Non-trivial $\operatorname{Hom}_R(M;N)$

Question

In K. Conrad's handout on dual modules(https://kconrad.math.uconn.edu/blurbs/linmultialg/dualmod.pdf), there is the following theorem-exercise 2.11, which states:

Let $R$ be an integral domain with fraction field $K$. For two nonzero $R$-modules $M$ and $N$ in $K$, the $R$-module $\operatorname{Hom}_R(M, N)$ is isomorphic to $\{c ∈ K : cM ⊂ N\}$. If $M$ and $N$ admit a common denominator from $R$ then so does $\operatorname{Hom}_R(M, N)$. Equivalently, if $M^∨$ and $N^∨$ are nonzero then $\operatorname{Hom}_R(M, N) \neq 0$.

I'm struggling with proving the second-third part on my own, it seems that the proof of the similar statement when $N$ is equal to the base ring $R$ (that is given in the notes) doesn't work, or at least I don't see how.

Any hint, full solution, link are weLcome.

Answer 1

Let me restate the second statement in a clearer way (to reflect better the fact that it is equivalent to the last statement):

For $M$ and $N$ be nonzero modules in $K$, each has a common denominator, then $\text{Hom}_R(M, N)$ is also nonzero.

We only have to prove the second statement. To do this, let $d$ be a common denominator of $M$ and $d'$ be a common denominator of $N$. Since $N$ is nonzero, there exists $c\in R, c\neq 0$ such that $c/d'\in N$. Denote $k=\frac{cd}{d'}\in K$. The map $x\mapsto kx$ belongs to $\text{Hom}_R(M, N)$ since it takes $\frac{m}{d}\in M$ to $\frac{cm}{d'}=m\frac{c}{d'}\in N$.