Lemma 10.52.6. A ring $R$ is Artinian if and only if it has finite length as a module over itself. Any such ring $R$ is both Artinian and Noetherian, any prime ideal of $R$ is a maximal ideal, and $R$ is equal to the (finite) product of its localizations at its maximal ideals.

**Proof.**
If $R$ has finite length over itself then it satisfies both the ascending chain condition and the descending chain condition for ideals. Hence it is both Noetherian and Artinian. Any Artinian ring is equal to product of its localizations at maximal ideals by Lemmas 10.52.3, 10.52.4, and 10.52.5.

Suppose that $R$ is Artinian. We will show $R$ has finite length over itself. It suffices to exhibit a chain of submodules whose successive quotients have finite length. By what we said above we may assume that $R$ is local, with maximal ideal $\mathfrak m$. By Lemma 10.52.4 we have $\mathfrak m^ n =0$ for some $n$. Consider the sequence $0 = \mathfrak m^ n \subset \mathfrak m^{n-1} \subset \ldots \subset \mathfrak m \subset R$. By Lemma 10.51.6 the length of each subquotient $\mathfrak m^ j/\mathfrak m^{j + 1}$ is the dimension of this as a vector space over $\kappa (\mathfrak m)$. This has to be finite since otherwise we would have an infinite descending chain of sub vector spaces which would correspond to an infinite descending chain of ideals in $R$. $\square$

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