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.



When is a localization of a commutative ring finitely generated as an algebra?
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.


