Tensor product and injective maps

Question

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.

Answer 1

Hint: the fundamental fact here is that $\mathbb{Z}^n$ is free.

The canonical map $\def\Z{\mathbb{Z}}\def\Q{\mathbb{Q}}\Z\to\Q$ is injective and you can identify $\Q$ with $\Z\otimes\Q$. But the tensor product commutes with direct sums, so $\Z^n\to\Z^n\otimes\Q$ is injective too.

It's just a check seeing that the diagram $$ \begin{array}{c} \Z^n & \xrightarrow{\Phi} & \Z^n \\ \downarrow && \downarrow \\ \Z^n\otimes\Q & \xrightarrow{\Phi_{\Q}} & \Z^n\otimes\Q \end{array} $$ is commutative.

There are other cases in which tensoring by $\Q$ reflects monomophisms (it always preserves them because $\Q$ is torsion-free, hence flat), but at a first stage a direct proof using freeness of $\Z^n$ seems better.

Answer 2

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.

Answer 3

Write down the short exact sequence

$$ 0 \to \ker \Phi \to \Bbb{Z} \to \Bbb{Z} \to \operatorname{coker} \Phi \to 0.$$

The functor $-\otimes_{\Bbb{Z}} \Bbb{Q}$ is exact since $\Bbb{Q}$ is a flat $\Bbb{Z}$-module. Thus if the tensored map is injective this means that $(\ker \Phi) \otimes_{\Bbb{Z}} \Bbb{Q} = 0$. Now $\ker \Phi$ is a submodule of a free module over a PID and hence is finitely generated. This means $\ker \Phi = T \oplus F$, torsion part plus a free part of finite rank. The torsion part has to be zero as $\Bbb{Z}^n$ is torsion free. So $\ker \Phi \cong \Bbb{Z}^k$ for some $0 \leq k \leq n$. But now

$$0 = \ker \Phi \otimes_{\Bbb{Z}}\Bbb{Q} = (\Bbb{Z}^k) \otimes_{\Bbb{Z}} \Bbb{Q} = \Bbb{Q}^k $$

which implies that $k= 0$, i.e. $\ker \Phi = 0$.