Noetherian ring

Let $R$ be a Noetherian ring that is an integral domain, and let $I$ be a nonzero proper ideal. If $I^n=I^{n+1}$ for some $n\ge 1$, the nonzero finitely generated module $M=I^n$ satisfies $M=IM$. Choose generators $u_1,\dots ,u_r$ and write $u_i={\sum }_j{a}_{ij}{u}_j$ with $a_{ij}\in I$. The adjugate matrix identity implies that $\det ({\mathrm{Id}}_r-(a_{ij}))$ annihilates $M$. This determinant is $1-b$ for some $b\in I$, since every term except the constant term contains an entry from $I$.

The determinant trick has thus supplied $(1-b)M=0$. A nonzero element of $M$ and multiplicative cancellation in an integral domain force $b=1$, contradicting properness of $I$. Therefore all positive powers of an ideal are distinct in this setting. Since they have the same radical of an ideal, there are infinitely many ideals with radical $\sqrt{I}$.

Let $\varphi :R\to S$ be a surjective ring homomorphism and let $R$ be a Noetherian ring. For every ideal $J$ of $S$, its inverse image is a finitely generated ideal of $R$. If ${\varphi }^{-1}(J)=(a_1,\dots ,a_r)$, then $J=(\varphi (a_1),\dots ,\varphi (a_r))$. Thus $S$ is a Noetherian ring. In particular, every quotient ring of a Noetherian ring is a Noetherian ring. For a ring homomorphism that is not surjective, apply the same argument with its image as codomain.

A commutative Noetherian ring is Cohen-Macaulay when every localization at a prime has depth of a Noetherian local ring equal to its Krull dimension. A regular local ring is Cohen-Macaulay, and quotienting one by a nonzero divisor preserves Cohen-Macaulayness while lowering dimension and depth by one. In particular a hypersurface in a regular ambient variety satisfies the depth condition $(S_2)$ in the Serre criterion for normality.

For a surjective ring endomorphism $f$ of a Noetherian ring, the ascending kernel chain $\ker f\subseteq \ker f^2\subseteq \cdots$ stabilizes. If $\ker f^m=\ker f^{m+1}$ and $f(a)=0$, choose $b$ with $f^m(b)=a$. Then $b\in \ker f^{m+1}=\ker f^m$, so $a=0$. Thus the endomorphism is a ring automorphism.

An $R$-module is Noetherian when every submodule is finitely generated, equivalently when every ascending chain of submodules stabilizes. Quotients and finite direct sums of Noetherian modules are Noetherian.

For a short exact sequence $0\to N\to M\to M/N\to 0$, $M$ is a Noetherian module exactly when both the other terms are Noetherian modules. For the extension direction, an ascending chain $M_i$ has eventually constant intersections $M_i\cap N$ and images $(M_i+N)/N$. Equality of both forces equality of consecutive $M_i$: subtract an element with the same image and then use the common intersection.

For a surjective module endomorphism of a Noetherian module, the chain $\ker f\subseteq \ker f^2\subseteq \cdots$ stabilizes at some $N$. If $f(x)=0$, choose $y$ with $f^N(y)=x$. Then $y\in \ker f^{N+1}=\ker f^N$, giving $x=0$. Thus the endomorphism is also injective and is an isomorphism.

A prime filtration of a module over a commutative ring is a finite increasing chain from zero to the module with each nonzero successive quotient module isomorphic to $R/P$ for a prime ideal $P$. Every finitely generated module over a Noetherian ring has one. In each nonzero quotient, a maximal annihilator of a module element is prime supplies a cyclic submodule $R/P$; lift it and repeat. The ascending chain condition for a Noetherian module makes the process terminate. This is not necessarily a composition series: the prime ideals need not be maximal.

A Noetherian local ring is a Noetherian ring with a unique maximal ideal.

An ideal $I$ of a Noetherian local ring $(A,\mathfrak{m})$ is an ideal of definition when $A/I$ has Krull dimension zero. Equivalently, $I$ is $\mathfrak{m}$-primary, so $\sqrt{I}=\mathfrak{m}$ and ${\mathfrak{m}}^r\subseteq I$ for some positive integer $r$.

The ascending chain condition requires every chainof ideals to stabilize. A ring is Noetherian exactly when it satisfies this condition, equivalently when every ideal is finitely generated.

To prove a property of all elements of a partially ordered set satisfying the ascending chain condition, assume a counterexample exists and take a maximal counterexample. All strictly larger elements satisfy the property, allowing an inductive contradiction. Applied to ideals, this often replaces induction on a numerical size.

If a ring $R$ is finitely generated as an abelian group, then it is Noetherian. Every ideal is an additive subgroup of that finitely generated abelian group and is therefore generated by finitely many elements as an abelian group. The same elements generate it as an $R$-ideal because integer multiples are already scalar multiples by elements of $R$.

A polynomial ring over a Noetherian ring is Noetherian.

For an ideal $I$ in $R[x]$, let $J_n$ contain the coefficients of $x^n$ in its elements of degree at most $n$. These are ideals of $R$ and form an ascending chain, since multiplying by $x$ embeds one into the next. If $R$ is a Noetherian ring, the chain stabilizes at $N$, and each $J_n$ for $0\le n\le N$ has finitely many generators. Choose corresponding polynomials in $I$. Their union generates $I$: for an element of degree $d$, subtract a combination of the chosen polynomials for $J_d$ if $d<N$, or the polynomials for $J_N$ multiplied by $x^{d-N}$ otherwise. This cancels its leading coefficient and lowers its degree. Induction completes the proof without assuming that $R$ is an integral domain.

For a nonsquare integer $d$,The integers are Noetherian, the Hilbert basis theorem makes $\mathbb{Z}[X]$ Noetherian, and quotients preserve Noetherianity.

In $R=\mathbb{Z}[\sqrt{-5}]$, the idealhas additive index two. If $I=(a+b\sqrt{-5})$ were principal, multiplication by its generator would givewhich has no solution in integers. Thus $I$ is not principal.

The subring $R=K+XYK[X,Y]$ is not Noetherian. Its idealsform a strict ascending chain because multiplying by a nonconstant element of $R$ introduces at least a second power of $X$.

An Artinian ring satisfies the descending chain condition on ideals. A commutative ring is Artinian exactly when it is Noetherian and has Krull dimension zero.

If $C$ is a commutative algebra of finite dimension over a field $K$, every prime ideal is maximal and $|\mathrm{Spec}C|\le {\dim }_{K}C$. Indeed, the quotient by a prime is a finite-dimensional integral domain. Multiplication by any nonzero element is an injective linear endomorphism, hence surjective, so the quotient is a field. Distinct maximal ideals are pairwise comaximal. For any finite list ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$, construct $e_i\equiv 1(\mathrm{mod}{\mathfrak{m}}_i)$ and $e_i\equiv 0(\mathrm{mod}{\mathfrak{m}}_j)$ for $j\ne i$ by multiplying elements in ${\mathfrak{m}}_j$ congruent to one modulo ${\mathfrak{m}}_i$. These give a surjection $C\to {\prod }_{i}C/{\mathfrak{m}}_i$. Each factor has positive $K$-dimension, so $r\le {\dim }_{K}C$, excluding an infinite list.

Every nonzero commutative Artinian ring is a finite product of Artinian local rings. The primary components of zero have distinct maximal radicals and are pairwise comaximal ideals, so the Chinese remainder theorem gives the product. Each local factor has nilpotent maximal ideal and finite composition length. These factors and the components of zero are unique.

A family of ideals satisfies the descending chain condition when every descending sequence of ideals eventually becomes constant. A ring is Artinian precisely when all its ideals satisfy this condition.

A commutative Artinian ring has finitely many prime ideals, and every one is a maximal ideal. If $\mathfrak{p}$ is prime, the Artinian domain $R/\mathfrak{p}$ is a field: stabilization of $(x)\supseteq (x^2)\supseteq \cdots$ gives $x^n=x^{n+1}a$, and cancellation gives $xa=1$. Infinitely many distinct maximal ideals ${\mathfrak{m}}_i$ would give a strictly descending chainbecause distinct maximal ideals are comaximal.

The nilradical $N$ of a commutative Artinian ring is a nilpotent ideal. Its powers stabilize, say $N^r=N^{r+1}=I$. If $I\ne 0$, choose an ideal $J$ minimal subject to $IJ\ne 0$, and then $x\in J$ with $Ix\ne 0$. Minimality and $I^2=I$ give $Rx=Ix$, so $x=ax$ for some $a\in I$. Since $a$ is nilpotent, $1-a$ is a unit, contradicting $(1-a)x=0$ and $x\ne 0$.

Every commutative Artinian ring is Noetherian. Its nilradical $N$ is nilpotent by the nilradical of a commutative Artinian ring, and the Chinese remainder theorem identifies the reduced quotient $R/N$ with the finite product of the residue fields at the finitely many maximal ideals. Each $N^i/N^{i+1}$ is an Artinian module over that product of fields and hence a finite-dimensional vector space. The finite filtration by these layers makes every ideal finitely generated.

A commutative Artinian ring has finite composition length as a module over itself. It has finitely many maximal ideals, and its Jacobson radical $J$ is nilpotent. The Chinese remainder theorem makes $R/J$ a finite product of fields; each Artinian module $J^i/J^{i+1}$ has finite-dimensional components over those fields. Adding their lengths along the finite radical filtration proves the assertion, and in particular the ascending chain condition.

An Artinian local ring is an Artinian ring with a unique maximal ideal. Its maximal ideal is nilpotent.

Let $(R,\mathfrak{m})$ be an Artinian local ring. If ${\dim }_{R/\mathfrak{m}}\mathfrak{m}/{\mathfrak{m}}^2=1$, then every ideal of $R$ is principal. The Nakayama lemma gives $\mathfrak{m}=(x)$, and nilpotence gives a finite chainFor a nonzero ideal $J$, choose the largest $r$ for which $J\subseteq (x^r)$ and take $y\in J\setminus (x^{r+1})$. Writing $y=x^{r}u$, one has $u\notin \mathfrak{m}$, so $u$ is a unit and $J=(x^r)=(y)$.

An Artinian local ring is a finite set exactly when its residue field is finite. If ${\mathfrak{m}}^r=0$, each layer ${\mathfrak{m}}^i/{\mathfrak{m}}^{i+1}$ is a finite-dimensional vector space over that residue field.

A commutative Noetherian ring is Artinian if and only if every prime ideal is maximal. Equivalently, it is Artinian exactly when its Krull dimension is zero.