Prove that there is an $R$-module isomorphism between $Q \otimes_R N \cong N$, Q is a quotient field of $R$.

Question

Let $R$ be an integral domain with quotient field Q and $N$ be a unitary, divisible, torsion-free left $R$-module. Show that there is an $R$-module isomorphism so that $Q \otimes_R N \cong N$.

Here is what I know. Every element of $Q \otimes_R N \cong N$ is of the form $1/d \otimes n$, $d \in R$. I want to use the universal property of tensor product to produce a group homomorphism from $Q \otimes_R N \to N$, but I was struggling to find a midlinear map from $Q \times_R N \to N$. Not sure if I am on the right track. Any hint would be appreciated.

Answer 1

$N$ is divisible, so for each $n\in N$ and $0\ne d\in R$, there is $m\in N$ with $dm=n$. Then $$ \frac1d\otimes n=\frac1d\otimes dm=1\otimes m$$

Answer 2

Note that $Q$ is a localization of $R$, so $Q\otimes_R N$ is a localization of $N$. Can you use your hypothesis on $N$ to show the canonical map

$$N\longrightarrow N[S^{-1}]$$

is an isomorphism? When is it onto? When is it injective?

Hint. This post here for injectivity, and divisibility for surjetivity.