If $B$ is $A$-flat, is $\_ \otimes_A B: A\text{-}\mathrm{Mod} \to B\text{-}\mathrm{Mod}$ exact?

Question

Let $f: A \to B$ be a flat morphism of rings, i.e. $B$ is a flat $A$-module by $a \cdot b:=f(a)b$. Then certainly $\_ \otimes_A B: A\text{-}\mathrm{Mod} \to A\text{-}\mathrm{Mod}$ is exact (by the very definition of flatness), but is $\_ \otimes_A B: A\text{-}\mathrm{Mod} \to B\text{-}\mathrm{Mod}$ exact? I would say yes, because the injectivity of a $A$ and $B$-linear map depends only on the abelian group structure of the modules involved, but I am not sure.

Answer 1

Yes, that's exactly right. Given an exact sequence $$0\to K\to M\to N\to 0$$ of $A$-modules, the sequence $$0\to K\otimes_A B\to M\otimes_A B\to N\otimes_A B\to 0$$ is exact as a sequence of $A$-modules because $A$ is flat. But this implies it is also exact as a sequence of $B$-modules, since the kernels and images as $B$-modules are the same subsets as the kernels and images as $A$-modules. As you said, determining whether a sequence is exact only uses the underlying abelian group structures, not the full module structures. (In fact, it really only uses the $0$ elements of the underlying sets, not even the addition operation!)