# 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
* $R$ is noetherian and every prime ideal is maximal
* $R$ 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 $I$.$\Box$