# Commutative Algebra/Artinian rings

## Definition, first property

Definition 19.1:

A ring ${\displaystyle R}$ is called artinian if and only if each descending chain

${\displaystyle I_{1}\supseteq I_{2}\supseteq \cdots \supseteq I_{k}\supseteq \cdots }$

of ideals of ${\displaystyle R}$ eventually terminates.

Equivalently, ${\displaystyle R}$ is artinian if and only if it is artinian as an ${\displaystyle R}$-module over itself.

Proposition 19.2:

Let ${\displaystyle R}$ be an artinian integral domain. Then ${\displaystyle R}$ is a field.

Proof:

Let ${\displaystyle r\in R}$. Consider in ${\displaystyle R}$ the descending chain

${\displaystyle \langle r\rangle \supseteq \langle r^{2}\rangle \supseteq \langle r^{3}\rangle \supseteq \cdots \supseteq \langle r^{n}\rangle \supseteq \cdots }$.

Since ${\displaystyle R}$ is artinian, this chain eventually stabilizes; in particular, there exists ${\displaystyle n\in \mathbb {N} }$ such that

${\displaystyle \langle r^{n}\rangle =\langle r^{n+1}\rangle }$.

Then write ${\displaystyle r^{n}=sr^{n+1}}$, that is, ${\displaystyle r^{n}(1-sr)=0}$, that is (as we are in an integral domain) ${\displaystyle sr=1}$ and ${\displaystyle r}$ has an inverse.${\displaystyle \Box }$

Corollary 19.3:

Let ${\displaystyle R}$ be an artinian ring. Then each prime ideal of ${\displaystyle R}$ is maximal.

Proof:

If ${\displaystyle p\leq R}$ is a prime ideal, then ${\displaystyle R/p}$ is an artinian (theorem 12.9) integral domain, hence a field, hence ${\displaystyle p}$ is maximal.${\displaystyle \Box }$

## Characterisation

Theorem 19.4:

Let ${\displaystyle R}$ be a ring. We have:

${\displaystyle R}$ is artinian ${\displaystyle \Leftrightarrow }$ ${\displaystyle R}$ is noetherian and every prime ideal of ${\displaystyle R}$ is maximal.

Proof:

First assume that the zero ideal ${\displaystyle \langle 0\rangle }$ of ${\displaystyle R}$ can be written as a product of maximal ideals; i.e.

${\displaystyle \langle 0\rangle =m_{1}\cdots m_{n}}$

for certain maximal ideals ${\displaystyle m_{1},\ldots ,m_{n}\leq R}$. In this case, if either chain condition is satisfied, one may consider the normal series of ${\displaystyle R}$ considered as an ${\displaystyle R}$-module over itself given by

${\displaystyle R\geq m_{1}\geq m_{1}\cdot m_{2}\geq \cdots \geq m_{1}\cdot m_{2}\cdots m_{n}=\langle 0\rangle }$.

Consider the quotient modules ${\displaystyle m_{1}\cdots m_{k}/m_{1}\cdots m_{k+1}}$. This is a vector space over the field ${\displaystyle R/m_{k+1}}$; for, it is an ${\displaystyle R}$-module, and ${\displaystyle m_{k+1}}$ annihilates it.

Hence, in the presence of either chain condition, we have a finite vector space, and thus ${\displaystyle R}$ has a composition series (use theorem 12.9 and proceed from left to right to get a composition series). We shall now go on to prove that ${\displaystyle \langle 0\rangle }$ is a product of maximal ideals in cases

1. ${\displaystyle R}$ is noetherian and every prime ideal is maximal
2. ${\displaystyle R}$ is artinian.

1.: If ${\displaystyle R}$ is noetherian, every ideal (in particular ${\displaystyle \langle 0\rangle }$) contains a product of prime ideals, hence equals a product of prime ideals. All these are then maximal by assumption.

2.: If ${\displaystyle R}$ is artinian, we use the descending chain condition to show that if (for a contradiction) ${\displaystyle \langle 0\rangle }$ is not product of prime ideals, the set of ideals of ${\displaystyle R}$ that are product of prime ideals is inductive with respect to the reverse order of inclusion, and hence contains a minimal (w.r.t. inclusion) element ${\displaystyle I\neq \langle 0\rangle }$. We lead this to a contradiction.

We form ${\displaystyle A:=(\langle 0\rangle :I)}$. Since ${\displaystyle 1\notin A}$ as ${\displaystyle I\neq 0}$, ${\displaystyle A\neq R}$. Again using that ${\displaystyle R}$ is artinian, we pick ${\displaystyle B}$ minimal subject to the condition ${\displaystyle B>A}$. We set ${\displaystyle p:=(A:B)}$ and claim that ${\displaystyle p}$ is prime. Let indeed ${\displaystyle a\notin p}$ and ${\displaystyle b\notin p}$. We have

${\displaystyle A\subsetneq aB+A\subseteq B}$, hence, by minimality of ${\displaystyle B}$, ${\displaystyle aB+A=B}$

and similarly for ${\displaystyle b}$. Therefore

${\displaystyle abB+A=a(bB+A)+A=aB+A=B}$,

whence ${\displaystyle ab\notin p}$. We will soon see that ${\displaystyle p\neq R}$. Indeed, we have ${\displaystyle pB\leq A}$, hence ${\displaystyle IpB\subseteq \langle 0\rangle }$ and therefore

${\displaystyle (\langle 0\rangle :Ip)\geq B>A=(\langle 0\rangle :I)}$.

This shows ${\displaystyle p\neq R}$, and ${\displaystyle Ip\subsetneq I}$ contradicts the minimality of ${\displaystyle I}$.${\displaystyle \Box }$