Commutative Algebra/Artinian rings

Definition, first property

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

Proof:

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

$⟨r⟩\supseteq ⟨r^2⟩\supseteq ⟨r^3⟩\supseteq \cdots \supseteq ⟨r^n⟩\supseteq \cdots$.

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

$⟨r^n⟩=⟨r^{n+1}⟩$.

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

Proof:

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

Characterisation

Proof:

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

$⟨0⟩=m_1\cdots m_n$

for certain maximal ideals $m_1,\dots ,m_n\le R$. 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

$R\ge m_1\ge m_1\cdot m_2\ge \cdots \ge m_1\cdot m_2\cdots m_n=⟨0⟩$.

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 $⟨0⟩$ 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 $⟨0⟩$) 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) $⟨0⟩$ 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 $I\ne ⟨0⟩$. We lead this to a contradiction.

We form $A:=(⟨0⟩:I)$. Since $1\notin A$ as $I\ne 0$, $A\ne R$. 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

$A⊊aB+A\subseteq B$, 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 $p\ne R$. Indeed, we have $pB\le A$, hence $IpB\subseteq ⟨0⟩$ and therefore

$(⟨0⟩:Ip)\ge B>A=(⟨0⟩:I)$.

This shows $p\ne R$, and $Ip⊊I$ contradicts the minimality of $I$.$◻$