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Let $G$ be a given finite group. Is there a way to extend the field $K$ such that for the extension $L\geq M\geq K$ we have that $L/M$ is Galois and it's Galois group $Gal(L/M)$ is isomorphic to $G$ ? If there is, along which lines does the proof run ?

(I didn't manage to come up with a proof - nor with a counterexample to this question)

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Is $K \subset L$ just an arbitrary field extension? – JSchlather Dec 11 '12 at 14:31
Just to clarify, you are asking: given a field $K$, does there exist an extension $M$ of $K$ and a Galois extension $L/M$ such that $\mathrm{Gal}(L/M)$ is isomorphic to a given finite group? – Keenan Kidwell Dec 11 '12 at 15:02
$L = M = K\phantom{}$? – Hans Giebenrath Dec 11 '12 at 21:00
@KeenanKidwell exactly! Ok, by looking at the comments, it seems that I should have been more precise. sorry – user47580 Dec 11 '12 at 21:06
@JacobSchlather Well...yes, under the constraints I listed – user47580 Dec 11 '12 at 21:07
up vote 2 down vote accepted

You're asking if for a field $K$ does there exist an extension $M$ such that the inverse Galois problem is solved over $M$. The inverse Galois problem being showing that for every finite group $G$ there exists a Galois extension $L$ of $M$ such that $\mathrm{Gal}(L/M)$. This is a difficult question. In the case that $K$ is characteristic $0$ the answer is yes. You can set $M=\overline{K}(t)$ for which the problem is solved. There are some results over the case that characteristic is $p$ but I don't understand them, or the previous result honestly.

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Let $n$ be large enough that $G$ embeds in $S_n$. Let $L$ be the field of rational functions over $K$ in $n$ variables. Let $E$ be the subfield of $L$ consisting of symmetric rational functions. Then $L/E$ is Galois with Galois group $S_n$. Consider a copy of $G$ in the Galois group, and let $M$ be its fixed field. By the fundamental theorem, $L/M$ is Galois with Galois group isomorphic to $G$.

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