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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?

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    @Brandon I'm confused. What about the chain of $\{(2^n)\}_n$ in $\mathbf{Z}$?2011-11-14

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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.

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    That sounds about right to me. A clean proof might go something like: Each time we have $I_j \not= I_{j+1}$ it must be the case that $a_{j+1}$ has more irreducible factors than $a_{j}$ (if they have the same factors they will necessarily be associates by uniqueness of factorizations). However, the number of times we can increase the number of irreducible factors is bounded by the number of factors in $a$ (the generator of $I$). Thus the chain must stabilize at some point. Notice that if $a=0$, it does not have$a$factorization and the argument falls apart :)2011-11-14
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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.

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    $R/I$ is not necessarily a domain, so it is not necessarily a PID.2013-02-15