{a}))$ of $B$ generated by the image of $\mathfrak{a}$. One of the most useful general facts is that contraction preserves primeness: if $\mathfrak{b}$ is prime in $B$, then $\mathfrak{b}^c$ is prime in $A$, because the composite $A \xrightarrow{f} B \to B/\mathfrak{b}$ has kernel $\mathfrak{b}^c$, so $A/\mathfrak{b}^c$ embeds in the integral domain $B/\mathfrak{b}$. An ideal $\mathfrak{a}$ of $A$ is called **contracted** if $\mathfrak{a} = \mathfrak{a}^{ec}$. An ideal $\mathfrak{b}$ of $B$ is called **extended** if $\mathfrak{b} = \mathfrak{b}^{ce}$. In general one has inclusions $\mathfrak{a} \subset \mathfrak{a}^{ec}$ and $\mathfrak{b} \supset \mathfrak{b}^{ce}$ for every ideal on the respective side. The contracted and extended ideals are in bijective correspondence via $\mathfrak{a} \mapsto \mathfrak{a}^e$ and $\mathfrak{b} \mapsto \mathfrak{b}^c$. ### Explicit Formulas for the Localization Map For the localization map $\iota: R \to S^{-1}R$, $r \mapsto \frac{r}{1}$, we have especially concrete descriptions of both operations. We also write $S^{-1}\mathfrak{a}$ for the extension $\mathfrak{a}^e$ when the multiplicative set is clear. [definition: Extension and Contraction Under Localization] Let $S$ be a multiplicative subset of $R$. For the localization map $\iota: R \to S^{-1}R$: - **Extension**: For an ideal $\mathfrak{a}$ of $R$, \begin{align*} \mathfrak{a}^e = S^{-1}\mathfrak{a} = \left\{ \frac{a}{s} \in S^{-1}R \;\middle|\; a \in \mathfrak{a},\, s \in S \right\}. \end{align*} - **Contraction**: For an ideal $\mathfrak{b}$ of $S^{-1}R$, \begin{align*} \mathfrak{b}^c = \left\{ r \in R \;\middle|\; \frac{r}{1} \in \mathfrak{b} \right\}. \end{align*} [/definition] To see that the formula for extension is correct: the left-hand side $\mathfrak{a}^e$ is the ideal generated by $\{\frac{a}{1} : a \in \mathfrak{a}\}$. Any element of $S^{-1}R$ of the form $\frac{a}{s}$ with $a \in \mathfrak{a}$ equals $\frac{1}{s} \cdot \frac{a}{1}$, so it lies in the ideal generated by $\mathfrak{a}$. Conversely, the set of all such fractions forms an ideal of $S^{-1}R$ (closure under addition follows from $\frac{a}{s} + \frac{a'}{s'} = \frac{s'a + sa'}{ss'}$ with numerator in $\mathfrak{a}$, and absorption of ring elements is immediate), so the two sides agree. These formulas give useful derived identities. For the double-contraction-extension: \begin{align*} \mathfrak{a}^{ec} = \bigcup_{s \in S} (\mathfrak{a} : s), \end{align*} where $(\mathfrak{a} : s) = \{r \in R : rs \in \mathfrak{a}\}$ is the colon ideal. To verify this: if $r \in \bigcup_{s \in S}(\mathfrak{a}:s)$, then $rs = a$ for some $s \in S$, $a \in \mathfrak{a}$, which means $\frac{r}{1} = \frac{a}{s} \in \mathfrak{a}^e$, so $r \in \mathfrak{a}^{ec}$. Conversely, if $\frac{r}{1} \in \mathfrak{a}^e$, write $\frac{r}{1} = \frac{a}{s}$ with $a \in \mathfrak{a}$, $s \in S$; the definition of equality in $S^{-1}R$ gives $u(rs - a) = 0$ for some $u \in S$, so $rus = ua \in \mathfrak{a}$, placing $r$ in $(\mathfrak{a} : us)$ with $us \in S$. For the contraction-extension of ideals already in $S^{-1}R$: every ideal of $S^{-1}R$ is extended, because $\mathfrak{b}^{ce} = \mathfrak{b}$ for all $\mathfrak{b}$. The inclusion $\mathfrak{b}^{ce} \subset \mathfrak{b}$ is general. For the reverse: given $\frac{r}{s} \in \mathfrak{b}$, note $\frac{r}{1} = s \cdot \frac{r}{s} \in \mathfrak{b}$, so $r \in \mathfrak{b}^c$, and then $\frac{r}{s} \in (\mathfrak{b}^c)^e = \mathfrak{b}^{ce}$. ### The Prime Ideal Correspondence We now state the central theorem of this section, which describes exactly which prime ideals of $R$ survive to prime ideals of $S^{-1}R$ and establishes a bijection between them. Write $\operatorname{Spec}(R)$ for the set of prime ideals of $R$. [quotetheorem:2927] [citeproof:2927] This bijection is one of the most useful computational tools in commutative algebra. To see it concretely, take $R = \mathbb{Z}$ and $S = \mathbb{Z} \setminus (p)$ for a prime $p$. The prime ideals of $\mathbb{Z}$ disjoint from $S$ are those contained in $(p)$, namely $(0)$ and $(p)$ itself. The theorem says $\operatorname{Spec}(\mathbb{Z}_{(p)}) = \{(0), p\mathbb{Z}_{(p)}\}$, which one can verify directly: the units of $\mathbb{Z}_{(p)}$ are fractions $a/b$ with $p \nmid a$, so the non-units are exactly the multiples of $p$, forming the unique maximal ideal $p\mathbb{Z}_{(p)}$. For non-prime ideals, the situation is different: the bijection is between **prime** ideals only. The ideal $(6) \subset \mathbb{Z}$ is not prime, and $(6)^e = 6\mathbb{Z}_{(5)}$ is the unit ideal in $\mathbb{Z}_{(5)}$ (since $6$ is a unit there), even though $(6) \cap (\mathbb{Z} \setminus (5)) = \{6, 12, \ldots\} \neq \varnothing$. Part (3) of the theorem accounts for exactly this collapse: an ideal extends to the unit ideal precisely when it meets $S$. Geometrically, the bijection says that $\operatorname{Spec}(S^{-1}R)$ is an open subset of $\operatorname{Spec}(R)$: the primes of $S^{-1}R$ are the primes of $R$ that "survive" the localization, meaning they do not get killed by meeting $S$. Localizing at $\mathfrak{p}$ corresponds to keeping only the primes contained in $\mathfrak{p}$ — zooming in on the closed point $\mathfrak{p}$ and its specialisations. ### The Radical and Prime Ideals Before turning to local properties, we record a fundamental connection between the radical of an ideal and the prime ideals containing it. [definition: Radical of an Ideal] The **radical** of an ideal $I$ of $R$ is \begin{align*} \sqrt{I} = \{ r \in R : r^n \in I \text{ for some } n \geq 1 \}. \end{align*} [/definition] That $\sqrt{I}$ is indeed an ideal: if $x^n \in I$ and $y^\ell \in I$, then expanding $(x+y)^{n+\ell}$ shows each term $x^i y^j$ has either $i \geq n$ or $j \geq \ell$, so the entire sum lies in $I$. Closure under scalar multiples is straightforward. Note that $I \subset \sqrt{I}$ always, and the nilradical is the special case $\operatorname{nil}(R) = \sqrt{(0)}$. If $J \subset I$ are ideals of $R$, then $\sqrt{I/J} = \sqrt{I}/J$. [quotetheorem:2928] [citeproof:2928] [remark: Nilradical as Intersection of All Primes] Taking $I = (0)$, we get $\operatorname{nil}(R) = \bigcap_{\mathfrak{p} \in \operatorname{Spec}(R)} \mathfrak{p}$: the nilradical is the intersection of all prime ideals of $R$. [/remark] ## Local Rings, Localization at a Prime, and Local Properties ### Local Rings and the Residue Field How do we study $R$ near a single prime ideal $\mathfrak{p}$? The quotient $R/\mathfrak{p}$ is one answer, but it discards all information about other primes contained in $\mathfrak{p}$ and about elements that are not in $\mathfrak{p}$. A better tool is to invert everything outside $\mathfrak{p}$: by passing to $R_\mathfrak{p} = (R \setminus \mathfrak{p})^{-1}R$, we make all elements not in $\mathfrak{p}$ into units, forcing $\mathfrak{p}$ to become the unique obstruction to invertibility. The prime ideal correspondence has a striking consequence for this construction: when $S = R \setminus \mathfrak{p}$, the ring $S^{-1}R = R_\mathfrak{p}$ has primes corresponding bijectively to the primes of $R$ contained in $\mathfrak{p}$. In particular, every prime of $R_\mathfrak{p}$ is contained in $\mathfrak{p}^e = \mathfrak{p}R_\mathfrak{p}$, which means $\mathfrak{p}R_\mathfrak{p}$ is the unique maximal ideal of $R_\mathfrak{p}$. [definition: Local Ring] A ring $R$ is **local** if it has exactly one maximal ideal, customarily denoted $\mathfrak{m}$ and the pair written $(R, \mathfrak{m})$. [/definition] Thus $(R_\mathfrak{p}, \mathfrak{p}R_\mathfrak{p})$ is a local ring for every prime $\mathfrak{p}$. [example: Localization of the Integers at a Prime] For $R = \mathbb{Z}$ and $\mathfrak{p} = (2)$, the localization is \begin{align*} \mathbb{Z}_{(2)} = \left\{ \frac{a}{b} : a, b \in \mathbb{Z},\, 2 \nmid b \right\}, \end{align*} a subring of $\mathbb{Q}$. Its unique maximal ideal is $(2)\mathbb{Z}_{(2)} = \left\{ \frac{2a}{b} : a, b \in \mathbb{Z},\, 2 \nmid b \right\}$. An element $\frac{a}{b}$ with $2 \nmid a$ and $2 \nmid b$ is a unit in $\mathbb{Z}_{(2)}$, so the non-units are exactly the elements of the maximal ideal. The prime ideals of $\mathbb{Z}_{(2)}$ correspond to prime ideals of $\mathbb{Z}$ contained in $(2)$, which are $(0)$ and $(2)$. [/example] For any $R$-module $M$, the localization $M_\mathfrak{p}$ is a module over the local ring $R_\mathfrak{p}$. [remark: Contrasting $R_\mathfrak{p}$ and $R/\mathfrak{p}$] It is worth comparing the two natural quotients associated to $\mathfrak{p}$. The quotient $R/\mathfrak{p}$ kills $\mathfrak{p}$ and retains only the structure "at the residue level", while $R_\mathfrak{p}$ inverts everything outside $\mathfrak{p}$ and retains the local structure "near $\mathfrak{p}$". If $\mathfrak{q} \subset \mathfrak{p}$ are prime ideals, then $R_\mathfrak{p}/\mathfrak{q}R_\mathfrak{p}$ focuses attention on primes between $\mathfrak{q}$ and $\mathfrak{p}$, and one has a natural ring isomorphism \begin{align*} R_\mathfrak{p} / \mathfrak{q}R_\mathfrak{p} \xrightarrow{\;\sim\;} (R/\mathfrak{q})_\mathfrak{p}, \quad \frac{r}{s} + \mathfrak{q}R_\mathfrak{p} \mapsto \frac{r + \mathfrak{q}}{s + \mathfrak{q}}, \end{align*} because localization commutes with quotients. Taking $\mathfrak{q} = \mathfrak{p}$ gives the **residue field** at $\mathfrak{p}$: \begin{align*} \kappa(\mathfrak{p}) := R_\mathfrak{p} / \mathfrak{p}R_\mathfrak{p} \cong (R/\mathfrak{p})_\mathfrak{p} = \operatorname{Frac}(R/\mathfrak{p}). \end{align*} This is a field because $\mathfrak{p}R_\mathfrak{p}$ is the unique maximal ideal of the local ring $R_\mathfrak{p}$. [/remark] ### The Local–Global Principle A fundamental paradigm in commutative algebra is the **local–global principle**: many properties of $R$-modules can be verified by checking them at every localization. The first instance is the detection of zero. [quotetheorem:2852] [citeproof:2852] The same principle extends to maps. Injectivity and surjectivity are determined locally. [quotetheorem:2853] [citeproof:2853] ### Flatness Is a Local Property Flatness — the property that $- \otimes_R M$ preserves exact sequences — also passes through the local–global principle. [quotetheorem:2929] [citeproof:2929] [remark: Localizable and Local-to-Global Properties] A property $\mathcal{P}$ of modules is called **localizable** if $M$ having $\mathcal{P}$ implies $M_\mathfrak{p}$ has $\mathcal{P}$ for all $\mathfrak{p} \in \operatorname{Spec}(R)$. It is called **local-to-global** if the converse holds: $M$ has $\mathcal{P}$ whenever $M_\mathfrak{p}$ has $\mathcal{P}$ for all $\mathfrak{p}$. A property is **local** if it is both. The three theorems above show that being zero, injectivity, surjectivity, and flatness are all local properties. [/remark] ### Checking Local Properties at Maximal Ideals In the theorems above, condition (3) always only refers to maximal ideals, which is often more convenient in practice. This is not accidental: for any "reasonable" module property, checking at all maximal ideals is equivalent to checking at all prime ideals. A property $\mathcal{P}$ is called **reasonable** if the following holds: whenever $\varphi: R_1 \xrightarrow{\sim} R_2$ is a ring isomorphism, $M_1$ an $R_1$-module, $M_2$ an $R_2$-module, and $M_1 \cong M_2$ as $R_1$-modules (viewing $M_2$ as an $R_1$-module via $\varphi$), then $M_1$ has $\mathcal{P}$ if and only if $M_2$ does. For any local reasonable property $\mathcal{P}$, checking $M_\mathfrak{m}$ for all maximal $\mathfrak{m}$ suffices to conclude $M$ has $\mathcal{P}$. ### A Criterion for Noetherianness The local–global principle yields a useful criterion for when a ring is noetherian. [quotetheorem:2930] [remark: The Second Condition Is Necessary] There exist non-noetherian rings $A$ for which $A_\mathfrak{m}$ is noetherian for every maximal ideal $\mathfrak{m}$. The finiteness condition (2) cannot be dropped. [/remark] [proof] Let $\mathfrak{a}$ be any nonzero ideal of $R$. We show $\mathfrak{a}$ is finitely generated. It suffices to find a finitely generated ideal $\mathfrak{b} \subset \mathfrak{a}$ with $\mathfrak{a}R_\mathfrak{m} = \mathfrak{b}R_\mathfrak{m}$ for all maximal $\mathfrak{m}$: then the inclusion $\mathfrak{b} \hookrightarrow \mathfrak{a}$ localizes to a surjection at every $\mathfrak{m}$, hence is surjective by the local property for surjectivity, giving $\mathfrak{a} = \mathfrak{b}$. Fix a nonzero element $x \in \mathfrak{a}$ and partition $\operatorname{mSpec}(R)$ into three subsets: \begin{align*} M_1 &= \{ \mathfrak{m} : x \notin \mathfrak{m} \}, \\ M_2 &= \{ \mathfrak{m} : x \in \mathfrak{m},\, \mathfrak{a} \not\subset \mathfrak{m} \}, \\ M_3 &= \{ \mathfrak{m} : \mathfrak{a} \subset \mathfrak{m} \}. \end{align*} For $\mathfrak{m} \in M_1$: $x$ is a unit in $R_\mathfrak{m}$, so $\mathfrak{a}R_\mathfrak{m} = R_\mathfrak{m} = (x)R_\mathfrak{m}$. For $\mathfrak{m} \in M_2$: pick $x^{(\mathfrak{m})} \in \mathfrak{a} \setminus \mathfrak{m}$. This element is a unit in $R_\mathfrak{m}$, so $\mathfrak{a}R_\mathfrak{m} = R_\mathfrak{m} = (x^{(\mathfrak{m})})R_\mathfrak{m}$. For $\mathfrak{m} \in M_3$: $R_\mathfrak{m}$ is noetherian, so $\mathfrak{a}R_\mathfrak{m}$ is finitely generated by elements $\frac{a_1}{s_1}, \ldots, \frac{a_\ell}{s_\ell}$ with $a_i \in \mathfrak{a}$. Since multiplying by $s_i$ does not change the generated ideal in $R_\mathfrak{m}$, the ideal $\mathfrak{a}R_\mathfrak{m}$ is generated by $\frac{a_1}{1}, \ldots, \frac{a_\ell}{1}$. Set $\mathfrak{a}^{(\mathfrak{m})} = (a_1, \ldots, a_\ell) \subset \mathfrak{a}$, so $\mathfrak{a}^{(\mathfrak{m})}R_\mathfrak{m} = \mathfrak{a}R_\mathfrak{m}$. By condition (2), $M_2 \cup M_3$ (the set of maximal ideals containing $x$) is finite. Set \begin{align*} \mathfrak{b} = (x) + \sum_{\mathfrak{m} \in M_2} (x^{(\mathfrak{m})}) + \sum_{\mathfrak{m} \in M_3} \mathfrak{a}^{(\mathfrak{m})}. \end{align*} This is a finite sum of finitely generated ideals, so $\mathfrak{b}$ is finitely generated, $\mathfrak{b} \subset \mathfrak{a}$, and $\mathfrak{b}R_\mathfrak{m} = \mathfrak{a}R_\mathfrak{m}$ for all $\mathfrak{m}$, as required. [/proof] ### Local Freeness and Other Non-Local Properties Not every natural property is local. We close with two instructive examples. [example: Locally Free But Not Free] An $R$-module $M$ is **locally free** if $M_\mathfrak{p}$ is a free $R_\mathfrak{p}$-module for every $\mathfrak{p} \in \operatorname{Spec}(R)$. Freeness itself is not a local property. Take $R = \mathbb{C} \times \mathbb{C}$. In a product ring $\prod_{i=1}^n R_i$, the ideals are exactly the products $\prod_{i=1}^n I_i$ with $I_i$ an ideal of $R_i$, and the prime ideals are those where $I_{i_0}$ is prime for one index $i_0$ and $I_i = R_i$ for all other $i$. Thus \begin{align*} \operatorname{Spec}(\mathbb{C} \times \mathbb{C}) = \{ \mathfrak{p}_1, \mathfrak{p}_2 \} \quad \text{where } \mathfrak{p}_1 = \mathbb{C} \times \{0\},\; \mathfrak{p}_2 = \{0\} \times \mathbb{C}. \end{align*} For $\mathfrak{p}_1$, the multiplicative set is $S = \mathbb{C} \times (\mathbb{C} \setminus \{0\})$. The map $(x, y) \mapsto y$ sends every element of $S$ to a unit in $\mathbb{C}$, inducing an isomorphism $R_{\mathfrak{p}_1} \cong \mathbb{C}$ (the kernel, consisting of fractions with numerator of the form $(x, 0)$, is zero because $(0, 1) \cdot (x, 0) = 0$ and $(0, 1) \in S$). Similarly $R_{\mathfrak{p}_2} \cong \mathbb{C}$. So $R$ is locally a field and every $R$-module is locally free. Now let $M = \mathbb{C} \times \{0\}$, an ideal of $R$ viewed as an $R$-module. For any $(x, 0) \in M$, we have $(0, 1) \cdot (x, 0) = (0, 0)$, so $(0, 1)$ annihilates all of $M$. The only $R$-linearly independent subset of $M$ is the empty set, which does not span $M$ (since $M \neq 0$). So $M$ is not free, despite being locally free. [/example] [example: Reducedness Is Local; Integrality Is Not] A ring $R$ is **reduced** if $0$ is its only nilpotent element, i.e., $\sqrt{(0)} = (0)$. Reducedness is a local property: if $R_\mathfrak{p}$ is reduced for all $\mathfrak{p} \in \operatorname{Spec}(R)$, then $R$ is reduced. However, being an **integral domain** is not a local property. The ring $\mathbb{C} \times \mathbb{C}$ shows this: it is locally a field (in particular locally an integral domain), yet it is not itself an integral domain since $(1, 0) \cdot (0, 1) = (0, 0)$. [/example] ## Localization via a Quotient of a Polynomial Ring The fraction-based construction of $S^{-1}R$ — as equivalence classes of pairs $(r, s)$ — is explicit and concrete, well-suited for computing with ideals and modules. But there is a second construction of the same ring, starting from an entirely different direction: instead of quotienting a set of pairs, one quotients a polynomial ring. The two constructions yield isomorphic rings by the universal property, and comparing them illuminates why the universal property is so powerful. ### The Polynomial Ring Construction Let $U \subset R$ be a subset and let $S$ be its multiplicative closure. The idea is to adjoin a formal inverse for each element of $U$ simultaneously by introducing polynomial variables. [definition: Localization as Polynomial Quotient] Let $R$ be a commutative ring and $U \subset R$ a subset. Consider the polynomial $R$-algebra \begin{align*} R[\{T_u\}_{u \in U}] \end{align*} in one variable $T_u$ for each $u \in U$. Let $I_U$ be the ideal generated by the elements $\{uT_u - 1\}_{u \in U}$, and define \begin{align*} R_U := R[\{T_u\}_{u \in U}] / I_U. \end{align*} Write $\bar{u}$ and $\overline{T_u}$ for the images of $u \in U$ and $T_u$ in $R_U$. By construction, $\bar{u} \cdot \overline{T_u} = 1$ in $R_U$, so every element of $U$ is invertible in $R_U$. [/definition] In $R_U$, the generator $\overline{T_u}$ plays the role of $u^{-1}$: the relation $u T_u - 1 \in I_U$ is precisely the declaration that $T_u$ is a multiplicative inverse of $u$. The ring $R_U$ is therefore the most general $R$-algebra in which every $u \in U$ has been made invertible. [remark: Two Constructions, One Object] Both $S^{-1}R$ (via fractions) and $R_U$ (via polynomial quotient) are constructed as quotients of a large object. To form $S^{-1}R$, one took the set of all pairs $(r, s) \in R \times S$ and imposed an equivalence relation; equivalently, one quotiented a free $R$-module by a huge submodule. To form $R_U$, one took a polynomial $R$-algebra in infinitely many variables and quotiented by a huge ideal. In both cases, the quotient was shown to satisfy a universal property, and the universal property determines the object up to unique isomorphism. This pattern — construct a huge object by generators and relations, verify the universal property, then use the universal property to extract concrete information — recurs throughout algebra. [/remark] ### Verification via the Universal Property The claim is that $R_U \cong S^{-1}R$ as $R$-algebras. The proof proceeds by verifying that $(R_U, \iota)$ satisfies the same universal property as $(S^{-1}R, \iota_{S^{-1}R})$, so the two must be isomorphic by the uniqueness clause of the universal property. [quotetheorem:2931] [citeproof:2931] Since $(R_U, \iota)$ satisfies the same universal property as $(S^{-1}R, \iota_{S^{-1}R})$, the uniqueness clause gives a ring isomorphism \begin{align*} R_U \xrightarrow{\;\sim\;} S^{-1}R \end{align*} which is an $R$-algebra isomorphism. Explicitly, a monomial $r \prod_{i=1}^{\ell} T_{u_i} + I_U$ in $R_U$ maps to the fraction $\tfrac{r}{u_1 \cdots u_\ell} \in S^{-1}R$, and the inverse map sends $\tfrac{r}{u_1 \cdots u_\ell}$ to $r \prod_{i=1}^{\ell} \overline{T_{u_i}}$. ### The Single-Element Case: $R_u = R[T]/(uT - 1)$ An important special case occurs when $U = \{u\}$ consists of a single element. Then $S = \{u^n \mid n \geq 0\}$ and the localization $S^{-1}R$ is written $R_u$. The polynomial ring is now an ordinary polynomial ring $R[T]$ in a single variable, and the ideal $I_U$ is the principal ideal $(uT - 1)$. [quotetheorem:2932] This makes the structure of $R_u$ completely concrete: its elements are polynomials in a single variable $T$, subject only to the relation $uT = 1$. For example, take $R = \mathbb{Z}$ and $u = 2$. Then $\mathbb{Z}_2 \cong \mathbb{Z}[T]/(2T - 1)$. In this ring $T$ plays the role of $1/2$, so any element can be written as $a_0 + a_1 T + \cdots + a_n T^n$ subject to $2T = 1$, which reduces everything to expressions of the form $a/2^n$ with $a \in \mathbb{Z}$, as expected. [example: Explicit Computation in $R[T]/(uT-1)$] Let $R = \mathbb{Z}$ and $u = 6$. By the theorem, $\mathbb{Z}_6 \cong \mathbb{Z}[T]/(6T - 1)$, whose elements are polynomials in $T$ with integer coefficients, subject to $6T = 1$. The element $\tfrac{1}{36} = \tfrac{1}{6^2}$ corresponds to $T^2 + (6T-1)$. To see that this is consistent, note that in $\mathbb{Z}[T]/(6T-1)$ one has $(6T)^2 = 1$, so $36 T^2 = 1$, confirming that $T^2$ is the inverse of $36$. More generally, a fraction $\tfrac{a}{6^n}$ with $a \in \mathbb{Z}$ corresponds to the polynomial class $aT^n + (6T-1)$. The ring $\mathbb{Z}_6 = \{a/6^n : a \in \mathbb{Z},\, n \geq 0\}$ is the subring of $\mathbb{Q}$ generated by $\mathbb{Z}$ and $1/6$, i.e., $\mathbb{Z}[1/6]$. [/example] ### Comparing the Two Constructions The fraction construction and the polynomial quotient construction each have advantages. Understanding when to use each clarifies the broader structure of localization. The **fraction construction** is better for direct computation: one can work concretely with numerators and denominators, and the explicit formulas for extension and contraction of ideals (developed in the preceding sections) arise naturally from the fraction description. For instance, the extended ideal $\iota(I) \cdot S^{-1}R = \{\tfrac{a}{s} : a \in I,\, s \in S\} = S^{-1}I$ is most naturally expressed in terms of fractions. The **polynomial quotient construction** is better for structural and categorical arguments. It makes the universal property automatic: any ring homomorphism out of $R[\{T_u\}_{u \in U}]$ that sends each $T_u$ to the inverse of $u$ descends to $R_U$. It also emphasizes that $R_U$ is a quotient of an explicit ring by an explicit ideal, which can be useful in Gröbner basis computations and other algorithmic contexts. The same interplay between "large quotient object" and "explicit description" appeared earlier in the course when constructing the tensor product $M \otimes_R N$: one formed the quotient of a huge free module by a huge submodule, proved the universal property, and then separately developed tools (base change, right-exactness, examples) to compute $M \otimes_R N$ in specific situations. The parallel is exact: universal properties are the right framework for proving uniqueness and functoriality, while explicit descriptions are the right tools for computation. [remark: The Universal Property as a Bridge] The universal property of $S^{-1}R$ characterizes morphisms **from** $S^{-1}R$ to other rings: any such morphism must send $u^{-1}$ to the inverse of $f(u)$. This makes the universal property powerful for studying how $S^{-1}R$ maps into other objects. But it gives almost no information about the **internal** structure of $S^{-1}R$ — which elements exist, how they multiply, what its ideals look like. For that, the fraction description is indispensable. The extension-contraction theory for the localization map $R \to S^{-1}R$ developed in the preceding sections depends essentially on the fraction description: one needs to know which elements of $R$ lie in the preimage of a given ideal of $S^{-1}R$, and this requires knowing how elements of $S^{-1}R$ look. The polynomial quotient construction, by itself, would make this calculation much harder. [/remark] The two constructions together — the explicit fraction model and the quotient of a polynomial ring — give a complete picture of the localization. The fraction model grounds the theory in computation; the polynomial quotient model grounds it in the universal property. Both perspectives are used throughout commutative algebra, and recognizing when to switch between them is part of the working algebraist's toolkit. Localization reveals which properties survive near a prime ideal and when switching between constructions is justified. Nakayama's lemma answers the fundamental question: when does supporting evidence actually imply a module vanishes? # 5. Nakayama's Lemma The results of Chapter 4 give us powerful tools for deciding when modules are finitely generated and when rings are Noetherian. But there is a subtler question that frequently arises in practice: when can we conclude that a module is *zero*? Or more usefully, when can we conclude that a submodule is the entire module? Nakayama's Lemma — deceptively simple in statement, extraordinary in application — answers exactly this question, and it does so by exploiting the structure of the Jacobson radical. This chapter develops the necessary ingredients: a module-theoretic version of the Cayley-Hamilton theorem, the Jacobson radical itself, and then the lemma and its immediate corollaries. ## The Cayley-Hamilton Theorem for Modules Suppose we have an endomorphism $f: M \to M$ of a module and an ideal $\mathfrak{a}$ such that $f$ maps $M$ into $\mathfrak{a}M$. Can we extract a polynomial relation that $f$ satisfies, with coefficients drawn from $\mathfrak{a}$? Over a field with a free module this is exactly classical Cayley-Hamilton, but here the module need not be free, and the ring need not be a field. The question is whether the argument survives in this generality — and it does, provided $M$ is finitely generated. The key insight is that one can encode the action of $f$ on a finite generating set by a matrix with entries in $\mathfrak{a}$, and then run the classical adjugate argument inside the endomorphism ring. [quotetheorem:2933] [citeproof:2933] [remark: Analogy with Classical Cayley-Hamilton] When $R = k$ is a field, $M = k^n$, and $f$ is multiplication by a matrix $A$, this reduces to the classical Cayley-Hamilton theorem: $A$ satisfies its characteristic polynomial $\chi_A(T) = \det(TI_n - A)$. The module-theoretic version is more general because the ideal $\mathfrak{a}$ need not be all of $R$, and $M$ need not be free. [/remark] It is worth noting what finite generation buys us here. The matrix $P$ in the proof is constructed by writing each $f(m_i)$ as an $\mathfrak{a}$-linear combination of the generators $m_1, \dots, m_n$. Without a finite generating set, no such matrix exists, and the entire adjugate argument collapses — there is no finite-dimensional object to take a determinant of. For infinite-dimensional modules, polynomial relations of this kind can fail entirely. The immediate payoff of the theorem comes from the special case $f = \operatorname{id}_M$, which gives the Determinant Trick below and sets up the path to Nakayama's Lemma. A particularly clean consequence arises when we take $f = \operatorname{id}_M$. [quotetheorem:2934] [citeproof:2934] This is sometimes called the "determinant trick" because it manufactures a single element $a \in \mathfrak{a}$ that acts as the identity on all of $M$. On its own, this says only that $\mathfrak{a}$ contains an idempotent-like element for $M$; the power comes when we combine it with a condition on $\mathfrak{a}$ that forces $1 - a$ to be invertible. Note how essential the hypothesis $\mathfrak{a}M = M$ is. If we only assume $\mathfrak{a}M \subseteq M$ (which is always true), we learn nothing. The hypothesis says every element of $M$ is already in the image of multiplication by $\mathfrak{a}$, a very strong condition. In the integers, for example, taking $\mathfrak{a} = (2)$ and $M = \mathbb{Z}$, we have $\mathfrak{a}M = 2\mathbb{Z} \neq \mathbb{Z}$, so the hypothesis fails and the trick says nothing. For the trick to have force, one needs a situation where $\mathfrak{a}M = M$ genuinely holds — and the next step is to understand which ideals $\mathfrak{a}$ convert this into the invertibility of $1 - a$. ## The Jacobson Radical To get from the determinant trick to a useful vanishing statement, we need to know when $1 - a$ is a unit. This is precisely what the Jacobson radical controls. [definition: Jacobson Radical] The **Jacobson radical** $\operatorname{Jac}(R)$ of a ring $R$ is the intersection of all maximal ideals of $R$: \begin{align*} \operatorname{Jac}(R) = \bigcap_{\mathfrak{m} \text{ maximal}} \mathfrak{m}. \end{align*} [/definition] [example: Jacobson Radical of Local Rings and $\mathbb{Z}$] Two important cases illustrate the definition at opposite extremes. **Local rings.** A ring $(R, \mathfrak{m})$ is local if it has a unique maximal ideal $\mathfrak{m}$. Then $\operatorname{Jac}(R) = \mathfrak{m}$ by definition. **The integers.** The maximal ideals of $\mathbb{Z}$ are the principal ideals $(p)$ for each prime $p$. An integer $n$ lies in $\operatorname{Jac}(\mathbb{Z})$ only if $p \mid n$ for every prime $p$. The only such integer is $0$, so $\operatorname{Jac}(\mathbb{Z}) = \{0\}$. [/example] The Jacobson radical has an elegant internal characterisation that does not require listing all maximal ideals explicitly. [quotetheorem:2860] [citeproof:2860] [remark: Alternative Characterisation] In some sources, $\operatorname{Jac}(R)$ is characterised as the largest ideal $I$ of $R$ such that $1 - a$ is a unit for every $a \in I$. The proposition above shows that these definitions agree. [/remark] The significance of this characterisation is that it gives us direct access to the units: elements of the Jacobson radical are precisely those that can be subtracted from $1$ to always produce a unit. ## Nakayama's Lemma When can we conclude that a finitely generated module is zero? The determinant trick produces an element $a \in \mathfrak{a}$ with $(1-a)M = 0$. For this to force $M = 0$, we need $1 - a$ to be a unit — and the Jacobson radical is precisely the set of elements of $R$ with this property. Nakayama's Lemma is the resulting conclusion: if $\mathfrak{a}$ sits inside the Jacobson radical and $\mathfrak{a}M = M$, then $M$ must be zero. [quotetheorem:2935] [citeproof:2935] [motivation] Why does the hypothesis $\mathfrak{a} \subseteq \operatorname{Jac}(R)$ matter? Without it, the conclusion can fail even when $\mathfrak{a}M = M$ is satisfied. Take $R = \mathbb{Z}$, $\mathfrak{a} = (5)$, and $M = \mathbb{Z}/5\mathbb{Z}$. Then $\mathfrak{a}M = (5)(\mathbb{Z}/5\mathbb{Z}) = \mathbb{Z}/5\mathbb{Z} = M$, so the hypothesis $\mathfrak{a}M = M$ holds. But $M = \mathbb{Z}/5\mathbb{Z} \neq 0$. What goes wrong? The Jacobson radical of $\mathbb{Z}$ is $\{0\}$ (since every non-zero integer misses some maximal ideal $(p)$), so $(5) \not\subseteq \operatorname{Jac}(\mathbb{Z})$. Correspondingly, the determinant trick produces $a = 5k$ for some $k$, but $1 - a$ need not be a unit in $\mathbb{Z}$: for instance, $a = 5$ gives $1 - a = -4$, which is not a unit. The Jacobson radical condition is what buys us the invertibility of $1 - a$, which is the keystone of the argument. The most common application is in local rings: for a local ring $(R, \mathfrak{m})$, we have $\operatorname{Jac}(R) = \mathfrak{m}$, so the lemma says: if $M$ is a finitely generated $R$-module and $\mathfrak{m}M = M$, then $M = 0$. This is enormously useful in local algebra because it gives a criterion for vanishing of modules without any explicit computation of their elements. [/motivation] The immediate corollary captures the form in which Nakayama's Lemma is most frequently applied: it tells us when a submodule is the whole module. [quotetheorem:2862] [citeproof:2862] [explanation: Reading Nakayama's Lemma] The submodule form is the version one encounters most in practice. In a local ring $(R, \mathfrak{m})$, it says: if $N$ is a submodule of a finitely generated module $M$ and $N$ generates $M$ modulo $\mathfrak{m}$ (i.e., $\mathfrak{m}M + N = M$), then $N$ already equals $M$. A particularly important special case arises when we want to find a generating set for a module $M$ over a local ring. Suppose $m_1, \dots, m_r \in M$ have images $\bar{m}_1, \dots, \bar{m}_r$ that span the $R/\mathfrak{m}$-vector space $M/\mathfrak{m}M$. Let $N$ be the submodule of $M$ generated by $m_1, \dots, m_r$. Then $N + \mathfrak{m}M = M$, and by the submodule form of Nakayama's Lemma, $N = M$. This means: to generate a finitely generated module over a local ring, it suffices to find elements whose images form a basis of $M/\mathfrak{m}M$. One can think of $M/\mathfrak{m}M$ as the "fibre" of $M$ at the closed point, and elements that generate this fibre lift to a generating set for the whole module. Finite generation is not merely a technical convenience — it is essential. The $p$-adic integers $\mathbb{Z}_p$ form a local ring with maximal ideal $\mathfrak{m} = (p)$. Consider the $\mathbb{Z}_p$-module $M = \mathbb{Q}_p$. Then $\mathfrak{m}M = p\mathbb{Q}_p = \mathbb{Q}_p = M$, since multiplication by $p$ is an automorphism of $\mathbb{Q}_p$. So the hypothesis $\mathfrak{m}M = M$ holds, and $\mathfrak{m} = \operatorname{Jac}(\mathbb{Z}_p)$. Yet $M = \mathbb{Q}_p \neq 0$. What saves us from a contradiction? The module $\mathbb{Q}_p$ is not finitely generated over $\mathbb{Z}_p$: any finite set of $p$-adic rationals generates a fractional ideal, never all of $\mathbb{Q}_p$. Without finite generation, the determinant trick never fires, and Nakayama's Lemma simply does not apply. [/explanation] The example below shows the submodule form at work in a concrete setting and illustrates how checking the fibre $M/\mathfrak{m}M$ suffices to determine an entire generating set. [example: Generating Sets Over Local Rings] Let $(R, \mathfrak{m})$ be a local ring and $M$ a finitely generated $R$-module. Suppose $M/\mathfrak{m}M$ is a one-dimensional $R/\mathfrak{m}$-vector space, spanned by the image of some $m \in M$. We show $M = Rm$. Let $N = Rm$. Then $N + \mathfrak{m}M \supseteq Rm + \mathfrak{m}M$. Since $\bar{m}$ spans $M/\mathfrak{m}M$, every $\bar{x} \in M/\mathfrak{m}M$ satisfies $\bar{x} = \bar{r}\bar{m}$ for some $r \in R$, meaning $x - rm \in \mathfrak{m}M$, so $x \in Rm + \mathfrak{m}M = N + \mathfrak{m}M$. Thus $N + \mathfrak{m}M = M$, and the submodule form of Nakayama's Lemma gives $N = M$. In particular, if $R$ is a local ring and $M$ is a finitely generated $R$-module with $M/\mathfrak{m}M \cong R/\mathfrak{m}$ as a one-dimensional vector space, then $M$ is a cyclic $R$-module. [/example] Nakayama's lemma is most powerful for finitely generated modules and generating sets over local rings. Integral extensions ask the inverse problem: when is one ring entirely determined by another? # 6. Integral and Finite Extensions (Part I) Chapter 5 developed the machinery of localisation, which allows us to focus on the behaviour of a ring at a prime ideal. Chapter 6 turns to a different but equally fundamental structural question: when is one ring "controlled" by another? Integral extensions formalise the idea that every element of an overring satisfies a monic polynomial equation with coefficients in the base ring. This is the commutative algebra analogue of algebraic extensions in field theory, but now the base is an arbitrary commutative ring rather than a field, which forces a more delicate analysis. The chapter introduces integral and finite extensions, establishes their equivalence under a finiteness hypothesis, and develops the integrality properties that govern quotients, localisations, and fields. ## Integral Elements Why require a monic polynomial? If we merely asked for any polynomial with coefficients in $R$ to vanish at $x$, we would get the notion of algebraic dependence, which is well-behaved when $R$ is a field but badly so otherwise. The monic condition is what gives integrality its arithmetic force: it ensures that the leading coefficient is 1, so no cancellation with the base ring can occur, and the resulting algebraic relationship is genuinely constraining. [definition: Integral Element] Let $A$ be an $R$-algebra. An element $x \in A$ is **integral over** $R$ (or **$R$-integral**) if there exists a monic polynomial $f \in R[T]$ such that $f(x) = 0$. [/definition] The monic condition is what distinguishes integrality from the weaker notion of algebraic dependence. Recall that $x \in A$ is $R$-algebraic if there is any polynomial (not necessarily monic) $f \in R[T]$ with $f(x) = 0$. Since a monic polynomial is in particular a polynomial, every $R$-integral element is $R$-algebraic. The converse fails when $R$ is not a field. [example: Integrality vs Algebraicity] Three instructive cases illustrate the gap between integrality and algebraicity. First, if $K$ is a field and $A$ is a $K$-algebra, then $x \in A$ is $K$-integral if and only if $x$ is $K$-algebraic. This is because any polynomial $f \in K[T]$ with $f(x) = 0$ can be divided by its leading coefficient (which is a unit in $K$) to produce a monic polynomial with the same root. Second, the set of $\mathbb{Z}$-integral elements in $\mathbb{Q}$ is exactly $\mathbb{Z}$. However, every element of $\mathbb{Q}$ is $\mathbb{Z}$-algebraic: for $a, b \in \mathbb{Z}$ with $b \neq 0$, the element $a/b$ is a root of the polynomial $bT - a \in \mathbb{Z}[T]$, which is not monic (unless $b = \pm 1$). To check that the only $\mathbb{Z}$-integral rationals are integers: if $a/b$ in lowest terms satisfies a monic polynomial $T^n + r_1 T^{n-1} + \cdots + r_n = 0$ with $r_i \in \mathbb{Z}$, multiplying through by $b^n$ gives $a^n = -b(r_1 a^{n-1} + \cdots + r_n b^{n-1})$, so $b \mid a^n$, and since $\gcd(a,b) = 1$ we get $b \mid 1$, meaning $b = \pm 1$ and the element is an integer. Third, the $\mathbb{Z}$-integral elements in $\mathbb{Q}(\sqrt{2})$ form the ring $\mathbb{Z}[\sqrt{2}]$, while in $\mathbb{Q}(\sqrt{5})$ they form the strictly larger ring $\mathbb{Z}\!\left[\frac{1+\sqrt{5}}{2}\right] \supsetneq \mathbb{Z}[\sqrt{5}]$. The element $\frac{1+\sqrt{5}}{2}$ is integral because it satisfies $T^2 - T - 1 = 0$, a monic polynomial over $\mathbb{Z}$, yet it does not lie in $\mathbb{Z}[\sqrt{5}]$. [/example] The field case shows that over a field, integrality reduces to algebraicity. The integer examples reveal the arithmetic content of the monic condition. ## The Faithful Module Criterion To make integrality amenable to algebraic manipulation, we need a criterion that translates the polynomial condition into a module-theoretic property. The key ingredient is faithfulness. [definition: Faithful Module] An $R$-module $M$ is **faithful** if the structural ring homomorphism $R \to \operatorname{End}(M)$ is injective, that is, for every $0 \neq r \in R$ there exists $m \in M$ such that $rm \neq 0$. [/definition] The condition says that $R$ acts non-trivially on $M$: no nonzero element of $R$ annihilates everything in $M$. A useful observation is that if $R \subset A$ are rings, then $A$ is automatically a faithful $R$-module. Indeed, if $0 \neq r \in R$, then $r \cdot 1_A = r \neq 0$ since the inclusion $R \hookrightarrow A$ is injective. [quotetheorem:2936] [citeproof:2936] This criterion is the key technical tool for proving that integrality is preserved under ring operations. Notice what faithfulness does: without it, the Cayley–Hamilton argument only shows that the polynomial $(x^n + r_1 x^{n-1} + \cdots + r_n)$ annihilates every element of $M$, which is too weak — we need the polynomial itself to be zero in $A$. Faithfulness is precisely what allows the jump from "annihilates $M$" to "equals zero". Without finite generation, the Cayley–Hamilton theorem does not apply, since it requires the endomorphism to act on a module with a finite generating set. The criterion connects forward to the equivalence theorem: a finite $R$-module is faithful and finitely generated, which is exactly what the criterion demands. ## Finite and Integral Algebras If single elements can be integral, what happens when we ask for the entire algebra to be integral? And is there a relationship between this global integrality and the more quantitative condition of being finitely generated as a module? The two conditions are genuinely different in general, but they coincide when combined with a finiteness hypothesis. [definition: Finite and Integral Algebras] Let $A$ be an $R$-algebra. 1. $A$ is **integral** over $R$ if every $a \in A$ is $R$-integral. 2. $A$ is **finite** over $R$ if $A$ is a finitely generated $R$-module. [/definition] The word "finite" here refers to finite generation as a module, not as an algebra. These two notions, together with finite generation as an integral algebra, are equivalent. [quotetheorem:2937] [citeproof:2937] This theorem says that the two ways of making an $R$-algebra "small"—generating it by finitely many integral elements, or having it be finitely generated as a module—coincide. The polynomial ring $R[T]$ is a finitely generated $R$-algebra but not a finite $R$-algebra (it has infinite $R$-module rank), so integrality of the generators is genuinely necessary. ## The Integral Closure Is the sum of two integral elements always integral? The definition makes this far from obvious: $x$ satisfies a monic polynomial of degree $n$ and $y$ satisfies one of degree $m$, but $x + y$ might satisfy a polynomial of degree $nm$ or more, and there is no transparent way to write it down. The answer is yes, but the cleanest proof goes through the equivalence theorem rather than by direct polynomial manipulation. [quotetheorem:2865] [citeproof:2865] The significance of this result is that the collection of all elements of $B$ satisfying an integral condition over $A$ forms a coherent algebraic object — a ring — not just a scattered set. This ring is the natural "integrally saturated" analogue of the base ring $A$ within $B$: it contains everything that is algebraically bound to $A$ via monic polynomials. The proof also makes clear why the equivalence theorem is the right tool: the claim is inherently about two elements together, and the finite module structure of the algebra they generate is what makes the argument work. [definition: Integral Closure] Let $A \subset B$ be rings. 1. The **integral closure** of $A$ in $B$ is $\overline{A} = \{b \in B : b \text{ is integral over } A\}$. 2. $A$ is **integrally closed in $B$** if $\overline{A} = A$. When $A$ is an integral domain: 1. The **integral closure** of $A$ is its integral closure in $\operatorname{Frac}(A)$. 2. $A$ is **integrally closed** if $A$ is integrally closed in $\operatorname{Frac}(A)$. [/definition] By the previous proposition, $\overline{A}$ is always an $A$-subalgebra of $B$. [example: Failure of Integral Closure] The ring $\mathbb{Z}[\sqrt{5}]$ is not integrally closed. The element $\alpha = \frac{1+\sqrt{5}}{2}$ belongs to $\mathbb{Q}(\sqrt{5}) = \operatorname{Frac}(\mathbb{Z}[\sqrt{5}])$ but not to $\mathbb{Z}[\sqrt{5}]$ itself, yet $\alpha$ satisfies $T^2 - T - 1 = 0$, which is monic with coefficients in $\mathbb{Z} \subset \mathbb{Z}[\sqrt{5}]$. So $\alpha$ is $\mathbb{Z}[\sqrt{5}]$-integral but $\alpha \notin \mathbb{Z}[\sqrt{5}]$. [/example] The golden ratio appearing as a witness to failure of integral closure illustrates why algebraic number theory requires working with the full ring of integers $\mathcal{O}_K$ rather than just $\mathbb{Z}[\sqrt{d}]$. ## Transitivity and Stability Does integrality compose in towers? If $C$ is integral over $B$ and $B$ is integral over $A$, is $C$ necessarily integral over $A$? This is not formal: given $c \in C$, we can write down a monic polynomial over $B$ that $c$ satisfies, but its coefficients live in $B$, not in $A$. To get a monic polynomial over $A$, we need to eliminate the $B$-coefficients, which requires the transitivity theorem. A second natural question is stability: if $B$ is integral over $A$, does integrality survive passage to quotients and localisations? These operations change the rings involved, so preservation of integrality is a structural coherence property. [quotetheorem:2866] [citeproof:2866] The transitivity theorem is indispensable for the definition of integral closure to be well-behaved: it ensures that $\overline{A}$ is already integrally closed — if $x$ is integral over $\overline{A}$, then by transitivity, $x$ is integral over $A$, so $x \in \overline{A}$. This makes taking integral closures an idempotent operation, not an infinite process that requires iteration. [quotetheorem:2867] [citeproof:2867] The stability theorem matters because integral extensions are supposed to capture a genuine relationship between rings, not just a property of a specific presentation. Stability under quotients says that if we impose additional relations (pass to a quotient), the overring remains controlled by the base. Stability under localisation says that zooming in on a prime preserves this control. Together, they mean that integrality is a local-global property compatible with the standard tools of commutative algebra. Integrality is essential here: without it, localising a module-finite extension can break the module structure entirely. [remark: Integral Closure Is Integrally Closed] For rings $A \subset B$, the integral closure $\overline{A}$ of $A$ in $B$ is itself integrally closed in $B$. If $x \in B$ is integral over $\overline{A}$, then by transitivity of integrality ($A \subset \overline{A} \subset \overline{A}[x]$, both integral extensions), $x$ is $A$-integral, so $x \in \overline{A}$. This confirms that taking the integral closure is an idempotent operation. [/remark] ## Integrality, Fields, and Maximal Ideals The most profound structural consequence of integrality is how it interacts with maximality. Before proving this, we need a lemma about units and fields. [quotetheorem:2868] [citeproof:2868] This lemma connects integrality to prime spectrum behaviour through the following key corollary. [quotetheorem:2869] [citeproof:2869] This result is the key to understanding the Spec correspondence in integral extensions. Recall that $\operatorname{Spec}(B) \to \operatorname{Spec}(A)$ denotes the map sending $\mathfrak{q} \mapsto \mathfrak{q} \cap A$. The theorem says that this map sends maximal ideals to maximal ideals and, conversely, maximal ideals in $B$ contract to maximal ideals in $A$. In other words, the closed points of $\operatorname{Spec}(B)$ map onto the closed points of $\operatorname{Spec}(A)$ — the prime spectrum of $B$ lies above that of $A$ without collapsing dimension in a way that disrupts maximality. Integrality is essential for this: for the non-integral extension $\mathbb{Z} \subset \mathbb{Q}$, the unique maximal ideal $(0)$ of $\mathbb{Q}$ contracts to the non-maximal prime $(0)$ of $\mathbb{Z}$, showing the statement fails without integrality. ## Unique Factorisation Domains Are Integrally Closed Not every domain is integrally closed, as the example $\mathbb{Z}[\sqrt{5}]$ shows. However, the following theorem provides a large natural class of integrally closed rings. [quotetheorem:2938] [citeproof:2938] [remark: Fields and Polynomial Rings] As a consequence, all fields, all PIDs (in particular $\mathbb{Z}$), and all polynomial rings $k[T_1, \ldots, T_n]$ over a field $k$ are integrally closed. The ring $\mathbb{Z}[\sqrt{5}]$ is not a UFD (for instance, $4 = 2 \cdot 2 = (1 + \sqrt{5})(1 - \sqrt{5})$ gives two distinct factorisations), consistent with it failing to be integrally closed. [/remark] Integral extensions exhibit rigidity absent in general ring homomorphisms, and UFDs are integrally closed. Noether normalization and the Nullstellensatz connect this algebraic rigidity to the geometry of spectra. # 7. Noether's Normalization Theorem and Hilbert's Nullstellensatz The two theorems at the heart of this chapter are among the most striking connections between algebra and geometry in the entire subject. Noether's normalization theorem says that every finitely generated algebra over a field can be "sandwiched" between a polynomial ring and itself in a finite extension — that after a clever linear change of variables, any such algebra looks like it sits finitely over a pure polynomial ring. Hilbert's Nullstellensatz then uses this structure to establish a perfect dictionary between algebra and geometry: radical ideals in $k[T_1,\dots,T_n]$ correspond bijectively to algebraic subsets of $\Omega^n$, and — most strikingly — for an algebraically closed field, the maximal ideals of $\Omega[T_1,\dots,T_n]$ are in bijection with the points of $\Omega^n$. These results are the algebraic foundation of classical algebraic geometry. ## Algebraic Independence and the Normalization Statement Given a finitely generated $k$-algebra $A = k[x_1,\dots,x_m]$, what structure can we extract? The generators satisfy all manner of polynomial relations, making $A$ hard to analyse directly. The key question is: can we always find a subalgebra of $A$ that looks like a clean polynomial ring — no relations, no torsion — and over which $A$ is a finite extension? If so, we can leverage the theory of finite integral extensions developed in Chapter 6 to study arbitrary finitely generated algebras. The answer is yes, and the right notion to make this precise is algebraic independence. [definition: Algebraic Independence] Let $A$ be a $k$-algebra. Elements $x_1,\dots,x_n \in A$ are **$k$-algebraically independent** if $p(x_1,\dots,x_n) \neq 0$ for every nonzero polynomial $p \in k[T_1,\dots,T_n]$. [/definition] Algebraic independence is the correct analogue, in the setting of algebras, of linear independence in vector spaces. Equivalently, $x_1,\dots,x_n$ are $k$-algebraically independent if and only if the evaluation map $k[T_1,\dots,T_n] \to A$ sending $T_i \mapsto x_i$ is injective. When this holds, the $k$-subalgebra $k[x_1,\dots,x_n]$ is isomorphic to a polynomial ring — there are no algebraic relations among the $x_i$ with coefficients in $k$. The following example illustrates what the normalization theorem achieves in a concrete case. [example: Laurent Polynomial Ring] Let $A = k[T, T^{-1}]$ where $k$ is an infinite field. The inclusion $k[T] \subset k[T, T^{-1}]$ is not a finite extension: $T^{-1}$ is not integral over $k[T]$. To see this, suppose $T^{-1}$ satisfies a monic polynomial over $k[T]$ of degree $n$, meaning $(T^{-1})^n \in \operatorname{span}_{k[T]}\{1, T^{-1}, \dots, (T^{-1})^{n-1}\}$. Multiplying through by $T^n$ gives $1 \in \operatorname{span}_{k[T]}\{T, T^2, \dots, T^n\}$, which is impossible since every element of the right side has positive degree. Noether's theorem predicts that there must be a different subalgebra $A' \subset A$ isomorphic to a polynomial ring over which $A$ is finite. Choose any nonzero scalar $c \in k$ and set $y = T^{-1} - cT$. Since $T^{-1} = y + cT$, we see that $T$ and $y$ together generate $A$ as a $k$-algebra, so $A = k[y][T]$. To check finiteness, we need to show $T$ is integral over $k[y] = k[T^{-1} - cT]$. Starting from the tautological relation $T^{-1} \cdot T - 1 = 0$ in $A$ and substituting $T^{-1} = y + cT$: \begin{align*} (y + cT) \cdot T - 1 &= 0 \\ cT^2 + yT - 1 &= 0. \end{align*} Since $c \neq 0$ we may divide by $c$, giving $T^2 + c^{-1}yT - c^{-1} = 0$, a monic polynomial over $k[y]$ satisfied by $T$. Thus $T$ is integral over $k[y]$, and $A = k[y][T]$ is a finite extension of $k[y] \cong k[T]$. [/example] The general statement now follows. [quotetheorem:2939] [citeproof:2939] [remark: Matrix Interpretation and Necessity of Hypotheses] The proof shows that the generators $(x_1,\dots,x_n)$ are obtained from the original generators $(t_1,\dots,t_\ell)$ of $A$ by an upper-triangular matrix $P$ with $1$s on the diagonal (an invertible matrix in $M_{\ell \times \ell}(k)$). More precisely, one can show that there is a nonzero polynomial $h$ in the entries above the diagonal of $P$ such that any $P$ with $h(P) \neq 0$ gives a valid normalization. The set of "bad" choices of $P$ is the zero set of $h$, and so a generic $P$ works. The hypothesis that $A$ is finitely generated is essential: the theorem fails for $k$-algebras that are not finitely generated. For instance, take $A = k(T)$, the field of rational functions. This is a $k$-algebra, and it contains no polynomial subring $k[x_1,\dots,x_n]$ with $n \geq 1$ over which $A$ is finite — were $k(T)$ finite over some $k[x_1,\dots,x_n]$ with $n \geq 1$, then $k[x_1,\dots,x_n]$ would be a field by the going-up / lying-over argument from Chapter 6, contradicting the fact that polynomial rings are never fields. And $A$ is not finite over $k$ itself, since $k(T)$ is an infinite-dimensional $k$-vector space. So no normalization exists. Over finite fields that are too small, it may happen that no $P$ with $h(P) \neq 0$ exists, because the finite field has too few elements to avoid the zero set of $h$. In that case a polynomial change of variables (rather than a linear one) replaces the linear substitution above — the bound on the degree of the polynomial change depends on the size of the field and the degree of $h$. [/remark] ## Zariski's Lemma and Its Consequences Normalization immediately raises a sharp question about fields: if $K$ is a field that happens to be finitely generated as a $k$-algebra, is $K$ a finite extension of $k$ in the classical sense? At first glance this seems unreasonable — a field finitely generated as an algebra is allowed to contain transcendental elements. But the existence of a transcendental element forces the polynomial ring $k[x]$ to embed in $K$, and $k[x]$ is not a field, which creates a conflict once $K$ is required to be integral over its normalization. Zariski's lemma makes this tension precise and resolves it decisively. [quotetheorem:2872] [citeproof:2872] This seemingly simple statement has an important interpretation: a field that is finitely generated as an algebra over $k$ must in fact be a finite extension of $k$ in the classical sense. This bridges the two meanings of "finitely generated": as an algebra (arbitrary polynomial combinations of generators) versus as a module (finite $k$-linear combinations). ## Algebraic Subsets and the Ideal-Variety Correspondence Polynomial equations cut out geometric shapes in affine space: $T_1^2 + T_2^2 - 1 = 0$ traces a circle, $T_1 T_2 = 0$ traces two coordinate axes, a single nonzero constant has no solutions at all. What can we say in general about the solution sets of systems of polynomial equations $\{f_1 = 0, \dots, f_t = 0\}$ in $\Omega^n$? And can we recover the equations from the solution set, or is information lost? Making these questions precise requires two maps — one from equations to geometry and one from geometry back to equations — and the Nullstellensatz will show when they are inverse to each other. Fix fields $k \subset \Omega$ where $\Omega$ is algebraically closed. [definition: Algebraic Subsets and Vanishing Ideal] Let $n \geq 0$. 1. For a subset $S \subset k[T_1,\dots,T_n]$, the **$k$-algebraic subset** defined by $S$ is \begin{align*} V(S) = \{\underline{x} \in \Omega^n : f(\underline{x}) = 0 \text{ for all } f \in S\}. \end{align*} 2. For a subset $X \subset \Omega^n$, the **vanishing ideal** of $X$ is \begin{align*} I(X) = \{f \in k[T_1,\dots,T_n] : f(\underline{x}) = 0 \text{ for all } \underline{x} \in X\}. \end{align*} [/definition] An **algebraic subset** of $\Omega^n$ (without the $k$-prefix) is a $\Omega$-algebraic subset, i.e., one defined by polynomials in $\Omega[T_1,\dots,T_n]$. Every $k$-algebraic subset is algebraic. The $k$-prefix carries real information: for example, $\{i\} \subset \mathbb{C}^1$ is an algebraic subset, but not an $\mathbb{R}$-algebraic subset — any real polynomial vanishing at $i$ must also vanish at $-i$ (since complex roots of real polynomials come in conjugate pairs), so every $\mathbb{R}$-algebraic set containing $i$ must also contain $-i$. Note also that replacing $S$ by the ideal $\mathfrak{a}$ it generates does not change $V(S)$: $V(\mathfrak{a}) = V(S)$. The maps $V(\cdot)$ and $I(\cdot)$ satisfy several formal properties that establish them as a Galois connection: 1. Both $I(\cdot)$ and $V(\cdot)$ are inclusion-reversing: $X \subset Y$ implies $I(X) \supset I(Y)$, and $S \subset T$ implies $V(S) \supset V(T)$. 2. For $S \subset k[T_1,\dots,T_n]$, one has $S \subset I(V(S))$: every polynomial in $S$ vanishes on $V(S)$ by definition. 3. For $X \subset \Omega^n$, one has $X \subset V(I(X))$: every point of $X$ satisfies every polynomial in $I(X)$ by definition. 4. For a $k$-algebraic set $X = V(\mathfrak{a})$, one has $X = V(I(X))$: this follows from (3) and the fact that $I(V(\mathfrak{a})) \supset \mathfrak{a}$, which by inclusion-reversal gives $V(I(X)) \subset V(\mathfrak{a}) = X$. 5. For any $X \subset \Omega^n$, the ideal $I(X)$ is radical: if $f^\ell \in I(X)$ then $f^\ell(\underline{x}) = 0$ for all $\underline{x} \in X$, and since $\Omega$ is an integral domain, $f(\underline{x}) = 0$. Property (4) says the map $X \mapsto I(X)$ is injective on $k$-algebraic sets. The strong Nullstellensatz (proved below) will give its right inverse and complete the bijection. ## Hilbert's Nullstellensatz The maps $V(\cdot)$ and $I(\cdot)$ set up a correspondence between ideals and algebraic sets, but the round trip $\mathfrak{a} \mapsto V(\mathfrak{a}) \mapsto I(V(\mathfrak{a}))$ does not simply return $\mathfrak{a}$. We always have $\mathfrak{a} \subset I(V(\mathfrak{a}))$, but the containment can be strict: the ideal $(T^2)$ and the ideal $(T)$ in $k[T]$ define the same algebraic set $\{0\}$, so $I(V((T^2))) = I(\{0\}) = (T) \supsetneq (T^2)$. What the round trip actually recovers is the radical $\sqrt{\mathfrak{a}}$. This is Hilbert's Nullstellensatz — and why it requires algebraic closure: over $\mathbb{R}$, the ideal $(T^2+1)$ is proper but $V(T^2+1) = \varnothing$, which would force $I(V((T^2+1))) = I(\varnothing) = k[T]$, breaking the correspondence entirely. [quotetheorem:2940] [citeproof:2940] The weak Nullstellensatz has a clean Hilbert-style formulation: if $f_1,\dots,f_t \in k[T_1,\dots,T_n]$ have no common zero in $\Omega^n$, then there exist $p_1,\dots,p_t \in k[T_1,\dots,T_n]$ such that $\sum_{i=1}^t p_i f_i = 1$. This is the algebraic certificate that a system of polynomial equations is inconsistent. Algebraic closure is indispensable here. Over the reals, the polynomial $T^2 + 1 \in \mathbb{R}[T]$ generates a proper ideal (it has no real roots, so it is not a unit), yet $V(T^2+1) = \varnothing \subset \mathbb{R}^1$. The weak Nullstellensatz would then claim $1 \in (T^2+1)$, but that is false: $T^2+1$ does not divide $1$ in $\mathbb{R}[T]$. The proof breaks because the quotient $\mathbb{R}[T]/(T^2+1) \cong \mathbb{C}$ is finite over $\mathbb{R}$ but does not embed into $\mathbb{R}$ itself — algebraic closure is what forces a $k$-algebra homomorphism into $\Omega$ to exist. The strong Nullstellensatz sharpens the picture by pinpointing exactly which ideal $I(V(\mathfrak{a}))$ is: always the radical $\sqrt{\mathfrak{a}}$, regardless of $\mathfrak{a}$. In particular, if $\mathfrak{a}$ is already radical then $V$ and $I$ are perfectly inverse to each other on the pair $(\mathfrak{a}, V(\mathfrak{a}))$. This motivates the next result. ## The Bijection Between Radical Ideals and Algebraic Sets The strong Nullstellensatz completes the Galois correspondence begun with the properties of $I(\cdot)$ and $V(\cdot)$. [quotetheorem:2941] [citeproof:2941] This is a profound result: it says that the geometry of $k$-algebraic subsets of $\Omega^n$ is entirely captured by the algebra of radical ideals of $k[T_1,\dots,T_n]$, and vice versa. Inclusion of algebraic sets corresponds to reverse inclusion of radical ideals. What the bijection does not see is the scheme-theoretic information encoded in non-radical ideals. The ideals $(T)$ and $(T^2)$ in $k[T]$ both correspond to the same algebraic set $\{0\}$ under the Nullstellensatz correspondence — only the radical $(T)$ appears on the ideal side. The "multiplicity" or "thickness" of the zero at the origin, which $(T^2)$ records, is invisible to classical algebraic geometry. Scheme theory remedies this by replacing algebraic sets with $\operatorname{Spec}(k[T]/\mathfrak{a})$ for arbitrary $\mathfrak{a}$, thereby retaining the full ideal and not just its radical. The correspondence extends naturally to the projective setting, where $\Omega^n$ is replaced by projective space $\mathbb{P}^n(\Omega)$ and polynomial rings are replaced by graded rings, with the caveat that the irrelevant ideal $(T_0,\dots,T_n)$ must be handled separately. But even in the affine setting, the bijection is already the central organising principle of classical algebraic geometry. Several important special cases deserve attention. [remark: Points, Maximal Ideals, and the Correspondence] 1. Every prime ideal is radical (if $x^n \in \mathfrak{p}$ then $x \in \mathfrak{p}$). So prime ideals correspond to a special class of algebraic sets, which we will identify shortly. 2. For any field $k$, the map $k^n \to \operatorname{mspec}(k[T_1,\dots,T_n])$ sending $\underline{x} = (x_1,\dots,x_n) \mapsto \mathfrak{m}_{\underline{x}} = (T_1 - x_1, \dots, T_n - x_n)$ is injective. The ideal $\mathfrak{m}_{\underline{x}}$ is maximal because it is the kernel of the surjective evaluation homomorphism $k[T_1,\dots,T_n] \to k$, $T_i \mapsto x_i$. Injectivity holds because if $\underline{x} \neq \underline{y}$ then $x_i \neq y_i$ for some $i$, which forces $(T_i - y_i) - (T_i - x_i) = x_i - y_i$ to lie in $\mathfrak{m}_{\underline{x}} + \mathfrak{m}_{\underline{y}}$, but $x_i - y_i$ is a nonzero scalar, so $1 \in \mathfrak{m}_{\underline{x}} + \mathfrak{m}_{\underline{y}}$, giving $\mathfrak{m}_{\underline{x}} \neq \mathfrak{m}_{\underline{y}}$. In general this map is not surjective: for instance $(T^2 + 1)$ is a maximal ideal of $\mathbb{R}[T]$ (being the kernel of $\mathbb{R}[T] \to \mathbb{C}$, $T \mapsto i$) but does not have the form $(T - x)$ for any $x \in \mathbb{R}$. 3. For every $\underline{x} \in k^n$, one has $\mathfrak{m}_{\underline{x}} = I(\{\underline{x}\})$: the generators of $\mathfrak{m}_{\underline{x}}$ vanish at $\underline{x}$, giving $\mathfrak{m}_{\underline{x}} \subset I(\{\underline{x}\})$, and since $\mathfrak{m}_{\underline{x}}$ is maximal and $I(\{\underline{x}\})$ is proper ($1$ does not vanish at $\underline{x}$), we get equality. 4. Since the bijection of the Nullstellensatz correspondence reverses inclusion, the **minimal** nonempty $k$-algebraic subsets of $\Omega^n$ correspond to the **maximal** radical ideals, i.e., the maximal ideals of $k[T_1,\dots,T_n]$. [/remark] The case $k = \Omega$ (the algebraically closed field itself) gives the cleanest statement. [quotetheorem:2875] [citeproof:2875] This corollary is the algebraic geometry motto: **points are maximal ideals**. It explains why, in modern algebraic geometry, one defines the "points" of a scheme to be the prime ideals of the corresponding ring — generalizing the classical case where, over an algebraically closed field, the closed points (maximal ideals) are precisely the geometric points. For $k \subsetneq \Omega$ not algebraically closed, the maximal ideals of $k[T_1,\dots,T_n]$ correspond to minimal nonempty $k$-algebraic subsets of $\Omega^n$, which need not be singletons. For example: - In $\mathbb{C}^1$, the subset $\{i, -i\}$ is a minimal $\mathbb{R}$-algebraic set, corresponding to the maximal ideal $(T^2 + 1)$ of $\mathbb{R}[T]$. - In characteristic $0$ (or more generally when $k \subset \Omega$ is separable), if $\{\underline{x}\} \subset \Omega^n$ is $k$-algebraic then $\underline{x} \in k^n$, so the minimal sets that do not come from $k^n$ always have more than one element. - In characteristic $p$, the situation can be more subtle: for $k = \mathbb{F}_p(X)$ and $\Omega = k^{\mathrm{alg}}$, the singleton $\{X^{1/p}\}$ is $k$-algebraic since $V(T^p - X) = \{X^{1/p}\}$ (as $T^p - X = (T - X^{1/p})^p$ over $\Omega$), yet $X^{1/p} \notin k$. ## Irreducible Algebraic Sets and Prime Ideals The bijection between radical ideals and algebraic sets raises a further question: can every algebraic set be broken into simpler, "atomic" pieces? The hyperbola $V(T_1 T_2) = V(T_1) \cup V(T_2)$ splits into two coordinate axes, each of which cannot be split further. These indecomposable pieces are the irreducible algebraic sets, and identifying them is the geometric counterpart of factoring an ideal into primes. What makes the Nullstellensatz so powerful here is that the algebraic characterisation of irreducibility is clean: the atoms of the geometry are exactly the algebraic sets whose vanishing ideals are prime. [definition: Irreducible Algebraic Set] An $\Omega$-algebraic set $X \subset \Omega^n$ is **irreducible** if it cannot be written as $X = X_1 \cup X_2$ with $X_1, X_2$ proper algebraic subsets of $X$. [/definition] Irreducibility is the geometric counterpart of primeness in the ideal-variety dictionary. [quotetheorem:2942] [citeproof:2942] Together, the results of this chapter give a complete algebraic framework for classical algebraic geometry over an algebraically closed field: geometric objects (algebraic sets) are classified by algebra (radical ideals), their irreducibility is detected by primeness, and their points correspond to maximal ideals. To see the decomposition into irreducible components concretely, consider $X = V(T_1 T_2(T_1 - T_2)) \subset \Omega^2$. This is the union of three lines: $X_1 = V(T_1) = \{(0, t) : t \in \Omega\}$, $X_2 = V(T_2) = \{(t, 0) : t \in \Omega\}$, and $X_3 = V(T_1 - T_2) = \{(t,t) : t \in \Omega\}$. Each $X_i$ is irreducible — its vanishing ideal $(T_1)$, $(T_2)$, $(T_1 - T_2)$ is prime — and $X = X_1 \cup X_2 \cup X_3$ is the decomposition of $X$ into its irreducible components. Algebraically, the radical of the corresponding ideal is $\sqrt{(T_1 T_2(T_1-T_2))} = (T_1) \cap (T_2) \cap (T_1 - T_2)$, and this intersection of prime ideals is the primary decomposition of the radical ideal $I(X)$. In general, every algebraic set can be written uniquely as a finite union of irreducible algebraic sets (the irreducible components), with none containing another, and this decomposition corresponds exactly to the primary decomposition of the radical ideal. The deeper study of this dictionary — through Krull dimension, primary decomposition, and the geometry of spectra — forms the substance of the remainder of the course. The Nullstellensatz reveals that radical ideals correspond bijectively to varieties, establishing the geometric meaning of prime ideals. Refining integrality to track how specific prime ideals relate under extensions sharpens this correspondence. # 8. Integral and Finite Extensions (Part II) The theory of integral extensions, developed in Chapters 6 and 7, concerned elements satisfying monic polynomial equations over the entire base ring $A$. This chapter refines that theory by asking: what if the polynomial coefficients are constrained to lie in a specified ideal $\mathfrak{a} \subset A$ rather than in all of $A$? The resulting notion of $\mathfrak{a}$-integrality turns out to interact beautifully with the integral closure $\overline{A}$ of $A$ in $B$: the $\mathfrak{a}$-integral elements of $B$ are precisely the elements of the radical $\sqrt{\mathfrak{a}\overline{A}}$. This identification, together with a characterisation of $\mathfrak{a}$-integrality via minimal polynomials over fraction fields, completes the foundational picture of integral extensions. ## Integrality over an Ideal Ordinary integrality over $A$ asks only that the polynomial coefficients land somewhere in $A$. But suppose we want to study how prime ideals behave under ring extensions — say, whether a prime $\mathfrak{p} \subset A$ contracts from a prime of $B$. Ordinary integrality cannot see this: an element can be integral over $A$ while its integrality witness has coefficients scattered throughout $A$, with no special relationship to $\mathfrak{p}$. What we need is a refined notion where the polynomial coefficients are forced to lie inside a specified ideal $\mathfrak{a} \subset A$. [definition: Integral Over an Ideal] Let $A \subset B$ be rings and $\mathfrak{a}$ an ideal of $A$. 1. An element $x \in B$ is **integral over $\mathfrak{a}$** (or **$\mathfrak{a}$-integral**) if there exists a monic polynomial \begin{align*} f = T^n + a_1 T^{n-1} + \cdots + a_n \in A[T], \quad a_1, \dots, a_n \in \mathfrak{a}, \end{align*} with $n \geq 1$, such that $f(x) = 0$. 2. The **integral closure of $\mathfrak{a}$ in $B$** is the set $\{x \in B : x \text{ is } \mathfrak{a}\text{-integral}\}$. [/definition] When $\mathfrak{a} = A$, the condition reduces to ordinary integrality over $A$, since every coefficient is automatically in $A$. For a proper ideal $\mathfrak{a} \subsetneq A$, the constraint is genuinely stronger: the element must satisfy a monic polynomial whose non-leading coefficients are all "small" in the sense of belonging to $\mathfrak{a}$. There is an important structural difference between the integral closure of $A$ in $B$ and the integral closure of $\mathfrak{a}$ in $B$. The former contains $1_A$ (since $1_A$ satisfies $T - 1 = 0$) and forms a subring of $B$. The latter does not necessarily contain $1_A$ — if $1_A$ were $\mathfrak{a}$-integral, it would satisfy $1 + a_1 + \cdots + a_n = 0$ for some $a_i \in \mathfrak{a}$, which would mean $-1 \in \mathfrak{a}$, forcing $\mathfrak{a} = A$. So for a proper ideal $\mathfrak{a}$, the integral closure of $\mathfrak{a}$ in $B$ is not a subring of $B$ in general. It is, however, an ideal of $\overline{A}$, as the following theorem makes precise. ## The Radical Characterisation The set of $\mathfrak{a}$-integral elements is not a subring — but does it have any algebraic structure at all? From the definition alone, closure under sums and products is not obvious. The answer identifies the $\mathfrak{a}$-integral elements with a familiar ideal-theoretic object. [quotetheorem:2877] [citeproof:2877] The theorem says that the $\mathfrak{a}$-integral elements form the preimage of the radical $\sqrt{\mathfrak{a}\overline{A}}$ under the inclusion $\overline{A} \hookrightarrow B$. Since $\sqrt{\mathfrak{a}\overline{A}}$ is an ideal of $\overline{A}$, it follows in particular that the set of $\mathfrak{a}$-integral elements is closed under addition and multiplication — something that is not at all obvious from the definition. An immediate consequence simplifies the condition by replacing $\mathfrak{a}$ with its own radical. [quotetheorem:2943] [citeproof:2943] This corollary is a useful reduction: when testing whether $b$ is integral over $\mathfrak{a}$, one may replace $\mathfrak{a}$ by its radical without changing the answer. In particular, $b$ is $\mathfrak{a}$-integral iff $b$ is $(a)$-integral for every $a \in \mathfrak{a}$, and the two conditions $b^n \in \mathfrak{a}$ and $b^m \in \sqrt{\mathfrak{a}}$ are equivalent for appropriate choices of $n$ and $m$. ## Integrality over Ideals in Integrally Closed Domains The radical characterisation requires computing the full integral closure $\overline{A}$ of $A$ in $B$, which can be difficult. When $A$ is already integrally closed — so $\overline{A} = A$ in $\operatorname{Frac}(A)$ — can one bypass $\overline{A}$ entirely and characterise $\mathfrak{a}$-integrality directly via the minimal polynomial? The answer is yes. [quotetheorem:2879] [citeproof:2879] [remark: Significance of Integrally Closed Domains] The hypothesis that $A$ is integrally closed is essential. In the proof, it is used to identify the $\mathfrak{a}$-integral elements of $\operatorname{Frac}(A)$ itself: because $A$ is its own integral closure in $\operatorname{Frac}(A)$, the integral closure of $\mathfrak{a}$ in $\operatorname{Frac}(A)$ is $\sqrt{\mathfrak{a} \cdot A} = \sqrt{\mathfrak{a}}$, and so the integrally closed condition guarantees that $\mathfrak{a}$-integral elements of $\operatorname{Frac}(A)$ actually live in $A$. Without this, the coefficients of the minimal polynomial could be $\mathfrak{a}$-integral fractions that are not in $A$. Important classes of integrally closed domains include: every field (since any element of a field extension satisfying a monic polynomial over a field already lies in the field), every UFD (since prime factorisation prevents non-integer rationals from being integral), the ring of integers $\mathbb{Z}$, and more generally the ring of integers $\mathcal{O}_K$ of any number field $K$. The latter example connects the theorem to classical algebraic number theory: the norm of an algebraic integer (the constant term of its minimal polynomial, up to sign) lies in $\mathbb{Z}$. [/remark] [example: Integral Closure of an Ideal in a Number Ring] Let $A = \mathbb{Z}$, $B = \mathbb{Z}[i]$ (the Gaussian integers), and $\mathfrak{a} = (2) \subset \mathbb{Z}$. We determine which Gaussian integers are $(2)$-integral. An element $\alpha = a + bi \in \mathbb{Z}[i]$ is $(2)$-integral if it satisfies a monic polynomial $T^n + c_1 T^{n-1} + \cdots + c_n$ with $c_j \in (2) = 2\mathbb{Z}$. Taking $n = 2$: this asks for $c_1, c_2 \in 2\mathbb{Z}$ with $\alpha^2 + c_1 \alpha + c_2 = 0$. By the theorem, the $(2)$-integral elements in $\mathbb{Z}[i]$ are exactly $\sqrt{(2)\overline{\mathbb{Z}}} = \sqrt{(2)\mathbb{Z}[i]}$. Now $(2) = -i(1+i)^2$ in $\mathbb{Z}[i]$ (since $(1+i)^2 = 2i$, so $2 = (1+i)^2 / i = -i(1+i)^2$), and $-i$ is a unit. Thus $(2)\mathbb{Z}[i] = ((1+i)^2)$, and its radical is $(1+i)$ (since $\mathbb{Z}[i]$ is a UFD, radical of a principal ideal $(p^k)$ is $(p)$). Therefore, the $(2)$-integral elements in $\mathbb{Z}[i]$ are the multiples of $(1+i)$, i.e., the elements of $(1+i)\mathbb{Z}[i] = \{a + bi : a \equiv b \pmod{2}\}$. To verify: $1+i$ satisfies $(T - (1+i))(T - (1-i)) = T^2 - 2T + 2$, which has coefficients $-2, 2 \in (2)$, confirming $1+i$ is $(2)$-integral. Meanwhile, $1 \in \mathbb{Z}[i]$ does not lie in $(1+i)\mathbb{Z}[i]$ (since $|N(1)| = 1$ is not divisible by $|N(1+i)| = 2$), confirming that $1$ is not $(2)$-integral, as expected for a proper ideal. [/example] The results of this chapter will be applied in the study of going-up and going-down theorems for prime ideals, where the integrality condition over a prime $\mathfrak{p}$ controls whether prime ideals in extensions can be "moved" up and down the chain. The minimal polynomial characterisation is especially powerful when $A = \mathbb{Z}$ or $A = \mathcal{O}_K$, where it provides a concrete arithmetic criterion for integrality over an ideal purely in terms of the norm. Integral extensions produce a web of ideal relationships that encode both algebra and number-theoretic data. Cohen-Seidenberg theorems decode this web by showing how prime ideals lie above one another. # 9. Cohen-Seidenberg Theorems (Going Up/Down) The Cohen–Seidenberg theorems — known informally as the Going Up and Going Down theorems — describe how prime ideals behave under integral extensions of rings. Given an integral extension $A \subset B$, there is a natural map $\iota^*: \operatorname{Spec}(B) \to \operatorname{Spec}(A)$ sending each prime $\mathfrak{q}$ of $B$ to $\mathfrak{q} \cap A$. The central questions are: which primes of $A$ arise as contractions of primes of $B$, and how do inclusion relations among primes of $A$ lift to inclusion relations among primes of $B$? This chapter answers both questions through three interlocking results — Incomparability, Lying Over, and Going Up — and then proves Going Down under the additional hypothesis that $A$ is an integrally closed domain. ## The Contraction Map on Prime Spectra When $A \subset B$ is an integral extension, one might ask: does every prime of $A$ come from a prime of $B$? And if so, does the process respect the ordering of primes? These are not automatic — for a general ring homomorphism the contraction map can fail to be surjective and can collapse inclusion relations arbitrarily. The special structure of integral extensions is what makes controlled answers possible. Let $A \subset B$ be an integral extension with inclusion map $\iota: A \hookrightarrow B$. Every prime ideal $\mathfrak{q} \in \operatorname{Spec}(B)$ contracts to a prime ideal $\mathfrak{q} \cap A \in \operatorname{Spec}(A)$, giving a map \begin{align*} \iota^*: \operatorname{Spec}(B) &\to \operatorname{Spec}(A), \\ \mathfrak{q} &\mapsto \mathfrak{q} \cap A. \end{align*} The three theorems of this chapter — Incomparability, Lying Over, and Going Up — together give a complete picture of this map: its fibers are nonempty and antichain-ordered. The same results hold for any ring homomorphism $f: A \to B$ making $B$ an integral $A$-algebra; the general case reduces to the inclusion case by factoring through $A/\ker(f) \hookrightarrow B$. Before stating the theorems, we introduce a localisation that will appear in all three proofs. For $A \subset B$ and $\mathfrak{p} \in \operatorname{Spec}(A)$, set $S = A \setminus \mathfrak{p}$. Since $A \subset B$, the set $S$ is multiplicative in $B$, and we define \begin{align*} B_\mathfrak{p} := S^{-1}B = (A \setminus \mathfrak{p})^{-1}B. \end{align*} Note that $B_\mathfrak{p}$ is not a localisation of $B$ at a prime ideal of $B$; it is a localisation of $B$ at a multiplicative subset drawn from the subring $A$. By a previous result (Proposition 6.7), the extension $A_\mathfrak{p} \subset B_\mathfrak{p}$ is again integral whenever $A \subset B$ is. ## Incomparability The first theorem says that the fibers of $\iota^*$ carry no nontrivial inclusion relations. [quotetheorem:2880] [citeproof:2880] Incomparability says that the fiber $(\iota^*)^{-1}(\mathfrak{p})$ is an antichain in $\operatorname{Spec}(B)$: no two distinct primes lying over the same prime of $A$ can be comparable. This is a sharp restriction — it means that a chain of primes in $B$ must map injectively to a chain in $A$ under $\iota^*$, so the Krull dimension of $B$ can be no larger than that of $A$. ## Lying Over Incomparability constrains fibers, but it does not tell us whether fibers are nonempty. Lying Over fills this gap. [quotetheorem:2944] [citeproof:2944] Together, Incomparability and Lying Over characterise the image and the fiber structure of $\iota^*$: the map is surjective, and each fiber is an antichain. It is worth emphasising what Lying Over does not say: it does not specify how many primes of $B$ lie over a given prime of $A$, nor does it say anything about inclusion relations among those primes beyond the antichain condition. In the example $\mathbb{Z} \subset \mathbb{Z}[i]$, the prime $(5)$ has two primes lying over it — $(2+i)$ and $(2-i)$ — while the prime $(2)$ has only one — $(1+i)$ (with multiplicity two, but as an ideal $(1+i)$ is the unique prime above $(2)$). Integrality tells us a prime exists; arithmetic tells us how many. Lying Over also requires integrality: for the non-integral extension $\mathbb{Z} \subset \mathbb{Z}[1/2]$, the prime $(2) \in \operatorname{Spec}(\mathbb{Z})$ has empty fiber under the contraction map, since $(2) \cdot \mathbb{Z}[1/2] = \mathbb{Z}[1/2]$ is the whole ring and cannot be contained in any prime. [example: Lying Over for Integer Extensions] Consider $A = \mathbb{Z} \subset B = \mathbb{Z}[i]$. The extension is integral since $i^2 + 1 = 0$. For the prime $\mathfrak{p} = (2) \in \operatorname{Spec}(\mathbb{Z})$, Lying Over guarantees some $\mathfrak{q} \in \operatorname{Spec}(\mathbb{Z}[i])$ with $\mathfrak{q} \cap \mathbb{Z} = (2)$. Indeed, $(2) = (1+i)^2$ in $\mathbb{Z}[i]$ (since $(1+i)(1-i) = 2$ and $1-i = -i(1+i)$), so $(1+i)$ is a prime of $\mathbb{Z}[i]$ contracting to $(2)$. For $\mathfrak{p} = (5)$, note $5 = (2+i)(2-i)$ in $\mathbb{Z}[i]$, so both $(2+i)$ and $(2-i)$ lie over $(5)$. By Incomparability, $(2+i)$ and $(2-i)$ cannot be comparable, consistent with the fact that $(2+i) \neq (2-i)$. [/example] ## Going Up While Lying Over shows every prime of $A$ has a prime of $B$ above it, Going Up shows that chains of primes can be lifted. [quotetheorem:2945] [citeproof:2945] By induction, Going Up extends to arbitrary finite chains: if $\mathfrak{p}_1 \subset \mathfrak{p}_2 \subset \cdots \subset \mathfrak{p}_n$ is a chain in $\operatorname{Spec}(A)$ and $\mathfrak{q}_1$ lies over $\mathfrak{p}_1$, then there exists a chain $\mathfrak{q}_1 \subset \mathfrak{q}_2 \subset \cdots \subset \mathfrak{q}_n$ in $\operatorname{Spec}(B)$ with each $\mathfrak{q}_i \cap A = \mathfrak{p}_i$. Integrality is essential for Going Up. For a non-integral extension, the lifting property can fail: consider the localisation $A = \mathbb{Z} \subset B = \mathbb{Z}_{(2)}$ (integers localised at $(2)$). The chain $(0) \subset (2)$ lies in $\operatorname{Spec}(\mathbb{Z})$, and $(0) \in \operatorname{Spec}(\mathbb{Z}_{(2)})$ lies over $(0)$. But the only prime of $\mathbb{Z}_{(2)}$ above $(2)$ is $(2)\mathbb{Z}_{(2)}$, and the extension $\mathbb{Z} \subset \mathbb{Z}_{(2)}$ is not integral (the element $1/2$ satisfies no monic polynomial over $\mathbb{Z}$). More dramatically, in the polynomial extension $\mathbb{Z} \subset \mathbb{Z}[x]$, the prime $(2) \subset \mathbb{Z}$ has no canonical lift to a prime containing a prescribed prime over $(0)$: the lying-over primes $(2, f(x))$ for irreducible $f$ are incomparable, and chain-lifting depends on the choice of $f$. The going up property for a fixed starting prime requires integrality to make the quotient extension $B/\mathfrak{q}_1$ integral over $A/\mathfrak{p}_1$, which is the key step. [remark: Dimension Equality] Combining Incomparability and Going Up gives $\dim B = \dim A$: chains in $B$ inject into chains in $A$ by Incomparability (so $\dim B \leq \dim A$), and chains in $A$ lift to chains in $B$ by Going Up (so $\dim A \leq \dim B$). [/remark] [example: Going Up for Rings of Integers] In the extension $\mathbb{Z} \subset \mathbb{Z}[i]$, consider the chain $(0) \subset (5)$ in $\operatorname{Spec}(\mathbb{Z})$ and the prime $\mathfrak{q}_1 = (0) \in \operatorname{Spec}(\mathbb{Z}[i])$ lying over $(0)$. Going Up guarantees a prime $\mathfrak{q}_2$ with $(0) \subset \mathfrak{q}_2$ and $\mathfrak{q}_2 \cap \mathbb{Z} = (5)$. Indeed, we can take $\mathfrak{q}_2 = (2+i)$: this is prime in $\mathbb{Z}[i]$, contains $(0)$, and contracts to $(5)$ since $N(2+i) = 5$. [/example] ## Going Down Going Up lifts chains in the upward direction. The analogue for going down — given $\mathfrak{p}_1 \supset \mathfrak{p}_2$ and $\mathfrak{q}_1$ over $\mathfrak{p}_1$, find $\mathfrak{q}_2 \subset \mathfrak{q}_1$ over $\mathfrak{p}_2$ — requires stronger hypotheses. Integrality alone is not sufficient; one needs $A$ to be an integrally closed domain. [quotetheorem:2883] [citeproof:2883] The proof of Going Down is considerably more delicate than Going Up. The key ingredients are: the description of the radical of $\mathfrak{p}_2 B$ via integral closure (from Chapter 8), the integrally closed hypothesis on $A$ which controls the minimal polynomial, and a careful argument in the fraction fields. [remark: Necessity of Hypotheses] Both the domain condition on $B$ and the integrally closed condition on $A$ are necessary for Going Down. For a counterexample without the domain condition: let $A = \mathbb{Z}$ and $B = \mathbb{Z}[x]/(x(x-1)) \cong \mathbb{Z} \times \mathbb{Z}$, which is integral over $\mathbb{Z}$ but not a domain. The prime $(2,0)$ of $B$ lies over $(2)$ in $\mathbb{Z}$, but the zero ideal $(0,0)$ of $B$ lies over $(0)$, and $(2,0) \not\supset (0,0)$. For a counterexample without the integrally closed hypothesis, one can exhibit extensions of non-normal rings where going down fails. [/remark] [example: Going Down for Integer Extensions] Let $A = \mathbb{Z}$ and $B = \mathbb{Z}[\sqrt{-5}]$. Both are domains, $\mathbb{Z}$ is integrally closed (as a PID), and $B$ is integral over $A$ since $\sqrt{-5}$ satisfies $x^2 + 5 = 0$. Consider the chain $\mathfrak{p}_1 = (3) \supset \mathfrak{p}_2 = (0)$ in $\operatorname{Spec}(\mathbb{Z})$. We need a prime $\mathfrak{q}_1 \in \operatorname{Spec}(B)$ lying over $(3)$. In $B = \mathbb{Z}[\sqrt{-5}]$, the norm form is $N(a + b\sqrt{-5}) = a^2 + 5b^2$. We check whether $3$ is irreducible in $B$: for $3 = \alpha \beta$ we need $N(\alpha)N(\beta) = 9$, so either both have norm $3$ or one has norm $1$. But $a^2 + 5b^2 = 3$ has no integer solutions (checking $b = 0$ gives $a^2 = 3$, impossible; $b \neq 0$ gives $a^2 + 5b^2 \geq 5$). Therefore $3$ is irreducible in $B$, and the ideal $(3, \sqrt{-5} + 1)$ is a prime of $B$ lying over $(3)$: one can verify that $B/(3, \sqrt{-5}+1) \cong \mathbb{F}_3$ since $(\sqrt{-5}+1)$ mod $3$ satisfies $(x-1)^2 + 5 = x^2 - 2x + 6 \equiv x^2 + x \equiv x(x+1) \pmod{3}$, giving residue field $\mathbb{F}_3$. Set $\mathfrak{q}_1 = (3, \sqrt{-5} + 1)$. Going Down guarantees $\mathfrak{q}_2 \subset \mathfrak{q}_1$ lying over $(0)$. Since $B$ is a domain, its zero ideal $(0)$ is prime, $(0) \subset \mathfrak{q}_1$, and $(0) \cap \mathbb{Z} = (0) = \mathfrak{p}_2$. So $\mathfrak{q}_2 = (0)$ witnesses Going Down. [/example] ## Chain Lifting and Dimension If chains of primes in $A$ can always be lifted to $B$, what does this imply for the dimension theory of the two rings? We have already seen the dimension inequality $\dim B \leq \dim A$ from Incomparability. Going Up provides the matching lower bound. [quotetheorem:2946] [citeproof:2946] Going Down adds to this picture: under the additional hypotheses that $B$ is a domain and $A$ is integrally closed, chains can also be lifted downward, ensuring that the dimension formula is witnessed not just by existence of long chains but by the ability to extend a given partial chain in either direction. [explanation: The Role of Integral Closure in Going Down] The proof of Going Down uses the integrally closed hypothesis at a specific moment: when applying Proposition 8.2 to conclude that the coefficients of the minimal polynomial equation for $s$ lie in $A$. Without this, one can only conclude the coefficients lie in the integral closure of $A$, which might be larger than $A$ itself, and the subsequent argument breaks down. This is more than a technical convenience. Integrally closed domains — most prominently, normal domains and in particular Dedekind domains and UFDs — are the natural setting for Going Down. In algebraic number theory, where one routinely studies extensions $\mathcal{O}_K \subset \mathcal{O}_L$ of rings of integers, both rings are Dedekind (hence normal) domains, and Going Down applies freely. This accounts for the clean structure of prime decomposition in extensions of number fields. [/explanation] Prime ideals in integral extensions satisfy going-up, going-down, and incomparability, creating a clean structure paralleling prime decomposition in number fields. Primary decomposition factors ideals into prime power components, dual to this prime structure. # 10. Primary Decompositions Primary decomposition is the commutative algebraist's answer to a classical question in number theory: how do we factor an integer into prime powers? In $\mathbb{Z}$, writing $n = p_1^{e_1} \cdots p_k^{e_k}$ translates at the level of ideals to $(n) = (p_1^{e_1}) \cap \cdots \cap (p_k^{e_k})$, an intersection of primary ideals. This chapter develops the general theory of primary decomposition, showing that in any Noetherian ring every ideal admits such a representation. The main theorems — proved by Lasker and Noether — give partial uniqueness: the prime ideals that appear as radicals of the primary components are intrinsic to the original ideal, and so are the primary components at the minimal primes. The chapter closes with the geometric interpretation, connecting primary decomposition to the decomposition of algebraic varieties. ## Prime, Radical, and Primary Ideals What makes an ideal behave like a prime power rather than an arbitrary intersection? The answer turns on what happens to zero divisors in the quotient ring — and distinguishing three levels of behaviour (prime, radical, primary) is the key to making the question precise. Recall that an element $a \in R$ is a zero divisor if there exists $b \neq 0$ in $R$ with $ab = 0$, and nilpotent if $a^n = 0$ for some $n \geq 1$. Every nilpotent element is a zero divisor (take $b = a^{n-1}$ when $a^n = 0$ and $a^{n-1} \neq 0$), but not conversely. [definition: Prime, Radical, and Primary Ideals] Let $I$ be an ideal of $R$. Then: 1. $I$ is **prime** if $R/I \neq 0$ and the only zero divisor in $R/I$ is $0$. 2. $I$ is **radical** if the only nilpotent element in $R/I$ is $0$. 3. $I$ is **primary** if $R/I \neq 0$ and every zero divisor in $R/I$ is nilpotent. [/definition] Since every nilpotent is a zero divisor, the three conditions satisfy the chain of implications: prime $\implies$ primary $\implies$ radical. The ring $R$ itself (corresponding to the ideal $I = R$) is a radical ideal — the quotient $R/R = 0$ has no nonzero nilpotents vacuously — but $R$ is neither prime nor primary because $R/R = 0$, failing the condition $R/I \neq 0$. The introductory example of $\mathbb{Z}/(p^n)$ makes the interplay concrete. For a prime $p$ and $n \geq 1$, every zero divisor in $\mathbb{Z}/(p^n)$ is an element $a + \mathbb{Z}$ with $p \mid a$; such an element satisfies $(a + \mathbb{Z})^n = 0$, so it is nilpotent. This means $(p^n)$ is a primary ideal of $\mathbb{Z}$. On the other hand, if $x \in \mathbb{Z}$ has a prime divisor $p$ with $p^2 \nmid x$, then $p + \mathbb{Z}$ is a zero divisor in $\mathbb{Z}/(x)$ but is not nilpotent, so $(x)$ fails to be primary. [example: Ideals in Z] Let $R = \mathbb{Z}$. The zero ideal $(0)$ is prime (hence also radical and primary). For $0 \neq x \in \mathbb{Z}$: 1. $(x)$ is prime if and only if $|x|$ is a prime number. 2. $(x)$ is radical if and only if $x$ is square-free. 3. $(x)$ is primary if and only if $x = p^n$ for some prime number $p$ and $n \geq 1$. Thus $(6)$ is radical (since $6 = 2 \cdot 3$ is square-free) but not primary, while $(9) = (3^2)$ is primary but not radical (since $3 + \mathbb{Z}$ is nilpotent in $\mathbb{Z}/(9)$, but $\mathbb{Z}/(9)$ has nilpotents). [/example] The example makes clear why the definition is the right generalisation: being primary is the correct abstraction of "being a prime power". ## The Associated Prime of a Primary Ideal If $I$ is a primary ideal, the zero divisors in $R/I$ are exactly the nilpotents, so the set of nilpotents forms a prime ideal in $R/I$ (it is closed under products and the complementary condition: if $ab$ is nilpotent then by primality of the radical either $a$ or $b$ must be nilpotent). This means $\sqrt{I}$ is prime. [definition: p-Primary Ideal] A primary ideal $I$ of $R$ is **$\mathfrak{p}$-primary** if $\sqrt{I} = \mathfrak{p}$ for some prime ideal $\mathfrak{p} \in \operatorname{Spec}(R)$. [/definition] The following proposition assembles the key structural facts. [quotetheorem:2885] [citeproof:2885] The minimality conditions in (4) ensure the decomposition is irredundant: we cannot remove any component without changing the intersection, and no two components share the same associated prime. ## Primary Decomposition and Its Uniqueness Minimality narrows down the decomposition, but it does not make it unique — the embedded components can vary. The first uniqueness theorem says the prime ideals that appear are intrinsic. [definition: Primary Decomposition] A **primary decomposition** of an ideal $I$ is an expression \begin{align*} I = \mathfrak{q}_1 \cap \cdots \cap \mathfrak{q}_n \end{align*} where each $\mathfrak{q}_i$ is a primary ideal. The decomposition is **minimal** if the radicals $\mathfrak{p}_i = \sqrt{\mathfrak{q}_i}$ are pairwise distinct and no $\mathfrak{q}_i$ contains $\bigcap_{j \neq i} \mathfrak{q}_j$. [/definition] [example: Primary Decomposition in Z] In $\mathbb{Z}$, the primary decomposition of $(90)$ is \begin{align*} (90) = (2) \cap (3^2) \cap (5) = (2) \cap (9) \cap (5). \end{align*} The three components have radicals $(2)$, $(3)$, $(5)$, which are the prime ideals dividing $(90)$. This recovers the prime factorisation $90 = 2 \cdot 3^2 \cdot 5$. The decomposition is minimal: removing any component enlarges the intersection, and the radicals are distinct. [/example] The relationship between primary ideals and powers of prime ideals is more subtle than the $\mathbb{Z}$ example suggests. Two important facts prevent the naive expectation "$\mathfrak{q}$ primary $\iff$ $\mathfrak{q} = \mathfrak{p}^n$": **A primary ideal need not be a power of a prime ideal.** Let $R = k[X, Y]$ (where $k$ is a field) and $\mathfrak{q} = (X, Y^2)$. Then $R/\mathfrak{q} \cong k[Y]/(Y^2)$, in which every zero divisor is a $k$-multiple of $Y$, hence nilpotent. So $\mathfrak{q}$ is primary. Computing the radical: \begin{align*} \sqrt{\mathfrak{q}} = \sqrt{(X) + (Y)^2} = \sqrt{\sqrt{(X)} + \sqrt{(Y^2)}} = \sqrt{(X) + (Y)} = (X, Y), \end{align*} where the final equality uses the fact that $(X, Y)$ is maximal. If $\mathfrak{q} = \mathfrak{p}^n$ for some prime $\mathfrak{p}$, then $\mathfrak{p} = (X, Y) =: \mathfrak{m}$. But $\mathfrak{m}^2 = (X^2, XY, Y^2) \subsetneq (X, Y^2) = \mathfrak{q} \subsetneq \mathfrak{m}$, so $\mathfrak{q}$ is strictly between $\mathfrak{m}^2$ and $\mathfrak{m}$ and cannot be a power of $\mathfrak{m}$. **A power of a prime ideal need not be primary.** Let $R = k[X, Y, Z]/(XY - Z^2)$, and write $\bar{X}, \bar{Y}, \bar{Z}$ for the images. Then $\mathfrak{p} = (\bar{X}, \bar{Z})$ is prime because $R/\mathfrak{p} \cong k[Y]$ is an integral domain. Consider $\mathfrak{p}^2 = (\bar{X}^2, \bar{X}\bar{Z}, \bar{Z}^2)$. The relation $\bar{X}\bar{Y} = \bar{Z}^2$ shows $\bar{X}\bar{Y} \in \mathfrak{p}^2$. However $\bar{X} \notin \mathfrak{p}^2$, so $\bar{Y}$ is a zero divisor in $R/\mathfrak{p}^2$. Yet $\bar{Y} \notin \mathfrak{p}$ (since $R/\mathfrak{p} \cong k[Y]$ and $\bar{Y}$ maps to $Y \neq 0$), so no power of $\bar{Y}$ lies in $\mathfrak{p} = \sqrt{\mathfrak{p}^2}$, meaning $\bar{Y}$ is not nilpotent in $R/\mathfrak{p}^2$. Thus $\mathfrak{p}^2$ is not primary. These examples show that primary decomposition requires genuinely primary ideals, not just prime powers. ## The First Uniqueness Theorem A minimal primary decomposition is not unique — the embedded components can be replaced. What, if anything, is forced? The question is whether the prime ideals $\mathfrak{p}_i = \sqrt{\mathfrak{q}_i}$ that label the components depend on the decomposition or on $I$ alone. [quotetheorem:2886] [citeproof:2886] The set $\{\mathfrak{p}_1, \ldots, \mathfrak{p}_n\}$ is called the set of **associated primes** of $I$, written $\operatorname{Ass}(I)$ or $\operatorname{Ass}(R/I)$. The theorem does not say the decomposition itself is unique — only its associated primes are. Minimality is essential: without it, redundant components could introduce extra ideals into the set $\{(I:x)\} \cap \operatorname{Spec}(R)$ that do not reflect genuine primary structure. The associated primes are the bridge to the second uniqueness theorem, which identifies which primary components are also forced by $I$. ## Isolated and Embedded Primes Among the associated primes of $I$, those that are minimal with respect to inclusion play a distinguished role. [definition: Isolated and Embedded Primes] An associated prime ideal $\mathfrak{p}$ of $I$ is **isolated** if it is minimal with respect to inclusion in $\operatorname{Ass}(I)$. An associated prime that is not isolated is **embedded**. [/definition] The terminology is geometric: an isolated prime corresponds to an irreducible component of the variety $V(I)$, while an embedded prime corresponds to a sub-variety "embedded" inside one of these components. [quotetheorem:2887] The second uniqueness theorem says we get no choice for the isolated primary components: they are forced by $I$. Only the embedded components — those lying over embedded primes — can vary between different minimal decompositions. The following worked example exhibits both phenomena simultaneously. [example: Non-uniqueness of Embedded Components] Let $R = k[X, Y]$ for a field $k$, and $I = (X^2, XY)$. Two distinct minimal primary decompositions are: \begin{align*} I &= (X) \cap (X, Y)^2, \\ I &= (X) \cap (X^2, Y). \end{align*} We verify that both are valid: For the first: The ideal $(X)$ is prime. The radical of $(X, Y)^2$ is $(X, Y)$, which is maximal, so $(X, Y)^2$ is $(X, Y)$-primary. To check the intersection: take $f \in (X) \cap (X, Y)^2$. Write $f = aX = bX^2 + cXY + dY^2$ with $a, b, c, d \in k[X, Y]$. The last term forces $d = eX$ for some $e$ (since $f \in (X)$), giving $X(a - bX - cY - eY^2) = 0$, so $a \in (X, Y)$ and $f \in X(X,Y) = I$. For the second: The radical of $(X^2, Y)$ is also $(X, Y)$ (since $X^2 \in (X^2, Y)$ and $Y \in (X^2, Y)$ forces $X, Y \in \sqrt{(X^2, Y)}$), and $(X, Y)$ is maximal, so $(X^2, Y)$ is $(X, Y)$-primary. Moreover $(X, Y)^2 \neq (X^2, Y)$ because $Y \notin (X, Y)^2$. To verify the intersection: take $f \in (X) \cap (X^2, Y)$. Write $f = aX = bX^2 + cY$, so $c = dX$ for some $d$, and $X(a - bX - dY) = 0$, giving $a \in (X, Y)$, so $f \in X(X,Y) = I$. Both decompositions are minimal. The associated primes are $(X) \subset (X, Y)$, so $(X)$ is the unique isolated prime and $(X, Y)$ is the unique embedded prime. The isolated primary component $(X)$ is the same in both decompositions (as the second uniqueness theorem guarantees), while the embedded component at $(X, Y)$ differs: $(X, Y)^2$ in the first and $(X^2, Y)$ in the second. [/example] ## Radical Ideals and the Absence of Embedded Primes The simplest case of primary decomposition occurs for radical ideals, where the theory simplifies dramatically. [remark: Radical Ideals Have Unique Decompositions] Let $I = \mathfrak{q}_1 \cap \cdots \cap \mathfrak{q}_n$ be a minimal primary decomposition, and let $\mathfrak{p}_1, \ldots, \mathfrak{p}_t$ be the isolated primes. Then $\sqrt{I} = \mathfrak{p}_1 \cap \cdots \cap \mathfrak{p}_t$, and this is a minimal primary decomposition of $\sqrt{I}$. If $I$ is radical — that is, $I = \sqrt{I}$ — then the primary decomposition of $I$ is unique: $I = \mathfrak{p}_1 \cap \cdots \cap \mathfrak{p}_t$ where the $\mathfrak{p}_i$ are the associated primes of $I$, and all associated primes are isolated. In particular, a radical ideal has no embedded primes, and its primary components are exactly its associated prime ideals. In a Noetherian ring, giving a radical ideal is therefore equivalent to giving a finite collection of prime ideals $\mathfrak{p}_1, \ldots, \mathfrak{p}_t$ with no inclusions $\mathfrak{p}_i \subseteq \mathfrak{p}_j$ for $i \neq j$. [/remark] Passing from $I$ to $\sqrt{I}$ in a Noetherian ring has a clean interpretation: it retains exactly the isolated primes and discards all other information about the primary components. In the polynomial ring $R = k[T_1, \ldots, T_n]$ over an algebraically closed field $\Omega \supset k$, the Nullstellensatz gives $\sqrt{I} = I(V(I))$, so the algebraic set $V(I)$ encodes exactly $\sqrt{I}$, which is exactly the data of the isolated primes of $I$. ## Existence and Non-Existence of Primary Decompositions Noetherian rings guarantee primary decompositions, but what fails when the Noetherian hypothesis is dropped? The obstacle is not algebraic complexity per se — it is that the isolated primes of a primary decomposition must be finite in number, and a ring with infinitely many minimal prime ideals cannot accommodate this. [remark: Non-Existence Outside the Noetherian Case] For some rings $R$, not every ideal $I$ has a primary decomposition. When $I$ does have one, its set of isolated primes is both finite (it is a subset of the finitely many associated primes) and equal to the set of minimal prime ideals among those containing $I$. Therefore, if one constructs a ring having infinitely many minimal prime ideals, the zero ideal $(0)$ cannot have a primary decomposition. The ring $C([0,1])$ of continuous functions $[0,1] \to \mathbb{R}$ is a concrete example: its maximal ideals $\mathfrak{m}_x = \{f : f(x) = 0\}$, indexed by points $x \in [0, 1]$, form an uncountable family, and $(0)$ has no primary decomposition. [/remark] This remark shows that primary decomposition is genuinely tied to finiteness — it belongs to the Noetherian world, and the general theory does not extend cleanly beyond it. Primary decomposition belongs entirely to the Noetherian world. Direct and inverse limits, and completions, extend this finiteness language to infinite systems and topological refinements. # 11. Direct and Inverse Limits, Completions The previous chapters built a rich theory of rings and modules, culminating in the structure of Noetherian rings, localisation, and primary decomposition. This chapter introduces two fundamental categorical constructions — direct limits and inverse limits — that capture the idea of forming an "ultimate" object from a coherent family of approximations. The chapter then applies these ideas to one of the most important constructions in algebra and number theory: the $\mathfrak{a}$-adic completion of a ring, which replaces a ring by a complete metric space encoding all information modulo powers of an ideal. The $p$-adic integers and formal power series rings are the two canonical examples. ## Directed Sets and Systems How do you compare objects from different levels of an approximation process? If $\mathbb{Z}/p\mathbb{Z}$, $\mathbb{Z}/p^2\mathbb{Z}$, $\mathbb{Z}/p^3\mathbb{Z}$ are successive snapshots of arithmetic modulo higher powers of $p$, they do not sit inside a common ambient ring in any obvious way — yet they clearly form a coherent family. The obstacle is that we need a notion of a partially ordered index set rich enough to guarantee that any two objects in the family have a common "upper bound" to which they can both be compared. [definition: Directed Set] A **directed set** $(I, \leq)$ is a partially ordered set such that for every $a, b \in I$ there exists $c \in I$ with $a \leq c$ and $b \leq c$. [/definition] The key property is that any two elements have a common upper bound — this ensures we can always find an object in the system that "dominates" any finite collection. [definition: Directed and Inverse Systems] Let $\mathcal{C}$ be one of the following categories: Sets, Groups, Rings, $R$-modules, or $R$-algebras for a fixed ring $R$. Let $(I, \leq)$ be a directed set. A **directed system** over $I$ is a pair $\bigl((X_i)_{i \in I},\, (f_{ij})_{i \leq j}\bigr)$, where each $X_i$ is an object of $\mathcal{C}$ and each $f_{ij} : X_i \to X_j$ is a morphism, satisfying: - $f_{ii} = \operatorname{id}_{X_i}$ for all $i \in I$, - $f_{jk} \circ f_{ij} = f_{ik}$ for all $i \leq j \leq k$. An **inverse system** over $I$ is a pair $\bigl((Y_i)_{i \in I},\, (h_{ij})_{i \leq j}\bigr)$, where each $Y_i$ is an object of $\mathcal{C}$ and each $h_{ij} : Y_j \to Y_i$ is a morphism (note the reversal of direction), satisfying: - $h_{ii} = \operatorname{id}_{Y_i}$ for all $i \in I$, - $h_{ij} \circ h_{jk} = h_{ik}$ for all $i \leq j \leq k$. [/definition] The compatibility conditions say precisely that the morphisms form a coherent system: composing two consecutive maps gives the same result as taking the direct map between the endpoints. [example: The Finite Fields and $p$-adic Approximations] Let $I = \mathbb{Z}$ with the usual ordering and $p$ a prime. **A directed system.** For $a \mid b$ there is a ring embedding $\mathbb{F}_{p^a} \hookrightarrow \mathbb{F}_{p^b}$ (since $\mathbb{F}_{p^a}$ is the unique subfield of $\mathbb{F}_{p^b}$ of order $p^a$ when $a \mid b$). Set $X_i = \mathbb{F}_{p^{i!}}$, and for $i \leq j$ let $f_{ij} : \mathbb{F}_{p^{i!}} \to \mathbb{F}_{p^{j!}}$ be the composition of the successive embeddings $f_{i,(i+1)}, f_{(i+1),(i+2)}, \dots, f_{(j-1),j}$. The identity and transitivity conditions hold because composition of embeddings is associative, so $\bigl((X_i), (f_{ij})\bigr)$ is a directed system. **An inverse system.** Set $Y_i = \mathbb{Z}/p^i\mathbb{Z}$, and for $i \leq j$ let $h_{ij} : \mathbb{Z}/p^j\mathbb{Z} \to \mathbb{Z}/p^i\mathbb{Z}$ be the natural projection (reduce modulo $p^i$). Since $p^i \mid p^j$, the map is well-defined. The identity condition is immediate, and transitivity $h_{ij} \circ h_{jk} = h_{ik}$ says reducing modulo $p^j$ then modulo $p^i$ is the same as reducing directly modulo $p^i$, which holds since $p^i \mid p^j \mid p^k$. Thus $\bigl((Y_i), (h_{ij})\bigr)$ is an inverse system. [/example] ## Direct and Inverse Limits A directed system presents a family of objects with maps pointing forward; an inverse system presents a family with maps pointing backward. In neither case is there an obvious single object that represents the whole family. The question is: can we build, canonically, one object that "absorbs" all the information in the system — and if so, what does it look like concretely? [definition: Direct Limit and Inverse Limit] Let $(I, \leq)$ be a directed set. **Direct limit.** Let $D = \bigl((X_i)_{i \in I}, (f_{ij})_{i \leq j}\bigr)$ be a directed system. The **direct limit** \begin{align*} \varinjlim X_i \end{align*} of $D$ is $\bigl(\coprod_{i \in I} X_i\bigr)/{\sim}$, where $\sim$ is the smallest equivalence relation such that $x_i \sim f_{ij}(x_i)$ for all $i \leq j$ and all $x_i \in X_i$. Equivalently, for $x_i \in X_i$ and $x_j \in X_j$, one has $x_i \sim x_j$ if and only if there exists $k \in I$ with $k \geq i$ and $k \geq j$ such that $f_{ik}(x_i) = f_{jk}(x_j)$. **Inverse limit.** Let $E = \bigl((Y_i)_{i \in I}, (h_{ij})_{i \leq j}\bigr)$ be an inverse system. The **inverse limit** \begin{align*} \varprojlim Y_i \end{align*} of $E$ is the subobject \begin{align*} \left\{ \underline{y} \in \prod_{i \in I} Y_i : y_i = h_{ij}(y_j) \text{ for all } i \leq j \right\}. \end{align*} [/definition] The direct limit identifies elements that eventually agree under the transition maps; the inverse limit picks out coherent sequences — tuples in the product whose components are compatible under all the projection maps. Both constructions come with canonical morphisms. The direct limit $\varinjlim X_i$ is equipped with a natural morphism $X_j \to \varinjlim X_i$ for each $j \in I$, sending $x_j$ to its equivalence class. The inverse limit $\varprojlim Y_i$ is equipped with a natural morphism $\varprojlim Y_i \to Y_j$ for each $j$, given by projecting $\underline{y}$ onto its $j$-th coordinate. The notations $\varinjlim$ and $\varprojlim$ suppress the transition morphisms $f_{ij}$ and $h_{ij}$; these must be inferred from context. [remark: The Geometric Intuition for Direct Limits] The direct limit is the algebraic analogue of "forming a union of objects that are not literally subsets of one large set, but are identified through maps." A classic example from algebraic geometry is the **stalk** of a sheaf: at a point $x$ in a topological space, the stalk is the direct limit of the sections over all open neighbourhoods $U_i$ of $x$, where the directed set is the set of open neighbourhoods ordered by reverse inclusion ($i \leq j$ if $U_j \subset U_i$), and the maps are restrictions. Elements of the stalk represent "germs" of functions at $x$. [/remark] ## Universal Properties Both the direct and inverse limit constructions look somewhat ad hoc from the explicit descriptions: a quotient of a disjoint union, or a subset of a product. Why should these specific constructions be the "right" ones? The answer is that each is characterised by a universal property — any other object receiving compatible maps from the system factors uniquely through the limit. This pins down the limit up to unique isomorphism and makes it the unavoidable target for any coherent system of maps. [example: Universal Properties of the Limits] Let $D = (X_i, f_{ij})$ be a directed system and $E = (Y_i, h_{ij})$ be an inverse system over a directed set $(I, \leq)$. **Universal property of the direct limit.** For any object $A$ and any system of morphisms $(g_i : X_i \to A)_{i \in I}$ satisfying $g_j = g_j \circ f_{ij}$ for all $i \leq j$ (compatibility: the morphisms agree on overlaps), there exists a unique morphism $g : \varinjlim X_i \to A$ such that each $g_i$ factors as \begin{align*} X_j \to \varinjlim X_i \xrightarrow{\;g\;} A, \end{align*} where the left map is the canonical map from $X_j$ into the direct limit. A simple instance in the category of sets: a function from the disjoint union $\coprod_i X_i$ to a set $A$ is the same as a collection of functions $(X_i \to A)_{i \in I}$ — this is the degenerate case where $\leq$ has no relations beyond $i \leq i$, so no compatibility condition is imposed. **Universal property of the inverse limit.** For any object $B$ and any system of morphisms $(g_i : B \to Y_i)_{i \in I}$ satisfying $g_i = h_{ij} \circ g_j$ for all $i \leq j$, there exists a unique morphism $g : B \to \varprojlim Y_i$ such that each $g_j$ factors as \begin{align*} B \xrightarrow{\;g\;} \varprojlim Y_i \to Y_j, \end{align*} where the right morphism is the canonical projection from the inverse limit to $Y_j$. A concrete instance: the quotient maps $g_i : \mathbb{Z} \to \mathbb{Z}/p^i\mathbb{Z}$ are compatible with the projections $h_{ij}$ in the sense that $g_i = h_{ij} \circ g_j$ for $i \leq j$ (reducing modulo $p^j$ and then modulo $p^i$ gives the same result as reducing directly modulo $p^i$). By the universal property, these induce a unique ring homomorphism $g : \mathbb{Z} \to \varprojlim \mathbb{Z}/p^i\mathbb{Z}$ given by $g(x) = (x + p^i\mathbb{Z})_{i=1}^\infty$. [/example] ## The Direct Limit as an Algebraic Closure The field $\mathbb{F}_p$ has no algebraic closure sitting inside it — we must enlarge it. But any algebraic closure must contain a root of every polynomial over $\mathbb{F}_p$, hence a copy of every finite field $\mathbb{F}_{p^n}$. The difficulty is that these finite fields do not naturally lie inside a single ambient field; they are related only by embeddings. The direct limit turns this web of embeddings into an actual field, which turns out to be the algebraic closure $\overline{\mathbb{F}_p}$. [example: The Direct Limit of Finite Fields Is an Algebraic Closure of $\mathbb{F}_p$] Consider the directed system $\bigl((\mathbb{F}_{p^{i!}})_{i \geq 1},\, (f_{ij})_{i \leq j}\bigr)$ from the first example, where $i \leq j$ and $f_{ij}$ is the chain of embeddings. The direct limit $\varinjlim \mathbb{F}_{p^{i!}}$ is a field. It is a field because the direct limit of a directed system of fields (with injective transition maps in a directed set) is again an integral domain, and since every element comes from some $\mathbb{F}_{p^{i!}}$ it is invertible. There is a natural ring homomorphism $\mathbb{F}_p \to \varinjlim \mathbb{F}_{p^{i!}}$, which is injective since $\mathbb{F}_p$ is a field. **The direct limit is algebraic over $\mathbb{F}_p$.** Take any $[x] \in \varinjlim \mathbb{F}_{p^{i!}}$; it is represented by some $x \in \mathbb{F}_{p^{i!}}$ for some $i \geq 1$. Then $x^{p^{i!}} - x = 0$, since every element of $\mathbb{F}_{p^{i!}}$ satisfies this equation (the field $\mathbb{F}_{p^{i!}}$ is exactly the set of roots of $T^{p^{i!}} - T$ over $\mathbb{F}_p$). Hence $[x]^{p^{i!}} - [x] = 0$ in the direct limit, showing $[x]$ is algebraic over $\mathbb{F}_p$. **The direct limit is algebraically closed.** Let $[h] \in (\varinjlim \mathbb{F}_{p^{i!}})[T]$ be a polynomial with coefficients in the direct limit. Each coefficient comes from some finite field, so $[h]$ is the image of some $h \in \mathbb{F}_{p^{j!}}[T]$ for some $j \geq 1$ under the canonical map (applying $[\cdot]$ to each coefficient). The splitting field of $h$ over $\mathbb{F}_{p^{j!}}$ is isomorphic to $\mathbb{F}_{p^\ell}$ for some $\ell$ with $j! \mid \ell$. This $\mathbb{F}_{p^\ell}$ embeds in $\mathbb{F}_{p^{\ell!}}$, so $h$ splits over $\mathbb{F}_{p^{\ell!}}$. The key point is that the image of $[h]$ under two different embeddings $f_{j\ell}$ splits into linear factors regardless of which embedding is chosen: any two field embeddings $\sigma, \tau : \mathbb{F}_{p^a} \to \mathbb{F}_{p^b}$ satisfy $\tau(x) = \sigma(x^{p^s}) = (\sigma(x))^{p^s}$ for some $s \geq 0$, because $\mathbb{F}_{p^b}$ contains a unique copy of $\mathbb{F}_{p^a}$ (the roots of $T^{p^a} - T$). Therefore if $\sigma(h) = \prod_{i=1}^\ell (T - \alpha_i)$ splits, then $\tau(h) = \prod_{i=1}^\ell (T - \alpha_i^{p^s})$ also splits. Thus $[h]$ splits over $\varinjlim \mathbb{F}_{p^{i!}}$, proving algebraic closure. [/example] ## The Ring of $p$-adic Integers Ordinary integers let us do arithmetic exactly, but at the cost of carrying information about all primes simultaneously. A persistent question in number theory is: what does arithmetic look like when you care only about divisibility by powers of a fixed prime $p$, and you want to work to any precision you choose? The inverse system $(\mathbb{Z}/p^i\mathbb{Z})_{i \geq 1}$ encodes exactly this — each term captures $p$-divisibility information to one more level of precision — and its inverse limit assembles all these approximations into a single ring. [example: The Ring of $p$-adic Integers] The inverse limit $\mathbb{Z}_p := \varprojlim \mathbb{Z}/p^i\mathbb{Z}$ is the **ring of $p$-adic integers**. An element of $\mathbb{Z}_p$ is a coherent sequence \begin{align*} (a_1 + p\mathbb{Z},\; a_2 + p^2\mathbb{Z},\; a_3 + p^3\mathbb{Z},\; \dots) \end{align*} where $a_i \equiv a_{i-1} \pmod{p^{i-1}}$ for all $i \geq 2$. Working with $p = 5$: each element of $\mathbb{Z}_5$ can be thought of as an infinite base-$5$ expansion $\sum_{k=0}^\infty d_k \cdot 5^k$ where $0 \leq d_k < 5$. More precisely, a sequence of integers $(d_i)_{i=0}^\infty$ with $0 \leq d_i < 5$ defines the element \begin{align*} x = \left(\sum_{k=0}^{i-1} d_k \cdot 5^k + 5^i\mathbb{Z}\right)_{i=1}^\infty \in \mathbb{Z}_5. \end{align*} The natural map $\mathbb{Z}_5 \to \mathbb{Z}/5^i\mathbb{Z}$ reveals the $i$ least significant base-$5$ digits (the "rightmost $i$ digits" in the expansion). The element $1 \in \mathbb{Z}_5$ corresponds to the sequence $(1 + 5^i\mathbb{Z})_{i=1}^\infty$. The element $-1 \in \mathbb{Z}_5$ is more interesting. Since $-1 \equiv 4 \pmod{5}$, $-1 \equiv 24 \pmod{25}$, and so on, we get the sequence \begin{align*} (4 + 5\mathbb{Z},\; 24 + 5^2\mathbb{Z},\; 124 + 5^3\mathbb{Z},\; 624 + 5^4\mathbb{Z},\; \dots). \end{align*} In base $5$, these are $4_5, 44_5, 444_5, 4444_5, \dots$ — an infinite string of $4$s to the left, just as $-1$ in decimal is the "infinite string of $9$s" $\dots 9999$. The ring homomorphism $g : \mathbb{Z} \to \mathbb{Z}_p$ given by the universal property embeds $\mathbb{Z}$ as a subring of $\mathbb{Z}_p$, but $\mathbb{Z}_p$ is strictly larger: elements like $-1$ described above do not come from $\mathbb{Z}$ in any obvious finite sense. [/example] ## The $\mathfrak{a}$-adic Completion The $p$-adic integers encode arithmetic modulo successive powers of a prime. What if instead of $(p) \subset \mathbb{Z}$ we start with an arbitrary ideal $\mathfrak{a}$ in an arbitrary ring $R$? The powers $\mathfrak{a}^i$ again form a descending chain of ideals, and the quotients $R/\mathfrak{a}^i$ again form an inverse system with the natural projections. The resulting inverse limit is the $\mathfrak{a}$-adic completion, and it encompasses $p$-adic integers, formal power series rings, and many other completions in a single framework. [definition: $\mathfrak{a}$-adic Completion of a Ring] Let $R$ be a ring and $\mathfrak{a}$ an ideal of $R$. The **$\mathfrak{a}$-adic completion** of $R$ is \begin{align*} \hat{R} = \varprojlim R/\mathfrak{a}^i, \end{align*} the inverse limit of the directed system $\bigl((R/\mathfrak{a}^i)_{i \geq 1},\, (f_{ij})_{i \leq j}\bigr)$, where $(I, \leq) = (\mathbb{N}, \leq)$ and $f_{ij} : R/\mathfrak{a}^j \to R/\mathfrak{a}^i$ is the natural projection (which is well-defined since $\mathfrak{a}^i \supset \mathfrak{a}^j$ for $i \leq j$). [/definition] The definition is most transparent through its three principal instances, which together cover the two main families that appear throughout algebra and number theory. [example: Three Fundamental Completions] The $\mathfrak{a}$-adic completion construction recovers several classical objects. **The $p$-adic integers.** For $R = \mathbb{Z}$ and $\mathfrak{a} = (p)$, one has $\mathfrak{a}^i = (p^i)$ and $R/\mathfrak{a}^i = \mathbb{Z}/p^i\mathbb{Z}$, so $\hat{R} = \mathbb{Z}_p$ as constructed above. **Formal power series in one variable.** For $R = k[T]$, $k$ a field, and $\mathfrak{a} = (T)$, the ideal $\mathfrak{a}^i = (T^i)$ consists of polynomials divisible by $T^i$. The projection $k[T]/(T^j) \to k[T]/(T^i)$ truncates a polynomial at degree $i-1$. A coherent sequence in the inverse limit is precisely a formal power series $\sum_{k=0}^\infty a_k T^k \in k[[T]]$: the $j$-th term of the sequence is the truncation $\sum_{k=0}^{j-1} a_k T^k$. Thus $\hat{R} = k[[T]]$. **Formal power series in several variables.** For $R = k[T_1, \dots, T_n]$ and $\mathfrak{a} = (T_1, \dots, T_n)$, the ideal $\mathfrak{a}^i$ is the span over $k$ of all monomials $T_1^{e_1} \cdots T_n^{e_n}$ with $e_1 + \dots + e_n \geq i$. An element of the inverse limit is a coherent sequence of polynomials modulo $\mathfrak{a}^i$, which corresponds precisely to a formal power series $\sum_{\alpha} c_\alpha T_1^{\alpha_1} \cdots T_n^{\alpha_n} \in k[[T_1, \dots, T_n]]$. Thus $\hat{R} = k[[T_1, \dots, T_n]]$. [/example] ## Completions of Modules When studying a module $M$ over a ring $R$, completing $R$ to $\hat{R}$ raises an immediate question: does the module $M$ have a natural $\hat{R}$-module analogue, and how does it relate to $\hat{R} \otimes_R M$? To answer this, we must first define what completion means for a module. The submodules $\mathfrak{a}^i M$ give a filtration of $M$ by successively smaller submodules, and we can form the inverse limit of the quotients $M/\mathfrak{a}^i M$ just as we did for the ring. [definition: Filtration and Completion of a Module] Let $M$ be an $R$-module and $\mathfrak{a}$ an ideal of $R$. A **filtration** of $M$ is a sequence of submodules $M = M_0 \supset M_1 \supset M_2 \supset \cdots$. The **completion** of $M$ with respect to the filtration $(M_n)_{n \geq 0}$ is \begin{align*} \varprojlim M/M_n, \end{align*} the inverse limit of the system $(M/M_n)_{n \geq 0}$ with the natural projections $M/M_n \to M/M_m$ for $m \leq n$. The **$\mathfrak{a}$-adic completion** of $M$ is the completion with respect to the filtration $(\mathfrak{a}^i M)_{i \geq 0}$: \begin{align*} \hat{M} = \varprojlim M/\mathfrak{a}^i M. \end{align*} [/definition] The $\mathfrak{a}$-adic completion $\hat{M}$ is naturally an $\hat{R}$-module. The module structure is defined componentwise: for \begin{align*} (r_i + \mathfrak{a}^i)_{i \geq 0} \in \hat{R} \quad \text{and} \quad (m_i + \mathfrak{a}^i M)_{i \geq 0} \in \hat{M}, \end{align*} the product is $(r_i m_i + \mathfrak{a}^i M)_{i \geq 0}$. This is well-defined: if $r \equiv r' \pmod{\mathfrak{a}^i}$ and $m \equiv m' \pmod{\mathfrak{a}^i M}$, then $rm \equiv r'm' \pmod{\mathfrak{a}^i M}$, because $rm - r'm' = (r - r')m + r'(m - m')$ with $r - r' \in \mathfrak{a}^i$ and $m - m' \in \mathfrak{a}^i M$, so both terms lie in $\mathfrak{a}^i M$. ## Properties of Completions over Noetherian Rings Completion is a powerful operation, but it can destroy good properties: a non-Noetherian ring may complete to a wild object, and the completion functor on modules can fail to be exact. The Noetherian hypothesis tames all of this. When $R$ is Noetherian, the $\mathfrak{a}$-adic completion $\hat{R}$ remains Noetherian, the completion functor on finitely generated modules is exact, and $\hat{R} \otimes_R M$ is canonically isomorphic to $\hat{M}$ for every finitely generated $M$. [quotetheorem:2888] [citeproof:2888] [explanation: Significance of the Theorem] Parts (2) and (3) together have an important consequence for the study of finitely generated modules. When $R$ is Noetherian, the $\mathfrak{a}$-adic completion functor $M \mapsto \hat{M}$ is exact on the category of finitely generated $R$-modules: exactness of $\hat{R} \otimes_R (-)$ (part 2) combined with the identification $\hat{R} \otimes_R M \cong \hat{M}$ for finitely generated $M$ (part 3) gives exactness of $M \mapsto \hat{M}$. This is a powerful tool: completing a short exact sequence of finitely generated modules yields another short exact sequence. Part (1) says that completing a Noetherian ring yields another Noetherian ring, which means the theory of Noetherian modules — primary decomposition, associated primes, and the rest — is available to study $\hat{R}$-modules as well. [/explanation] ## Formal Power Series Rings Are Noetherian Hilbert's basis theorem tells us that polynomial rings over Noetherian rings are Noetherian, using an argument with leading coefficients that is intrinsically about finite-degree objects. Formal power series have no leading term and no finite degree bound, so that argument fails outright. Is $R[[T_1, \dots, T_n]]$ Noetherian when $R$ is? The answer is yes — but the proof requires a completely different strategy, passing through the completion theorem rather than any direct combinatorial argument. [quotetheorem:2889] [citeproof:2889] This result is noteworthy because $R[[T_1, \dots, T_n]]$ is not a finitely generated $R$-algebra (it is uncountable as a set when $R$ is infinite), so Hilbert's basis theorem does not directly apply. The completion theorem is essential. [remark: Comparison with Polynomial Rings] Recall that Hilbert's basis theorem establishes noetherianity of $R[T_1, \dots, T_n]$ for Noetherian $R$ by an argument with leading terms of polynomials. The analogous statement for formal power series requires a different proof, going through the completion machinery. The key structure to remember is: \begin{align*} R[T_1, \dots, T_n] &\hookrightarrow R[[T_1, \dots, T_n]] = \widehat{R[T_1, \dots, T_n]}, \end{align*} where the completion is taken with respect to $\mathfrak{m} = (T_1, \dots, T_n)$. Both rings are Noetherian when $R$ is, but for entirely different reasons. [/remark] Inverse limits like adic completions become Noetherian for entirely different reasons — they inherit finiteness from their defining sequences. Graded rings and filtrations decompose structures by level, turning infinite algebra into layer-by-layer analysis. # 12. Filtrations, Graded Rings Graded rings and filtrations are two complementary tools for studying how an algebraic structure decomposes or approximates itself at different "levels of complexity." A graded ring breaks apart into homogeneous pieces indexed by degree, with multiplication respecting the grading. A filtration instead provides a descending chain of submodules that becomes increasingly fine, and an ideal $\mathfrak{a}$ controls how fast this chain descends. The two ideas are linked by the associated graded construction, which converts a filtered ring or module into a graded one and thereby opens the door to powerful noetherianness arguments. This chapter culminates in the Artin-Rees lemma, a technical result about how filtrations interact with submodules — a lemma that will be indispensable in the proof of the Krull intersection theorem. ## Graded Rings and Graded Modules The simplest motivating example is the polynomial ring $k[T_1, \dots, T_r]$: every polynomial decomposes uniquely into a sum of homogeneous polynomials of each degree, and multiplying a homogeneous polynomial of degree $m$ by one of degree $n$ yields one of degree $m + n$. Graded rings axiomatise precisely this structure. [definition: Graded Ring] A **graded ring** is a ring $A$ together with a family $(A_n)_{n \geq 0}$ of additive subgroups of $A$ such that \begin{align*} A = \bigoplus_{n=0}^{\infty} A_n \end{align*} (internal direct sum as abelian groups) and $A_m A_n \subseteq A_{m+n}$ for all $m, n \geq 0$. An element $x \in A_n$ is called **homogeneous of degree $n$**. [/definition] The degree-zero piece $A_0$ is automatically a subring of $A$. To see this, note that $A_0$ is closed under multiplication since $A_0 A_0 \subseteq A_{0+0} = A_0$. Moreover, $1_A \in A_0$: write $1_A = \sum_{i=0}^{m} y_i$ with $y_i \in A_i$. For any $z_n \in A_n$, the product $1_A z_n = z_n$ lies in $A_n$, while $\sum_i y_i z_n$ has its $A_n$-component equal to $y_0 z_n$ (since $y_i z_n \in A_{n+i}$ and the direct sum decomposition is unique). Hence $y_0 z_n = z_n$ for all $z_n$, and since every element of $A$ is a finite sum of homogeneous pieces, $y_0 = 1_A$. In particular, each $A_n$ is an $A_0$-module. [example: Polynomial Ring Is Graded] Let $k$ be a field. The polynomial ring $k[T_1, \dots, T_r]$ is a graded ring with $A_n$ the set consisting of $0$ together with all homogeneous polynomials of degree $n$ (polynomials in which every monomial has total degree exactly $n$). Every polynomial decomposes uniquely into its homogeneous components, and the product of a homogeneous polynomial of degree $m$ with one of degree $n$ is homogeneous of degree $m+n$. [/example] With the notion of a graded ring in hand, the natural module-theoretic counterpart is immediate. [definition: Graded Module and Homomorphism] Let $A = \bigoplus_{n=0}^{\infty} A_n$ be a graded ring. 1. A **graded $A$-module** is an $A$-module $M = \bigoplus_{n=0}^{\infty} M_n$, where each $M_n$ is an additive subgroup of $M$, such that $A_m M_n \subseteq M_{m+n}$ for all $m, n \geq 0$. In particular, each $M_n$ is an $A_0$-module. 2. A **homomorphism of graded $A$-modules** is an $A$-module homomorphism $f: \bigoplus_{n \geq 0} M_n \to \bigoplus_{n \geq 0} N_n$ such that $f(M_n) \subseteq N_n$ for all $n \geq 0$. 3. An element $x \in M$ is **homogeneous of degree $n$** if $x \in M_n$. Any $y \in M$ can be written uniquely as $y = \sum_n y_n$ with $y_n \in M_n$ and all but finitely many $y_n$ equal to zero; the nonzero $y_n$ are the **homogeneous components** of $y$. 4. Set $A_+ = \bigoplus_{n=1}^{\infty} A_n$. This is an ideal of $A$, since it is the kernel of the natural projection $A \to A_0$. [/definition] The ideal $A_+$ measures "everything of positive degree" and plays a central role in what follows: it is the obstruction to an element of $A$ living in the base ring $A_0$. ### Noetherianness of Graded Rings When is a graded ring noetherian? The answer is clean and parallels the structure of finitely generated algebras. [quotetheorem:2947] [citeproof:2947] The key step — projecting the equation $y = \sum r_i x_i$ onto the homogeneous component $A_n$ — is a characteristic feature of arguments in graded algebra. It reduces a global statement to a homogeneous one, where induction on degree becomes available. ## Filtrations and the Associated Graded Ring A filtration is a different way to impose a hierarchical structure on a module: rather than decomposing the module into graded pieces, a filtration provides a nested decreasing sequence of submodules that become "smaller" (closer to $0$) as the index grows. [definition: Filtration and Stable Filtration] Let $R$ be a ring, $\mathfrak{a}$ an ideal of $R$, and $M$ an $R$-module. 1. A **filtration** of $M$ is a sequence $(M_n)_{n \geq 0}$ of submodules of $M$ with $M_0 = M$ and $M_n \supseteq M_{n+1}$ for all $n \geq 0$. 2. The filtration $(M_n)_{n \geq 0}$ is an **$\mathfrak{a}$-filtration** if $\mathfrak{a} M_n \subseteq M_{n+1}$ for all $n \geq 0$. 3. An $\mathfrak{a}$-filtration $(M_n)_{n \geq 0}$ is **$\mathfrak{a}$-stable** (or simply stable) if $\mathfrak{a} M_n = M_{n+1}$ for all sufficiently large $n$. [/definition] The prototypical example is the $\mathfrak{a}$-adic filtration. [example: The Adic Filtration] For any $R$-module $M$ and ideal $\mathfrak{a} \trianglelefteq R$, the sequence $(\mathfrak{a}^n M)_{n \geq 0}$ is a stable $\mathfrak{a}$-filtration. It is an $\mathfrak{a}$-filtration because $\mathfrak{a} \cdot \mathfrak{a}^n M = \mathfrak{a}^{n+1} M$, and it is stable since equality $\mathfrak{a} \cdot \mathfrak{a}^n M = \mathfrak{a}^{n+1} M$ holds for all $n \geq 0$. [/example] The associated graded ring encodes the "successive quotients" of the $\mathfrak{a}$-adic filtration on $R$ itself. [definition: Associated Graded Ring] Let $\mathfrak{a}$ be an ideal of $R$. The **associated graded ring** of $R$ with respect to $\mathfrak{a}$ is \begin{align*} G_{\mathfrak{a}}(R) = \bigoplus_{n=0}^{\infty} \mathfrak{a}^n / \mathfrak{a}^{n+1} \end{align*} where $\mathfrak{a}^0 = R$. [/definition] [remark: Multiplication in the Associated Graded Ring] This is indeed a graded ring. For $x \in \mathfrak{a}^n$ and $y \in \mathfrak{a}^m$, let $\bar{x}$ and $\bar{y}$ denote their images in $\mathfrak{a}^n / \mathfrak{a}^{n+1}$ and $\mathfrak{a}^m / \mathfrak{a}^{m+1}$ respectively. The product $\bar{x} \cdot \bar{y}$ is defined to be the image of $xy$ in $\mathfrak{a}^{n+m} / \mathfrak{a}^{n+m+1}$. One checks that this is well-defined (it does not depend on the choice of representatives $x, y$) using the fact that $xy' - xy = x(y'-y) \in \mathfrak{a}^n \cdot \mathfrak{a}^{m+1} \subseteq \mathfrak{a}^{n+m+1}$, and similarly for the first variable. [/remark] For a module with a filtration, there is an analogous graded module construction. [definition: Graded Module of a Filtration] Let $M$ be an $R$-module with $\mathfrak{a}$-filtration $(M_n)_{n \geq 0}$. Define \begin{align*} G(M) = \bigoplus_{n=0}^{\infty} M_n / M_{n+1} \end{align*} and write $G_n(M) = M_n / M_{n+1}$ for the $n$-th graded piece. [/definition] [remark: Module Structure on G(M)] $G(M)$ is a graded $G_{\mathfrak{a}}(R)$-module in a natural way: for $x \in \mathfrak{a}^n$ with image $\bar{x} \in \mathfrak{a}^n/\mathfrak{a}^{n+1}$ and $m \in M_k$ with image $\bar{m} \in M_k/M_{k+1}$, define $\bar{x} \cdot \bar{m}$ to be the image of $xm$ in $M_{k+n}/M_{k+n+1}$. This is well-defined because $(M_n)$ is an $\mathfrak{a}$-filtration, so $\mathfrak{a}^n M_k \subseteq M_{k+n}$. [/remark] ### Noetherianness of the Associated Graded Ring Passing to the associated graded ring looks like it might lose information — we are replacing exact multiplication in $\mathfrak{a}^n$ with multiplication in the quotient $\mathfrak{a}^n/\mathfrak{a}^{n+1}$. Does this process destroy the noetherianness of $R$? If it did, graded methods would be of limited use. The answer is reassuring: the associated graded construction preserves noetherianness. [quotetheorem:2948] [citeproof:2948] The theorem tells us that passing to the associated graded ring does not destroy the finiteness properties of the original ring — a reassuring fact that makes graded techniques genuinely applicable. ## Equivalent Filtrations and the Rees Algebra Two filtrations of the same module may define the same "approximation" in a suitable sense. Making this precise requires a notion of equivalence. [definition: Equivalent Filtrations] Let $(M_n)_{n \geq 0}$ and $(M'_n)_{n \geq 0}$ be filtrations of an $R$-module $M$. They are **equivalent** if there exists $n_0 \geq 0$ such that \begin{align*} M_{n + n_0} \subseteq M'_n \quad \text{and} \quad M'_{n + n_0} \subseteq M_n \end{align*} for all $n \geq 0$. [/definition] This is indeed an equivalence relation. Intuitively, two filtrations are equivalent if they are "offset" from each other by a bounded amount — neither one can get too far ahead of the other. [quotetheorem:2892] [citeproof:2892] This equivalence result is more than a curiosity: it means that stable filtrations are interchangeable for topological purposes (e.g., for the $\mathfrak{a}$-adic topology on a module), even though their individual terms may differ. ### The Rees Algebra To study filtrations via noetherianness arguments, one encodes the entire filtration into a single graded ring, the Rees algebra. Given an ideal $\mathfrak{a}$ of $R$ and an $\mathfrak{a}$-filtration $(M_n)_{n \geq 0}$ of an $R$-module $M$, define \begin{align*} R^* := \bigoplus_{n=0}^{\infty} \mathfrak{a}^n \quad (\mathfrak{a}^0 = R) \end{align*} and \begin{align*} M^* := \bigoplus_{n=0}^{\infty} M_n. \end{align*} For $x \in \mathfrak{a}^n$ and $y \in \mathfrak{a}^\ell$ (viewed as elements of the $n$-th and $\ell$-th summands of $R^*$), their product in $R^*$ is $xy \in \mathfrak{a}^{n+\ell}$ in the $(n+\ell)$-th summand. This makes $R^*$ a graded ring. For $x \in \mathfrak{a}^n$ in the $n$-th summand of $R^*$ and $m \in M_\ell$ in the $\ell$-th summand of $M^*$, the action is $xm \in \mathfrak{a}^n M_\ell \subseteq M_{n+\ell}$ in the $(n+\ell)$-th summand. This makes $M^*$ a graded $R^*$-module. When $R$ is noetherian, the ideal $\mathfrak{a}$ is generated by some $x_1, \dots, x_r$, and $R^*$ is generated as an $R$-algebra by $x_1, \dots, x_r$ (viewed as elements of the first summand $\mathfrak{a}$ of $R^*$). By Hilbert's basis theorem, $R^*$ is noetherian whenever $R$ is. The following lemma characterises stability of a filtration in terms of a finiteness property of $M^*$. [quotetheorem:2949] [citeproof:2949] This lemma is the technical engine behind the Artin-Rees lemma: stability of a filtration is equivalent to finite generation of the associated Rees module, and the noetherianness of $R^*$ makes that finite generation automatically pass to submodules. ## The Artin-Rees Lemma The Artin-Rees lemma asks a natural question: if $(M_n)_{n \geq 0}$ is a stable filtration of $M$, is the induced filtration $(N \cap M_n)_{n \geq 0}$ on a submodule $N \subseteq M$ also stable? The answer is yes, and this has far-reaching consequences. [quotetheorem:2950] [citeproof:2950] The Artin-Rees lemma is short to state, but the proof path runs through almost everything developed in this chapter: graded rings, the Rees algebra, the equivalence of stability with finite generation, and Hilbert's basis theorem in the background. The payoff is significant: in the next chapter, it will be used to prove the Krull intersection theorem, which gives a definitive answer to when the intersection $\bigcap_{n=1}^{\infty} \mathfrak{a}^n M$ is zero. [remark: Explicit Form of Artin-Rees] The conclusion that $(N \cap M_n)_{n \geq 0}$ is a stable $\mathfrak{a}$-filtration means precisely that there exists $n_0 \geq 0$ such that \begin{align*} N \cap \mathfrak{a}^n M = \mathfrak{a}^{n - n_0}(N \cap \mathfrak{a}^{n_0} M) \end{align*} for all $n \geq n_0$. In particular, $(N \cap \mathfrak{a}^n M)_{n \geq 0}$ is equivalent to the $\mathfrak{a}$-adic filtration $(\mathfrak{a}^n N)_{n \geq 0}$ of $N$, since all stable $\mathfrak{a}$-filtrations are mutually equivalent. [/remark] One consequence worth noting immediately: the filtration induced on a submodule is equivalent to the $\mathfrak{a}$-adic filtration of that submodule. This says, in a precise sense, that the $\mathfrak{a}$-adic topology on $M$ restricts to the $\mathfrak{a}$-adic topology on $N$ — not in a literally obvious sense, but up to the equivalence of filtrations. Filtrations and gradings transform complicated rings into more transparent ones via associated graded rings. Dimension theory is the culmination: it measures ring complexity via heights and regularity, synthesizing all prior machinery. # 13. Dimension Theory Dimension theory is the culmination of the course. After building up the machinery of Noetherian rings, primary decomposition, localisation, integral extensions, and graded rings, the course now asks a deceptively simple question: what is the "size" of a ring? The answer — the Krull dimension — turns out to be a surprisingly subtle invariant, and proving that it equals two other, completely different-looking quantities is the central achievement of this chapter. The proof of the Dimension Theorem for Noetherian local rings is one of the most satisfying arguments in commutative algebra: three numbers, defined in three different ways, are forced to be equal by a beautiful chain of inequalities. ## The Krull Dimension What does it mean for a ring to have "dimension $n$"? For the polynomial ring $k[T_1, \dots, T_n]$, the answer should be $n$ — one needs $n$ independent parameters to describe a point of $k^n$. But a purely algebraic definition must avoid geometry entirely and work for any ring. The difficulty is that "size" has to be defined without a metric or vector space structure: the right answer turns out to come from the longest chain of prime ideals one can fit inside the ring. [definition: Chain of Prime Ideals] Let $R$ be a ring. A **chain** of distinct prime ideals of $R$ is a strictly ascending sequence \begin{align*} \mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_d \end{align*} with each $\mathfrak{p}_i \in \operatorname{Spec}(R)$. The **length** of this chain is $d$ (so a chain of $d+1$ prime ideals has length $d$). [/definition] With the notion of a chain in hand, the Krull dimension is simply a supremum: [definition: Krull Dimension and Height] Let $R$ be a ring. - The **height** $\operatorname{ht}(\mathfrak{p})$ of a prime ideal $\mathfrak{p} \in \operatorname{Spec}(R)$ is the supremum of the lengths of all chains of prime ideals contained in $\mathfrak{p}$: \begin{align*} \operatorname{ht}(\mathfrak{p}) = \sup\{d \ge 0 : \exists\, \mathfrak{p}_0 \subsetneq \cdots \subsetneq \mathfrak{p}_d = \mathfrak{p}\}. \end{align*} - The **Krull dimension** (or simply **dimension**) of $R$ is \begin{align*} \dim(R) = \sup\{\operatorname{ht}(\mathfrak{p}) : \mathfrak{p} \in \operatorname{Spec}(R)\} = \sup\{\operatorname{ht}(\mathfrak{m}) : \mathfrak{m} \in \operatorname{mSpec}(R)\}. \end{align*} For an ideal $I$ of $R$, the **height** of $I$ is \begin{align*} \operatorname{ht}(I) = \inf\{\operatorname{ht}(\mathfrak{p}) : I \subset \mathfrak{p} \in \operatorname{Spec}(R)\}. \end{align*} [/definition] The equality $\dim(R) = \sup_{\mathfrak{m}}\operatorname{ht}(\mathfrak{m})$ holds because any chain of prime ideals that does not end with a maximal ideal can always be extended by one step. The supremum over all prime ideals therefore equals the supremum over maximal ideals. A key computational reduction: for every $\mathfrak{p} \in \operatorname{Spec}(R)$, one has $\operatorname{ht}(\mathfrak{p}) = \dim(R_\mathfrak{p})$, and consequently \begin{align*} \dim(R) = \sup\{\dim(R_\mathfrak{m}) : \mathfrak{m} \in \operatorname{mSpec}(R)\}. \end{align*} This means the global dimension of a ring is determined by the dimensions of its localisations at maximal ideals, which are all local rings. The Dimension Theorem (proved later in this chapter) will then give us complete control over the dimension of a Noetherian local ring. [remark: Module Length] The notion of **length** of a module $M$, defined as the supremum of the lengths of chains $M_0 \subsetneq \cdots \subsetneq M_n$ of submodules, plays an auxiliary but important role. Recall from earlier in the course that for a finitely generated module over an Artinian ring, the length is finite. When $A_0 = k$ is a field and $M$ is a $k$-vector space, $\ell(M) = \dim_k(M)$. [/remark] ## Transcendence Degree and Dimension of Finitely Generated Algebras How large can the Krull dimension of a finitely generated $k$-algebra be? For $k[T_1, \dots, T_d]$ the answer is $d$, and one might hope this is always determined by how many "truly independent" elements the algebra contains — that is, by a field-theoretic invariant, not a chain-combinatorial one. The notion that makes this precise is transcendence degree, which counts the size of a maximal algebraically independent subset of the fraction field. Let $k \subset L$ be a field extension. Recall that a subset $A \subset L$ is **algebraically independent** over $k$ if no nonzero polynomial in $k[T_1, \dots, T_n]$ vanishes when evaluated at any finite tuple from $A$. [definition: Transcendence Basis] A subset $A \subset L$ is a **transcendence basis** for $L$ over $k$ if any of the following three equivalent conditions holds: 1. $A$ is algebraically independent over $k$, and $L$ is algebraic over $k(A)$. 2. $A$ is algebraically independent over $k$, and $A \cup \{\beta\}$ is algebraically dependent over $k$ for every $\beta \in L$. 3. $L$ is algebraic over $k(A)$, but not over $k(A \setminus \{\alpha\})$ for any $\alpha \in A$. [/definition] The equivalence of these three characterisations is entirely analogous to the three characterisations of a linear basis for a vector space — maximally independent or minimally spanning. The proof that transcendence bases exist and have a unique cardinality runs parallel to the analogous facts for vector space bases. [quotetheorem:2895] For a $k$-algebra $A$ that is also an integral domain, one sets $\operatorname{trdeg}_k(A) = \operatorname{trdeg}_k(\operatorname{Frac}(A))$, viewing the transcendence degree as a property of the field of fractions. The bridge between transcendence degree and Krull dimension is built through integral extensions. [quotetheorem:2896] [citeproof:2896] The theorem does not say that every prime of $A$ has a unique lift to $B$: over a given $\mathfrak{p} \in \operatorname{Spec}(A)$ there can be several primes of $B$ lying above it. What Incomparability does say is that two distinct primes lying over the same prime cannot be comparable, so contracting a chain in $B$ always gives a chain of the same length in $A$. [example: Dimension of Polynomial Rings] For the polynomial ring $k[T_1, \dots, T_d]$ over a field $k$, Example Sheet 3 establishes $\dim(k[T_1, \dots, T_d]) = d$. This follows by exhibiting the chain $(0) \subsetneq (T_1) \subsetneq (T_1, T_2) \subsetneq \cdots \subsetneq (T_1, \dots, T_d)$ of length $d$, and Krull's Height Theorem (proved below) shows no prime of height greater than $d$ can exist. [/example] Combining Noether's Normalisation Theorem with the dimension-preserving property of integral extensions gives: [quotetheorem:2951] [citeproof:2951] This result is geometrically compelling: the dimension of an affine variety over an algebraically closed field $k$ equals the transcendence degree of its function field over $k$, which in turn equals the number of independent parameters needed to describe a general point of the variety. ## Hilbert Functions and the Poincaré Series Krull dimension counts the length of chains of primes. Can one measure the same quantity by watching how fast a graded ring grows? If the ring grows like $n^{d-1}$ in degree $n$, one expects dimension $d$. The obstacle is making this precise: one must encode the growth data into a power series, show the series is rational, and identify the pole order at $T = 1$ as an invariant. This is the Hilbert–Serre approach, and it will later become one of the three equal quantities in the Dimension Theorem. Throughout this section, let $A = \bigoplus_{n \ge 0} A_n$ be a Noetherian graded ring with $A_0$ Artinian, generated as an $A_0$-algebra by elements $x_1, \dots, x_s$ with $x_i \in A_{k_i}$, $k_i > 0$. Let $M = \bigoplus_{n \ge 0} M_n$ be a nonzero finitely generated graded $A$-module. Since $A_0$ is Artinian and each $M_n$ is a finitely generated $A_0$-module (which follows from $M$ being finitely generated over the Noetherian ring $A$), the module $M_n$ has **finite length** $\ell(M_n) < \infty$ over $A_0$ for every $n \ge 0$. [definition: Poincaré Series] The **Poincaré series** of $M$ is the formal power series \begin{align*} P(M, T) = \sum_{n=0}^{\infty} \ell(M_n) T^n \in \mathbb{Z}[[T]]. \end{align*} [/definition] The Hilbert–Serre theorem asserts that this power series is "almost" rational: [quotetheorem:2952] [citeproof:2952] Since $P(M, T)$ is the Taylor expansion of the rational function $R(T) = f(T)/\prod_{i=1}^{s}(1-T^{k_i})$ at $T = 0$, and the radius of convergence is at least $1$, one can ask about the behaviour of $R(T)$ near $T = 1$. [definition: Pole Order d(M)] Write $d(M)$ for the **order of the pole** of $R(T)$ at $T = 1$ (equivalently, the order of vanishing of the denominator minus the order of vanishing of $f$ at $T = 1$). Note that $d(M) \ge 0$: if $d(M) < 0$, then $R(1) = 0$, forcing $\ell(M_n) = 0$ for all $n \ge 0$ and hence $M = 0$, a contradiction. [/definition] [example: Polynomial Ring] Take $A = k[T_1, \dots, T_s] = \bigoplus_{n \ge 0} A_n$ graded by total degree, with $A_0 = k$ and generators $T_1, \dots, T_s \in A_1$ (so $k_1 = \cdots = k_s = 1$). By Stars and Bars, the number of degree-$n$ monomials is $\binom{n+s-1}{n}$, so $\ell(A_n) = \binom{n+s-1}{n}$. The Poincaré series is \begin{align*} P(A, T) &= \sum_{n \ge 0} \binom{n+s-1}{n} T^n = \left(\sum_{\ell \ge 0} T^\ell\right)^s = \frac{1}{(1-T)^s}, \end{align*} where the second equality holds because the coefficient of $T^n$ in $(1/(1-T))^s$ counts the number of ways to write $n$ as an ordered sum of $s$ nonneg integers, which is $\binom{n+s-1}{n}$. Thus $d(A) = s$ and $f(T) = 1$. [/example] The key structural result is the Hilbert polynomial, which shows that the growth of $\ell(M_n)$ is polynomial for large $n$: [quotetheorem:2953] [citeproof:2953] The hypothesis $k_1 = \cdots = k_s = 1$ cannot be dropped: if generators have different degrees, one can still extract a rational function with a pole at $T = 1$, but the function $\ell(M_n)$ is only eventually polynomial along arithmetic progressions, not along all integers. The degree of $\operatorname{HP}_M$ being $d(M) - 1$ (rather than $d(M)$) reflects the fact that $\ell(M_n)$ counts the size of a single graded piece, while $d(M)$ measures cumulative growth via the partial sums $\ell(A/\mathfrak{m}^n)$. [definition: Hilbert Function and Hilbert Polynomial] The function $n \mapsto \ell(M_n)$ is the **Hilbert function** of $M$. The polynomial $\operatorname{HP}_M$ is the **Hilbert polynomial** of $M$. [/definition] [remark: Conventions on Degree] By convention, the zero polynomial has degree $-1$, and $\binom{n}{-1} = 0$ for $n \ge 0$ with $\binom{-1}{-1} = 1$. These conventions ensure the Hilbert polynomial statement remains valid even when $d(M) = 0$ (where $\operatorname{HP}_M = 0$, indicating $M_n = 0$ for large $n$). [/remark] ## The Dimension Theorem for Noetherian Local Rings Three completely different ways of measuring the size of a Noetherian local ring have been lurking in the background: chains of primes, pole orders of Poincaré series, and minimum generators of $\mathfrak{m}$-primary ideals. Why should these have anything to do with each other? A priori there is no obvious reason — one is combinatorial, one is analytic, and one is purely about generators. The Dimension Theorem asserts they are always equal, and the proof is a chain of three inequalities, each requiring a separate argument. Fix a Noetherian local ring $(A, \mathfrak{m})$. The three invariants are: 1. $\dim(A)$: the Krull dimension of $A$. 2. $\delta(A) = \min\{\delta(\mathfrak{q}) : \mathfrak{q} \text{ is an } \mathfrak{m}\text{-primary ideal of } A\}$, where $\delta(\mathfrak{q})$ denotes the minimum number of generators of $\mathfrak{q}$. 3. $d(G_\mathfrak{m}(A))$: the order of the pole at $T = 1$ of the Poincaré series $P(G_\mathfrak{m}(A), T) = \sum_{n \ge 0} \ell(\mathfrak{m}^n/\mathfrak{m}^{n+1}) T^n$, where $G_\mathfrak{m}(A) = \bigoplus_{n \ge 0} \mathfrak{m}^n/\mathfrak{m}^{n+1}$ is the associated graded ring. Before stating the theorem, one needs a basic fact about $\mathfrak{m}$-primary ideals: [quotetheorem:2900] [citeproof:2900] Part (1) shows that $\mathfrak{m}$-primary ideals are precisely those sandwiched between $\mathfrak{m}$ and some power $\mathfrak{m}^t$; in particular, every power $\mathfrak{m}^t$ is itself $\mathfrak{m}$-primary. Part (2) is crucial for the Dimension Theorem: it allows one to compute lengths $\ell(A/\mathfrak{q}^n)$, which will be the finite quantities whose growth rate encodes the dimension. [quotetheorem:860] [citeproof:860] The theorem is remarkable: $\dim(A)$ is a purely combinatorial invariant (lengths of chains), $d(G_\mathfrak{m}(A))$ is an analytic/algebraic invariant (pole order of a power series), and $\delta(A)$ is a generator-counting invariant (minimum generators of an $\mathfrak{m}$-primary ideal). Their equality is not at all obvious. ### The Three Inequalities The proof of the Dimension Theorem is assembled from three propositions. The first requires an elementary but useful lemma about partial sums of polynomials. [quotetheorem:860] [citeproof:860] The first inequality $\delta(A) \ge d(G_\mathfrak{m}(A))$ comes from estimating the growth of $\ell(A/\mathfrak{q}^n)$. Fix an $\mathfrak{m}$-primary ideal $\mathfrak{q}_0$ achieving the minimum $\delta(A) = \delta(\mathfrak{q}_0)$. The associated graded ring $G_\mathfrak{q}(A)$ is a Noetherian graded ring generated by $\delta(\mathfrak{q})$ elements of degree $1$ over the Artinian ring $A/\mathfrak{q}$. By the Hilbert polynomial theorem, $\ell(\mathfrak{q}^n/\mathfrak{q}^{n+1})$ is eventually a polynomial of degree at most $\delta(\mathfrak{q}) - 1$. By the summation lemma, $\ell(A/\mathfrak{q}^n) = \sum_{i=0}^{n-1} \ell(\mathfrak{q}^i/\mathfrak{q}^{i+1})$ is eventually a polynomial of degree at most $\delta(\mathfrak{q})$. The inclusion $\mathfrak{m}^t \subset \mathfrak{q}_0 \subset \mathfrak{m}$ for some $t \ge 1$ gives sandwiching inequalities $\ell(A/\mathfrak{m}^n) \le \ell(A/\mathfrak{q}_0^n) \le \ell(A/\mathfrak{m}^{tn})$ for all $n \ge 0$, which forces $\ell(A/\mathfrak{q}_0^n)$ and $\ell(A/\mathfrak{m}^n)$ to have the same degree as polynomials. Thus $\deg(\ell(A/\mathfrak{m}^n)) = d(G_\mathfrak{m}(A))$. [quotetheorem:2168] [citeproof:2168] The second inequality $d(G_\mathfrak{m}(A)) \ge \dim(A)$ is proved by induction, using the following key proposition about how $d$ behaves when one passes to a quotient by a non-zero-divisor: [quotetheorem:2124] [citeproof:2124] The non-zero-divisor hypothesis is essential here: if $x$ is a zero-divisor then multiplication by $x$ is not injective, the isomorphism $A \xrightarrow{\sim} xA$ fails, and the leading terms need not cancel. In geometric terms, dividing by a non-zero-divisor corresponds to restricting to a hypersurface that genuinely cuts the variety, reducing dimension by exactly one. [quotetheorem:2136] [citeproof:2136] The third inequality $\dim(A) \ge \delta(A)$ asserts that one can always find an $\mathfrak{m}$-primary ideal generated by exactly $\dim(A)$ elements: [quotetheorem:915] [citeproof:915] The inductive prime-avoidance construction does not say that $x_1, \dots, x_d$ form a regular sequence; they are merely chosen to avoid the minimal primes at each stage. Whether $x_i$ is a non-zero-divisor on $A/(x_1, \dots, x_{i-1})$ is a stronger condition, and rings where such a regular sequence exists are called Cohen–Macaulay rings, an important refinement studied in further courses. ## Krull's Height Theorem How many equations does it take to cut out a subvariety of codimension $r$? Geometrically the answer is "at least $r$", and algebraically this should mean that a prime ideal of height $r$ cannot arise as a minimal prime over a set of fewer than $r$ generators. Without the Dimension Theorem this bound would be very hard to prove; with it, the argument is short. [quotetheorem:2156] Recall that a prime $\mathfrak{p}$ is **minimal over** $\mathfrak{a}$ if no prime lies strictly between $\mathfrak{a}$ and $\mathfrak{p}$: [definition: Minimal Prime Over an Ideal] A **minimal prime** over an ideal $I \trianglelefteq R$ is a prime ideal $\mathfrak{p} \supset I$ such that no prime $\mathfrak{q}$ satisfies $I \subset \mathfrak{q} \subsetneq \mathfrak{p}$. [/definition] [proof] Localise at $\mathfrak{p}$ to get the local ring $A_\mathfrak{p}$. If any prime of $A_\mathfrak{p}$ contains $\mathfrak{a} A_\mathfrak{p}$, its contraction to $A$ is a prime between $\mathfrak{a}$ and $\mathfrak{p}$; by minimality of $\mathfrak{p}$, this contraction must be $\mathfrak{p}$ itself, so the prime in $A_\mathfrak{p}$ is $\mathfrak{p} A_\mathfrak{p}$. Thus $\sqrt{\mathfrak{a} A_\mathfrak{p}} = \mathfrak{p} A_\mathfrak{p}$, so $\mathfrak{a} A_\mathfrak{p}$ is $\mathfrak{p} A_\mathfrak{p}$-primary (by Proposition 10.1). The ideal $\mathfrak{a} A_\mathfrak{p}$ is generated by the $r$ elements $x_1/1, \dots, x_r/1$, so $\delta(A_\mathfrak{p}) \le \delta(\mathfrak{a} A_\mathfrak{p}) \le r$. By the Dimension Theorem, $\operatorname{ht}(\mathfrak{p}) = \dim(A_\mathfrak{p}) = \delta(A_\mathfrak{p}) \le r$. [/proof] The theorem bounds height from above but says nothing about lower bounds: a prime minimal over an $r$-generator ideal can have height anywhere from $0$ to $r$. There is no converse: having height $r$ does not force the prime to be minimal over some $r$-generator ideal, and in a non-Cohen–Macaulay ring the height can be strictly less than $r$ even when the ideal genuinely "needs" $r$ generators. [example: A Principal Ideal Has Height At Most One] The special case $r = 1$: if $x \in A$ is any element of a Noetherian ring and $\mathfrak{p}$ is a minimal prime over $(x)$, then $\operatorname{ht}(\mathfrak{p}) \le 1$. This means the principal ideal theorem: a single equation can cut codimension at most $1$. More generally, $r$ equations can cut codimension at most $r$. In the polynomial ring $k[T_1, \dots, T_n]$, every prime $\mathfrak{p} = (T_1, \dots, T_r)$ for $r \le n$ achieves equality, showing the bound is sharp. [/example] [remark: Dimension and Generators] Krull's Height Theorem, combined with the Dimension Theorem, shows that $\dim(A) \le$ (minimum number of generators of $\mathfrak{m}$) for any Noetherian local ring $(A, \mathfrak{m})$. Rings where equality holds — where $\mathfrak{m}$ can be generated by exactly $\dim(A)$ elements — are called **regular local rings**. These are among the most important objects in commutative algebra and algebraic geometry, as they correspond to smooth points on algebraic varieties. The polynomial ring $k[T_1, \dots, T_n]$ localised at any maximal ideal gives an example of a regular local ring of dimension $n$. [/remark] ## References