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$\DeclareMathOperator{\Spec}{Spec}$ I am self-studying commutative algebra and I would like to have my proof verified. In what follows, all rings are assumed to be commutative and have a multiplicative unit. The claim I intend to prove is the following:

Let $k$ be a field, and let $R$ be a commutative $k$-algebra, finite dimensional over $k$. Then $R$ has only finitely many prime ideals, each of which is maximal, and the canonical ring maps induces an isomorphism $R\to R_P$ $$R\cong \prod_{P\in\Spec R}R_P$$ of rings.

My proof is posted as an answer. Any critique/suggestion/comment/better proof is welcome. Thanks in advance.

Ken
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  • Somehow the tag [proof-verification] is automatically changed into [solution-verification] and I can't seem to change it. Hopefully someone will edit the tag... – Ken Jul 02 '20 at 09:04
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    Your proof looks great. Do you know what an Artinian ring is? Every finite dimensional commutative algebra over a field is Artinian. This follows by dimension reasons via the exact argument you give in your post. And indeed, any Artinian ring has dimension 0 (all primes are maximal), finite spectrum, and the structure theorem for Artinian rings implies your localization claim. (Of course, there is much overlap between the ideas you use above and the ideas used to prove these facts about Artin rings. But it is worth noting that there is a more general setting for what you've done here.) – Alex Wertheim Jul 02 '20 at 09:55
  • @Alex Wertheim I just googled Artinian rings and realized that, as you mentioned, a lot of the ideas in the proof are applicable to a more general setting. I appreciate you for telling me about this. And of course, thank you very much for taking your time! – Ken Jul 02 '20 at 10:37
  • A reference for artinian rings: https://stacks.math.columbia.edu/tag/00J4 – Ken Jul 02 '20 at 10:40
  • If you searched, you should have found this which I think confirms what you're doing in the first half. Secondly, please don't post "does everything look ok with my proof?" posts. If you really must submit it for inspection, do it as a self-solution to your own post (in which you actually formulate a real question.) be sure to thoroughly search the site for exiting solutions, though. – rschwieb Jul 02 '20 at 15:29
  • @rschwieb You are right in that I should have searched more. As for your second point, I will make change the format of the question and post my proof as an answer shortly. Thank you for your comments. – Ken Jul 03 '20 at 00:42
  • @ken thanks! This is excellent now. – rschwieb Jul 03 '20 at 11:13

1 Answers1

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$\DeclareMathOperator\Spec{Spec}$ $\DeclareMathOperator\Ann{Ann}$ As pointed out in the comments by Alex, there is a more general setting in which the claim holds. By definition, Artinian rings refers to rings satisfying the descending chain condition(DCC) on ideals. Note that every finite-dimensional algebra over a field is Artinian for dimensional reasons.

Lemma. Every Artinian ring is Noetherian.

(proof.) Let $R$ be an Artinian ring, and suppose it is not Noetherian. Let $I\subset R$ be an ideal minimal among those that are not finitely generated. We claim that $P=\Ann_{R}(I)$ is a prime ideal. Given any $a\in R$, we can form an exact sequence$$0\to K\to I\xrightarrow{a\cdot}aI\to0,$$where $K$ is the kernel of the multiplication map $a\cdot:I\to aI$. If $aI\neq I$, then $aI$ is finitely generated, so $K$ cannot be finitely generated. By minimality of $I$, this means $K=I$. Thus $aI=0$. So we have either $aI=I$ or $aI=0$. This is enough to conclude that $P$ is prime.

It follows that $R/P$ is an artinian domain, i.e., a field. As an $R/P$-vector space, $I$ is infinite-dimensional. Thus $I$ contains a proper infinite-dimensional subspace, which is an ideal properly contained in $I$ that is not finitely generated, a contradiction.$\square$

Theorem. (Structure Theorem for Artinian Rings) If $R$ is an Artinian ring, then it has only finitely many primes, each of which is maximal, and $R$ is isomorphic to $\prod_{P\in\Spec R}R_P$ as rings.

(proof.) Since every Artinian domain is a field, every prime in $R$ is maximal. To show that there are only finitely many primes, suppose there existed infinitely many primes $P_{1},P_{2},\cdots$. Since each $P_{i}$ is maximal, $\{P_{i}\}_{i}$ is pairwise comaximal, so by CRT we have $R/\bigcap_{i=1}^{n}P_{i}\cong R/P_{1}\times\cdots\times R/P_{n}$. In particular, for each $n$ there is $r\in R$ such that $r\in\bigcap_{i=1}^{n}P_{i}\setminus P_{n}$, so the chain $\{\bigcap_{i=1}^{n}P_{i}\}_{n=1}^{\infty}$ violates the DCC. Hence $\Spec R$ is finite.

Next, let $J$ denote the Jacobson radical of $R$. By DCC we have $J^{n}=J^{n+1}$ for some $n$. Since Artinian rings are Noetherian, $J^{n}$ is finitely generated, and hence Nakayama's lemma shows $J^{n}=0$. Therefore by CRT we have $$ R\cong\prod_{P\in\Spec R}\frac{R}{P^{n}}. $$ By taking $n$ large enough, we may assume $P^{n}=P^{n+1}$ for all $P\in\Spec R$ (because of DCC and because $\Spec R$ is finite). Then $(PR_{P})^{n}=(PR_{P})^{n+1}=(PR_{P})(PR_{P}^{n})$. Since $(R_{P},PR_{P})$ is local and $(PR_{P})^{n}$ is finitely generated as an $R_{P}$-module (because $R_{P}$ is Noetherian since $R$ is Noetherian), Nakayma's lemma shows that $(PR_{P})^{n}=0$. Therefore $$ R_{P}=\frac{R_{P}}{(PR_{P})^{n}}=\frac{R_{P}}{P^{n}R_{P}}\cong\left(\frac{R}{P^{n}}\right)_{P} $$ as rings. Now since $P$ is maximal, given any $u\in R\setminus P$ we can find $v\in R\setminus P$ such that $uv+1\in P$. Then $(uv+1)^{n}\in P^{n}$. Expanding the LHS we find that $ua+1\in P^{n}$ for some $a\in R$, so $u$ acts as a unit in $R/P^{n}$. Hence $(R/P^{n})_{P}\cong R/P^{n}$ as rings. Thus $R_{P}\cong R/P^{n}$ as rings. The claim follows. $\square$

Ken
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