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I'm trying to find an example showing that subrings of noetherian rings are not necessarily noetherian. So I just searched the net and this site, and came upon this:

"A common example showing that a subring of a Noetherian ring is not necessarily Noetherian is to take a polynomial ring over a field $k$ in infinitely many indeterminates, $k[x_1,x_2,\dots]$. The quotient field is then Noetherian obviously, but the subring $k[x_1,x_2,\dots]$ is not since there is an infinite ascending chain of ideals which never stabilizes."

I haven't had the chance to read my textbook the past week so I'm a bit behind on my understanding of noetherian rings. I understand that they are rings in which every chain of its ideals stabilizes. I also understand that a polynomial ring $R[x_1, x_2, ...]$ of noetherian ring $R$ is also noetherian. So what I don't get in the example is when they say polynomial ring, do they mean a ring that has infinite indeterminates or it could just be any polynomial. Another question is what are quotient fields and why is it obviously noetherian?

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If you have not had time to read your book, you should really do that first. – Tobias Kildetoft Oct 25 '12 at 8:02
@QiaochuYuan The example describes it as a subring of its quotient field. Where is the problem with that? – Tobias Kildetoft Oct 25 '12 at 8:07
@Tobias: my mistake; I got confused about the meaning of "quotient field" (I'm much more used to "fraction field"). – Qiaochu Yuan Oct 25 '12 at 17:11
up vote 5 down vote accepted

$k[x_1,x_2,\dotsc]$ is not noetherian since the chain of ideals $(x_1) \subseteq (x_1,x_2) \subseteq \dotsc$ is not stationary. Fields are noetherian since there are only two ideals. Thus a counterexample is $k[x_1,x_2,\dotsc] \subseteq k(x_1,x_2,\dotsc)$. But you can do this with any other non-noetherian integral domain, such as the ring of holomorphic functions on a region.

By the way, Hilbert's Basis Theorem says that if $R$ is a noetherian ring and $n \in \mathbb{N}$, then $R[x_1,\dotsc,x_n]$ is noetherian. We don't need this here, but this shows that we really need infinitely many indeterminates in order to get a non-noetherian polynomial ring.

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Got it, thank you guys :) And Tobias, you are correct to criticize the reading part, I will read asap. – Kiyoshi Oct 25 '12 at 8:44

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