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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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Careful! $\mathbf{Z}$ is not Artinian, for example. It's important that $I$ be non-zero, here. –  Dylan Moreland Nov 14 '11 at 15:11
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This corresponds to the statement that if a descending chain of ideals in $R$ has a non-zero intersection, then it must be finite. –  Thomas Andrews Nov 14 '11 at 15:32
    
@BrandonCarter , you've essentially assumed what you are trying to prove. Where have you used that $I\neq 0$? –  Thomas Andrews Nov 14 '11 at 15:33
    
@Dylan Moreland, Yes of course, silly mistake. –  user50948 Nov 14 '11 at 15:36
    
@Brandon I'm confused. What about the chain of $\{(2^n)\}_n$ in $\mathbf{Z}$? –  Dylan Moreland Nov 14 '11 at 15:47

2 Answers 2

up vote 7 down vote accepted

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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Ok. Thank you for your response! Is the following understood correctly?: $a$ has a unique factorization $a = c_1c_2c_3 \cdots c_k$ into irreducible (prime) elements of $R$ (since $R$ is a PID). So $a$ can have only finitely many proper divisors. Since $a_j$ divides $a$ for all $j$ then not all $(a_j)$ can be different since this would give an infinite set of proper divisors. So either $a_j = a_n$ for some $n$ or $a_j$ is an associate with $a$. If $a_j$ and $a$ are associates then $(a_k) = (a)$ so we are done, if $(a_j) = (a_n)$ then we are also done. –  user50948 Nov 14 '11 at 16:38
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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 :) –  Bill Cook Nov 14 '11 at 21:43

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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I think you meant to say that in a PID, every nonzero prime ideal is maximal. –  Manny Reyes Dec 10 '12 at 15:41
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This is correct (modulo Manny's comment) but I think it is overkill to use the characterization of Artinian rings here. In fact the ring $R/I$ has only finitely many ideals, and I think it is instructive to show this directly. (Hint: $I = (x)$ for some nonzero $x$. Now factor $x$ into primes....) –  Pete L. Clark Dec 10 '12 at 16:57
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$R/I$ is not necessarily a domain, so it is not necessarily a PID. –  user26857 Feb 15 '13 at 11:55

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