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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2$\begingroup$ 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 $2$. $\endgroup$– Tom GoodwillieCommented Oct 29, 2011 at 0:45
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2$\begingroup$ 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... $\endgroup$– WandererCommented Oct 29, 2011 at 12:35
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2$\begingroup$ @Wanderer: I think one 'problem' with your question was really the superficial fact that it was phrased in a typical homework/excercise style. It is true that the context you later gave does not add much to the question itself. Yet, from observing a lot of question, I can asure you that it can make a considerable difference whether some personal motivation is given or not. In some sense one can consider this as strange, but in my experience it is like this. $\endgroup$– user9072Commented Oct 29, 2011 at 14:04
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6$\begingroup$ 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. $\endgroup$– Greg KuperbergCommented Oct 29, 2011 at 15:56
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3$\begingroup$ @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.) $\endgroup$– Greg KuperbergCommented Oct 29, 2011 at 19:42
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1 Answer
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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.
answered Oct 29, 2011 at 0:49
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$\begingroup$ By the Galois group of $K_3/K_1$ you actually mean the Galois group of the Galois closure $K_4/K_1$? $\endgroup$– WandererCommented Oct 29, 2011 at 11:21
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$\begingroup$ Yes in the first instance I said it slightly wrong. I will fix it. $\endgroup$ Commented Oct 29, 2011 at 15:51