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The question is asking how to prove (not necessarily in high detail but concisely) that the Galois Group (the group of Q-automorphisms of F where Q is the base field and F is a field extension of the base field) of a radical field extension (an extension of a field K that is obtained by adjoining a sequence of nth roots - radicals - of elements) is always soluble/solvable (a group which has a normal series such that each normal factor is abelian).

Thank you

p.s. I am only 17 so please don't make your answer to complicated - cheers.

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  • $\begingroup$ Please make the question body self-contained so the title is not necessary to understand the question. Also, writing up what definition of radical extension and solvable group you are familiar with will help people give useful answers. $\endgroup$ – Tobias Kildetoft Apr 12 '16 at 8:22
  • $\begingroup$ Doing that now. $\endgroup$ – Daniele1234 Apr 12 '16 at 8:25
  • $\begingroup$ Done. random stuff to get 15 characters. $\endgroup$ – Daniele1234 Apr 12 '16 at 8:32
  • $\begingroup$ Geoff Robertson has written a sketch solution, but my immediate reaction is that this is a standard textbook result, and you would do better to read a book about it rather than trying to do it as an exercise. Lots of of people recommend Ian Stewart's book on Galois Theory for example. $\endgroup$ – Derek Holt Apr 12 '16 at 10:52
  • $\begingroup$ @Derek Holt : It's Geoff Robinson! I certainly agree that it's better to read a proper text book. I responded because the question has a history on Mathoverflow too- although Daniele's age seems to fluctuate somewhat! $\endgroup$ – Geoff Robinson Apr 12 '16 at 11:07
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Here is a rough sketch: You use induction on the degree of the field extension, and it is useful to allow the base field to vary. So we need to consider the case that $F = L[a^{\frac{1}{n}}]$ for some $a$, and some intermediate field $L$, itself a radical extension. Set $b = a^{\frac{1}{n}}$, a particular fixed choice of $n$-th root of $a \in L$. Any field automorphism $\alpha$ of $F$ which fixes $L$ ( ie $\alpha \in {\rm Gal}(F/L) )$ must send $b$ to another $n$-th root of $a$ and this must have the form $\omega b$ for some $n$-th root of unity $\omega$ and we have $\omega \in F$. It follows from this that ${\rm Gal}(F/L)$ is Abelian ( it is isomorphic to a group roots of unity under multiplication, which is certainly Abelian). Now the basic idea is to show that ${\rm Gal}(F/Q)/{\rm Gal}(L/Q)$ is isomorphic to ${\rm Gal}(F/L)$, and to use inductive assumption that ${\rm Gal}(L/Q)$ is solvable to conclude that ${\rm Gal}(F/Q)$ is solvable. However, there are quite a lot of technicalities needed to make the induction work, and it is usual to work with separable normal intermediate extensions, to ensure that they are left invariant under enough automorphisms ( for example, if you don't make some assumptions, it might be that an automorphism of $F$ which fixes $Q$ does not send elements of $L$ back into $L$: eg, think about the case that $F = Q[\omega,2^{\frac{1}{3}}]$ where $\omega$ is a complex cube root of unity and the cube root of $2$ is the real one. Take $L$ to be $F \cap \mathbb{R}$. Then any automorphism of $F$ which does not fix the real cube root of $2$ must send that cube root somewhere outside $L$).

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