Prove that if an integral domain $R$ is an injective $R$-module then $R$ is a field.
Choose a non-zero element $x$ of $R$ and consider the map $f: R \rightarrow R$ given by $f(t)=tx$.
This yields an exact sequence $0 \rightarrow R \rightarrow R \rightarrow R/\langle x \rangle \rightarrow 0$ where the first map is $f$.
By assumption $R$ is injective and thus $R \cong R \oplus R/\langle x \rangle$.
Now the claim is that $1 \in \langle x \rangle$. Otherwise $(1,0 + \langle x \rangle)$, $(0,1+ \langle x \rangle)$ are zero divisors of $R \oplus R/\langle x \rangle$ which is impossible since $R$ is an integral domain.
Thus $1 \in \langle x \rangle$ so $x$ is invertible and we're done. Is this OK?