7
$\begingroup$

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!

1 Answers 1

6

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, ยง3, Cor. to Prop.2, page 246.