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.

Let $K$ be a field and $x, y$ be independent variables. How can I show that $K(x, y)/K$ is not a simple extension?

share|improve this question
1  
Suppose to the contrary that it is simple and generated by some $t$. Write $x$ and $y$ as rational functions in $t$. Then show there is some relation between them, which contradicts independence. –  Qiaochu Yuan Jun 15 '12 at 13:41

2 Answers 2

Consider a simple extension $K(f)/K$ with $f\in K(x,y) \setminus K$.
Since $K$ is algebraically closed in $K(x,y)$, $f$ is transcedental over $K$ so that $\operatorname{trdeg}_K K(f)=1$.
However $\operatorname{trdeg}_K K(x,y)=2$ so that necessarily $K(f)\subsetneq K(x,y)$ and thus $K(x,y)/K$ is not simple.

share|improve this answer
3  
Georges, it seems to me that one has to prove statements generalizing this one in order to establish the basic properties of transcendence degree, so you'll forgive me if this argument strikes me as a bit circular... –  Qiaochu Yuan Jun 15 '12 at 13:47
    
Dear Qiaochu: 1) I have browsed the proofs in Bourbaki and they don't use the OP's result. Of course they use and prove very general results but nobody thinks the question here is the most general possible result on transcendental extensions. 2) My answer clearly isn't from first principles. There is a choice for each question between giving completely elementary calculations or just quoting a general result. 3) I don't think there is anything circular involved since I'm not writing a systematic treatise. (to be continued ) –  Georges Elencwajg Jun 15 '12 at 14:10
    
(continued) 4) I'd be very happy to see you develop your brief comment into an explicit elementary answer and will happily upvote it then. –  Georges Elencwajg Jun 15 '12 at 14:11
    
My apologies. I admit I haven't actually gone through proofs of the basic properties of transcendence degree myself. As for an explicit elementary answer, the problem looks like homework to me and so I am a little reluctant... –  Qiaochu Yuan Jun 15 '12 at 14:35
    
Dear @Qiaochu: no problem. Actually you raise a very interesting question: at what level of sophistication should we answer? The gut feeling is certainly "at the level of the question". But sometimes, probably by laziness or lack of time the temptation becomes irrestible to just quote a result. It will amuse you if I confess that I actually tried (maybe not long enough) to give an elegant, simple and elementary solution to the question here but that I failed to find one! So I resigned myself to using a sledgehammer... –  Georges Elencwajg Jun 15 '12 at 14:59

This is a simple and elementary proof, I think.

Assume that $K(x,y)=K(t)$ for some $t\in K(x,y)$.
Since $x\in K(t)$, there exist (coprime and nonzero) polynomials $u(z),v(z)\in K[z]$ such that $x=\frac{u(t)}{v(t)}$.
Consider now the polynomial $f(z)=xv(z)-u(z)\in K(x)[z]$: clearly it is not $0$ and $f(t)=0$, which means that $t$ is algebraic over $K(x)$, that is, $K(x)[t]$ is an algebraic extension of $K(x)$. But then $K(x,y)=K(t)=K(x,t)=K(x)(t)=K(x)[t]$ is algebraic over $K(x)$, a contradiction.

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.