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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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Please write your questions so that the complete question is in the body. Your previous question was editing by someone else, but you can do it yourself... (And there is no need for abreviations like «comm.»: you are not charged by the character!) –  Mariano Suárez-Alvarez Feb 3 '11 at 3:52
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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. math.buffalo.edu/~badzioch/MTH620/Homework_files/hw1.pdf –  David Speyer Feb 13 '11 at 19:27

1 Answer 1

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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I think I may understand. So (f\tens id)(4 \tens 1) = 2*2 \tens 1 = 2 \tens 2 = 2 \tens 0 = (f \tens id)(1 \tens 0). That look about right? –  user6560 Feb 3 '11 at 4:11
    
@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... –  Mariano Suárez-Alvarez Feb 3 '11 at 7:13

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