# Tensor product of modules and being torsion-free

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 ?

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.