# Commutative Algebra/Artinian rings

## Definition, first property

Definition 19.1:

A ring $R$ is called artinian if and only if each descending chain

$I_{1}\supseteq I_{2}\supseteq \cdots \supseteq I_{k}\supseteq \cdots$ of ideals of $R$ eventually terminates.

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

Proposition 19.2:

Let $R$ be an artinian integral domain. Then $R$ is a field.

Proof:

Let $r\in R$ . Consider in $R$ the descending chain

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

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

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

Then write $r^{n}=sr^{n+1}$ , that is, $r^{n}(1-sr)=0$ , that is (as we are in an integral domain) $sr=1$ and $r$ has an inverse.$\Box$ Corollary 19.3:

Let $R$ be an artinian ring. Then each prime ideal of $R$ is maximal.

Proof:

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

Theorem 19.4:

Let $R$ be a ring. We have:

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

Proof:

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

$\langle 0\rangle =m_{1}\cdots m_{n}$ for certain maximal ideals $m_{1},\ldots ,m_{n}\leq R$ . In this case, if either chain condition is satisfied, one may consider the normal series of $R$ considered as an $R$ -module over itself given by

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

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

Hence, in the presence of either chain condition, we have a finite vector space, and thus $R$ 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 $\langle 0\rangle$ is a product of maximal ideals in cases

1. $R$ is noetherian and every prime ideal is maximal
2. $R$ is artinian.

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

2.: If $R$ is artinian, we use the descending chain condition to show that if (for a contradiction) $\langle 0\rangle$ is not product of prime ideals, the set of ideals of $R$ 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 $I\neq \langle 0\rangle$ . We lead this to a contradiction.

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

$A\subsetneq aB+A\subseteq B$ , hence, by minimality of $B$ , $aB+A=B$ and similarly for $b$ . Therefore

$abB+A=a(bB+A)+A=aB+A=B$ ,

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

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

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