Tensoring a monomorphism of free modules with an identity map Suppose $R$ is a commutative ring, $f\colon F_1\to F_2$ is a homomorphism of free modules, and $M$ is an $R$-module.
If $f$ is a surjective homomorphism, then $f\otimes_R \mathrm{id}_M$ is surjective.
Is it true that if $f$ is a monomorphism, then $f\otimes_R \mathrm{id}_M$ is a monomorphism? I can't think of any counterexamples.
 A: Take $f:\mathbb Z\to\mathbb Z$ be multiplication by $2$, and $M=\mathbb Z/2\mathbb Z$. This example captures precisely why it does not work.
N.B. If you have an injective map $f:P\to Q$ with $Q$ is free, you can complete it to a short exact sequence $$0\longrightarrow P\xrightarrow{\;f\;}Q\longrightarrow Q/f(P)\longrightarrow 0$$ If $M$ is another module, then there is a short exact sequence $$0\longrightarrow\mathrm{Tor}_1^R(Q/f(P),M)\longrightarrow P\otimes_RM\xrightarrow{\;f\otimes\mathrm{id}_M\;} Q\otimes_RM\longrightarrow Q/f(P)\otimes_RM\longrightarrow 0$$ where $\mathrm{Tor}_1^R(Q/f(P),M)$ is a certain $R$-module one learns to construct when one learns about homological algebra; it measures how far $f\otimes\mathrm{id}_M$ is from being a monomorphism. In the example above we have $$\mathrm{Tor}_1^R(Q/f(P),M)\cong \mathbb Z/2\mathbb Z,$$ as one can easily check as soon as one knows how this is done, and in fact this is how I found the example: I looked for a $\mathrm{Tor}$ I knew was not zero.
