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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!

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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 $k[x,1/g(x)]$?

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  • $\begingroup$ Thank you, this is helpful- is this sort of a proof by contradiction, where we could suppose that $f_1(x),...,f_(x)$ are a finite generating set for $k(x)$? Just a few points I would like to clarify for my understanding- when you take the product of the denominators of each $f_j(x)$, are we first guaranteeing that each polynomial has a single denominator to help simplify this product? $\endgroup$ – MathStudent1324 Jan 17 '18 at 7:48
  • $\begingroup$ You can take all denominators appearing anywhere among those $m$ polynomials; or for each $f_i$ the common denominator of all the coeffs of $f_i$ — it doesn’t matter. You should be able to prove this directly. $\endgroup$ – Lubin Jan 17 '18 at 19:19

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