Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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.

share|improve this answer
    
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 trascendental 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 $M = K(u_1,\cdots,u_m)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.