It is well known that $\mathbb{Z}/n\mathbb{Z}$ is injective as a module over itself. And so because $\mathbb{Z}/n\mathbb{Z}$ is a PID (even if it isn't an integral domain, each ideal is generated by a single element), this implies that over itself $\mathbb{Z}/n\mathbb{Z}$ is divisible.
That being said, even though I get each step in theory, in practice something must be escaping my understanding as consider
$$10 \equiv 4a [12]$$
Because $\mathbb{Z}/12\mathbb{Z}$ is injective over itself, there should exist an $a$ solving this. However, $\forall a\in\mathbb{Z}/12\mathbb{Z}$ the RHS and LHS are different when after reducing the congruence to mod 3, and so this congruence is impossible to solve.
What am I missing/What did missunderstand? My reading is leading me to believe that my mistake is in PID $\Rightarrow$ injective iff divisble, because the ring $\mathbb{Z}/n\mathbb{Z}$ isn't integral. But I don't think my proof uses integrality of R...