Lemma 10.53.5. Any ring with finitely many maximal ideals and locally nilpotent Jacobson radical is the product of its localizations at its maximal ideals. Also, all primes are maximal.

**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.
$\square$

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