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Let $A$ be a commutative ring with identity. If $A$ has finite number of prime ideals $p_1,...p_n$ and moreover $\prod_{i=1}^n p_i^{k_i} = 0$ for some $k_i$. Are the prime ideals necessarily maximal?

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No, but the counterexample is trivial. Take any integral domain with finitely many prime ideals which is not a field. For example, the localization $\mathbb{Z}_{(p)}$ of the integers at a prime p. The zero ideal is non-maximal and prime so, trivially, $\prod_{i=1}^np_i=0$. Maybe this isn't exactly what you were meaning to ask?

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  • $\begingroup$ Thank you! Actually that was what I meant to ask. The reason behind the question is that according to our teacher in commutative algebra, this should be true. From this we were supposed to conclude that the ring factors into a product of localized rings around $p_i$ using the Chinese Remainder Theorem. That is $A = \prod_{i=1}^n A_{p_i}$. But maybe the conclusion holds anyway. It should be a generalisation of the structure theorem for Artinian rings. $\endgroup$ – Maria Nov 19 '10 at 20:37
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    $\begingroup$ If the ring has only finitely many maximal ideals $m_i$ satisfying $\prod_im_i^{k_i}=0$ then the conclusion that every prime ideal is maximal would hold. Maybe your teacher was thinking of that? $\endgroup$ – George Lowther Nov 19 '10 at 21:00
  • $\begingroup$ It is certainly possible that that was what he was thinking, however he explicitly stated that it should be prime ideals. But it would make a lot of sense if it were maximal ideals, then the proof for Artin rings would apply almost word by word. $\endgroup$ – Maria Nov 19 '10 at 21:15
  • $\begingroup$ @user3620: If you think that my response answers your question, then it can be accepted. $\endgroup$ – George Lowther Nov 19 '10 at 23:13
  • $\begingroup$ I wouldn't ask, but it certainly seems to be a complete and correct answer and, maybe, being a new user, you don't know about accepted answers? $\endgroup$ – George Lowther Nov 19 '10 at 23:16

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