$R/Ra$ is an injective module over itself

Let $R$ be a PID, $a\in R$ be a nonzero nonunit in $R$. Prove that $R/Ra$ is an injective module over itself.

If $R$ is a PID, every $R$- divisible module is injective, but the question concerns with $R/Ra$-module, so I have no idea to solve this problem. Help me.

Use Baer's criterion: Ideals in $R/aR$ correspond to divisors of $a$, so you have to show the following:

For any $d \in R$ with $d|a$ any $R/aR$-homomorphism $f:dR/aR \to R/aR$ extends to a $R/aR$-homomorphism $R/aR \to R/aR$.

A homomorphism $f:dR/aR \to R/aR$ is given by the image of $d$ and since we have $\frac{a}{d}d=0$ in $dR/aR$, the image $f(d)$ must also satisfy $\frac{a}{d}f(d)=0$, equivalent to $d|f(d)$. Hence we can extend $f$ to $$R/aR \to R/aR ~,~ 1 \mapsto \frac{f(d)}{d}$$

You could also embed this into a more general context: If $M$ is an injective $R$-Module and $I$ an ideal of $R$, then $$M[I] := \{m \in M | Im=0 \}$$ is an injective $R/I$-module (This is a one-liner, just write down the defining diagram of an injective object and do some abstract nonsense).

In your setting, we know that $Q(R)/R$ is an injective $R$-module, hence $R/aR \cong (Q(R)/R)[aR]$ is an injective $R/aR$-module (The elements in $Q(R)/R$, which are annihilated by $a$ are precisely spanned by $\frac{1}{a}$, hence isomorphic to the cyclic module $R/aR$).