# Ring with finitely many prime ideals with an extra condition. Are they maximal?

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?
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. – user3620 Nov 19 '10 at 20:37
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? – George Lowther Nov 19 '10 at 21:00