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.


A non-empty subset of a commutative ring is indeed a submodule iff it is an ideal, so your intuition is spot on.

Then use

  • 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 $a \mid b$.
| cite | improve this answer | |
  • $\begingroup$ Wait, what does $J/I$ mean when $J$ is an ideal? $\endgroup$ – Jack M Oct 28 '16 at 17:27
  • $\begingroup$ Same thing as $R/I$, that is, $J/I = \{ a + I : a \in J \} \subseteq R/I$. $\endgroup$ – Andreas Caranti Oct 28 '16 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.