The Stacks project

Proof. The descending chain condition for ideals obviously holds. $\square$

Proof. Suppose that $\mathfrak m_ i$, $i = 1, 2, 3, \ldots $ are pairwise distinct maximal ideals. Then $\mathfrak m_1 \supset \mathfrak m_1\cap \mathfrak m_2 \supset \mathfrak m_1 \cap \mathfrak m_2 \cap \mathfrak m_3 \supset \ldots $ is an infinite descending sequence (because by the Chinese remainder theorem all the maps $R \to \oplus _{i = 1}^ n R/\mathfrak m_ i$ are surjective). $\square$

Proof. Let $I \subset R$ be the Jacobson radical. Note that $I \supset I^2 \supset I^3 \supset \ldots $ is a descending sequence. Thus $I^ n = I^{n + 1}$ for some $n$. Set $J = \{ x\in R \mid xI^ n = 0\} $. We have to show $J = R$. If not, choose an ideal $J' \not= J$, $J \subset J'$ minimal (possible by the Artinian property). Then $J' = J + Rx$ for some $x \in R$. By NAK, Lemma 10.20.1, we have $IJ' \subset J$. Hence $xI^{n + 1} \subset xI \cdot I^ n \subset J \cdot I^ n = 0$. Since $I^{n + 1} = I^ n$ we conclude $x\in J$. Contradiction. $\square$

Proof. Let $R$ be a ring with finitely many maximal ideals $\mathfrak m_1, \ldots , \mathfrak m_ n$. Let $I = \bigcap _{i = 1}^ n \mathfrak m_ i$ be the Jacobson radical of $R$. Assume $I$ is locally nilpotent. Let $\mathfrak p$ be a prime ideal of $R$. Since every prime contains every nilpotent element of $R$ we see $ \mathfrak p \supset \mathfrak m_1 \cap \ldots \cap \mathfrak m_ n$. Since $\mathfrak m_1 \cap \ldots \cap \mathfrak m_ n \supset \mathfrak m_1 \ldots \mathfrak m_ n$ we conclude $\mathfrak p \supset \mathfrak m_1 \ldots \mathfrak m_ n$. Hence $\mathfrak p \supset \mathfrak m_ i$ for some $i$, and so $\mathfrak p = \mathfrak m_ i$. By the Chinese remainder theorem (Lemma 10.15.4) we have $R/I \cong \bigoplus R/\mathfrak m_ i$ which is a product of fields. Hence by Lemma 10.32.6 there are idempotents $e_ i$, $i = 1, \ldots , n$ with $e_ i \bmod \mathfrak m_ j = \delta _{ij}$. Hence $R = \prod Re_ i$, and each $Re_ i$ is a ring with exactly one maximal ideal. $\square$

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 10.53.3, 10.53.4, and 10.53.5.

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 10.53.4 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 10.52.6 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 $R$. $\square$