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A finite dimensional commutative algebra is a finite product of commutative local algebras. why? In fact, every commutative semiperfect ring is a basic ring and isomorphic to a finite product of local rings, but I do not how to prove it.

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I've changed algebra tag to abstract-algebra, since we don't use algebra tag anymore, see meta for details. If you can think of a more suitable tag, please, retag your post. –  Martin Sleziak Nov 2 '12 at 12:25
Thank you very much, Martin Sleziak –  Aimin Xu Nov 2 '12 at 12:42
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2 Answers 2

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A slightly different proof than that of rschwieb. Let $R$ be a finite dimensional commutative artinian algebra.

  1. $R$ has only finitely many maximal ideals. Let $n$ be the dimension of $R$. If $m_1, \dots, m_r$ are distinct maximal ideals of $R$, then by Chinese Remainder Theorem, the canonical map $$ R\to R/m_1\oplus \cdots \oplus R/m_r$$ is surjective. This imples implies that $r\le n$. So $R$ has only finitely many maximal ideals. Let $m_1, \dots, m_q$ be the maximal ideals of $R$.

  2. For some $N\ge 1$, $(m_1m_2\dots m_q)^N=0$. The decreasing sequence of ideals $(m_1m_2\dots m_q)^N$ is stationary. For some $N\ge 1$, we have $$ (m_1m_2\dots m_q)^N=(m_1m_2\dots m_q).(m_1m_2\dots m_q)^N.$$ By Nakayama's lemma, this implies that $(m_1m_2\dots m_q)^N=0$.

  3. End of proof. Let $N$ be as above. Again by CRT, we have $$ R\simeq \oplus_{1\le i\le q} R/m_i^N.$$ Now $R/m_i^N$ is local because its only maximal ideal is the image of $m_i$ in $R/m_i^N$.

Remark. The same proof works similarly if $R$ is not an algebra over a field, but has finite length.

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In fact, every commutative semiperfect ring is a basic ring and isomorphic to a finite product of local rings, but I can not give a proof. –  Aimin Xu Nov 3 '12 at 2:23
@AiminXu: which definition of semiperfect do you use ? –  user18119 Nov 3 '12 at 9:45

Because it is Artinian, and commutative Artinian rings factor this way. Here is a sketch.

For each idempotent $e$ in a commutative Artinian ring $R$, the ideal $eR$ is actually a subring with identity $e$, and also an algebra, if $R$ is. Consequently $(1-e)R$ is also a subring, since $1-e$ is also idempotent, and $eR\oplus (1-e)R=R$.

Now imagine breaking these idempotents into smaller ones: $e=f+g$ with $fg=0$. Under these conditions, $eR=fR\oplus gR$, and the original ideal is broken down to two smaller ideals.

Using the Artinian hypothesis, you can show that this process of breaking down cannot continue indefinitely, and you must arrive at idempotents such that $eR$ cannot be broken down any further. This turns out to be equivalent to $eR$ having only the trivial idempotents $0$ and $e$. Now a commutative Artinian ring with only trivial idempotents is local.

So what is the end result? $R$ decomposes into a finite product of $e_iR$, all of which are finite and local (algebras).

(I thought I had an elementary proof handy but it's not coming to mind, so here's an off the cuff version.) Lemma A commutative Artinian ring with only trivial idempotents is local.

Proof sketch: This is trivially true when $R$ is semisimple (a.k.a semisimple Artinian) because commutative semisimple rings are products of fields, and "no nontrivial idempotents" means it consists of exactly one field.

$R$ is local iff $R/\mathrm{Nil}(R)$ is local. $R/\mathrm{Nil}(R)$ is semisimple. Any idempotent of $R/\mathrm{Nil}(R)$ lifts to one in $R$.

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There are several lemmas hiding in there, so if you would like a few hints on those, it's fine to ask. –  rschwieb Nov 2 '12 at 12:52
For example, this 'Now a commutative Artinian ring with only trivial idempotents is local.' seems a bigger bite. –  Berci Nov 2 '12 at 13:00
@Berci Included some content on that, now. –  rschwieb Nov 2 '12 at 13:27

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