Prove or counterexample: If A is a commutative ring and $A_p$ is a finitely generated algebra over A for all prime ideal p of A, then A is a product of local rings.
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Here is a counterexample. Let p and q be distinct prime numbers, let S denote the complement of {p,q} in the set of all primes, and let A denote Z[1/S]. Then the prime ideals of A are pA, qA, and {0}, the first two of which are maximal. Also, since A is domain (being a subring of Q) with two maximal ideals, it's not a product of local rings. On the other hand, the local rings at the various prime ideals are A[1/q], A[1/p], and A[1/p,1/q], which are clearly finitely generated as A-algebras.
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$\begingroup$ Cool - I figured something had to go wrong without stronger finiteness conditions. $\endgroup$ Commented Nov 22, 2009 at 10:14
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In the case that $A$ is noetherian and we replace finitely generated as an algebra with finitely generated as a module we can argue as follows. We have for any choice of minimal prime ideal $P$ of $\mathrm{Spec} A$ that $A_P$ is flat and finitely presented hence projective. It is also an artinian local ring. As it is projective the corresponding sheaf is locally free on $\mathrm{Spec} A$. But $$\mathrm{Supp} A_P = \mathrm{Spec} A_P$$ so that if $\mathrm{Spec} A$ were connected it would have to agree with the spectrum of $A_P$ since a locally free sheaf has constant rank on connected components. Thus $A$ is in fact artinian and local. $$ $$ If $A$ does not have connected spectrum then we can rewrite $A$ as $A_P \times A_1$. Where $A_P$ doesn't split up at all. We then just keep playing the same game (next with $A_1$) which terminates since $A$ is noetherian so has finitely many connected components and we in fact see that $A$ is a product of artinian local rings.
The stronger hypotheses I have are probably only somewhat necessary. I don't see a problem (except one needs to remove the word artinian) if one just assumes that the localizations are finitely presented as modules - there is the issue of whether or not this terminates but if you have an infinite product you also pick up extra prime ideals via choice and I have no idea what localizing at one of those would look like so I am not quite sure if this causes problems.
answered Nov 22, 2009 at 9:00
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$\begingroup$ cool for the argument in the case of f.g. as modules $\endgroup$– TJCMCommented Nov 24, 2009 at 14:57
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I think it's correct. Since the localization A_p is finitely generated over A for any prime ideal p, A has only finite maximal ideals. Now A is a semi-local ring, which is also a a product of local rings. The following link maybe helpful http://www.mathreference.com/ring-jr,sloc.html
