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If M and N are modules over some commutative ring A and $\mathfrak{a} \subset \operatorname{Ann(M)} \cap \operatorname{Ann(N)}$ is an ideal, is it true that $M \otimes_A N \cong M \otimes_{A/\mathfrak{a}} N$ as A-modules?

I think that I can interpret $M \otimes_A N$ as an $A/\mathfrak{a}$-module since $\mathfrak{a} \subset \operatorname{Ann}(M \otimes_A N)$, but I don't know if that gets me any further. Thanks!

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1 Answer 1

up vote 5 down vote accepted

To be absolutely precise , let us denote by $M'$ the the $A/\mathfrak a$-module whose underlying abelian group is that of $M$ and whose multiplication is the obvious one: $\bar a \cdot m=a\cdot m $ and similarly for $N'$.
The result you want is then

$$ M \otimes _A N = ( M' \otimes_{A/\mathfrak{a}} N')_A \quad \text {(isomorphism of} \; A-\text {modules)}$$ where, given an $A/\mathfrak a$-module $P$, the notation $P_A$ means the $A$- module gotten from $P$ by the restriction of scalars $A\to A/\mathfrak{a}$.
A reference is Bourbaki, Algebra, Chap. 2, §3, Cor. to Prop.2, page 246.

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