Artinian Rings and Length

By The ideal generated by a subset and principal ideals, the principal ideals $(a)\supseteq (a^2)\supseteq (a^3)\supseteq \cdots$ form a descending chain of ideals. Since $D$ is Artinian, this chain stabilizes, so $(a^n)=(a^{n+1})$ for some integer $n\ge 1$.

Because $a^n\in (a^{n+1})$, there is $b\in D$ with $a^n=a^{n+1}b$, hence $a^n(1-ab)=0$. As $D$ is an integral domain, every power of the nonzero element $a$ is nonzero, so $a^n\mathrm{\ne }0$ and therefore $1-ab=0$. Thus $ab=1$, and the chosen nonzero element $a$ is a unit.

Every nonzero element of $D$ is a unit, so Field makes $D$ a field.

By Correspondence theorem: ideals of $R\mathrm{/}I$ correspond to ideals of $R$ containing $I$, ideals of $R\mathrm{/}\mathfrak{p}$ correspond to ideals of $R$ containing $\mathfrak{p}$. Hence every descending chain of ideals in $R\mathrm{/}\mathfrak{p}$ lifts to a descending chain of ideals in $R$, so $R\mathrm{/}\mathfrak{p}$ is Artinian. Because $\mathfrak{p}$ is prime, $R\mathrm{/}P$ is an integral domain if and only if $P$ is a prime ideal says that $R\mathrm{/}\mathfrak{p}$ is an integral domain.

The quotient $R\mathrm{/}\mathfrak{p}$ is therefore an Artinian integral domain, so An Artinian integral domain is a field makes it a field. Then $R\mathrm{/}M$ is a field if and only if $M$ is a maximal ideal forces $\mathfrak{p}$ to be maximal.

Hence every prime ideal of an Artinian ring is maximal.

If $R$ has no maximal ideals then the conclusion is immediate. Otherwise let $\mathcal{F}$ be the set of all finite nonempty intersections of maximal ideals of $R$. This set is nonempty, so [L1] gives a member minimal under inclusion; write it as $I={\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_n$ for maximal ideals ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_n$.

Let $\mathfrak{m}$ be any maximal ideal of $R$. Then $I\cap \mathfrak{m}$ is again a finite nonempty intersection of maximal ideals, so $I\cap \mathfrak{m}\in \mathcal{F}$ and $I\cap \mathfrak{m}\subseteq I$. By the minimality from step 1.1, one has $I=I\cap \mathfrak{m}$, hence $I\subseteq \mathfrak{m}$. If $\mathfrak{m}$ were distinct from every ${\mathfrak{m}}_i$, then for each $1\le i\le n$ we could choose $x_i\in {\mathfrak{m}}_i\setminus \mathfrak{m}$ and put $x=x_1\cdots x_n$. Now $x\in I\subseteq \mathfrak{m}$, but every maximal ideal is prime by Every maximal ideal of a commutative ring is prime, so some factor $x_i$ would lie in $\mathfrak{m}$, contradicting the choice of the $x_i$. Therefore every maximal ideal of $R$ is one of ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_n$.

Therefore an Artinian ring has only finitely many maximal ideals.

Put $N=\mathrm{Nil}(R)$. The chain $N\supseteq N^2\supseteq N^3\supseteq \cdots$ is a descending chain of ideals, so it stabilizes: $N^n=N^{n+1}$ for some $n\ge 1$. If already $N^n=0$, then $N$ is nilpotent and there is nothing more to prove.

Assume instead that $N^n\mathrm{\ne }0$. Let $\mathcal{F}=\{\mathfrak{a}⊴R:\mathfrak{a}{N}^n\mathrm{\ne }0\}\mathrm{.}$ This family is nonempty because $(N^n)N^n=N^{2n}=N^n\mathrm{\ne }0$. By [L1], choose a minimal member $\mathfrak{a}$ of $\mathcal{F}$, and then choose $a\in \mathfrak{a}$ with $a{N}^n\mathrm{\ne }0$. Since $(a)\subseteq \mathfrak{a}$ and $(a)N^n\mathrm{\ne }0$, minimality gives $\mathfrak{a}=(a)$. Also $a{N}^n\subseteq \mathfrak{a}$, and because $N^n=N^{n+t}$ for every $t\ge 0$, one has $(a{N}^n)N^n=a{N}^{2n}=a{N}^n\mathrm{\ne }0$. So $a{N}^n\in \mathcal{F}$, whence minimality again gives $a{N}^n=\mathfrak{a}=(a)$. Therefore $a=ax$ for some $x\in N^n$.

The element $x$ lies in $N$, so The nilradical and reduced rings says that $x$ is nilpotent; say $x^m=0$. Iterating the identity $a=ax$ gives $a=a{x}^m=0$, contradicting $a{N}^n\mathrm{\ne }0$. Thus the assumption in step 2.1 is false, and the alternative left open in step 1.1 must hold: $N^n=0$.

Hence the nilradical of an Artinian ring is a nilpotent ideal.

First take $r=2$. Choose $u\in I_1$ and $v\in I_2$ with $u+v=1$. If $x\in I_1\cap I_2$, then $x=xu+xv\in I_1{I}_2$, so $I_1\cap I_2=I_1{I}_2$. For classes $a+I_1$ and $b+I_2$, the element $av+bu$ satisfies $av+bu\equiv a(\mathrm{m}\mathrm{o}\mathrm{d}{I}_1)$ and $av+bu\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}{I}_2)$, so the canonical map $R\to R\mathrm{/}{I}_1\times R\mathrm{/}{I}_2$ is surjective with kernel $I_1\cap I_2=I_1{I}_2$.

Now assume $r\ge 2$. Fix $i$ and put $J_i={\prod }_{j\mathrm{\ne }i}{I}_j$. For each $j\mathrm{\ne }i$, choose $u_j\in I_i$ and $v_j\in I_j$ with $u_j+v_j=1$. Expanding ${\prod }_{j\mathrm{\ne }i}(u_j+v_j)=1$ shows that $1-{\prod }_{j\mathrm{\ne }i}{v}_j\in I_i$ and ${\prod }_{j\mathrm{\ne }i}{v}_j\in J_i$, so $I_i+J_i=R$. Choose $e_i\in J_i$ with $e_i\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{I}_i)$. Then $e_i\in I_j$ for every $j\mathrm{\ne }i$. Given residue classes $a_i+I_i$, the element $x:={\sum }_{i=1}^r{a}_i{e}_i$ satisfies $x\equiv a_i(\mathrm{m}\mathrm{o}\mathrm{d}{I}_i)$ for every $i$. Hence the canonical map $R\to {\prod }_{i=1}^{r}R\mathrm{/}{I}_i$ is surjective.

The kernel of the canonical map is plainly ${\bigcap }_{i=1}^r{I}_i$. To compare this with the product, induct on $r$. The case $r=1$ is tautological and the case $r=2$ is step 1.1. For $r>2$, let $K={\bigcap }_{i=2}^r{I}_i$. By the induction hypothesis, $K={\prod }_{i=2}^r{I}_i$. Step 2.1 with $i=1$ gives $I_1+K=R$, so applying the two-ideal case to $I_1$ and $K$ yields ${\bigcap }_{i=1}^r{I}_i=I_1\cap K=I_{1}K={\prod }_{i=1}^r{I}_i$. Therefore $R\mathrm{/}{\prod }_{i=1}^r{I}_i\cong {\prod }_{i=1}^{r}R\mathrm{/}{I}_i$.

Let $N=\mathrm{Nil}(R)$. By An Artinian ring has only finitely many maximal ideals, the maximal ideals of $R$ are ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$ for some $r\ge 1$ unless $R=0$, in which case the conclusion is immediate. By The nilradical is the intersection of all prime ideals and Every prime ideal of an Artinian ring is maximal, one has $N={\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_r$. Distinct maximal ideals are comaximal, so Chinese remainder theorem for pairwise comaximal ideals gives $R\mathrm{/}N\cong {\prod }_{i=1}^{r}R\mathrm{/}{\mathfrak{m}}_i$. Each factor is a field by $R\mathrm{/}M$ is a field if and only if $M$ is a maximal ideal, and a field is Noetherian because its only ideals are $0$ and itself. Repeated use of A product of two Noetherian rings is Noetherian therefore shows that $B:=R\mathrm{/}N$ is Noetherian.

By The nilradical of an Artinian ring is a nilpotent ideal, choose $t\ge 1$ with $N^t=0$. For each $0\le j<t$, put $M_j=N^j\mathrm{/}{N}^{j+1}$. Because $N{M}_j=0$, the action of $R$ on $M_j$ factors through $B=R\mathrm{/}N$, so $M_j$ is a $B$-module. Transport the standard idempotents of the product ring in step 1.1 to elements $e_1,\dots ,e_r\in B$. Then $M_j=e_1{M}_j\oplus \cdots \oplus e_r{M}_j$, and each summand is naturally a vector space over the field $R\mathrm{/}{\mathfrak{m}}_i$. If some $e_i{M}_j$ had no finite spanning set, recursively choose $v_1,v_2,\dots$ with $v_{n+1}\mathrm{\notin }\mathrm{span}(v_1,\dots ,v_n)$; then $\mathrm{span}(v_1,v_2,\dots )⊋\mathrm{span}(v_2,v_3,\dots )⊋\cdots$ would be a strict descending chain of $R$-submodules of $M_j$. But submodules of $M_j$ correspond to submodules of the ideal $N^j$ containing $N^{j+1}$, so this would give a strict descending chain of ideals in the Artinian ring $R$, impossible. Hence every $e_i{M}_j$ has a finite basis, and $M_j$ is finitely generated as a $B$-module.

Since $B$ is Noetherian, Finitely generated modules over a left Noetherian ring are Noetherian shows that each $M_j$ is Noetherian as a $B$-module, hence as an $R$-module. The filtration $0=N^t\subseteq N^{t-1}\subseteq \cdots \subseteq N\subseteq R$ has successive quotients $M_{t-1},\dots ,M_0$ and $R\mathrm{/}N=B$, all Noetherian. Repeatedly applying Noetherian and Artinian conditions are each exact in short exact sequences to the short exact sequences $0\to N^{j+1}\to N^j\to M_j\to 0$ shows that the regular module ${}_{R}R$ is Noetherian. Therefore Left and right Noetherian rings makes $R$ a Noetherian ring.

If ${}_{R}R$ has finite length, then by Composition series and length of a module it has a composition series. The forward implication of A module has a composition series if and only if it is Noetherian and Artinian, the converse using dependent choice therefore makes ${}_{R}R$ Artinian, and Left and right Artinian rings says exactly that $R$ is an Artinian ring.

Suppose now that $R$ is Artinian. Then Left and right Artinian rings says that the regular module ${}_{R}R$ is Artinian. Under the Axiom of Choice assumed in the Statement, Every commutative Artinian ring is Noetherian makes $R$ Noetherian, so Left and right Noetherian rings makes ${}_{R}R$ Noetherian. Since the same choice assumption also suffices for the dependent-choice use recorded in A module has a composition series if and only if it is Noetherian and Artinian, the converse using dependent choice, that theorem gives a composition series for ${}_{R}R$.

By Composition series and length of a module, a module has finite length exactly when it has a composition series. So step 1.1 proves the forward implication, and step 1.2 proves the reverse implication.

Therefore a commutative ring is Artinian exactly when its regular module has finite length.

Because $R$ is local, $\mathfrak{m}$ is its only maximal ideal. By Every prime ideal of an Artinian ring is maximal, every prime ideal of $R$ is maximal, so $\mathfrak{m}$ is also the only prime ideal. Therefore The nilradical is the intersection of all prime ideals gives $\mathrm{Nil}(R)=\mathfrak{m}$. Now The nilradical of an Artinian ring is a nilpotent ideal yields an integer $n\ge 1$ with ${\mathfrak{m}}^n=0$.

By Every commutative Artinian ring is Noetherian, the ring $R$ is Noetherian. Hence A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member makes each ideal ${\mathfrak{m}}^i$ finitely generated; fix generators $a_{i1},\dots ,a_{i{s}_i}$ of ${\mathfrak{m}}^i$. Also choose generators $x_1,\dots ,x_t$ of $M$. Then for every $0\le i<n$, the quotient ${\mathfrak{m}}^{i}M\mathrm{/}{\mathfrak{m}}^{i+1}M$ is spanned over the residue field $k=R\mathrm{/}\mathfrak{m}$ by the finitely many classes of the elements $a_{iq}{x}_j$. Indeed every element of ${\mathfrak{m}}^{i}M$ is a finite sum ${\sum }_q{a}_{iq}{m}_q$, and each $m_q$ is a finite $R$-linear combination of the $x_j$; modulo ${\mathfrak{m}}^{i+1}M$, only the residue classes of the coefficients in $k$ matter because multiplication by an element of $\mathfrak{m}$ lands in ${\mathfrak{m}}^{i+1}M$. Deleting redundant spanning vectors yields a basis $y_1,\dots ,y_d$ of ${\mathfrak{m}}^{i}M\mathrm{/}{\mathfrak{m}}^{i+1}M$, and the partial spans $0<k{y}_1<k{y}_1+k{y}_2<\cdots <k{y}_1+\cdots +k{y}_d$ form a composition series. So every quotient ${\mathfrak{m}}^{i}M\mathrm{/}{\mathfrak{m}}^{i+1}M$ has finite length.

The filtration $M\supseteq \mathfrak{m}M\supseteq \cdots \supseteq {\mathfrak{m}}^{n}M=0$ is finite by step 1.1. Applying Module length is additive in short exact sequences successively to $0\to {\mathfrak{m}}^{i+1}M\to {\mathfrak{m}}^{i}M\to {\mathfrak{m}}^{i}M\mathrm{/}{\mathfrak{m}}^{i+1}M\to 0$ shows that $M$ has finite length. Taking $M=R$ recovers the ring case from the first sentence of the theorem.

If $R=0$, then the theorem has no further content. Assume from now on that $R\mathrm{\ne }0$. Let ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$ be the maximal ideals of $R$, finite by An Artinian ring has only finitely many maximal ideals, and put $N=\mathrm{Nil}(R)$. By The nilradical is the intersection of all prime ideals and Every prime ideal of an Artinian ring is maximal, one has $N={\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_r$. Choose $n\ge 1$ with $N^n=0$ by The nilradical of an Artinian ring is a nilpotent ideal. Distinct maximal ideals are comaximal, and if $i\mathrm{\ne }j$ then choosing $a\in {\mathfrak{m}}_i$ and $b\in {\mathfrak{m}}_j$ with $a+b=1$ gives, from the binomial expansion of $(a+b{)}^{2n-1}$, the inclusion $1\in {\mathfrak{m}}_i^n+{\mathfrak{m}}_j^n$. So the ideals ${\mathfrak{m}}_i^n$ are pairwise comaximal. Also ${\prod }_{i=1}^r{\mathfrak{m}}_i^n={\left({\prod }_{i=1}^r{\mathfrak{m}}_i\right)}^n\subseteq N^n=0$. Hence Chinese remainder theorem for pairwise comaximal ideals yields $R\cong {\prod }_{i=1}^{r}R\mathrm{/}{\mathfrak{m}}_i^n$.

For each $i$, set $A_i=R\mathrm{/}{\mathfrak{m}}_i^n$. This quotient is local. Indeed Prime ideals of a quotient ring are exactly the prime ideals containing the ideal identifies prime ideals of $A_i$ with prime ideals $\mathfrak{p}$ of $R$ containing ${\mathfrak{m}}_i^n$. Since $N^n=0\subseteq {\mathfrak{m}}_i^n\subseteq \mathfrak{p}$, the prime $\mathfrak{p}$ contains $N$ and therefore equals one of the maximal ideals ${\mathfrak{m}}_j$. If $j\mathrm{\ne }i$, choose $a\in {\mathfrak{m}}_i$ and $b\in {\mathfrak{m}}_j$ with $a+b=1$; the same binomial argument as in step 1.1 gives $1\in {\mathfrak{m}}_i^n+{\mathfrak{m}}_j$. So ${\mathfrak{m}}_i^n⊈{\mathfrak{m}}_j$, a contradiction. Thus $\mathfrak{p}={\mathfrak{m}}_i$, and $A_i$ has the unique maximal ideal ${\mathfrak{m}}_i\mathrm{/}{\mathfrak{m}}_i^n$.

Let ${\lambda }_i:R\to R_{{\mathfrak{m}}_i}$ be localization and $q_i:R\to A_i$ the quotient map. Since $A_i$ is local with maximal ideal ${\mathfrak{m}}_i\mathrm{/}{\mathfrak{m}}_i^n$, every $s\mathrm{\notin }{\mathfrak{m}}_i$ maps to a unit of $A_i$, so Universal property of localisation: maps that invert $S$ factor uniquely through $S^{-1}R$ gives a unique homomorphism ${\psi }_i:R_{{\mathfrak{m}}_i}\to A_i$ with $q_i={\psi }_i{\lambda }_i$. Conversely, for each $j\mathrm{\ne }i$ choose $u_j\in {\mathfrak{m}}_j\setminus {\mathfrak{m}}_i$ and put $u={\prod }_{j\mathrm{\ne }i}{u}_j^n$. Then $u$ becomes a unit in $R_{{\mathfrak{m}}_i}$, while $u{\mathfrak{m}}_i^n\subseteq {\prod }_{j=1}^r{\mathfrak{m}}_j^n=0$ by step 1.1. Hence ${\lambda }_i$ kills ${\mathfrak{m}}_i^n$, and A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring gives a unique homomorphism ${\bar{\lambda }}_i:A_i\to R_{{\mathfrak{m}}_i}$ with ${\lambda }_i={\bar{\lambda }}_i{q}_i$. Both composites ${\psi }_i{\bar{\lambda }}_i$ and ${\bar{\lambda }}_i{\psi }_i$ agree with the identity after precomposing with the relevant universal map, so uniqueness in the same two universal properties forces them to be identities. Therefore $A_i\cong R_{{\mathfrak{m}}_i}$.

Combining steps 1.1 and 3.1 gives $R\cong {\prod }_{i=1}^r{R}_{{\mathfrak{m}}_i}$ canonically, and also $R\cong {\prod }_{i=1}^{r}R\mathrm{/}{\mathfrak{m}}_i^n$. The second form depends on the chosen nilpotence exponent $n$, while the first is the canonical product of localizations.

By Every prime ideal of an Artinian ring is maximal, every prime ideal of $R$ is maximal. By An Artinian ring has only finitely many maximal ideals, the set of maximal ideals is finite.

Conversely every maximal ideal of a commutative ring is prime by Every maximal ideal of a commutative ring is prime. Therefore the prime ideals of $R$ are exactly its maximal ideals, and there are only finitely many of them.

If $R$ is Artinian, then Every prime ideal of an Artinian ring is maximal says that every prime ideal is maximal. The Noetherian hypothesis in the statement is then automatic from Every commutative Artinian ring is Noetherian.

Conversely, assume every prime ideal of $R$ is maximal. By A Noetherian ring has finitely many minimal prime ideals, the minimal primes over $(0)$ are ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_r$ for some $r\ge 1$ unless $R=0$, in which case the conclusion is immediate. By hypothesis, each ${\mathfrak{p}}_i$ is maximal. Every prime ideal contains a minimal prime over $(0)$, so every prime ideal is one of the ${\mathfrak{p}}_i$. Hence The nilradical is the intersection of all prime ideals gives $N:=\mathrm{Nil}(R)={\mathfrak{p}}_1\cap \cdots \cap {\mathfrak{p}}_r$. Because distinct maximal ideals are comaximal, Chinese remainder theorem for pairwise comaximal ideals yields $R\mathrm{/}N\cong {\prod }_{i=1}^{r}R\mathrm{/}{\mathfrak{p}}_i$, and each factor is a field by $R\mathrm{/}M$ is a field if and only if $M$ is a maximal ideal.

By The nilradical of a Noetherian ring is nilpotent, choose $t\ge 1$ with $N^t=0$. Since $R$ is Noetherian, A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member makes each ideal $N^j$ finitely generated, so each quotient $N^j\mathrm{/}{N}^{j+1}$ is a finitely generated $R$-module. Because $N$ annihilates that quotient, it is a finitely generated $R\mathrm{/}N$-module. Under the product decomposition of step 1.2, this means a finite product of finitely generated vector spaces over the fields $R\mathrm{/}{\mathfrak{p}}_i$, so each quotient $N^j\mathrm{/}{N}^{j+1}$ has finite length. The same is true for $R\mathrm{/}N$ itself. Repeated use of Module length is additive in short exact sequences along $0\to N^{j+1}\to N^j\to N^j\mathrm{/}{N}^{j+1}\to 0$ therefore shows that $R$ has finite length as a module over itself.

By A commutative ring is Artinian exactly when it has finite length as a module over itself, a commutative ring has finite length as a module over itself exactly when it is Artinian. So step 2.1 proves that $R$ is Artinian. Together with step 1.1, this gives the asserted equivalence.

Let $I\subseteq R$ be an ideal. By Correspondence theorem: ideals of $R\mathrm{/}I$ correspond to ideals of $R$ containing $I$, descending chains of ideals in $R\mathrm{/}I$ correspond exactly to descending chains of ideals of $R$ containing $I$. Since those stabilize in the Artinian ring $R$, the quotient $R\mathrm{/}I$ is Artinian.

If $R=0$, then every localization is again $0$ and there is nothing to prove. Otherwise An Artinian ring is canonically the finite product of its localizations at its maximal ideals gives maximal ideals ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$ such that $R\cong {\prod }_{i=1}^r{R}_{{\mathfrak{m}}_i}$. Localization of a finite product acts factorwise, so it is enough to localize an Artinian local ring $(A,\mathfrak{m})$. If $S\cap \mathfrak{m}\mathrm{\ne }\mathrm{\varnothing }$, choose $s\in S\cap \mathfrak{m}$. By An Artinian local ring has nilpotent maximal ideal, and its finite modules have finite length, $s^n=0$ for some $n$, so $(s\mathrm{/}1{)}^n=0$ in $S^{-1}A$ while $s\mathrm{/}1$ is a unit; therefore $1=0$ and $S^{-1}A$ is the zero ring. If $S\cap \mathfrak{m}=\mathrm{\varnothing }$, then every element of $S$ is a unit of $A$: otherwise $(s)$ would be a proper ideal and therefore lie in the unique maximal ideal $\mathfrak{m}$, contradicting $s\mathrm{\notin }\mathfrak{m}$. So the localization map $A\to S^{-1}A$ is an isomorphism. Hence a localization of $R$ is exactly the product of those local factors whose maximal ideals avoid $S$, with the other factors collapsing to zero.

A finite product of Artinian rings is Artinian, because a descending chain of ideals in the product is coordinatewise a descending chain in each factor and therefore stabilizes once every coordinate chain does. Step 1.1 handles quotients, and step 2.1 handles localizations.