$A \subset B$ is an extension of commutative rings s.t. $B$ is a f.g. free $A$-module of rank $n$, so I have $A^n \stackrel{\sim}{\longrightarrow} B$ as $A$-modules. Let $\mathfrak a$ be an ideal of $A$.
I want to show that $(A/\mathfrak a)^n\stackrel{\sim}{\longrightarrow} B/\mathfrak a B$ as $A/\mathfrak a$-modules.
Can I do this by tensoring through by $A/\mathfrak a$? The problem is that I'm not sure if tensoring by $A/\mathfrak a$ preserves exact sequences of $A$-modules.
Many thanks.
Yes you can, because tensoring is exact on the right, i.e. if $P\to Q\to R\to 0$ is an exact sequence of $A$-modules for some ring $A$, then for every $A$-module $M$ the sequence $M\otimes P\to M\otimes Q\to M\otimes R\to 0$ is exact as well. So in particular this holds when $P=0$, i.e. when $Q\to R$ is an isomorphism. Note that $A/\mathfrak a$ doesn't need to be flat, so tensoring with it is not exact on the left in general.