We want a number system where $x^2 + 1 = 0$ has a solution. The real numbers do not provide one: for any $a \in \mathbb{R}$, we have $a^2 \geq 0$, so $a^2 + 1 \geq 1 > 0$. The equation is simply impossible in $\mathbb{R}$. But we know complex numbers exist and solve it — the question is how to *build* $\mathbb{C}$ from $\mathbb{R}$ in a principled way, without pulling $i$ out of thin air and asking the reader to accept it. The answer is a quotient ring: form the polynomial ring $\mathbb{R}[x]$ and then "declare" $x^2 + 1 = 0$ by passing to $\mathbb{R}[x]/(x^2 + 1)$. This construction — imposing an algebraic relation by taking a quotient — is what the theory of rings is built to do. The same pattern recurs everywhere. We want arithmetic modulo $n$: form $\mathbb{Z}$ and pass to $\mathbb{Z}/(n)$. We want a field with $p^2$ elements: start from $\mathbb{F}_p[x]$, pick an irreducible quadratic, and quotient by it. We want to understand whether a polynomial equation is solvable: study the rings and fields generated by its roots. Rings are the algebraic structures that carry both addition and multiplication, and their quotients are the machines for building new number systems by fiat. [example: The Polynomial Ring Modulo x² + 1] We construct the complex numbers from the reals without postulating $i$. Consider the polynomial ring $\mathbb{R}[x]$, whose elements are polynomials $a_0 + a_1 x + \cdots + a_n x^n$ with $a_i \in \mathbb{R}$. The polynomial $x^2 + 1$ is irreducible over $\mathbb{R}$ — it has no real roots. Form the quotient $\mathbb{R}[x]/(x^2+1)$. Every element of this quotient is a coset $f(x) + (x^2 + 1)$. Since we can divide any polynomial by $x^2 + 1$ with remainder of degree at most $1$, every coset has a unique representative of the form $a + bx$ with $a, b \in \mathbb{R}$. Addition is componentwise: \begin{align*} (a + bx) + (c + dx) &= (a+c) + (b+d)x. \end{align*} Multiplication uses the relation $x^2 \equiv -1 \pmod{x^2+1}$: \begin{align*} (a + bx)(c + dx) &= ac + (ad + bc)x + bd \cdot x^2 \\ &= ac + (ad+bc)x + bd(-1) \\ &= (ac - bd) + (ad + bc)x. \end{align*} This is exactly the multiplication rule for complex numbers under the identification $x \leftrightarrow i$. The element $x + (x^2+1)$ satisfies $(x + (x^2+1))^2 + 1 = (x^2 + 1) + (x^2+1) = 0 + (x^2+1)$, which is the zero element of the quotient. So $x^2 + 1 = 0$ has a solution in $\mathbb{R}[x]/(x^2+1)$, and that quotient ring is isomorphic to $\mathbb{C}$. [/example] This is the ring-theoretic miracle: you can solve equations by building the solution into the coefficient system. But to understand why this works — and when it doesn't — we need to develop the theory from the ground up. ## Definition What makes a ring a ring? The minimal requirements for doing algebra: you need to add and subtract freely (an abelian group under addition), you need to multiply (associatively, with an identity), and multiplication must distribute over addition. That is it. Commutativity of multiplication is a bonus, not a requirement. [definition: Ring] A **ring** is a set $R$ equipped with two binary operations, addition $+: R \times R \to R$ and multiplication $\cdot: R \times R \to R$, satisfying the following axioms: **(R1)** $(R, +)$ is an abelian [group](/page/Group). That is: $+$ is associative and commutative, there is a zero element $0_R \in R$ with $a + 0_R = a$ for all $a \in R$, and every $a \in R$ has an additive inverse $-a$ with $a + (-a) = 0_R$. **(R2)** Multiplication is associative: $(ab)c = a(bc)$ for all $a, b, c \in R$. **(R3)** There is a multiplicative identity $1_R \in R$ with $1_R \cdot a = a \cdot 1_R = a$ for all $a \in R$. **(R4)** Multiplication distributes over addition from both sides: \begin{align*} a(b + c) &= ab + ac, \\ (a + b)c &= ac + bc, \end{align*} for all $a, b, c \in R$. [/definition] Notice what is *not* required: multiplication need not be commutative ($ab = ba$ is not assumed), and nonzero elements need not have multiplicative inverses. These omissions are intentional — they admit important examples like matrix rings and group rings that would be excluded by stronger axioms. The axioms above capture the bare minimum needed to do algebra with two operations. From these four axioms alone, we can already derive some essential properties that might seem obvious but genuinely require proof. [quotetheorem:2572] The claim is not circular — it follows from the distributive law, not from some general principle about zero. Here is the argument for $0_R \cdot a = 0_R$. Since $0_R + 0_R = 0_R$ (the zero element satisfies this by definition of an additive identity), we compute: \begin{align*} 0_R \cdot a &= (0_R + 0_R) \cdot a = 0_R \cdot a + 0_R \cdot a. \end{align*} Subtract $0_R \cdot a$ from both sides (using the additive inverse) to get $0_R = 0_R \cdot a$. The argument for $a \cdot 0_R = 0_R$ is symmetric. This tells us that the multiplicative identity $1_R$ and the additive identity $0_R$ cannot be equal unless $R = \{0\}$: if $1_R = 0_R$, then for any $a \in R$, $a = 1_R \cdot a = 0_R \cdot a = 0_R$. A ring with $1_R = 0_R$ is called the **zero ring**; it has exactly one element. [quotetheorem:2573] These follow from distributivity and the zero annihilation theorem. For the first identity: \begin{align*} ab + (-a)b &= (a + (-a))b = 0_R \cdot b = 0_R. \end{align*} So $(-a)b$ is the additive inverse of $ab$, which is $-(ab)$. The second identity is symmetric. For the third: $(-a)(-b) = -(a(-b)) = -(-(ab)) = ab$, where the last step uses the fact that the inverse of the inverse is the original element. These are not just bookkeeping — the rule $(-1)(-1) = 1$ is what ensures that the integers behave correctly, and it is a theorem, not an axiom. Now we name the important special cases. In commutative rings, the order of multiplication is irrelevant, which simplifies algebra enormously: the theory of polynomials, divisibility, and factorization all rely on commutativity. [definition: Commutative Ring] A ring $R$ is **commutative** if $ab = ba$ for all $a, b \in R$. [/definition] Most of the rings appearing in number theory and algebraic geometry are commutative: $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, polynomial rings, and function rings. Matrix rings over a field are the prototypical noncommutative rings and are essential in representation theory and geometry. What happens when every nonzero element has a multiplicative inverse? The arithmetic becomes much simpler — you can divide freely, and the ring behaves like a number system. [definition: Division Ring] A ring $R$ is a **division ring** (or **skew field**) if $R \neq \{0\}$ and every nonzero element $a \in R$ has a multiplicative inverse $a^{-1} \in R$ satisfying $a \cdot a^{-1} = a^{-1} \cdot a = 1_R$. [/definition] [definition: Field] A **field** is a commutative division ring. [/definition] The distinction between a field and a division ring is commutativity of multiplication. Every field is a division ring, but not conversely: the quaternions $\mathbb{H}$ are a division ring but not a field, since quaternion multiplication is noncommutative. ## Examples Diverse examples are essential here, because the ring axioms admit structures that look wildly different from one another. We work through several families, computing something nontrivial in each to show what the axioms really mean in practice. The integers $\mathbb{Z}$ are the original motivation for the axioms. Addition, subtraction, and multiplication work as expected; there is no division in general ($2$ has no multiplicative inverse in $\mathbb{Z}$). This is a commutative ring. [example: The Ring Z/6Z Has Zero Divisors] The ring $\mathbb{Z}/6\mathbb{Z} = \{\bar{0}, \bar{1}, \bar{2}, \bar{3}, \bar{4}, \bar{5}\}$ consists of residue classes modulo $6$, with addition and multiplication inherited from $\mathbb{Z}$. Addition and multiplication are well-defined on cosets. Notice that $\bar{2} \neq \bar{0}$ and $\bar{3} \neq \bar{0}$, yet: \begin{align*} \bar{2} \cdot \bar{3} = \overline{2 \cdot 3} = \overline{6} = \bar{0}. \end{align*} Two nonzero elements multiply to zero. This is called a **zero divisor** — and it prevents $\mathbb{Z}/6\mathbb{Z}$ from being an integral domain or a field. By contrast, consider $\mathbb{Z}/5\mathbb{Z}$. Since $5$ is prime, if $\bar{a} \cdot \bar{b} = \bar{0}$ in $\mathbb{Z}/5\mathbb{Z}$, then $5 \mid ab$, which (since $5$ is prime) forces $5 \mid a$ or $5 \mid b$, i.e., $\bar{a} = \bar{0}$ or $\bar{b} = \bar{0}$. So $\mathbb{Z}/5\mathbb{Z}$ has no zero divisors. Moreover, every nonzero class in $\mathbb{Z}/5\mathbb{Z}$ is a unit: $\bar{1}^{-1} = \bar{1}$, $\bar{2}^{-1} = \bar{3}$, $\bar{3}^{-1} = \bar{2}$, $\bar{4}^{-1} = \bar{4}$. So $\mathbb{Z}/5\mathbb{Z}$ is a field. The contrast makes a point we will sharpen later: $\mathbb{Z}/(p)$ is a field if and only if $p$ is prime. [/example] [example: Matrix Rings Are Noncommutative] Let $k$ be a field. The set $M_2(k)$ of $2 \times 2$ matrices with entries in $k$ forms a ring under matrix addition and multiplication. This ring is not commutative. Take: \begin{align*} A &= \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}. \end{align*} Compute: \begin{align*} AB &= \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}, \quad BA = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}. \end{align*} Since $AB \neq BA$, $M_2(k)$ is not commutative. Matrix rings are also rich in zero divisors: $A \cdot C = 0$ whenever the columns of $C$ lie in the null space of $A$. [/example] [example: The Quaternions Are a Division Ring] The quaternions $\mathbb{H}$ are the real vector space with basis $\{1, i, j, k\}$ and multiplication given by: \begin{align*} i^2 = j^2 = k^2 = -1, \quad ij = k, \quad ji = -k, \quad jk = i, \quad kj = -i, \quad ki = j, \quad ik = -j. \end{align*} A general quaternion is $q = a + bi + cj + dk$ with $a, b, c, d \in \mathbb{R}$. Define the conjugate $\bar{q} = a - bi - cj - dk$ and the norm $|q|^2 = a^2 + b^2 + c^2 + d^2$. A direct computation shows $q\bar{q} = |q|^2$: the cross terms cancel in pairs (e.g., the $i$-component contributes $-ab + ba = 0$), while the diagonal terms give $a^2 + b^2 + c^2 + d^2$. For any nonzero $q \in \mathbb{H}$, we have $|q|^2 > 0$, so the inverse is: \begin{align*} q^{-1} = \frac{\bar{q}}{|q|^2} = \frac{a - bi - cj - dk}{a^2 + b^2 + c^2 + d^2}. \end{align*} One verifies $q \cdot q^{-1} = \frac{q\bar{q}}{|q|^2} = 1$. So every nonzero quaternion is invertible: $\mathbb{H}$ is a division ring. It is not a field because $ij = k \neq -k = ji$. [/example] [example: The Polynomial Ring R[x]] For any ring $R$, the polynomial ring $R[x]$ consists of all formal sums $\sum_{j=0}^n a_j x^j$ with $a_j \in R$ and finitely many nonzero coefficients, with the usual polynomial addition and multiplication by the convolution formula: \begin{align*} \left(\sum_{i} a_i x^i\right)\left(\sum_{j} b_j x^j\right) = \sum_{k} \left(\sum_{i+j=k} a_i b_j\right) x^k. \end{align*} This is a ring with identity $1_{R[x]} = 1_R$. If $R$ is commutative, so is $R[x]$. Polynomial rings are where most of algebraic geometry and number theory lives. [/example] [example: Function Rings] Let $X$ be a topological space. The set $C(X) = C(X; \mathbb{R})$ of continuous functions $f: X \to \mathbb{R}$ forms a ring under pointwise operations: $(f + g)(x) := f(x) + g(x)$ and $(fg)(x) := f(x)g(x)$. The zero element is the constant function $0$, and the identity is the constant function $1$. Function rings see the topology of $X$: two compact Hausdorff spaces $X$ and $Y$ are homeomorphic if and only if $C(X) \cong C(Y)$ as rings. The ideals of $C(X)$ correspond to closed subsets of $X$, making function rings the starting point for the algebraic study of geometry. [/example] ## Ring Homomorphisms Having defined the objects, we need the maps between them. The right notion of "structure-preserving map" between rings must respect both operations simultaneously. [definition: Ring Homomorphism] Let $R$ and $S$ be rings. A **ring homomorphism** is a function $\varphi: R \to S$ satisfying, for all $a, b \in R$: \begin{align*} \varphi(a + b) &= \varphi(a) + \varphi(b), \\ \varphi(ab) &= \varphi(a)\varphi(b), \\ \varphi(1_R) &= 1_S. \end{align*} A ring homomorphism that is also a bijection is a **ring isomorphism**. The **kernel** of $\varphi$ is $\ker \varphi = \{a \in R : \varphi(a) = 0_S\}$. [/definition] The condition $\varphi(1_R) = 1_S$ is not automatic from the first two conditions — without it, the map $\varphi: \mathbb{Z} \to \mathbb{Z}$ given by $\varphi(n) = 0$ would be a "homomorphism" in the weaker sense but clearly fails to be a meaningful ring map. The kernel of a ring homomorphism has a property that is much stronger than being an additive subgroup: it absorbs multiplication from both sides. If $a \in \ker \varphi$ and $r \in R$, then $\varphi(ra) = \varphi(r)\varphi(a) = \varphi(r) \cdot 0_S = 0_S$, so $ra \in \ker \varphi$. Similarly $ar \in \ker \varphi$. This forces the concept of ideal. Now see what goes wrong if you try to form a quotient ring $R/H$ for an ordinary additive subgroup $H$ that does not have this extra closure. You want to define multiplication of cosets: $(r + H)(s + H) := rs + H$. For this to be well-defined, if $r' \in r + H$ and $s' \in s + H$, we need $r's' \in rs + H$. Write $r' = r + h_1$ and $s' = s + h_2$ with $h_1, h_2 \in H$. Then: \begin{align*} r's' &= (r + h_1)(s + h_2) = rs + rh_2 + h_1 s + h_1 h_2. \end{align*} We need $rh_2 + h_1 s + h_1 h_2 \in H$. The term $h_1 h_2$ lies in $H$ if $H$ is closed under multiplication, but $rh_2$ requires $H$ to absorb left multiplication by arbitrary $r \in R$, and $h_1 s$ requires $H$ to absorb right multiplication by arbitrary $s \in R$. This is exactly the definition of a two-sided ideal. ## Ideals and Quotient Rings This is the core construction of ring theory, just as normal subgroups and quotient groups are the core of group theory. Ideals are the subsets you can "kill" to form a new ring; quotient rings are the result. [definition: Ideal] Let $R$ be a ring. A subset $I \subset R$ is a **left ideal** if: - $I$ is an additive subgroup of $(R, +)$: $0_R \in I$, and $a - b \in I$ whenever $a, b \in I$. - $I$ is closed under left multiplication by $R$: for all $r \in R$ and $a \in I$, $ra \in I$. Similarly, $I$ is a **right ideal** if $ar \in I$ for all $r \in R$, $a \in I$. An ideal that is both a left ideal and a right ideal is called a **two-sided ideal**, or simply an **ideal**. In a commutative ring, left, right, and two-sided ideals coincide. We write $I \trianglelefteq R$ to denote that $I$ is a (two-sided) ideal of $R$. [/definition] Two examples anchor the definition: $\{0_R\}$ and $R$ itself are always ideals (the trivial ideals). All other ideals are called proper. Given an ideal $I \trianglelefteq R$, one can collapse $I$ to zero by forming a new ring out of cosets. This is the same move made in group theory when passing to quotient groups, and it is just as powerful: quotient rings let you enforce algebraic relations by fiat. Want a ring where $x^2 + 1 = 0$? Take $\mathbb{R}[x]$ and kill the ideal $(x^2 + 1)$. The result is $\mathbb{C}$. [definition: Quotient Ring] Let $R$ be a ring and $I \trianglelefteq R$ a two-sided ideal. The **quotient ring** $R/I$ is the set of cosets $\{r + I : r \in R\}$, equipped with: \begin{align*} (r + I) + (s + I) &:= (r + s) + I, \\ (r + I)(s + I) &:= rs + I. \end{align*} The zero element of $R/I$ is $0_R + I = I$, and the identity is $1_R + I$. [/definition] The preceding discussion shows that the multiplication is well-defined precisely because $I$ is a two-sided ideal. The quotient ring $R/I$ captures the arithmetic of $R$ subject to the relation $a = 0$ for all $a \in I$ — equivalently, any two elements differing by an element of $I$ become equal in the quotient. The bridge between quotient rings and homomorphisms is the First Isomorphism Theorem: it says that every homomorphic image of $R$ is isomorphic to a quotient of $R$, and conversely. [quotetheorem:851] The First Isomorphism Theorem is the bridge between quotient rings (built abstractly from ideals) and homomorphic images (built from maps). Any time you want to understand $\operatorname{im} \varphi$, you can instead study $R/\ker\varphi$, which is often more tractable because you know $R$ and you know the ideal $\ker\varphi$. [example: R[x]/(x²+1) Is Isomorphic to C] We prove rigorously that $\mathbb{R}[x]/(x^2+1) \cong \mathbb{C}$. Define the evaluation homomorphism: \begin{align*} \varphi: \mathbb{R}[x] &\to \mathbb{C}, \quad f(x) \mapsto f(i). \end{align*} This is a ring homomorphism: $\varphi(f + g) = (f+g)(i) = f(i) + g(i)$ and $\varphi(fg) = (fg)(i) = f(i)g(i)$, and $\varphi(1) = 1$. The image: every complex number $a + bi$ equals $\varphi(a + bx)$, so $\varphi$ is surjective. The kernel: $\varphi(f) = 0$ iff $f(i) = 0$, i.e., iff $i$ is a root of $f$. Since $f$ has real coefficients, $\bar{i} = -i$ is also a root, so $(x-i)(x+i) = x^2+1$ divides $f(x)$ in $\mathbb{C}[x]$. Since $x^2+1$ has real coefficients, this division takes place in $\mathbb{R}[x]$: $\ker\varphi = (x^2+1)$. By the First Isomorphism Theorem: \begin{align*} \mathbb{R}[x]/(x^2+1) \cong \operatorname{im}\varphi = \mathbb{C}. \end{align*} [/example] [example: Z/(p) Is a Field When p Is Prime] Fix a prime $p$ and consider the quotient ring $\mathbb{Z}/(p) = \mathbb{Z}/p\mathbb{Z}$. We claim this is a field. Let $\bar{a} \in \mathbb{Z}/p\mathbb{Z}$ be nonzero, i.e., $p \nmid a$. Since $p$ is prime and $\gcd(a, p) = 1$, Bézout's theorem gives integers $u, v$ with $au + pv = 1$. Reducing modulo $p$: $\bar{a} \cdot \bar{u} = \bar{1}$ in $\mathbb{Z}/p\mathbb{Z}$. So $\bar{a}$ is a unit with inverse $\bar{u}$. Every nonzero element is a unit, so $\mathbb{Z}/p\mathbb{Z}$ is a field — the **prime field** $\mathbb{F}_p$. Now compare with $\mathbb{Z}/6\mathbb{Z}$: since $6 = 2 \cdot 3$ is not prime, $\gcd(2, 6) = 2 \neq 1$, and $\bar{2}$ has no inverse. Specifically, $\bar{2} \cdot \bar{3} = \bar{0}$, so $\bar{2}$ is a zero divisor — far from being a unit. [/example] ## Prime and Maximal Ideals Quotient rings come in different flavors depending on the ideal. Two special classes of ideals tell you exactly what algebraic structure the quotient inherits: prime ideals give integral domains, maximal ideals give fields. These characterizations are the structural payoff of the entire ideal theory — they let you read off the quotient's properties from the ideal without computing in the quotient directly. An ideal is prime when it mimics the defining property of a prime number: $p \mid ab$ implies $p \mid a$ or $p \mid b$. [definition: Prime Ideal] Let $R$ be a commutative ring and $\mathfrak{p} \trianglelefteq R$ a proper ideal ($\mathfrak{p} \neq R$). We say $\mathfrak{p}$ is a **prime ideal** if for all $a, b \in R$: \begin{align*} ab \in \mathfrak{p} \implies a \in \mathfrak{p} \text{ or } b \in \mathfrak{p}. \end{align*} [/definition] An ideal is maximal when it is as large as possible without being the whole ring — there is no room for a proper ideal between it and $R$. [definition: Maximal Ideal] Let $R$ be a commutative ring and $\mathfrak{m} \trianglelefteq R$ a proper ideal. We say $\mathfrak{m}$ is a **maximal ideal** if there is no ideal $J$ with $\mathfrak{m} \subsetneq J \subsetneq R$; equivalently, $\mathfrak{m}$ is maximal among proper ideals under inclusion. [/definition] The power of these definitions comes from the following characterization, which connects ideal-theoretic properties directly to the structure of the quotient ring. [quotetheorem:853] [quotetheorem:852] Instead of examining the quotient ring directly, you can read off its algebraic type — domain or field — from the combinatorial structure of the ideal in $R$. For part (1): $R/I$ has no zero divisors if and only if whenever $(a + I)(b + I) = 0 + I$, we have $a + I = 0 + I$ or $b + I = 0 + I$. But $(a+I)(b+I) = ab + I = 0 + I$ means $ab \in I$, and $a + I = 0+I$ means $a \in I$. So "no zero divisors in $R/I$" translates directly to the condition "$ab \in I \implies a \in I$ or $b \in I$" — exactly the prime ideal condition. For part (2): $R/I$ is a field if and only if every nonzero element $a + I$ (i.e., $a \notin I$) has a multiplicative inverse, if and only if for every $a \notin I$, there exists $b \in R$ with $ab \equiv 1 \pmod{I}$, i.e., $ab - 1 \in I$. This means the ideal generated by $I$ and $a$ contains $1$, hence equals $R$. This holds for every $a \notin I$ precisely when $I$ is maximal. The last claim — every maximal ideal is prime — now follows: fields are integral domains (a unit cannot be a zero divisor), so if $\mathfrak{m}$ is maximal then $R/\mathfrak{m}$ is a field, hence a domain, hence $\mathfrak{m}$ is prime. [example: Prime and Maximal Ideals in Z and Z[x]] **The ideal $(p)$ in $\mathbb{Z}$.** If $p$ is prime, the quotient $\mathbb{Z}/(p) = \mathbb{F}_p$ is a field, so $(p)$ is maximal (hence also prime). If $p$ is composite, say $p = 6$: then $2 \cdot 3 \in (6)$ but $2 \notin (6)$ and $3 \notin (6)$, so $(6)$ is not prime (hence not maximal either). **The ideal $(x)$ in $\mathbb{Z}[x]$.** The ideal $(x)$ consists of polynomials with zero constant term: $(x) = \{f \in \mathbb{Z}[x] : f(0) = 0\}$. The quotient $\mathbb{Z}[x]/(x) \cong \mathbb{Z}$ (via the evaluation map $f(x) \mapsto f(0)$). Since $\mathbb{Z}$ is an integral domain, $(x)$ is prime. But $\mathbb{Z}$ is not a field, so $(x)$ is not maximal in $\mathbb{Z}[x]$. Indeed, the ideal $(x, 2) = \{f \in \mathbb{Z}[x] : f(0) \text{ is even}\}$ strictly contains $(x)$. The quotient $\mathbb{Z}[x]/(x, 2) \cong \mathbb{F}_2$ (send $f(x) \mapsto f(0) \bmod 2$), which is a field, so $(x, 2)$ is maximal. This illustrates the general hierarchy: $(x)$ is prime but not maximal in $\mathbb{Z}[x]$, while $(x, p)$ is maximal for any prime $p$. [/example] ## Units and Zero Divisors We now examine the multiplicative structure of a ring more carefully. Two classes of elements stand out: those with inverses (units), and those that "kill" other nonzero elements (zero divisors). These two classes are always disjoint, and their relative sizes determine the arithmetic character of the ring. Which elements of a ring can we "divide by"? In $\mathbb{Z}$, dividing by $2$ takes us out of the ring — $\frac{1}{2} \notin \mathbb{Z}$ — so $2$ is not invertible. But $-1$ is invertible: $(-1)(-1) = 1$. The invertible elements form a group under multiplication, and this group controls much of the ring's structure. [definition: Unit] Let $R$ be a ring. An element $a \in R$ is a **unit** if there exists $b \in R$ with $ab = ba = 1_R$. The set of units of $R$ forms a group under multiplication, denoted $R^\times$. [/definition] At the opposite extreme, a zero divisor is an element whose multiplication collapses information irreversibly. [definition: Zero Divisor] Let $R$ be a ring. A nonzero element $a \in R$ is a **zero divisor** if there exists a nonzero $b \in R$ with $ab = 0$ or $ba = 0$. [/definition] [quotetheorem:2576] The proof is immediate: if $a$ is a unit and $ab = 0$, multiply on the left by $a^{-1}$ to get $b = a^{-1} \cdot 0 = 0$. So units and zero divisors are disjoint among nonzero elements. Integral domains are the rings where the zero-divisor pathology is entirely absent. This is the setting where cancellation works and factorization theory is well-behaved. [definition: Integral Domain] A commutative ring $R \neq \{0\}$ is an **integral domain** if it has no zero divisors: $ab = 0$ implies $a = 0$ or $b = 0$. [/definition] [quotetheorem:2577] From $ac = bc$, subtract: $(a-b)c = 0$. Since $c \neq 0$ and $R$ has no zero divisors, $a - b = 0$, i.e., $a = b$. [example: Units and Zero Divisors in Z/nZ] In $\mathbb{Z}/n\mathbb{Z}$, the units and zero divisors are determined by the $\gcd$. An element $\bar{a} \in \mathbb{Z}/n\mathbb{Z}$ is a **unit** if and only if $\gcd(a, n) = 1$: Bézout gives $au + nv = 1$, so $\bar{a}\bar{u} = \bar{1}$. An element $\bar{a}$ with $\bar{a} \neq \bar{0}$ is a **zero divisor** if and only if $\gcd(a,n) > 1$: set $b = n/\gcd(a,n)$; then $\bar{b} \neq \bar{0}$ and $\bar{a}\bar{b} = \bar{0}$. For example, in $\mathbb{Z}/12\mathbb{Z}$: the units are $\{\bar{1}, \bar{5}, \bar{7}, \bar{11}\}$ (the four elements coprime to $12$); $|(\mathbb{Z}/12\mathbb{Z})^\times| = \varphi(12) = 4$. The zero divisors are $\{\bar{2}, \bar{3}, \bar{4}, \bar{6}, \bar{8}, \bar{9}, \bar{10}\}$ — every nonzero non-unit is a zero divisor. [/example] ## Generated Ideals and Principal Ideal Domains How do we build ideals in practice? The most common method is to generate an ideal from a set of elements, just as we generate subgroups from a set of generators. [definition: Generated Ideal] Let $R$ be a commutative ring and $S \subset R$. The **ideal generated by $S$** is the smallest ideal containing $S$: \begin{align*} (S) = \left\{ \sum_{i=1}^n r_i s_i : n \in \mathbb{N},\; r_i \in R,\; s_i \in S \right\}. \end{align*} When $S = \{a\}$, we write $(a) = aR = \{ra : r \in R\}$ and call it the **principal ideal generated by $a$**. When $S = \{a_1, \ldots, a_k\}$, we write $(a_1, \ldots, a_k)$. [/definition] The most tractable rings are those where every ideal is principal — generated by a single element. In such rings, the ideal structure is completely determined by divisibility, and factorization theory works cleanly. [definition: Principal Ideal Domain] An integral domain $R$ is a **principal ideal domain** (PID) if every ideal of $R$ is principal: for every $I \trianglelefteq R$, there exists $a \in R$ with $I = (a)$. [/definition] The integers $\mathbb{Z}$ are a PID. The proof uses the division algorithm: given a nonzero ideal $I \trianglelefteq \mathbb{Z}$, let $a$ be the smallest positive element of $I$. For any $b \in I$, write $b = qa + r$ with $0 \leq r < a$. Since $b, a \in I$, we have $r = b - qa \in I$. The minimality of $a$ forces $r = 0$, so $a \mid b$ and $I = (a)$. In a PID, two elements $a, b$ generate an ideal $(a, b)$ that is itself principal: $(a, b) = (d)$ for some $d \in R$. This $d$ is a greatest common divisor of $a$ and $b$, and it is an $R$-linear combination of $a$ and $b$: $d = ua + vb$ for some $u, v \in R$. In $\mathbb{Z}$, this is Bézout's theorem. In $k[x]$ for a field $k$, it is the polynomial Bézout identity — and this is what makes polynomial factorization into irreducibles unique (up to units), paralleling unique prime factorization in $\mathbb{Z}$. [example: Generated Ideals in Z[x]] In $\mathbb{Z}[x]$, the ideal $(2, x)$ consists of all polynomials $f \in \mathbb{Z}[x]$ whose constant term is even: \begin{align*} (2, x) = \{2g(x) + xh(x) : g, h \in \mathbb{Z}[x]\} = \{f \in \mathbb{Z}[x] : f(0) \in 2\mathbb{Z}\}. \end{align*} This ideal is *not* principal. If $(2, x) = (d)$ for some $d \in \mathbb{Z}[x]$, then $d \mid 2$ and $d \mid x$. Since $d \mid 2$ and $2$ has degree $0$, $d$ must be a constant; since $d \mid x$ and $x$ is irreducible, $d$ must be $\pm 1$. But $(1) = \mathbb{Z}[x] \neq (2, x)$ since $1 \notin (2, x)$ (we have $1(0) = 1$, which is odd). Contradiction. So $\mathbb{Z}[x]$ is not a PID. [/example] ## Characteristic Every ring $R$ receives a unique ring homomorphism from $\mathbb{Z}$: the map $n \mapsto n \cdot 1_R$ (where $n \cdot 1_R$ means $1_R$ added to itself $n$ times). Its kernel is an ideal of $\mathbb{Z}$, hence principal: $\ker(n \mapsto n \cdot 1_R) = (m)$ for some $m \geq 0$. That integer $m$ is the characteristic. [definition: Characteristic] The **characteristic** of a ring $R$, denoted $\operatorname{char}(R)$, is the unique nonnegative integer $m$ such that $\ker(\mathbb{Z} \to R) = (m)$. Explicitly, $\operatorname{char}(R) = m$ means $m \cdot 1_R = 0_R$ and $m$ is the smallest such positive integer (or $m = 0$ if no positive integer annihilates $1_R$). [/definition] If $\operatorname{char}(R) = 0$, then $\mathbb{Z}$ injects into $R$: $\mathbb{Z}$ appears as a subring. This is the case for $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, and $\mathbb{Z}$ itself. [quotetheorem:2578] If $\operatorname{char}(R) = m > 0$ and $m = ab$ with $1 < a, b < m$, then $(a \cdot 1_R)(b \cdot 1_R) = ab \cdot 1_R = m \cdot 1_R = 0_R$, making $a \cdot 1_R$ and $b \cdot 1_R$ zero divisors — contradicting the integral domain assumption. [example: The Frobenius Endomorphism] Let $R$ be a commutative ring with $\operatorname{char}(R) = p$ prime. The **Frobenius endomorphism** is the map $\varphi_p: R \to R$ defined by $\varphi_p(a) = a^p$. We claim this is a ring homomorphism. Since $\varphi_p(1) = 1$ and $\varphi_p(ab) = (ab)^p = a^p b^p$ (using commutativity), the only nontrivial condition is additivity. By the binomial theorem: \begin{align*} (a + b)^p = \sum_{k=0}^p \binom{p}{k} a^k b^{p-k}. \end{align*} For $0 < k < p$, the binomial coefficient $\binom{p}{k} = \frac{p!}{k!(p-k)!}$ is divisible by $p$ (since $p$ is prime and does not divide $k!$ or $(p-k)!$ when $0 < k < p$). In a ring of characteristic $p$, these terms vanish: \begin{align*} (a + b)^p = a^p + b^p. \end{align*} So $\varphi_p$ is a ring homomorphism. In characteristic $2$, this gives $(a+b)^2 = a^2 + b^2$ — there is no cross term, since $2 = 0$. In a finite field $\mathbb{F}_q$ with $q = p^n$, the Frobenius generates the Galois group $\operatorname{Gal}(\mathbb{F}_q/\mathbb{F}_p) \cong \mathbb{Z}/n\mathbb{Z}$. [/example] ## The Chinese Remainder Theorem We close with a classical result that shows ideals and quotient rings working in concert. The Chinese Remainder Theorem decomposes a quotient ring into a product when the ideal factors into coprime parts. Two ideals are coprime when their sum fills the whole ring — they are as far apart as possible. [definition: Comaximal Ideals] Two ideals $I$ and $J$ in a commutative ring $R$ are **comaximal** (or **coprime**) if $I + J = R$, i.e., there exist $a \in I$ and $b \in J$ with $a + b = 1_R$. [/definition] In $\mathbb{Z}$, the ideals $(m)$ and $(n)$ are comaximal if and only if $\gcd(m, n) = 1$: Bézout gives $um + vn = 1$, so $um \in (m)$ and $vn \in (n)$ sum to $1$. [quotetheorem:734] [example: Chinese Remainder Theorem for Z/15Z] We compute $\mathbb{Z}/15\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$. Since $\gcd(3, 5) = 1$, the ideals $(3)$ and $(5)$ are comaximal in $\mathbb{Z}$: Bézout gives $2 \cdot 3 + (-1) \cdot 5 = 1$. The isomorphism $\mathbb{Z}/15\mathbb{Z} \to \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$ sends $\bar{a} \mapsto (a \bmod 3,\; a \bmod 5)$. To compute the inverse, we need idempotents. Since $6 \equiv 0 \pmod{3}$ and $6 \equiv 1 \pmod{5}$, the element $\bar{6}$ projects onto the second factor. Since $10 \equiv 1 \pmod{3}$ and $10 \equiv 0 \pmod{5}$, the element $\overline{10}$ projects onto the first factor. The inverse map is: \begin{align*} (r, s) \mapsto 10r + 6s \bmod 15. \end{align*} For instance, the unique solution to $a \equiv 2 \pmod{3}$ and $a \equiv 4 \pmod{5}$ is: \begin{align*} a &= 10 \cdot 2 + 6 \cdot 4 = 20 + 24 = 44 \equiv 14 \pmod{15}. \end{align*} Check: $14 \equiv 2 \pmod{3}$ and $14 \equiv 4 \pmod{5}$. [/example] The Chinese Remainder Theorem has a deep structural reading: it says that $\mathbb{Z}/N\mathbb{Z}$ "splits" as a product whenever $N$ factors into coprime parts. This is the beginning of the decomposition theory for rings and modules — the product decomposition corresponds to the ring splitting into independent components that do not interact. ## References - M. Artin, *Algebra* (2nd ed., 2011). Chapters 11–14 cover rings, ideals, homomorphisms, and factorization with a geometric flavor. - D.S. Dummit and R.M. Foote, *Abstract Algebra* (3rd ed., 2004). Part II is the standard graduate reference for ring theory. - S. Lang, *Algebra* (Revised 3rd ed., 2002). Terse but comprehensive; Chapter II covers rings and ideals. - M.F. Atiyah and I.G. Macdonald, *Introduction to Commutative Algebra* (1969). The standard reference for prime/maximal ideals, localization, and the spectrum. - N. Jacobson, *Basic Algebra I* (2nd ed., 1985). Thorough treatment of noncommutative rings and the Jacobson radical.