# Subfields $k\subseteq F\subseteq k(x_1,\dots,x_n)$. Is then $F\cap k[x_1,\dots,x_n]=k(f_1,\dots,f_m)\cap k[x_1,\dots,x_n]$ for polynomials $f_i$?

Let $F\subseteq k(x_1,\dots,x_n)$ be a subfield with $k\subseteq F$. I know that $F=k(\psi_1,\dots,\psi_r)$ for rational functions $\psi_i\in k(x_1,\dots,x_n)$. I'm interested in the intersection $F\cap k[x_1,\dots,x_n]$. Can I always write this intersection as $k(f_1,\dots,f_m)\cap k[x_1,\dots,x_n]$ for polynomials $f_j\in k[x_1,\dots,x_n]$ (instead of rational functions)?

Background: Hilbert's 14th problem asks about the finite generation of this intersection. But in his original formulation, Hilbert starts out by taking polynomials $f_1,\dots,f_m$ in $n$ variables, and then asks if the intersection $k(f_1,\dots,f_m)\cap k[x_1,\dots,x_n]$ (the relatively integer functions) is finitely generated. The modern formulation of the problem takes any subfield $F$ as above and intersects it with the polynomial ring. So I thought there should be a "way back" from the modern formulation to Hilbert's original one. That is, if I can decide if $k(f_1,\dots,f_m)\cap k[x_1,\dots,x_n]$ is finitely generated for arbitrary polynomials $f_i$, I hope I should be able to decide if $F\cap k[x_1,\dots,x_n]$ is finitely generated for an arbitrary $F$. But I don't see a way to break it down to polynomials.

I guess that this is a very simple problem, but just can't wrap my head around it. I tried asking on MSE, but there hasn't been an answer yet, which may be because I didn't make it really clear what I wanted. Thanks for your help in advance!

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I'm not sure whether this is a simple problem. Here is a related result of Emmy Noether: if the transcendence degree of $F/k$ is $1$, and $F$ contains a nonconstant polynomial, then $F=k(f)$ for some polynomial $f$. Note that $n$ can be arbitrary here. But things get much more complicated when the transcendence degree is bigger, and I don't know what the truth should be. – Michael Zieve Sep 30 '13 at 4:47