$M \otimes_R K$ is uniquely divisible as $R$-module

Question

A module $M$ over a domain $R$ is called uniquely divisible if for every $r \in R \setminus\lbrace 0 \rbrace$ the endomorphism $$\phi_r : M \to M$$ $$m \mapsto r \cdot m$$ is a bijection.

Let $K$ be the field of fraction of $R$. I have to prove that for any $R$-module $M$ , $M \otimes_R K$ is uniquely divisible as $R$-module.

I have proved that $\phi_r$ is surjective, but how to prove the injectivity ?

Answer 1

Hint $1$: If $R$ is a domain, then $K=\mathrm{Frac}(R)$ is a flat $R$-module.

Hint $2$: If $N_r=\{m\in M\mid r\cdot m=0\}$, then $N_r\otimes_{R}K=0$ for $r\ne 0$.

Now for any $r\ne 0$, consider the exact sequence of $R$-modules:

$$0\to N_r\to M\to rM\to 0$$

Can you finish the argument from here?

Answer 2

The module $M\otimes_RK$ is a vector space over $K$. For any $\alpha\in K$, $\alpha\ne0$, and any vector space $V$ over $K$, the map $v\mapsto \alpha v$ is bijective.