Let $R$ be a commutative domain with field of fractions $F$. Prove that $F$ is an injective $R$-module.

Let $R$ be a commutative integral domain with field of fractions $F$. Prove that $F$ is an injective $R$-module.

I have tried to apply Baer's criterion: every $R$-module homomorphism from any ideal $I$ of $R$ can be extended to an $R$-module homomorphism from $R$ to $F$. But I couldn't. Thanks for any help.

It is true because $K$ is torsion-free and divisible. So I'll prove that, over an integral domain, a torsion-free, divisible module $E$ is injective.

Let $I$ be an ideal in $A$. We may suppose $I\ne 0$. Using Baer's criterion, we have to prove that any homomorphism $f\colon I\to E$ can be extended to $A$.

As $I\ne 0$, we can pick a non-zero element $x\in I$ and, as $E$ is divisible, we can write $f(x)=xe$ for a certain element $e\in E$. Now let $a\in I$; we have $$xf(a)=f(ax)=af(x)=axe$$ As $E$ is torsion-free, we can simplify by $x$ and obtain $\; f(a)=ae$ for all $a\in I$. Thus we can extend $f$ to $A$ just setting, for any $a\in A$, $\; f(a)=ae$.

Added: Over a P.I.D. (and even a Dedekind domain), one has more: an $A$-module (torsion-free or not) is injective if and only if it is divisible.

Thus for a Dedekind domain $A$, the $A$-module $K/A$ is injective.