TITLE: Tensor product and injective maps
QUESTION [8 upvotes]: during a class we met a map $\Phi : \mathbb{Z}^n\to \mathbb{Z}^n$. We saw that $\Phi\otimes 1_\mathbb{Q}$ was an isomorphism and then the Professor said that it implied that $\Phi$ was injective. Unfortunately, I am not sure about the reason for this and I wanted to ask you if you may help me to understand it. I have doubts because I was studying on my own commutative algebra and saw that I may have two modules $M'\subset M$, $N$ and an inclusion $i: M'\to M$, but $i\otimes 1: M'\otimes N\to M\otimes N$ may not be not injective ($M'=2\mathbb{Z}$, $M=\mathbb{Z}$, $N=\mathbb{Z}/2\mathbb{Z}$ and consider the element $2\otimes x=0 \in \mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z}$, $\neq 0 \in 2\mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z}$). So I understood that tensor product should be considered carefully in both direction: $\Phi\otimes 1_\mathbb{Q}$ injective => $\Phi$ injective; $\Phi$ injective => $\Phi\otimes 1_\mathbb{Q}$ injective. I was wondering whether there is an object that measure the problem in both direction. I thank you in advance if you may help with the first part of this question, the second or both. Thanks for the attention.

REPLY [4 votes]: Use the commutative diagram
$\begin{array}{c} \mathbb{Z}^n & \xrightarrow{\Phi} & \mathbb{Z}^m \\ \downarrow && \downarrow \\ \mathbb{Q}^n & \xrightarrow{\Phi_{\mathbb{Q}}} & \mathbb{Q}^m. \end{array}$
Since $\Phi_{\mathbb{Q}}$ is an injective and $\mathbb{Z}^n \to \mathbb{Q}^n$ is injective, it follows that $\Phi$ is injective.