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

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 $$r^n = s r^{n+1}$$, 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 $$\langle 0 \rangle$$ 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 $$m_1, \ldots, m_n \le R$$. 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
 * $$R \ge m_1 \ge m_1 \cdot m_2 \ge \cdots \ge m_1 \cdot m_2 \cdots m_n = \langle 0 \rangle$$.

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 $$pB \le A$$, hence $$IpB \subseteq \langle 0 \rangle$$ and therefore
 * $$(\langle 0 \rangle : Ip) \ge B > A = (\langle 0 \rangle : I)$$.

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