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Let $A$ be a commutative ring with identity. If $A$ is a ring with only a finite set of prime ideals $p_1...p_n$ and moreover $\prod_{i=1}^n p_i^{k_i}=0$ for some k_i. Is $A$ then isomorphic to $\prod_{i=1}^nA_{(p_i)}$?

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Exactly on how many examples did you try this on? :) –  Mariano Suárez-Alvarez Nov 21 '10 at 4:03
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ti4: Did you ask this question math.stackexchange.com/questions/10980/… ? –  Hailong Dao Nov 21 '10 at 4:03
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Did you mean to assume that $A$ was Artinian? –  Karl Schwede Nov 21 '10 at 4:28
    
Mariano: Perhaps I should've added that I thought it was wrong but could'nt find a convincing counterexample :) –  Pandamic Nov 21 '10 at 11:31
    
Hailong: Yes I did, it is almost the same question, only this is a little weaker perhaps. –  Pandamic Nov 21 '10 at 11:33
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up vote 7 down vote accepted

No. Let $A$ be a DVR. It has two prime ideals: the maximal ideal $p_1=\mathfrak m\subset A$ and $p_2=(0)\subset A$. So, $p_1p_2=0$, but $A$ is not a product (of two local rings).

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Perhaps I have missed something huge but in this example it seems to me that product will be the ring localized at m, i.e. itself multiplied by the trivial ring, is this not isomorphic to the original ring? –  Pandamic Nov 21 '10 at 11:30
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You probably need to think carefully about what it means to localize at the prime $(0)$. –  Todd Trimble Nov 21 '10 at 11:42
    
Oh, I was being stupid. Thank you –  Pandamic Nov 21 '10 at 12:06
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