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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1$\begingroup$ Exactly on how many examples did you try this on? :) $\endgroup$– Mariano Suárez-ÁlvarezCommented Nov 21, 2010 at 4:03
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1$\begingroup$ ti4: Did you ask this question math.stackexchange.com/questions/10980/… ? $\endgroup$– Hailong DaoCommented Nov 21, 2010 at 4:03
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1$\begingroup$ Did you mean to assume that $A$ was Artinian? $\endgroup$– Karl SchwedeCommented Nov 21, 2010 at 4:28
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$\begingroup$ Mariano: Perhaps I should've added that I thought it was wrong but could'nt find a convincing counterexample :) $\endgroup$– PandamicCommented Nov 21, 2010 at 11:31
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$\begingroup$ Hailong: Yes I did, it is almost the same question, only this is a little weaker perhaps. $\endgroup$– PandamicCommented Nov 21, 2010 at 11:33
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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).
answered Nov 21, 2010 at 3:54
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$\begingroup$ 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? $\endgroup$– PandamicCommented Nov 21, 2010 at 11:30
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1$\begingroup$ You probably need to think carefully about what it means to localize at the prime $(0)$. $\endgroup$ Commented Nov 21, 2010 at 11:42