Lemma 10.53.5 (00JA)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.53: Artinian rings
Lemma 10.53.5 (cite)

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. $\square$

Comments (7)

Comment #1360 by abcxyz on March 17, 2015 at 08:59

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 March 17, 2015 at 12:46

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.

Comment #7150 by Ryo Suzuki on March 28, 2022 at 03:13

In my opinion, this lemma should be proved by an observation about topological spaces.

Lemma. Let $X$ be a topological space. Assume that (1) $X$ has finitely many closed points. (2) A subset consiting of all closed point is dense in $X$ . Then every point of $X$ is closed. Hence $X$ is a finite discrete spadce.

Proof. $X = \bar{Y} = \bigcup_{y\in Y} \bar{\lbrace y\rbrace} = Y$ .

Lemma 00JA can be proved by this lemma and Lemma 00EM.

Comment #7300 by Johan on May 04, 2022 at 11:11

Going to leave as is for now.

Comment #8123 by anon on February 21, 2023 at 16:07

The claim that the factors in the product are localizations is not justified. See here for a justification: https://math.stackexchange.com/questions/4643653/in-lemma-10-53-5-of-stacks-about-commutative-artinian-rings-how-did-they-use/4643742#4643742

Comment #8124 by anon on February 21, 2023 at 17:24

More generally, if a ring R is a product of local rings, then the factors in this product can be obtained by localizing at R's maximal ideals.

Comment #8225 by Stacks Project on March 03, 2023 at 13:12

Dear Anon, OK, I agree we could do a bit better here. For example, the product decomposition at the end of the argument holds because the $e_i$ are orthogonal to each other and sum up to $1$ . Then we could add some lemmas that $Re$ is a localization of $R$ if $e$ is an idempotent. We could add a lemma that if a $R \to S$ is a localization and $S$ is local, then $S$ is the localization of $R$ at a prime ideal. We could add the lemma you stated in #8124. I don't want this to become a long unwieldy thing however, so maybe there is a short sweet proof of this someone could tell me?

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