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Let $A\to B$ be a morphism of (commutative) algebras and $M$ a $B$-module. The $A$-bilinear map $B\times M\to M$ given by $(b,m)\mapsto bm$ induces a surjective homomorphism $B\otimes_{A}M\to M$.

Give sufficient and necessary (or at least sufficient) conditions for this mapping to be an isomorphism of $B$-modules.

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For the context of my problem you can actually assume that $A$ is a finitely generated UFD over a field of characteristic $p>0$ (you can even more assume that $A$ is a polynomial ring over $K$), that $B$ is a domain, and that $B$ is faithfully flat as an $A$-module. –  David Apr 19 '11 at 7:22

1 Answer 1

up vote 2 down vote accepted

Frankild, Sather-Wagstaff, and Wiegand: As long as ${}_BM$ is finitely generated, $A \longrightarrow B$ is flat local, $\mathfrak{m}_A B = \mathfrak{m}_B$, and the extension of residue fields is an isomorphism, this happens if and only if $\mathrm{Ext}_A^i(B,M)$ is $A$-finitely generated for all $i \geq 1$.

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In fact they need the stronger hypothesis that $A \to B$ is unramified, i.e. that $\mathfrak m_B = \mathfrak m_A B$. They also restrict to the situation where $M$ is finitely generated as an $A$-mdoule. –  Emerton Apr 19 '11 at 13:02
Quite right, I was overhasty. –  Graham Leuschke Apr 19 '11 at 13:34
@Graham: Nice reference, I'll take a look. In the context of my problem $B$ is not Noetherian (nor local). $A$ is Noetherian nonetheless, and the expansion of every maximal ideal of $A$ is maximal in $B$ (I'm just gonna think about it). –  David Apr 19 '11 at 17:04
@Graham: Thanks a lot. The reference was very useful. I could solve my problem (and, even more, the solution gave something interesting). –  David Apr 22 '11 at 18:49
Outstanding. Glad to hear it. –  Graham Leuschke Apr 23 '11 at 0:05

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