Artinian rings and PID Let $R$ be a commutative ring with unity. Suppose that $R$ is a principal ideal domain, and $0\ne t\in R$. I'm trying to show that $R/Rt$ is an artinian $R$-module, and so is an artinian ring if $t$ is not a unit in $R$.I'm not sure how to begin. please help.
 A: Hint: Show that the ideals of $R/Rt$ are all principal, and in fact, are in bijection with the divisors of $t$ in $R$ (considered up to multiplication by a unit). Then use  that $R$ is a UFD (since it's a PID). Finally, note that the ideals of $R/Rt$ are also precisely its $R$-submodules.
A: It happens that I recently incorporated a more general result into my commutative algebra notes ($\S 18.3$ on the Krull-Akizuki Theorem).
Lemma: For a Noetherian integral domain $R$, the following are equivalent:
(i) $R$ has Krull dimension at most one: i.e., nonzero prime ideals are maximal.
(ii) For every nonzero ideal $I$ of $R$, $R/I$ is an Artinian ring.
(iii) For every nonzero ideal $I$ of $R$, $R/I$ is a finite length $R$-module.
Note that (ii) $\iff$ (iii) and (iii) $\implies$ (i) are almost immediate.  The OP asks about (i) $\implies$ (ii) in the special case that $R$ is a PID.  The proof of (i) $\implies$ (ii) is still pretty easy when $R$ is a Dedekind domain: e.g. one can use the Chinese Remainder Theorem and localization to reduce to the case of a PID and reason as Zev does in his answer.  In general though the proof I came up with of (i) $\implies$ (ii) was a bit more technical than I would like.  Maybe someone can suggest something simpler.
