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I'm a bit embarrassed to ask this, but I've gotten myself confused over what I think is a simple issue. Let $A$ be a local ring, $k$ its residue field, and $M,N$ finitely generated $A$-modules. An exercise in Atiyah and Macdonald asserts that if $M \otimes_A N = 0$, then either $M = 0$ or $N = 0$. They give a hint in which they use the notation $M_k = M \otimes_A k$: it says $M \otimes_A N = 0$ implies $(M \otimes_A N)_k = 0$ implies $M_k \otimes_k N_k = 0$. I don't fully understand what happened in the second step.

This motivates the following more general question: if $A$ is any commutative ring and $B$ a commutative $A$-algebra, and $M,N$ are $B$-modules, can we identify $M \otimes_A N \cong M \otimes_B N$ as, say, $A$-modules? I think this is either very obvious or very naive.

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$M\otimes_A N$ is an $A$ module, so $(M\otimes_A N)_k$ is a $k$-module (a vector space). But if $M\otimes_A N$ is the zero module, then tensoring with $k$ is still the $0$ module, so $(M\otimes_A N)_k = 0$. –  Arturo Magidin Apr 18 '11 at 19:30
Right, but what happened in the second step? As elgeorges pointed out, the more general question is false, so why does $(M \otimes_A N)_k = 0$ imply $M_k \otimes_k N_k = 0$? The first module can be identified with $M_k \otimes_A N_k$... –  Justin Campbell Apr 18 '11 at 20:27

2 Answers 2

up vote 1 down vote accepted

No, we may not in general make the identification evoked in your more general question.
For example take $A=\mathbb R , B=\mathbb C , M=N =\mathbb C$. Then $\mathbb C \otimes_{\mathbb C} \mathbb C =\mathbb C$ whereas $\mathbb C \otimes_{\mathbb R} \mathbb C=\mathbb C \times \mathbb C $.

But sometimes we may...

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Thanks! I should have thought about it for longer... but does this work when $A$ is local and $B$ is its residue field? This is not a localization map, so the answer linked to does not apply. –  Justin Campbell Apr 18 '11 at 20:28
Given a completely arbitrary $A$-algebra $B$ and arbitrary $A$-modules $M$ and $N$, the $B$-modules $ (M \otimes_A N)\otimes_A B$ and $(M \otimes_A B)\otimes_B (N \otimes_A B)$ are $B$-linearly isomorphic. This justifies the second step you mentioned in your question (where $B$ is the residue field $k$ ) . A reference is Bourbaki, Algebra, Chapter II, §5.1, Proposition 3. Strangely this result doesn't seem to be mentioned in the more popular references I have just checked so don't be too harsh on yourself for not having known this :) –  Georges Elencwajg Apr 18 '11 at 22:34

For the second step, you actually have equality: $(M \otimes _A N)_k = (M \otimes _A N) \otimes _A (k \otimes _A k) = M_k \otimes _A N_k = M_k \otimes _k N_k$, hence the conclusion.

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