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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.

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    Commenting so that this will be visible at the top: This question is Problem 1.5 on the homework for Math 620, at the University of Buffalo. As user "Student" points out, every question user6560 has asked is a homework question from that course. http://www.math.buffalo.edu/~badzioch/MTH620/Homework_files/hw1.pdf2011-02-13

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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.

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    @user: to check that $f\otimes\mathrm{id}$ is zero, you need to apply to to every element in its domain and see that the result is zero: you picked the element $4\otimes 1\in\mathbb Z\otimes\mathbb Z/2\mathbb Z$, and that element *is* zero, so it is not surprising that its image vanishes...2011-02-03