I am working through Chapter 15 of Dummit and Foote's Abstract Algebra text, and I am stumped on how to prove the following (Exercise 3):
Prove that the field $k(x)$ of rational functions over $k$ in the variable $x$ is not a finitely generated $k$-algebra. (Recall that $k(x)$ is the field of fractions of the polynomial ring $k[x]$. Note that $k(x)$ is a finitely generated field extension over $k$.)
This portion discusses Noetherian rings and provides the following theorem which I think I should apply (Theorem 2):
The following are equivalent:
(1) $R$ is a Noetherian ring.
(2) Every nonempty set of ideals of $R$ contains a maximal element under inclusion.
(3) Every ideal of $R$ is finitely generated.
I also have the following definition:
(1) The ring $R$ is a finitely generated $k$-algebra if $R$ is generated as ring by $k$ together with some finite set $r_1,…,r_n$ elements of $R$.
I believe that this is all of the information I need, but I'm unsure of how to piece it together.
Thanks in advance!
Best Answer
You can get it from the Nullstellensatz, but that's really overkill.
Let $f_1(x),\ldots,f_m(x)\in k(x)$. The subalgebra generated by these is contained in $k[x,1/g(x)]$ where $g$ is the product of the denominators of the $f_j$. But $k[x]$ has infinitely many irreducibles. Pick one $p(x)$ not dividing $g(x)$. Can $1/p(x)$ be an element of $k[x,1/g(x)]$?