# B is a flat A-module, and M is a finitely generated $A$-module then $σ:\operatorname{Hom}_A(M, N)⊗_AB→\operatorname{Hom}_B(M⊗_AB,N⊗_AB)$ is injective

Let A and B be rings, and let f : A → B be a homomorphism of rings; we consider B as an A-module with the structure induced by f. Let M and N be two A-modules. There is a map of $B$-modules $σ : \operatorname{Hom}_A(M, N) ⊗_A B → \operatorname{Hom}_B(M ⊗_A B, N ⊗_A B)$. If B is a flat A-module, and $M$ is a finitely generated $A$-module, then σ is injective.

I have no idea on construct the map $\sigma$ and prove the claim. Can anyone just give me some hints? Thank you in advance !

• I assume that $A$ and $B$ are commutative rings. Constructing $\sigma$ is easy: it sends $f \otimes b$ to the $B$-linear map which sends every pure tensor $m \otimes c$ to $f\left(m\right) \otimes cb$. Not sure why $\sigma$ is injective, though. – darij grinberg Dec 3 '15 at 21:09
• Can we use exact sequence? – Thomas Edison Dec 3 '15 at 21:38
• – Watson Nov 22 '18 at 19:14

## 1 Answer

Notice that

\begin{align*} Hom_B\left(M \otimes_A B, N\otimes_A B\right) \overset{f} \simeq Hom_A\left(M, Hom_B \left(B, N\otimes_A B \right)\right) \end{align*}

via Hom-Tensor Adjoint Property.

Now, \begin{align*} Hom_A\left(M, Hom_B \left(B, N\otimes_A B \right)\right) \overset{g} \simeq Hom_A\left(M, N\otimes_A B\right) \end{align*} Observe that the diagram below is commutative -

$$\require{AMScd} \begin{CD} Hom_A\left(M,N\right)\otimes_A B @>{\sigma}>> Hom_B\left(M \otimes_A B, N\otimes_A B\right)\\ @V{\alpha}VV @V{g \circ f}VV \\ Hom_A\left(M, N\otimes_A B\right)@>{Id}>>Hom_A\left(M, N\otimes_A B\right) \end{CD}$$

Since $Id$ and $g \circ f$ are isomorphisms, hence, it is enough to prove that the map $\alpha$ is injective.

As $M$ is a finitely generated $A$ module, let $F$ be a finite-rank free $A$-module that surjects onto $M$. We immediately obtain the exact sequence -

$$0 \to Hom_A\left(M,N\right) \overset{i_0}\to Hom_A(F,N)$$ Tensoring by the flat $A$-module $B$, we get $$0 \to Hom_A(M,N)\otimes_A B \overset{i}\to Hom_A(F,N)\otimes_A B$$ Upon replacing $N$ with $N\otimes_A B$, we obtain the exact sequence - $$0 \to Hom_A\left(M,N\otimes_A B\right) \overset{j}\to Hom_A\left(F,N\otimes_A B\right)$$ And yet again we have a commutative diagram -

$$\require{AMScd} \begin{CD} Hom_A\left(M,N\right)\otimes_A B @>{i}>> Hom_A(F,N)\otimes_A B\\ @V{\alpha}VV @V{h}VV \\ Hom_A\left(M,N\otimes_A B\right)@>{j}>>Hom_A\left(F,N\otimes_A B\right) \end{CD}$$

The maps $i$ and $j$ are injective maps, and the map $h$ is an isomorphism. Therefore, the map $\alpha$ is injective as well.