Finding all submodules of $K[x]/\langle Q^3\rangle$, $Q$ irreducible

This question on an Algebra sheet stumped me.

Let $K$ be a field and $Q$ an irreducible polynomial of $K[x]$. $\langle Q^3\rangle$ will denote the ideal generated by $Q^3$. Determine all submodules of $K[x]/\langle Q^3\rangle$.

I could solve the equivalent question over $\mathbb Z$: find all submodules of $\mathbb Z/p^3\mathbb Z$ for prime $p$, but that's because over $\mathbb Z$, all submodules of its quotient modules are generated by a single element (right?). I tried to generalize this to $K[x]$, maybe something like "over a PID, all submodules of quotient modules are generated by a single element", but I'm not sure if I'm going in the right direction. I think that only works over $\mathbb Z$ because over $\mathbb Z$, all submodules are ideals.

• the fact that $K[x]$ is a PID,
• the correspondence theorem (if $I$ is an ideal of the ring $R$, then every ideal of $R/I$ is of the form $J/I$, where $J$ is an ideal of $R$ containing $I$), and
• the fact that in a domain $(a) \supseteq (b)$ iff $a \mid b$.
• Wait, what does $J/I$ mean when $J$ is an ideal? – Jack M Oct 28 '16 at 17:27
• Same thing as $R/I$, that is, $J/I = \{ a + I : a \in J \} \subseteq R/I$. – Andreas Caranti Oct 28 '16 at 17:28