TITLE: Can someone explain this non-noetherian subring example?
QUESTION [6 upvotes]: 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?

REPLY [7 votes]: $k[x_1,x_2,\dotsc]$ 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.