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I have an assignment to show the known result that any finite group occurs as Galois group of $k(x_1,...,x_n)/F$ for some field $F$. This seems like an insurmountable task to be given in a first course on GT. Despite that, here's my idea for a solution, although its simplicity makes me question whether it's complete:

Any group G is isomorphic to a permutation group, so for some appropriate n we can associate bijectively G with a subgroup of $S_n$, which we also call G. The extension $k(x_1,...,x_n)/k(x_1,...,x_n)^{S_n}$ is Galois with Galois group $S_n$. We have $k(x_1,...,x_n)^{S_n} \subset k(x_1,...,x_n)^G \subset k(x_1,...,x_n)$ and that $Gal(k(x_1,...,x_n)/k(x_1,...,x_n)^G) = G$, qed.

This is my approach, basically. Very simple. Is it enough? Seems too simple to be true, yet I can't find anything wrong with this "proof".

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The proof is fine, but the question is off-topic. – Angelo May 5 '13 at 13:58
Why is it off topic? – Erik Vesterlund May 6 '13 at 13:21
Please read the FAQ. – Angelo May 7 '13 at 4:37

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