Ideal

An ideal is an additive subgroup of a ring that absorbs multiplication by ring elements.

The sum of two ideals $I,J$ of a ring isIt is the smallest ideal containing both $I$ and $J$: it is closed under subtraction and multiplication by ring elements, and any ideal containing $I$ and $J$ contains every such sum. The ideals are comaximal ideals when $I+J=R$, equivalently when $1=a+b$ for some $a\in I$ and $b\in J$.

For a commutative ring $R$ and an ideal $I$, the maps $J\mapsto J/I$ and $H\mapsto q^{-1}(H)$ give inverse inclusion-preserving bijections between ideals $J\supseteq I$ of $R$ and ideals $H$ of the quotient ring $R/I$, where $q:R\to R/I$ is the quotient map. The preimage is an ideal containing the kernel $I$. Conversely, $q^{-1}(q(J))=J$ for $J\supseteq I$, and surjectivity of $q$ gives $q(q^{-1}(H))=H$. These identities prove the claim.

An ideal in a polynomial ring generated by monomials. Its exponent vectors form an upward-closed subset of ${\mathbb{N}}^r$ under the componentwise order. The Dickson lemma makes the minimal exponent set finite. This description permits Maclagan's theorem to compare ideals by reverse inclusion independently of the coefficient field.

Upward-closed subsets of ${\mathbb{N}}^r$, including the empty subset, form a well-quasi-ordering by reverse inclusion. For $r=1$, tails correspond to the well-order $\mathbb{N}\cup \{\infty \}$. Inductively slice in the last coordinate. The slices increase and eventually stabilize: their union has a finite minimal basis by the Dickson lemma, all of which has appeared in one finite slice. Encode an upset by the word of its pre-stabilization slices and its final slice. The Higman lemma and finite product closure of well-quasi-orderings compare these encodings in the reverse-inclusion slice order. A subsequence map $f$ has $f(k)\ge k$, so containment of the matched later slice implies containment of the slice at $k$; the final-slice comparison handles the tail. The entire earlier upset therefore contains the later upset.

An ideal $I$ is pure when the quotient is a flat module. Equivalently every $a\in I$ satisfies $a=ab$ for some $b\in I$. The latter condition makes $I$ vanish after localizing at primes containing it, while at primes not containing it the quotient is zero. Pure ideals need not be finitely generated. They describe flat morphisms that are closed immersions on affine charts.

The ideal consisting only of zero. It is the principal ideal generated by zero.

For ideals $I,J$ in a commutative ring, the ideal quotient is $(I:J)=\{r\in R:rJ\subseteq I\}$. Closure under subtraction and multiplication by arbitrary ring elements proves it is an ideal. If $I\subseteq J=(a)$, then $I=a(I:J)$: write $x=ar\in I$ and observe that $rJ=xR\subseteq I$. Consequently, if also $(I:J)=(b)$, then $I=(ab)$ is a principal ideal.

The whole ring $R$, viewed as an ideal. Elements $f_1,\dots ,f_r$ generate the unit ideal exactly when coefficients $g_i\in R$ give ${\sum }_i{g}_i{f}_i=1$.

If ${\sum }_i{g}_i{f}_i=1$, expand its power $r(N-1)+1$. By the pigeonhole principle, each monomial contains some $f_i^N$. Collecting terms expresses $1$ as a linear combination of these powers. This lets one use a common denominator exponent in localization arguments.

In ${\mathbb{F}}_2[u,v]/(u,v{)}^2$, the ideals $(u)$, $(v)$ and $(u+v)$ are pairwise incomparable and nonprime. Their union is the ideal $(u,v)$: the three lines cover its two-dimensional vector space over ${\mathbb{F}}_2$. This shows why a primality hypothesis matters in prime avoidance.

For an ideal $I\subseteq R$, its conormal module is $I/I^2$, naturally an $R/I$-module. It records generators and relations to first order along the quotient $R/I$. If $I$ is the maximal ideal of a local ring, it is the cotangent space of a local ring over the residue field. For the augmentation ideal $(X_1-1,\dots ,X_r-1)$ of $A[X_1^{\pm 1},\dots ,X_r^{\pm 1}]$, reduction modulo $I^2$ gives $I/I^2\cong A^r$, with basis the classes of $X_i-1$.

A proper ideal $I$ is irreducible if $I=J\cap K$ implies $I=J$ or $I=K$. In a Noetherian ring, every irreducible ideal is a primary ideal; every proper ideal is a finite intersection of irreducible ideals.

In a Noetherian ring, an irreducible ideal is a primary ideal. In its quotient, if $xy=0$ and $x\ne 0$, choose $r$ after the annihilators of powers of $y$ stabilize. Then $(x)\cap (y^r)=0$, so irreducibility forces $y^r=0$. Combined with finite decomposition into irreducible ideals, this proves the Lasker–Noether theorem.

An ideal $I$ is radical when $f^m\in I$ for some positive integer $m$ implies $f\in I$. Over an algebraically closed field, the Hilbert Nullstellensatz identifies the vanishing ideal of $V(I)$ with the radical $\sqrt{I}$.

For an ideal $I$ and an element $a$ of a commutative ring,

An ideal $I$ is nilpotent when $I^r=0$ for some positive integer $r$.

For a nilpotent ideal $I$, its nilpotence index of an ideal is the least positive integer $n$ for which $I^n=0$. In $k[X]/(g^e)$, with $g$ an irreducible polynomial, the unique maximal ideal $(g)$ has index $e$. In a discrete valuation ring with $p=u{\pi }^e$, the maximal ideal of its reduction modulo $p$ has the same index. Comparing these indices under a quotient ring isomorphism recovers the ramification exponent in the Kummer-Dedekind theorem.

An Oka family of ideals is a family $\mathcal{G}$ such that $I+(a)\in \mathcal{G}$ and $(I:a)\in \mathcal{G}$ imply $I\in \mathcal{G}$.

Every ideal maximal outside an Oka family is prime. If $xy\in I$ while $x,y\notin I$, both $I+(x)$ and $(I:x)$ properly contain $I$, contradicting the Oka condition.

A two-sided ideal $I$ of a noncommutative ring satisfies $RI\subseteq I$ and $IR\subseteq I$.

A square-zero ideal is an ideal $I$ satisfying $I^2=0$, so every product of two elements of $I$ vanishes.

For a unital associative algebra $A$ and prescribed $A$-bimodule $M$, an extension is $0\to M\to E\to A\to 0$ where $E$ is unital, $E\to A$ is unital, $M$ is a square-zero ideal, and the induced actions are the prescribed ones. Equivalences induce the identity on $A$ and $M$. A unital linear section produces a normalized Hochschild cocycle $\mu (a,b)=s(a)s(b)-s(ab)$. Conversely, $(a,m)(b,n)=(ab,an+mb+\mu (a,b))$ on $A\oplus M$ defines the extension. Changing section changes $\mu$ by a coboundary, giving classification by $H{H}^2(A,M)$.

If $I^2=0$, then $1+I=\{1+a:a\in I\}$ is an abelian subgroup of the unit group. Multiplication satisfies $(1+a)(1+b)=1+a+b$, and $(1+a{)}^{-1}=1-a$, so $a\mapsto 1+a$ identifies the additive group of $I$ with $1+I$.

An ideal $I$ is primary when $ab\in I$ and $a\notin I$ imply $b^n\in I$ for some positive integer $n$.

If a prime ideal $P=(r)$ is generated by a non-zero-divisor, its $P$-primary ideals are exactly $(r^k)$ for $k\ge 1$, in any commutative ring. To prove $(r^k)$ is primary, if $ab\in (r^k)$ and $a\notin (r)$, the prime ideal condition gives $b=r{b}_1$; cancelling $r$ and inducting gives $b\in (r^k)$. Also $\sqrt{(r^k)}=(r)$. Conversely, for a $P$-primary ideal $Q$, choose the least $k$ with $r^k\in Q$. If $x\in Q\setminus (r^k)$, factor $x=r^{j}u$ as far as possible with $0\le j<k$. Then $u\notin (r)$, so the primary ideal condition forces $r^j\in Q$, a contradiction. Thus $Q=(r^k)$. The powers are strictly decreasing: equality $(r^k)=(r^{k+1})$ would let cancellation give $1=rc$. This also proves uniqueness of $k$.

In a Noetherian ring, an ideal $Q$ is $\mathfrak{m}$-primary for a maximal ideal $\mathfrak{m}$ if and only if ${\mathfrak{m}}^n\subseteq Q\subseteq \mathfrak{m}$ for some $n\ge 1$. For necessity, choose generators $x_1,\dots ,x_d$ of $\mathfrak{m}$ and exponents $e_i\ge 1$ with $x_i^{e_i}\in Q$. Every product of $1+{\sum }_i(e_i-1)$ generators lies in $Q$. If $\mathfrak{m}=0$, take $n=1$. Conversely the inclusions imply $\sqrt{Q}=\mathfrak{m}$, so $R/Q$ is a local ring with maximal ideal $\mathfrak{m}/Q$. Every element outside this maximal ideal is a unit, while every element inside is nilpotent. Therefore every zero divisor is nilpotent, which is the primary ideal condition. The reverse implication works for any commutative ring.

In the Noetherian local ring $A=k[[u,v]]$ with maximal ideal $\mathfrak{m}=(u,v)$, the embedding dimension is two but ${\mathfrak{m}}^r$ needs $r+1$ generators for every $r\ge 1$. The degree-$r$ monomials $u^r,u^{r-1}v,\dots ,v^r$ generate ${\mathfrak{m}}^r$, and their images form a basis of ${\mathfrak{m}}^r/{\mathfrak{m}}^{r+1}$, so no smaller generating list exists. These ideals are $\mathfrak{m}$-primary: if $ab\in {\mathfrak{m}}^r$ and $a\notin {\mathfrak{m}}^r$, then $b$ cannot be a unit; hence $b\in \mathfrak{m}$ and $b^r\in {\mathfrak{m}}^r$. Consequently embedding dimension bounds the minimum over all maximal-primary ideals, because the maximal ideal itself is eligible, but does not bound the number of generators of each such ideal.

The radical $\sqrt{I}=\{x:x^n\in I\text{ for some }n\ge 1\}$ of a primary ideal is prime. In a number ring this forces a nonzero primary ideal to have only one prime in its unique factorization.

A minimal primary decomposition of an ideal $I$ is an irredundant finite intersection $I={\bigcap }_i{Q}_i$ of primary ideals whose radicals are pairwise distinct.

For a finite minimal primary decomposition $I={\bigcap }_i{Q}_i$, these are the prime ideals $\sqrt{Q_i}$. This terminology applies without a Noetherian hypothesis when the decomposition exists. Its invariance follows from the intrinsic description as those prime ideals of the form $\sqrt{(I:x)}$, where $x\notin I$ and $(I:x)$ is a colon ideal. Indeed, $(I:x)={\bigcap }_i(Q_i:x)$, and each factor is either the whole ring or a primary ideal with radical of an ideal $\sqrt{Q_i}$. A finite intersection of prime ideals can be prime only if it equals one of them: otherwise multiplying one element outside the intersection from each factor contradicts the prime ideal condition. Conversely, irredundancy supplies $x\in {\bigcap }_{j\ne i}{Q}_j\setminus Q_i$, giving $\sqrt{(I:x)}=\sqrt{Q_i}$. In a Noetherian ring these agree with the associated primes of a module $R/I$. For arbitrary rings, the annihilator definition of an associated prime of a module must not be substituted without a further hypothesis.

The distinct radicals in a minimal primary decomposition are exactly the associated primes of a module $R/I$, and hence are invariant. The injection into the sum of primary quotients proves one inclusion; an element supported on one component, supplied by irredundancy, proves the other. This does not imply uniqueness of embedded primary components.

A component in a minimal primary decomposition is embedded if its radical of an ideal strictly contains another radical occurring in that decomposition. Such components need not be unique. For example,in $k[x,y]$ gives different $(x,y)$-primary ideals as embedded components, with the same isolated component $(x)$.

An isolated prime of a minimal primary decomposition is a minimal member of the set of radicals of its primary components.

In a Noetherian ring, the primary component belonging to each isolated prime $\mathfrak{p}$ is uniquely determined byEmbedded primary components need not be unique.

A set $S\subseteq I$ generates an ideal $I$ when every element of $I$ is a finite sum of ring multiples of elements of $S$.

A principal ideal is generated by one element: $(a)=\{ra:r\in R\}$.

The product $IJ$ consists of finite sums of products $xy$ with $x\in I$ and $y\in J$.

For an ideal $I$ in a commutative ring $R$, put $I^0=R$ and let $I^n$ for $n\ge 1$ be the product of ideals with $n$ factors equal to $I$. It consists of finite sums of products of $n$ elements of $I$. These ideals form a descending chain $R\supseteq I\supseteq I^2\supseteq \cdots$.

Every positive power has the same radical of an ideal as $I$:One inclusion follows from $I^n\subseteq I$. Conversely, if $x^m\in I$, then $x^{mn}\in I^n$, giving the other inclusion. Different ideals can therefore define the same radical ideal, even when their successive powers are all distinct.

An ideal $I$ is idempotent when $I^2=I$. Then $I^j=I$ for every $j\ge 1$, so a nonzero idempotent ideal gives a nonzero intersection of all its powers. A semigroup algebra with arbitrarily divisible positive exponents supplies examples in a non-Noetherian integral domain. If $I$ is finitely generated, the determinant trick applied to $I=I^2$ produces $r\in I$ such that $(1+r)I=0$.

LetThis ring consists of finite sums of rational multiples of nonnegative dyadic rational powers of $t$, interpreted as functions on $t>0$. Every pair of elements lies in one polynomial ring $\mathbb{Q}[u_n]$, so $R$ is an integral domain. The constant term defines a surjective ring homomorphism $R\to \mathbb{Q}$. Its kernel of a ring homomorphism $\mathfrak{m}$ is nonzero and is a maximal ideal, by the maximal ideal quotient criterion.

Every $f\in \mathfrak{m}$ has the form $u_{n}g(u_n)$ for some $n$ and some polynomial $g$. The identityexpresses it as a product of two elements of $\mathfrak{m}$. Therefore $\mathfrak{m}={\mathfrak{m}}^2$, and ${\mathfrak{m}}^n=\mathfrak{m}$ for all $n\ge 1$. This shows that strict descent of powers of an ideal can fail in an integral domain, even for a nonzero maximal ideal. By the determinant trick, such an ideal cannot be finitely generated, so this ring is not a Noetherian ring.

The set-theoretic intersection of any family of ideals is again an ideal.

Ideals $I$ and $J$ are comaximal when $I+J=R$.

A prime-power ideal has the form ${\mathfrak{p}}^m$ for a prime ideal $\mathfrak{p}$ and positive integer $m$.

For finitely many distinct prime ideals, one can prescribe finite valuation data simultaneously by the Chinese remainder theorem.

Reduction modulo an ideal $I$ is the quotient homomorphism $R\to R/I$.

For a two-sided ideal $I$ of a ring $R$, the quotient $R/I$ has addition and multiplication induced from $R$.

Two elements of a commutative ring are congruent modulo an ideal if their difference lies in that ideal, equivalently if their images in the quotient ring agree. This is an equivalence relation compatible with addition and multiplication. For $R={\mathbb{Z}}_p[[T]]$ and $I=p^{k}R$, it means every coefficient difference is divisible by $p^k$.

An ideal in a graded ring is homogeneous when it is generated by homogeneous elements, equivalently when every homogeneous component of each of its elements also belongs to the ideal.

The radical of an ideal $I$ isAn ideal equal to its radical is a radical ideal.

In a commutative ring, the radical of an ideal is monotone and idempotent: inclusion $I\subseteq J$ gives $\sqrt{I}\subseteq \sqrt{J}$, and $a^n\in \sqrt{I}$ means $a^{nm}\in I$ for some $m$, giving $\sqrt{\sqrt{I}}=\sqrt{I}$. Since $\sqrt{I+J}$ contains both $\sqrt{I}$ and $\sqrt{J}$, it contains their sum of ideals. Taking radicals of ideals gives $\sqrt{\sqrt{I}+\sqrt{J}}\subseteq \sqrt{I+J}$. The reverse inclusion follows from $I+J\subseteq \sqrt{I}+\sqrt{J}$.

In particular, $\sqrt{I}+\sqrt{J}=R$ implies $I+J=R$. One can also see this directly: write $1=a+b$ with $a^m\in I$ and $b^n\in J$, then expand $(a+b{)}^{m+n-1}$ by the binomial theorem. Each term belongs to $I$ or $J$, so $1\in I+J$.

For any ideal $I$ in a commutative ring, every prime ideal containing $I$ contains $\sqrt{I}$, because $a^n\in P$ implies $a\in P$. Conversely, if $a\notin \sqrt{I}$, the multiplicatively closed set $T=\{1,a,a^2,\dots \}$ is disjoint from $I$. By Zorn's lemma, there is an ideal $P\supseteq I$ maximal among those disjoint from $T$.

This $P$ is a prime ideal: if $bc\in P$ but $b,c\notin P$, maximality makes both $P+(b)$ and $P+(c)$ meet $T$. Multiplication of representatives from these two intersections puts an element of $T$ in $P$, a contradiction. Thus there is a prime ideal containing $I$ but avoiding $a$, establishing the identity. For $I=R$, the intersection is empty and means $R$. It follows that an ideal is a radical ideal exactly when it is an intersection of ideals that are all prime ideals.

In a Noetherian ring, every radical of an ideal is an intersection of finitely many prime ideals. A maximal counterexample ideal $J$ cannot be prime. Choose $a,b\notin J$ with $ab\in J$; the strictly larger ideals $J+(a)$ and $J+(b)$ satisfy the result, and $\sqrt{J}=\sqrt{J+(a)}\cap \sqrt{J+(b)}$. This contradicts maximality. The unit ideal uses an empty intersection.

The nilradical of a commutative ring is its ideal of nilpotent elements, equivalently the radical of an ideal applied to its zero ideal.

For any multiplicative subset $S$ of a commutative ring $R$, fractions with nilpotent numerators are nilpotent in $S^{-1}R$. Conversely, if $(a/s{)}^n=0$, the vanishing criterion in a module localization gives $t\in S$ with $t{a}^n=0$. Then $(ta{)}^n=t^{n-1}t{a}^n=0$ and $a/s=ta/(ts)$ has a nilpotent numerator. This proves the formula for the nilradical without any finiteness assumption.

In $R=k[t_1,t_2,\dots ]/(t_1^2,t_2^2,\dots )$, the nilradical is generated by all ${\bar{t}}_i$. Every element uses finitely many variables and is nilpotent, while each square-free product ${\bar{t}}_1\cdots {\bar{t}}_j$ is nonzero. Thus this nilradical is not a nilpotent ideal. A Noetherian ring cannot have this behavior: finitely many nilpotent generators give a nilpotent ideal.

For every commutative ring $R$,Every nilpotent element lies in every prime ideal. If $a$ is not nilpotent, an ideal maximal among those disjoint from $\{1,a,a^2,\dots \}$ is a prime ideal avoiding $a$, by the Zorn lemma and the maximality argument.