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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?

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    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.2012-06-15

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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.

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    Dear @Qiaochu: no $p$roblem. 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...2012-06-15
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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.