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 $G$ 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)

share|improve this question
    
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
1  
$L = M = K\phantom{}$? –  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
show 6 more comments

2 Answers

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.

share|improve this answer
add comment

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 $G$.

share|improve this answer
add comment

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.