# Finitely generated field extensions

If $F=K(u_1,\ldots,u_n)$ is a finitely generated extension of $K$ and $M$ is an intermediate field, then $M$ is a finitely generated extension of $K$.

I'm not exactly sure how to start this problem. Any help at all would be a tremendous help.

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Firstly, I encourage you to simply try this with the tower rule ( $[F:K] = [F:M][M:K]$ ). Remember that you can consider the transcendent degree separately. However, if that is insufficient, I direct you to this answer over at MO.

The accepted answer covers this question very nicely, including the cases for transcendence.

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I've edited your answer to get rid of the entire link - in general, the format is [link text](url). – Zev Chonoles Apr 22 '11 at 0:41
@Zev: Thanks for that! – mixedmath Apr 22 '11 at 0:42
@ mixedmath thanks for the help. You offered a lot of help. – user8771 Apr 22 '11 at 1:09

I messed up last time. What I showed was not correct, $F$ need not be finite dimensional over $K$ just because it is finitely generated, counter example would be any transcendental extension.

Let $T = \{u_1,\cdots,u_n\}$. WLOG, let $A = \{u_1,\cdots,u_m\}\subset M$ and $B = \{u_{m+1},\cdots,u_n\}\cap M = \varnothing$. That is we partition $T = A\cup B$ into two sets $A$ and $B$, where $A$ contains all $u_i\in M$ and $B$ contains the rest.

Since $M\subset K(T)$, all elements of $M$ are in the form of $f(u_1,\cdots,u_n)$, where $f\in K(x_1,\cdots,x_n)$, the field of rational functions with coefficients in $F$ with $n$ variables.

But since $M\cap B = \varnothing$, all elements of $M$ are in the form of $f(u_1,\cdots,u_m,0,\cdots,0)$ where from position $m+1$ to $n$ its all $0$. That is all elements of $M$ are the form of $f(u_1,\cdots,u_m)$ where $f\in K(x_1,\cdots,x_m)$. Thus $M\subset K(u_1,\cdots,u_m)$.

Also $M\supset K(u_1,\cdots,u_m)$ because $M\supset K$ and $M\supset A$. That is $M = K(u_1,\cdots,u_m)$.

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It's curious that nobody remarked so far that this answer is wrong. (What if none of the $u_i$s belong to $M$? It follows that $M=K$? I don't think so.) – user26857 Apr 4 '15 at 18:03