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Given a finite field extension $K\subset L$, prove there exists and intermediate field $M$ that is purely transcendental over $K$ (or $K=M$) and $[L:K]<\infty$.

I feel like in concept the answer to this is fairly straightforward. After all, since $L$ is a finite field extension of $K$, we know there is some subset $S\subset L$ such that $K(S) = L$. We then take some subset $S_0 \subset S$ such that all elements of $S_0$ are algebraically independent. Then $M := K(S_0)$ must be purely transcendental over K.

To my eye that proves this intermediate field exists, as it is certainly contained in $L$. As to the degree $[L:M]$, we simply take those elements of $S$ we had earlier ignored and have $L = M(S - S_0)$, which is another finite extension.

I'm personally not fully happy with these arguments because they don't seem watertight. Could anybody point out anything I'm missing or doing wrong?

EDIT

I believe the following sketch provides the correct proof. Indeed choosing our elements of the set $S$ is no problem: we choose as many transcendental elements such that they are all algebraically independent as we can. Since the extension is finitely generated this is no problem. Then note that any remaining elements of $S$, i.e. the elements in $S - S_0$ are either algebraic to begin with or algebraically dependent on the elements in $S_0$. Hence each element only brings about an extension of finite degree.

rwmak
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  • I don't understand your "intuitive argument", since for example the extension $\mathbb R \subset \mathbb C$ shows your argument fails : here $S = {i}$ but $\mathbb R[i] = \mathbb C$ is not purely transcendental over $\mathbb R$. (For cardinality reason $\overline{\mathbb C(X)} \cong \mathbb C$ so here can take $M = \mathbb C(X)$). I have no ideas for the general solution. –  Sep 10 '16 at 12:07
  • @N.H. I'm new to the concept of purely transcendental extensions, but how is can an extension of a single element even be algebraically independent? (after all, there's nothing to be dependent or independent on) – rwmak Sep 10 '16 at 12:44
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    $i^2 = -1$ means that ${i}$ is an algebraically dependent set. By the way, non-trivial transcendental extensions are always infinite, so $M=K$ is the only purely transcendental sub-extension of a finite extension. – Andrew Dudzik Sep 10 '16 at 13:06
  • @Slade Oh, thank you for that information. Could you point me to a proof for that assertion? I'm very curious now. – rwmak Sep 10 '16 at 17:06
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    @rwmak It is enough to show that if $x$ is transcendental over $K$, then $K(x)$ has infinite degree over $K$. But if it had finite degree, then the set ${1,x,x^2,\cdots}$ would contain a linearly dependent set ${1,x,\cdots , x^n}$. Then $x$ would satisfy a nontrivial polynomial of degree $\leq n$, and therefore be algebraic. – Andrew Dudzik Sep 10 '16 at 19:46
  • @Slade Thank you very much, that makes a lot of sense. – rwmak Sep 11 '16 at 06:16

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