How to show that $R/I$ is Artinian when R is PID

Question

I'm working through some of Hungerfords "Algebra", and having trouble with Excercise VIII 1.2.:

Show that if $I$ is a non-zero ideal in a principal ideal domain (PID) $R$, then the ring $R/I$ is both Noetherian and Artinian.

I know that $R$ is Noetherian since it is a PID (this follows from Lemma III. 3.6 ). To show that $R/I$ is Noetherian I have then noted that since $I$ is a submodule of $R$ (viewed as an $R$-module) and since $R$ is Noetherian it follows that $R/I$ is Noetherian (by Corollary VIII 1.6).

My problem is then how to show that $R/I$ is Artinian.

Can someone give me a hint?

Answer 1

We know ideals of $R/I$ are of the form "ideal of $R$ containing $I$" mod $I$. So a descending chain of ideals looks like $I_1/I \supseteq I_2/I \supseteq I_3/I \supseteq \cdots$ where $I_j$ are ideals of $R$ such that $I \subseteq I_j$.

Next, $R$ is a PID so there exists $a_j \in R$ such that $I_j=(a_j)$ and $a \in R$ such that $I=(a)$. Don't forget $a \not=0$ because $I \not= \{ 0 \}$.

What does $I \subseteq I_j$ say? What does $I_{j+1} \subseteq I_j$? Notice that $I \subseteq \cap_{j=0}^\infty I_j$. Use unique factorization into primes to see that an infinite chain of proper divisors is impossible.

Answer 2

We know that a ring $R$ is Artinian iff it is Noetherian and every prime ideal is maximal. Now since $R$ is PID, so $R/I$ is PID. On the other hand, in a PID ring every prime ideal is maximal.So the Noetherian ring $R/I$ is Artinain.