Homomorphism always exists between modules over an integral domain 
Let $R$ be an integral domain. Let $F$ be a free module over $R$, and let $M$ be an arbitrary nonzero $R$-module. Is it true that there always exists a nonzero module homomorphism from $M$ to $F$? 

I know that there always exists one from $F$ to $M$ by the universal property, but I don't know if it is true the other way around, and I can't think of any counter examples.
 A: Not necessarily. Consider $R=\mathbb{Z}$, $F=\mathbb{Z}$, and $M = \mathbb{Z}/2\mathbb{Z}$.  If $\varphi: M \to F$ is a module homomorphism, then $2\varphi(1)=\varphi(2\cdot 1) = \varphi(0)=0$ implies that $\varphi(1)=0$, and hence there are no non-zero module homomorphisms from $M$ into $F$.
A: You're implicitly assuming $F\ne0$.
Asking for a nonzero homomorphism $M\to F$ is the same as asking for a nonzero homomorphism $M\to R$, because $F$ is a direct sum of copies of $R$ and, if a homomorphism $M\to F$ is nonzero, then at least one of the projections on a summand is nonzero as well.
Now $R$ is torsionfree as a module over itself, whereas $R/aR$ is a nonzero torsion module whenever $a\ne0$ and $a$ is not invertible.
So what you're asking is true only when $R$ is a field.

If you drop the condition of $R$ being an integral domain, then it can be true without $R$ being a field. An example is $R=\mathbb{Z}/p^2\mathbb{Z}$, which is self-injective, local and Artinian.
If $M\ne0$, then take $x\in M$, $x\ne0$. Since $xR$ is Artinian, it has a simple submodule $S$, which embeds in $R$ (there is, up to isomorphisms, a single simple module). By injectivity of $R$, the embedding $S\to R$ extends to a homomorphism $M\to R$, which is nonzero because it doesn't vanish on $S$.
