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I was reading about solvability of quintics by radicals, but unfortunately there were no many examples and I am afraid that I do not understand the whole concept. How to show $x^5-3$ is solvable by radicals over $\mathbb{Q}$?

stella
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    What about inserting $3^{1/5}$ and observing that you get $0$? – celtschk Jul 06 '13 at 22:51
  • @celtschk The OP asked if it is solvable over $\mathbb{Q}$. I'm pretty sure that $\sqrt[5]{3}$ is not a rational number. In fact, I'm quite sure that $x^5-3$ is irreducible over $\mathbb{Q}$. – Fly by Night Jul 06 '13 at 23:31
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    But in the title and first line it is written 'solvable by radicals'. – Berci Jul 06 '13 at 23:35
  • Yes, by radicals. I am not sure, but I had in mind something like showing that the Galois Group is one of solvable subgroups of S5. Or it might be simpler in this case. – stella Jul 06 '13 at 23:36
  • @Berci ...then the next two words are "over $\mathbb{Q}$". – Fly by Night Jul 06 '13 at 23:47
  • @sansa1213 There seems to be some misunderstanding arising from your post. What exactly would you like us to show? – Fly by Night Jul 06 '13 at 23:48
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    There is a theorem that a polynomial can be solved by radicals if and only if its Galois group is a solvable group. I just was not sure that it works for this example, but the following answer shows that it works. Its Galois group is F20 - Frobenius group of order 20, which is solvable. – stella Jul 07 '13 at 00:01
  • @sansa1213 I've added to my post. You should now be able to find the roots explicitly. – Fly by Night Jul 07 '13 at 00:18
  • @FlybyNight, thanks a lot. – stella Jul 07 '13 at 00:23

1 Answers1

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The splitting field $L =\mathbb{Q}[x]/(x^5-3)$ has 20 elements, and the Galois Group is

$$\text{Gal}(L\backslash\mathbb{Q}) \cong \langle (1,2)(3,4),(2,3,4,5)\rangle$$

Addition:

You can find the roots by using de Moivre's theorem. They are $\sqrt[5]{3}(\cos(\frac{2\pi n}{5})+\operatorname{i}\sin(\frac{2\pi n}{5})).$

Well-known formulae tell us that $\sin(\frac{\pi}{5}) = \frac{1}{4}(\sqrt{10-2\sqrt{5}})$ and $\cos(\frac{\pi}{5}) = \frac{1}{4}(1+\sqrt{5}).$

Using the formula $\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha$ and $\cos(\alpha\pm\beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta$ will get you closed form expressions for all of the roots.

Fly by Night
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