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Let $K_1$ be a perfect field. Let $K_2/K_1$ and $K_3/K_2$ be quadratic extensions. Let $K_4/K_3$ be the Galois closure of $K_3$ over $K_1$. Is it true that either $K_3 = K_4$ or $K_4/K_3$ is quadratic such that the Galois group of $K_4$ over $K_1$ is isomorphic to $\mathcal{D}_4$, the dihedral group with 8 elements?

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And if you really want to know, I am trying to do some nasty calculations with Galois cohomology for non-Galois extensions; I will have to specialize to specific Galois groups of the Galois closure at some point, and I just hope that the only non-commutative case I will have to do is $\mathcal{D}_4$, and not the quaternion group, $\mathcal{A}_4$ or $\mathcal{S}_4$. But this doesn't add much to the question. –  Wanderer Oct 28 '11 at 23:34
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Hints: (1) The Galois group of the splitting field of a degree four extension is a subgroup of $S_4$. (2) Think of a group having a subgroup of index $2$ that in turn has a subgroup of index $2$. –  Tom Goodwillie Oct 29 '11 at 0:45
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Thanks for the link. About the question itself: the question was voted up five times. I can find the answer in a research article. Hence I can only conclude that this is an OK question for Mathoverflow... –  Wanderer Oct 29 '11 at 12:35
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Guys, could you just give the question the benefit of the doubt. It could be given as graduate-level homework, but so could a lot of things. And citing a paper from only 20 years ago as an answer --- that's a good way to answer questions in MO, and not a good way to argue that they should be closed. –  Greg Kuperberg Oct 29 '11 at 15:56
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@Franz I understand that if you work in number theory or finite group theory, then the question can be regarded as trivial. Or even that it should be called trivial. But I think it was Grothendieck's view that the best kind of mathematics is a sum of interesting trivialities. (I also understand that the result in question is really due to Galois, not Waterhouse, but still.) –  Greg Kuperberg Oct 29 '11 at 19:42

1 Answer 1

up vote 10 down vote accepted

Yes. Let $a_1$ be a generator of $K_3$ over $K_1$ and let $a_2$, $a_3$, and $a_4$ be the other three roots of the polynomial of $a_1$. Then say that $a_2$ is the other root of the polynomial of $a_1$ over $K_2$. Then The Galois group of $K_4/K_1$ acts on the partitioned set $\{\{a_1,a_2\},\{a_3,a_4\}\}$, and is therefore a transitive subgroup of $D_4$, either $D_4$ itself or $C_4$ or $C_2 \times C_2$.

I think that the following is a more general fact that can be proven in the same way. If $K_2/K_1$ is separable and its Galois closure has Galois group $G$, and if $K_3/K_2$ is separable and its Galois closure has Galois group $H$, then the Galois group of the Galois closure of $K_3/K_1$ is a subgroup of a wreath product of $G$ and $H$. If you take a longer chain of extensions you get an iterated wreath product.

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By the Galois group of $K_3/K_1$ you actually mean the Galois group of the Galois closure $K_4/K_1$? –  Wanderer Oct 29 '11 at 11:21
    
Yes in the first instance I said it slightly wrong. I will fix it. –  Greg Kuperberg Oct 29 '11 at 15:51

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