
Lemma 10.52.5. 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 10.14.4) we have $R/I \cong \bigoplus R/\mathfrak m_ i$ which is a product of fields. Hence by Lemma 10.31.6 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. $\square$

Comment #1360 by abcxyz on

If by "radical" you mean Jacobson radical, then I don't know how every prime contain this. If by "radical" you mean nilradical, then I don't know how this equals the intersection of maximal ideals.

Comment #1363 by Auguste Hoang Duc on

The radical is the Jacobson radical (http://stacks.math.columbia.edu/tag/0AMD). The ring is assumed to have locally nilpotent radical. This is an other way to say that the Jacobson radical is equal to the nilradical.

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