Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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 $i \geq 1$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.