A ring lets us add, subtract, and multiply, but quotients are more demanding than subrings. If we try to declare every element of a subset $I \subset R$ equal to zero, addition only asks that $I$ be an additive subgroup. Multiplication asks for a stronger condition: anything in $I$ must stay in $I$ after multiplication by arbitrary elements of the ambient ring. Ideals are the subsets that make this collapse compatible with the whole ring structure. [example: Why Even Integers Work] In $\mathbb{Z}$, write $a\equiv b \pmod{2\mathbb{Z}}$ to mean $a-b\in 2\mathbb{Z}$. Suppose $a-b\in 2\mathbb{Z}$ and $c-d\in 2\mathbb{Z}$. Then there are integers $m,n\in\mathbb{Z}$ such that \begin{align*} a-b=2m \end{align*} and \begin{align*} c-d=2n. \end{align*} For addition, \begin{align*} (a+c)-(b+d)=(a-b)+(c-d)=2m+2n=2(m+n), \end{align*} so $(a+c)-(b+d)\in 2\mathbb{Z}$. For multiplication, first rewrite the difference as \begin{align*} ac-bd=ac-ad+ad-bd=a(c-d)+d(a-b). \end{align*} Substituting $c-d=2n$ and $a-b=2m$ gives \begin{align*} ac-bd=a(2n)+d(2m)=2(an)+2(dm)=2(an+dm), \end{align*} so $ac-bd\in 2\mathbb{Z}$. The important point is that $a$ and $d$ need not be even; $2\mathbb{Z}$ stays closed under multiplication by arbitrary integers, which is exactly the absorption property isolated by ideals. [/example] The example is familiar, but it exposes the main point. A modulus is not just a collection of elements; it is a collection stable under every multiplication forced by the ring. Ideals are the ring-theoretic analogue of normal subgroups: they are exactly the pieces that can be collapsed in a quotient. Throughout this page, a ring means an associative ring with multiplicative identity $1_R$, and a ring homomorphism preserves the multiplicative identity. A commutative ring is always understood in this unital sense. When noncommutative or [Lie algebra](/page/Lie%20Algebra) variants appear later, the relevant hypotheses are stated explicitly. ## Definition ### Absorption by the Ambient Ring The quotient problem gives the primary definition. We want $a \equiv b \pmod I$ to imply $ra \equiv rb \pmod I$ and $ar \equiv br \pmod I$ for every $r \in R$. The next definition records exactly the additive and multiplicative closure needed for that congruence relation to respect ring operations. [definition: Ideal] Let $R$ be a ring. A subset $I \subset R$ is an ideal of $R$, written $I \trianglelefteq R$, if the following conditions hold: \begin{align*} I \text{ is an additive subgroup of } (R,+,0_R). \end{align*} \begin{align*} \text{for all } r \in R \text{ and } a \in I, \quad ra \in I \text{ and } ar \in I. \end{align*} [/definition] The first condition includes $0_R \in I$, additive inverses, and closure under addition, so the empty subset is excluded. The second condition is the absorption condition: $I$ absorbs multiplication from the entire ring, not only from its own elements. In noncommutative ring theory this is also called a two-sided ideal. In a commutative ring the two products $ra$ and $ar$ give the same requirement, so the unqualified word ideal has the same meaning. Noncommutative rings force us to separate the two sides of multiplication. A subset may be stable under multiplication from the left but not from the right, and such subsets naturally appear as submodules of the regular module. The next definition keeps only the left action because that is the structure needed for left-module arguments. [definition: Left Ideal] Let $R$ be a ring. A subset $I \subset R$ is a left ideal of $R$ if $I$ is an additive subgroup of $(R,+,0_R)$ and, for all $r \in R$ and $a \in I$, one has $ra \in I$. [/definition] A left ideal can be used when the ring acts on itself from the left. It need not support a quotient ring because right multiplication may fail to be compatible with the proposed congruence. To track the other regular action, we need the mirror-image definition. [definition: Right Ideal] Let $R$ be a ring. A subset $I \subset R$ is a right ideal of $R$ if $I$ is an additive subgroup of $(R,+,0_R)$ and, for all $r \in R$ and $a \in I$, one has $ar \in I$. [/definition] An ideal in the primary sense is both a left ideal and a right ideal. The distinction disappears in commutative algebra, but it becomes essential in matrix rings and operator algebras. Since quotient rings require a subset that is stable from both sides, the rest of the ring-theoretic discussion focuses on ideals in the primary two-sided sense. The zero subset and the whole ring always satisfy the axioms. The whole ring gives a degenerate quotient, so many structural questions ask for ideals that do not already contain $1_R$. The next term isolates the ideals that impose a genuine restriction on the ring. [definition: Proper Ideal] Let $R$ be a ring. A proper ideal of $R$ is an ideal $I \trianglelefteq R$ such that $I \subsetneq R$. [/definition] Properness is equivalent to $1_R \notin I$ when $R$ has a multiplicative identity. It is the condition that prevents $R/I$ from collapsing to the zero ring. Prime and maximal ideals are both proper by definition. [example: Ideals in $\mathbb{Z}$] Fix $n\in\mathbb{N}$. The set $n\mathbb{Z}=\{nk:k\in\mathbb{Z}\}$ contains $0=n\cdot 0$. If $a,b\in n\mathbb{Z}$, then $a=nm$ and $b=n\ell$ for some $m,\ell\in\mathbb{Z}$, so \begin{align*} a-b=nm-n\ell=n(m-\ell). \end{align*} Since $m-\ell\in\mathbb{Z}$, this shows $a-b\in n\mathbb{Z}$. If $r\in\mathbb{Z}$ and $a=nm\in n\mathbb{Z}$, then \begin{align*} ra=r(nm)=n(rm). \end{align*} Since $rm\in\mathbb{Z}$, this shows $ra\in n\mathbb{Z}$. Thus $n\mathbb{Z}$ is an ideal of $\mathbb{Z}$. Conversely, let $I\trianglelefteq\mathbb{Z}$. If $I=\{0\}$, then $I=0\mathbb{Z}$. Otherwise $I$ contains a nonzero integer $t$. If $t<0$, then $-t\in I$ because $I$ is an additive subgroup, so $I$ contains a positive integer. Let $n$ be the least positive integer in $I$. Since $n\in I$ and $I$ absorbs multiplication by every integer, every element $qn$ with $q\in\mathbb{Z}$ lies in $I$, so $n\mathbb{Z}\subset I$. For the reverse inclusion, take $x\in I$. By the [division algorithm](/theorems/725), write \begin{align*} x=qn+s \end{align*} with $q\in\mathbb{Z}$ and $0\leq s<n$. Since $x\in I$ and $qn\in I$, the additive subgroup property gives \begin{align*} s=x-qn\in I. \end{align*} If $s>0$, then $s$ is a positive element of $I$ smaller than $n$, contradicting the choice of $n$. Hence $s=0$, so $x=qn\in n\mathbb{Z}$. Therefore $I\subset n\mathbb{Z}$, and $I=n\mathbb{Z}$. The generator is unique in $\mathbb{N}\cup\{0\}$: if $n\mathbb{Z}=m\mathbb{Z}$ with $n,m\geq 0$, then either both ideals are $\{0\}$, giving $n=m=0$, or both contain positive integers and their least positive elements must be equal. Thus every ideal of $\mathbb{Z}$ is exactly one of the ideals $n\mathbb{Z}$ with $n\in\mathbb{N}\cup\{0\}$. [/example] This example explains why ideals generalise divisibility. Passing from $n$ to $n\mathbb{Z}$ replaces an integer by the set of all its multiples. Ring theory then asks which rings behave like $\mathbb{Z}$ and which rings have more complicated ideals. ### Ideals Generated by Elements A ring can contain many ideals, and listing their elements is rarely practical. The better question is what ideal is forced once certain elements are required to vanish. Principal and generated ideals answer that question by closing a chosen list under addition and multiplication by arbitrary ring elements. [definition: Principal Ideal] Let $R$ be a commutative ring and let $a \in R$. The principal ideal generated by $a$ is \begin{align*} (a)=\{ra:r\in R\}. \end{align*} [/definition] A single generator models one imposed equation, but most algebraic systems have several independent equations. If $a_1,\dots,a_n$ must all vanish, the ring also forces every combination with coefficients in $R$ to vanish. The next definition packages the smallest ideal containing a finite list of prescribed generators. [definition: Generated Ideal] Let $R$ be a commutative ring and let $a_1,\dots,a_n \in R$. The ideal generated by $a_1,\dots,a_n$ is \begin{align*} (a_1,\dots,a_n)=\left\{\sum_{i=1}^n r_i a_i:r_i\in R\right\}. \end{align*} [/definition] Generated ideals encode algebraic consequences. If the equations $a_1=0,\dots,a_n=0$ are imposed in a quotient, then every linear combination $\sum_i r_i a_i$ must vanish as well. This viewpoint is the starting point for polynomial equations and computational commutative algebra. [example: The Ideal $(6,10)$ in $\mathbb{Z}$] In $\mathbb{Z}$, the ideal $(6,10)$ consists of all integer linear combinations \begin{align*} 6a+10b \end{align*} with $a,b\in\mathbb{Z}$. For any such combination, \begin{align*} 6a+10b=2(3a)+2(5b)=2(3a+5b). \end{align*} Since $3a+5b\in\mathbb{Z}$, this shows $6a+10b\in 2\mathbb{Z}$. Hence $(6,10)\subset 2\mathbb{Z}$. For the reverse inclusion, first note that \begin{align*} 2=2\cdot 6+(-1)\cdot 10. \end{align*} Thus $2\in(6,10)$. If $t\in 2\mathbb{Z}$, then $t=2q$ for some $q\in\mathbb{Z}$, and multiplying the displayed expression by $q$ gives \begin{align*} t=2q=q(2\cdot 6+(-1)\cdot 10)=(2q)\cdot 6+(-q)\cdot 10. \end{align*} Because $2q,-q\in\mathbb{Z}$, this expression lies in $(6,10)$. Therefore $2\mathbb{Z}\subset(6,10)$, and the two inclusions give \begin{align*} (6,10)=2\mathbb{Z}. \end{align*} So the two generators $6$ and $10$ impose exactly the same ideal as their greatest common divisor $2$. [/example] The calculation shows how generated ideals recover greatest common divisors in $\mathbb{Z}$. In more general rings, the same notation may hide far richer behaviour because no single greatest common divisor may exist. ## Quotients and Kernels ### Quotient Rings To make congruence modulo $I$ into arithmetic, the class of a product must not depend on the representatives chosen. This is the obstruction that ordinary additive subgroups do not solve: replacing $a$ by $a+i$ must leave products unchanged modulo $I$, which requires multiplication by elements of $I$ to stay inside $I$. When that absorption condition holds, the following construction collapses exactly the elements of $I$ to zero. [definition: Quotient Ring by an Ideal] Let $R$ be a ring and let $I \trianglelefteq R$ be a two-sided ideal. The quotient ring $R/I$ is the set of additive cosets $a+I$ with operations \begin{align*} (a+I)+(b+I)=(a+b)+I \end{align*} and \begin{align*} (a+I)(b+I)=ab+I. \end{align*} [/definition] The formulas for addition and multiplication use representatives $a$ and $b$, so there is still a well-definedness problem: two names for the same coset must give the same output coset. The absorption property of an ideal is precisely what removes this ambiguity, and once it is removed the quotient comes equipped with a natural map from the original ring. [quotetheorem:8156] This theorem says that ideals are exactly the allowable zero-sets for quotienting. After $I$ is collapsed, every product involving an element of $I$ must also become zero. The reverse direction asks where ideals come from when the quotient map is not already known. ### Kernels A homomorphism may identify distinct elements of its domain, and the strongest form of identification occurs when an element becomes zero. The elements with this behaviour record exactly which relations the map has forced. This zero-fiber is the object that will connect homomorphisms back to ideals. [definition: Kernel of a Ring Homomorphism] Let $\varphi:R\to S$ be a ring homomorphism. The kernel of $\varphi$ is \begin{align*} \ker\varphi=\{r\in R:\varphi(r)=0_S\}. \end{align*} [/definition] A kernel is not merely an additive subgroup because multiplication in the domain can be transported through the homomorphism. If $r$ maps to zero, then multiplying $r$ before applying the map still gives zero. The next theorem turns that observation into the main recognition principle for ideals. [quotetheorem:8157] Together with quotient rings, kernels give the central slogan: ideals are precisely the subsets that can occur as kernels of ring homomorphisms. This slogan is useful because it replaces an internal closure check with a structural map. The next theorem makes the slogan precise by identifying the quotient by the kernel with the image. [quotetheorem:851] The theorem decomposes every ring homomorphism into a quotient followed by an isomorphism onto its image. This is why quotient rings are not optional machinery: they are built into every homomorphism. ## Evaluation and Polynomial Quotients Here $k$ denotes a field, and $k[x]$ denotes the [polynomial ring](/page/Polynomial%20Ring) in one variable $x$ with coefficients in $k$. More generally, $k[x,y]$ is the polynomial ring in two variables, and $k[x_1,\dots,x_n]$ is the polynomial ring in $n$ variables over $k$. This notation lets evaluation at a point and quotienting by polynomial ideals be treated as ring-theoretic operations rather than as ad hoc substitution rules. [example: Evaluation and Polynomial Quotients] Let $k$ be a field and define \begin{align*} \operatorname{ev}_0:k[x]\to k \end{align*} by $\operatorname{ev}_0(f)=f(0)$. If \begin{align*} f(x)=c_0+c_1x+\cdots+c_nx^n \end{align*} with $c_i\in k$, then \begin{align*} \operatorname{ev}_0(f)=f(0)=c_0+c_1\cdot 0+\cdots+c_n\cdot 0^n=c_0. \end{align*} Thus $f\in\ker(\operatorname{ev}_0)$ exactly when $c_0=0$. In that case \begin{align*} f(x)=c_1x+\cdots+c_nx^n=x(c_1+c_2x+\cdots+c_nx^{n-1}), \end{align*} so $f\in(x)$. Conversely, if $f\in(x)$, then $f=xg$ for some $g\in k[x]$, and \begin{align*} \operatorname{ev}_0(f)=f(0)=0\cdot g(0)=0. \end{align*} Therefore $\ker(\operatorname{ev}_0)=(x)$. The map is surjective because every $\lambda\in k$ is the value of the constant polynomial $\lambda$, so *[First Isomorphism Theorem for Rings](/theorems/851)* gives \begin{align*} k[x]/(x)\cong k. \end{align*} More generally, fix $a\in k$ and define $\operatorname{ev}_a(f)=f(a)$. For \begin{align*} f(x)=c_0+c_1x+\cdots+c_nx^n, \end{align*} we have \begin{align*} f(x)-f(a)=\sum_{i=1}^n c_i(x^i-a^i). \end{align*} For each $i\geq 1$, \begin{align*} x^i-a^i=(x-a)(x^{i-1}+x^{i-2}a+\cdots+xa^{i-2}+a^{i-1}), \end{align*} so $f(x)-f(a)$ is a multiple of $x-a$. If $f(a)=0$, this gives $f(x)=f(x)-f(a)\in(x-a)$. Conversely, if $f=(x-a)g$, then \begin{align*} \operatorname{ev}_a(f)=f(a)=(a-a)g(a)=0. \end{align*} Hence $\ker(\operatorname{ev}_a)=(x-a)$. Since every $\lambda\in k$ is again hit by the constant polynomial $\lambda$, the map $\operatorname{ev}_a$ is surjective, and *[First Isomorphism Theorem](/theorems/791) for Rings* gives \begin{align*} k[x]/(x-a)\cong k. \end{align*} Setting $x=a$ is therefore the same quotient operation as collapsing the ideal generated by $x-a$ to zero. [/example] Setting $x=a$ is therefore the same as quotienting by the ideal generated by $x-a$. This is a small example of a broad principle: equations become ideals, and solving or imposing equations becomes quotienting. ## Finiteness and Principal Behaviour ### Finite Generation An ideal is often given by generators, but an arbitrary ideal may require infinitely many pieces of data to describe. That creates a practical obstruction: membership, comparison, and computation become hard to control without a finite list of generators. The finiteness condition below singles out the ideals whose entire content is determined by finitely many elements. [definition: Finitely Generated Ideal] Let $R$ be a commutative ring. An ideal $I\trianglelefteq R$ is finitely generated if there exist $a_1,\dots,a_n\in R$ such that \begin{align*} I=(a_1, \dots, a_n). \end{align*} [/definition] Finite generation turns ideal membership into a question about representing an element as a finite combination of specified generators. Noetherian rings are built by requiring this finite control for every ideal. Before reaching that general theory, it is useful to study the extreme case where one generator always suffices. [definition: Principal Ideal Domain] Let $R$ be an integral domain. The ring $R$ is a [principal ideal domain](/page/Principal%20Ideal%20Domain) if every ideal $I\trianglelefteq R$ is principal. [/definition] In a principal ideal domain, an ideal may have many possible generators, so the notation $(a)$ does not identify a single distinguished element. This creates a problem for using principal ideals as divisibility data: changing the generator should not change the ideal, but it should also not introduce uncontrolled choices. The needed result identifies exactly how much freedom remains when two elements generate the same ideal. [quotetheorem:8158] The generator is not unique as an element. It is unique up to multiplication by a unit, just as $n\mathbb{Z}=(-n)\mathbb{Z}$ in the integers. This controlled ambiguity is what makes principal ideal domains computationally friendly. ### Failure of Principal Generation The principal case is powerful because it is special. As soon as polynomial rings have two independent variables, one generator may no longer describe a natural ideal. The next example shows the first basic failure. [example: A Nonprincipal Ideal] Let $k$ be a field and set $R=k[x,y]$. We show that the ideal $(x,y)$ cannot be generated by one element. Suppose, toward a contradiction, that $(x,y)=(f)$ for some $f\in k[x,y]$. Since $x,y\in(f)$, there are polynomials $g,h\in k[x,y]$ such that \begin{align*} x=fg \end{align*} and \begin{align*} y=fh. \end{align*} The polynomial $f$ is nonzero, because otherwise $(f)=\{0\}$ would not contain $x$. Using total degree and the identity $\deg(uv)=\deg u+\deg v$ for nonzero polynomials over a field, the equation $x=fg$ gives \begin{align*} 1=\deg x=\deg f+\deg g. \end{align*} Thus $\deg f$ is either $0$ or $1$. If $\deg f=1$, then $\deg g=0$ and, from $y=fh$, also $\deg h=0$. Hence $g=\alpha$ and $h=\beta$ for some nonzero $\alpha,\beta\in k$, so \begin{align*} x=\alpha f \end{align*} and \begin{align*} y=\beta f. \end{align*} Multiplying the first equation by $\beta$ and the second by $\alpha$ gives \begin{align*} \beta x=\alpha\beta f=\alpha y. \end{align*} This is impossible in $k[x,y]$, because the polynomial $\beta x-\alpha y$ has coefficient $\beta\neq 0$ on $x$ and coefficient $-\alpha\neq 0$ on $y$, so it is not the zero polynomial. Therefore $\deg f\neq 1$. Hence $\deg f=0$, so $f\in k^\times$ is a unit and $(f)=R$. This would force $(x,y)=R$, so $1\in(x,y)$. But every element of $(x,y)$ has the form \begin{align*} ax+by \end{align*} with $a,b\in k[x,y]$, and both $ax$ and $by$ have zero constant term; therefore $ax+by$ has zero constant term. Since $1$ has constant term $1$, we have $1\notin(x,y)$. This contradiction shows that $(x,y)$ is not principal. [/example] The example is small but decisive. The origin in the affine plane needs the two independent equations $x=0$ and $y=0$. This is the point where commutative algebra starts to differ from elementary divisibility theory. ## Prime and Maximal Ideals ### Detecting Domains To detect when a quotient has no nonzero zero divisors, one must translate a statement about zero products in $R/I$ back into $R$. A product becomes zero in the quotient exactly when it lies in the ideal, so the obstruction is whether a product can enter the ideal without either factor doing so. The following condition rules out that obstruction. [definition: Prime Ideal] Let $R$ be a commutative ring. A proper ideal $\mathfrak{p}\trianglelefteq R$ 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] The condition says that a product can fall into $\mathfrak{p}$ only because at least one factor already fell into $\mathfrak{p}$. In the quotient, this is exactly the question of whether a zero product forces one factor to be zero. Thus the ideal-theoretic condition should be equivalent to the quotient having domain-like multiplication. [quotetheorem:853] Prime ideals are therefore the points of view from which the ring has domain-like behaviour. They are central in algebraic geometry because irreducible geometric pieces correspond to prime ideals under suitable hypotheses. ### Detecting Fields A field quotient is stronger than a domain quotient: every nonzero class becomes invertible. If $R/I$ is to be a field, then there can be no meaningful intermediate quotient after $I$ has been collapsed, because a field has no nonzero proper ideals. This turns the search for field quotients into a maximality condition among proper ideals. [definition: Maximal Ideal] Let $R$ be a commutative ring. A proper ideal $\mathfrak{m}\trianglelefteq R$ is a maximal ideal if the only ideals $J\trianglelefteq R$ satisfying \begin{align*} \mathfrak{m}\subset J\subset R \end{align*} are $J=\mathfrak{m}$ and $J=R$. [/definition] Maximality forbids an intermediate quotient between $R/\mathfrak{m}$ and the zero ring. This is the same obstruction a field removes: any nonzero element generates the whole field as an ideal because it is invertible. The quotient criterion below identifies maximal ideals precisely as those whose collapse leaves a field. [quotetheorem:852] Every field is an integral domain, so every maximal ideal in a commutative ring is prime. The converse can fail, and the failure is geometrically meaningful. Positive-dimensional objects give prime ideals that are not maximal. [example: Prime but Not Maximal] Let $k$ be a field and work in $R=k[x,y]$. We first show that $(x)$ is prime. A polynomial $f\in k[x,y]$ lies in $(x)$ exactly when every term of $f$ is divisible by $x$, equivalently when substituting $x=0$ gives the zero polynomial in $k[y]$: \begin{align*} f\in(x)\quad\Longleftrightarrow\quad f(0,y)=0. \end{align*} Now suppose $fg\in(x)$. Then \begin{align*} 0=(fg)(0,y)=f(0,y)g(0,y). \end{align*} Since $k$ is a field, $k[y]$ is an integral domain, so the product $f(0,y)g(0,y)$ can be zero only if \begin{align*} f(0,y)=0 \end{align*} or \begin{align*} g(0,y)=0. \end{align*} By the displayed membership criterion, this means $f\in(x)$ or $g\in(x)$. Thus $(x)$ is prime. It is not maximal because there is a proper ideal strictly between $(x)$ and $k[x,y]$, namely $(x,y)$. The inclusion $(x)\subset (x,y)$ holds because $x\in(x,y)$. It is strict because $y\in(x,y)$, while $y\notin(x)$: if $y=xh$ for some $h\in k[x,y]$, then substituting $x=0$ would give \begin{align*} y=0\cdot h(0,y)=0, \end{align*} which is false in $k[y]$. Finally, $(x,y)\subsetneq k[x,y]$ because every element of $(x,y)$ has the form \begin{align*} ax+by \end{align*} with $a,b\in k[x,y]$, and both $ax$ and $by$ have constant term $0$, so $ax+by$ has constant term $0$. Since $1$ has constant term $1$, we have $1\notin(x,y)$. Therefore \begin{align*} (x)\subsetneq(x,y)\subsetneq k[x,y], \end{align*} so $(x)$ is prime but not maximal. [/example] The ideal $(x)$ cuts out the $y$-axis, while $(x,y)$ cuts out the origin. Prime ideals can describe irreducible pieces of positive dimension, not only points. Many arguments need a maximal ideal before they can pass from a ring to a field quotient, but a general ring need not present such an ideal constructively. The obstruction is that one may have to enlarge a proper ideal through a long chain without ever seeing a final largest choice by hand. The existence theorem guarantees that this enlargement process can be completed in every nonzero commutative ring. [quotetheorem:8159] This result is often proved with [Zorn's lemma](/theorems/1226). In Noetherian rings the search for maximal ideals is more concrete, but the general theorem explains why field quotients are available in great generality. ## Operations on Ideals ### Sums and Intersections Ideals form a partially ordered collection under inclusion, and algebra often requires combining constraints. Simply taking the union of two ideals usually fails to be an ideal, so it cannot serve as the result of imposing both sets of relations. The correct join is the smallest ideal containing both, built by allowing sums of elements from the two ideals. [definition: Sum of Ideals] Let $R$ be a commutative ring and let $I,J\trianglelefteq R$. The sum of $I$ and $J$ is \begin{align*} I+J=\{a+b:a\in I,\ b\in J\}. \end{align*} [/definition] The sum is the least ideal containing both $I$ and $J$. Quotienting by $I+J$ imposes the relations from $I$ and the relations from $J$ simultaneously. A different combination asks for elements that already satisfy both ideal memberships. [definition: Intersection of Ideals] Let $R$ be a commutative ring and let $I,J\trianglelefteq R$. The intersection of $I$ and $J$ is \begin{align*} I\cap J=\{a\in R:a\in I\text{ and }a\in J\}. \end{align*} [/definition] Intersections of ideals are ideals. They often encode functions that vanish on a union of geometric pieces. To encode multiplicative interaction between constraints, however, intersection is too large and the product construction is needed. ### Products and Comaximality Multiplying two systems of equations should record expressions that contain one factor from each ideal. The set of single products $ab$ with $a\in I$ and $b\in J$ is usually not additively closed, so it may fail to be an ideal. To get the ideal generated by these pairwise products, the construction must allow finite sums. [definition: Product of Ideals] Let $R$ be a commutative ring and let $I,J\trianglelefteq R$. The product ideal $IJ$ is \begin{align*} IJ=\left\{\sum_{i=1}^n a_i b_i:n\in\mathbb{N},\ a_i\in I,\ b_i\in J\right\}. \end{align*} [/definition] The product always satisfies $IJ\subset I\cap J$. Equality requires the two ideals to be independent in a precise algebraic sense. That independence is called comaximality, and it is the hypothesis that makes simultaneous congruences split. [definition: Comaximal Ideals] Let $R$ be a commutative ring. Ideals $I,J\trianglelefteq R$ are comaximal if \begin{align*} I+J=R. \end{align*} [/definition] Comaximality means that $1_R$ can be written as a sum of an element of $I$ and an element of $J$. Such an identity lets residue classes modulo $I$ and modulo $J$ be prescribed independently. The next theorem is the ideal-theoretic form of the [Chinese remainder theorem](/theorems/734). [quotetheorem:8160] For integers, comaximality is coprimality. The theorem says that arithmetic modulo a product of coprime moduli separates into independent arithmetic modulo each factor. [example: Splitting $\mathbb{Z}/6\mathbb{Z}$] In $\mathbb{Z}$, the ideals $2\mathbb{Z}$ and $3\mathbb{Z}$ are comaximal because \begin{align*} 1=2(-1)+3(1). \end{align*} Here $2(-1)\in 2\mathbb{Z}$ and $3(1)\in 3\mathbb{Z}$, so $1\in 2\mathbb{Z}+3\mathbb{Z}$. Since every integer $r$ satisfies $r=r\cdot 1\in 2\mathbb{Z}+3\mathbb{Z}$, we get $2\mathbb{Z}+3\mathbb{Z}=\mathbb{Z}$. Their product ideal is $6\mathbb{Z}$. Indeed, every generator of the product has the form \begin{align*} (2a)(3b)=6ab \end{align*} with $a,b\in\mathbb{Z}$, so every finite sum of such products lies in $6\mathbb{Z}$. Conversely, if $6q\in 6\mathbb{Z}$, then \begin{align*} 6q=(2q)(3), \end{align*} with $2q\in 2\mathbb{Z}$ and $3\in 3\mathbb{Z}$, so $6q\in (2\mathbb{Z})(3\mathbb{Z})$. Thus \begin{align*} (2\mathbb{Z})(3\mathbb{Z})=6\mathbb{Z}. \end{align*} By *Chinese [Remainder Theorem](/theorems/1707) for Two Ideals*, the map \begin{align*} \Phi:\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z} \end{align*} defined by \begin{align*} \Phi(n)=(n+2\mathbb{Z},n+3\mathbb{Z}) \end{align*} has kernel $6\mathbb{Z}$ and induces an isomorphism \begin{align*} \mathbb{Z}/6\mathbb{Z}\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}. \end{align*} Thus knowing an integer modulo $6$ is exactly the same as knowing it modulo $2$ and modulo $3$ simultaneously. [/example] The same mechanism works in polynomial rings and coordinate rings. Comaximal ideals represent constraints that do not interfere with each other. ## Ideals as Equations ### Vanishing Ideals In polynomial rings, ideals can be read as systems of equations. A polynomial $f$ represents the equation $f=0$, and an ideal contains all polynomial consequences of the equations inside it. The next definition moves from a set of points to the ideal of all polynomials vanishing there. [definition: Vanishing Ideal of a Subset] Let $k$ be a field, let $S\subset k^n$, and let $R=k[x_1,\dots,x_n]$. The vanishing ideal of $S$ is \begin{align*} I(S)=\{f\in R:f(a)=0\text{ for all }a\in S\}. \end{align*} [/definition] The set $I(S)$ is an ideal because sums of functions vanishing on $S$ vanish on $S$, and multiplying such a function by any polynomial still gives a function vanishing on $S$. This is the child notion often called the ideal of a variety when $S$ is an algebraic variety. The reverse construction starts with equations and asks for their common solutions. [definition: Zero Set of an Ideal] Let $k$ be a field, let $R=k[x_1,\dots,x_n]$, and let $I\trianglelefteq R$. The zero set of $I$ over $k$ is \begin{align*} V(I)=\{a\in k^n:f(a)=0\text{ for all }f\in I\}. \end{align*} [/definition] The passage $I\mapsto V(I)$ reverses inclusions: more equations mean fewer solutions. This reversal is a central organizing idea in algebraic geometry. The simplest example is the ideal of all polynomials that vanish at a point. [example: The Origin in the Affine Plane] Let $R=k[x,y]$. We compute the zero set of the ideal $(x,y)$, meaning the common zeros of every polynomial in $(x,y)$. Since $x\in(x,y)$ and $y\in(x,y)$, any point $(a,b)\in V(x,y)$ must satisfy \begin{align*} x(a,b)=a=0 \end{align*} and \begin{align*} y(a,b)=b=0. \end{align*} Hence $V(x,y)\subset\{(0,0)\}$. Conversely, every element of $(x,y)$ has the form $xf+yg$ with $f,g\in k[x,y]$, and evaluating at $(0,0)$ gives \begin{align*} (xf+yg)(0,0)=0\cdot f(0,0)+0\cdot g(0,0)=0. \end{align*} Thus $(0,0)$ is a common zero of all elements of $(x,y)$, so \begin{align*} V(x,y)=\{(0,0)\}. \end{align*} The quotient also records only the value at the origin. Define \begin{align*} \operatorname{ev}_{(0,0)}:k[x,y]\to k \end{align*} by $\operatorname{ev}_{(0,0)}(f)=f(0,0)$. If \begin{align*} f(x,y)=c+\sum_{i+j\geq 1} c_{ij}x^iy^j, \end{align*} then $f(0,0)=c$. The polynomial $f-c$ has no constant term, so each monomial in $f-c$ is divisible by $x$ or by $y$, and therefore $f-c\in(x,y)$. Hence $f\in\ker(\operatorname{ev}_{(0,0)})$ exactly when $c=0$, which is exactly when $f\in(x,y)$. The map is surjective because every $\lambda\in k$ is the value at $(0,0)$ of the constant polynomial $\lambda$, so \begin{align*} k[x,y]/(x,y)\cong k. \end{align*} Thus quotienting by $(x,y)$ identifies two polynomials precisely when they have the same value at the origin. [/example] This is the simplest geometric quotient: all functions agreeing at the origin become the same after quotienting by the ideal of functions vanishing there. But ideals contain more information than their ordinary zero sets. To separate classical vanishing from nilpotent thickening, we need radicals. ### Radicals and Nilpotents A polynomial may fail to lie in an ideal even though some power of it does. In the quotient ring, its class is then not zero, but it becomes zero after taking a power. This is the obstruction to recovering an ideal from ordinary vanishing data: zero sets cannot see the difference between an element that vanishes and an element that is only nilpotent modulo the ideal. The radical records exactly the elements whose failure to lie in the ideal disappears after taking a power. [definition: Radical of an Ideal] Let $R$ be a commutative ring and let $I\trianglelefteq R$. The radical of $I$ is \begin{align*} \sqrt{I}=\{a\in R:a^n\in I\text{ for some }n\in\mathbb{N}\}. \end{align*} [/definition] Radicals raise a geometric question: if two ideals define the same common zero set, how much of the ideal can be recovered from that set? Powers matter here, because a function whose power vanishes on a quotient behaves like a nilpotent error. Before the full recovery statement, there is a more basic vanishing question: when do given complex polynomials have no common zero at all? The certificate form of the Nullstellensatz answers that emptiness question by turning it into an algebraic membership test for $1$ in the generated ideal. [quotetheorem:3721] This theorem should be read as an emptiness certificate, not as the radical formula itself. It says that over $\mathbb{C}^n$, having no common zero is equivalent to the generated ideal being the whole polynomial ring. The radical viewpoint above explains why stronger forms of the Nullstellensatz recover $\sqrt{I}$, rather than $I$ itself, from an ordinary zero set. [example: Same Zeros, Different Quotients] In $k[x]$, both $(x)$ and $(x^2)$ vanish exactly at $0$. For $(x)$, if $a\in V(x)$, then the polynomial $x\in(x)$ must vanish at $a$, so \begin{align*} x(a)=a=0. \end{align*} Conversely, every element of $(x)$ has the form $xg(x)$ with $g\in k[x]$, and evaluating at $0$ gives \begin{align*} (xg)(0)=0\cdot g(0)=0. \end{align*} Thus $V(x)=\{0\}$. For $(x^2)$, if $a\in V(x^2)$, then $x^2\in(x^2)$ must vanish at $a$, so \begin{align*} x^2(a)=a^2=0. \end{align*} Since $k$ is a field, $a^2=0$ implies $a=0$. Conversely, every element of $(x^2)$ has the form $x^2g(x)$ with $g\in k[x]$, and \begin{align*} (x^2g)(0)=0^2g(0)=0. \end{align*} Hence $V(x^2)=\{0\}$ as well. The quotients are different. In $k[x]/(x)$, every polynomial can be written as \begin{align*} f(x)=c_0+xg(x) \end{align*} with $c_0\in k$ and $g\in k[x]$, so \begin{align*} f(x)+(x)=c_0+(x). \end{align*} Thus the quotient keeps only the constant term, giving $k[x]/(x)\cong k$. In $k[x]/(x^2)$, let \begin{align*} \epsilon=x+(x^2). \end{align*} Then \begin{align*} \epsilon^2=(x+(x^2))^2=x^2+(x^2)=0+(x^2). \end{align*} But $\epsilon\neq 0+(x^2)$: if $x\in(x^2)$, then $x=x^2h(x)$ for some $h\in k[x]$, which is impossible because the left side has degree $1$ while the nonzero right side has degree at least $2$. Thus $k[x]/(x^2)$ contains a nonzero nilpotent element. The ideal $(x^2)$ therefore records an infinitesimal thickening of the origin that the ordinary zero set cannot see. [/example] This example warns against identifying ideals with their zero sets. Ideals control quotient rings, and quotient rings can retain nilpotent information invisible to ordinary points. ## Ideals Beyond Commutative Rings ### One-Sided Matrix Ideals The ring-theoretic definition above is the central definition for this page, but noncommutative examples show why side conditions matter. In a noncommutative ring, two-sided ideals still control quotient rings, while left and right ideals control quotient modules over the regular module. Matrix multiplication can preserve a subspace from one side while destroying it from the other. The next example displays that asymmetry concretely. [example: A Left Ideal in a Matrix Ring] Let $R=M_2(k)$ and let \begin{align*} I=\{B\in M_2(k):B_{12}=0\text{ and }B_{22}=0\}, \end{align*} the set of matrices whose second column is zero. The zero matrix lies in $I$. If $B,C\in I$, then \begin{align*} (B-C)_{12}=B_{12}-C_{12}=0-0=0 \end{align*} and \begin{align*} (B-C)_{22}=B_{22}-C_{22}=0-0=0, \end{align*} so $B-C\in I$. Thus $I$ is an additive subgroup of $M_2(k)$. Now take any $A\in M_2(k)$ and $B\in I$. The entries in the second column of $AB$ are \begin{align*} (AB)_{12}=A_{11}B_{12}+A_{12}B_{22}=A_{11}\cdot 0+A_{12}\cdot 0=0 \end{align*} and \begin{align*} (AB)_{22}=A_{21}B_{12}+A_{22}B_{22}=A_{21}\cdot 0+A_{22}\cdot 0=0. \end{align*} Hence $AB\in I$, so $I$ is a left ideal. It is not a right ideal. Let $B\in I$ be the matrix with $B_{11}=1$ and all other entries $0$, and let $C\in M_2(k)$ be the matrix with $C_{12}=1$ and all other entries $0$. Then \begin{align*} (BC)_{12}=B_{11}C_{12}+B_{12}C_{22}=1\cdot 1+0\cdot 0=1. \end{align*} Thus $BC$ has nonzero second column, so $BC\notin I$. Multiplication on the left preserves the condition defining $I$, while multiplication on the right can destroy it. [/example] This example shows why two-sidedness is not cosmetic. Quotient rings require compatibility with both sides of multiplication. Other algebraic structures have analogous quotient conditions for their own operations. ### Lie Algebra Ideals Lie algebras have a bracket rather than an associative product, but quotienting raises the same question. If a subspace is to be collapsed to zero, bracketing it with any ambient element must remain inside that subspace. The next definition records the Lie-algebra version of absorption. [definition: Lie Algebra Ideal] Let $\mathfrak{g}$ be a Lie algebra over a field $F$. A vector subspace $\mathfrak{i}\subset\mathfrak{g}$ is an ideal of $\mathfrak{g}$ if \begin{align*} [x,y]\in\mathfrak{i} \end{align*} for all $x\in\mathfrak{g}$ and $y\in\mathfrak{i}$. [/definition] This is not the same object as a ring ideal, but it has the same quotient role. The obstruction is well-definedness: if representatives of two cosets are changed by elements of $\mathfrak{i}$, the bracket of the new representatives must change only by another element of $\mathfrak{i}$. Stability under bracketing is precisely the condition that removes this ambiguity, so it is what makes a [quotient Lie algebra](/theorems/8146) possible. [quotetheorem:3763] The parallel with quotient rings is the conceptual lesson. An ideal is the kind of subobject that can be collapsed while preserving the operations of the ambient algebraic structure. ## Beyond and Connected Topics Ideals lead directly to quotient rings, where the phrase "collapse an ideal to zero" becomes a construction used throughout algebra. The first isomorphism theorem explains why quotients and kernels are inseparable. In commutative algebra, prime ideals form the spectrum $\operatorname{Spec}(R)$ of a ring. This turns a ring into a geometric object and is the starting point for localization, dimension theory, schemes, and the material developed in [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). In algebraic geometry, ideals of polynomial rings encode equations. The vanishing ideal of a variety is a specialization of the general ideal concept, and the Nullstellensatz explains when algebraic and geometric data determine each other. In module theory, ideals are submodules of the regular module $R$. This point of view connects ideals to exact sequences, tensor products, annihilators, and the homological constructions developed in [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions). In Lie theory, ideals classify quotient Lie algebras and measure simplicity. They play the same role for brackets that normal subgroups play for groups and two-sided ideals play for rings. ## References Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Androma, [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions). Atiyah and Macdonald, *Introduction to Commutative Algebra* (1969). Dummit and Foote, *Abstract Algebra* (2004). Eisenbud, *Commutative Algebra with a View Toward Algebraic Geometry* (1995).