Noetherian ring

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.

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.

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 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.

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.