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Let $N$ be a $\mathbb Z$-module. Then, the map $(i\otimes_{\mathbb{Z}} 1_N):M'\otimes_{\mathbb{Z}} N \rightarrow M\otimes_{\mathbb{Z}} N$ is a monomorphism for every monomorphism $i:M'\rightarrow M$ iff $\mathbb{Z}$-module $N$ is torsion-free.

$(\Rightarrow)$:

(Assume that $N$ is finitely generated.)

Assume that exist nonzero $q\in \mathbb{Z}$ such that for every $n\in N$ we have $qn=0$. Then for every $N\ni n\neq 0$ and for every $\mathbb{Z}$-module $M$ and its submodule $M'$ we have

$M'\otimes\{n\}\cong M'\stackrel{i}{\rightarrow} M $

and $qM'\cong 0$.

Taking $M=M':=\mathbb{Z}$ we obtain $M'\cap qM =\mathbb{Z}\cap q\mathbb{Z}=q\mathbb{Z}\neq0$.

If $N$ is finitely generated then we can use fact that $\ker\left(M'\otimes \left(\mathbb{Z}/q\mathbb{Z}\right)\rightarrow M\otimes \left(\mathbb{Z}/q\mathbb{Z}\right)\right)\cong \left(M'\cap qM\right)/qM'$, but in this case we obtain $q\mathbb{Z}\cong 0$, which is contradiction.

Is it good?

$(\Leftarrow)$

Propably I should use this theorem, but I have no idea how to continue it. Any hints ?

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I would suggest the following proof which one can find it in Rotman's book (An Introduction to homological algebra p134). I also assume $_{\mathbb{Z}}N$ is f.g as you used it like that in your proof.

Let $N$ be a f.g. $\mathbb{Z}$-module. If $N$ is torsion-free, then it is free, by using the fundamental theorem of finitely generated modules over a PID. Thus, $-\otimes_{\mathbb{Z}}N$ takes monomorphisms to monomorphisms.

Conversely, we have $0\to {}_{\mathbb{Z}}\mathbb{Z}\to {}_{\mathbb{Z}}\mathbb{Q}$, but by the condition we obtain the exact sequence $0\to\mathbb{Z}\otimes_{\mathbb{Z}}N\to\mathbb{Q}\otimes_{\mathbb{Z}}N$. Hence, $N\simeq\mathbb{Z}\otimes_{\mathbb{Z}}N$ is a $\mathbb{Z}$-submodule of a torsion-free module $\mathbb{Q}\otimes_{\mathbb{Z}}N$, since $\mathbb{Q}\otimes_{\mathbb{Z}}N$ is a $\mathbb{Q}$-vector space. But any submodule of a torsion-free module is torsion-free, which implies what we wanted.

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