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Let $a = \sqrt{\sqrt2 + \sqrt3}$ Is $\mathbb{Q}(a)$ a normal extension of $\mathbb{Q}$?

I thought the answer was no because the minimum polynomial of $a$ is $x^8 -10x^4 + 1$ so $[\mathbb{Q}(a):\mathbb{Q}] = 8$ and I read a similar question here where they say that is a normal extension iff the order of the Galois group is 4, but wouldn't this mean the order of the Galois group is 8? Also, wouldn't an extension by $a$ not include anything imaginary and therefore only half of the other roots. I would almost definitely conclude that this extension is not normal, but part b of the question is to describe the Galois group $\operatorname{Gal}(\mathbb{Q}(a)/\mathbb{Q})$ up to isomorphism, which would mean that the extension would have to be normal

user558377
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1 Answers1

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The normal closure of $\Bbb Q(\sqrt{\sqrt2+\sqrt3})$ contains a square root $\alpha$ of $\sqrt2-\sqrt3$. As $\alpha^2<0$, $\alpha\notin\Bbb R$. As $\Bbb Q(\sqrt{\sqrt2+\sqrt3})\subseteq\Bbb R$, then $\Bbb Q(\sqrt{\sqrt2+\sqrt3})$ is not a normal extension of $\Bbb Q$.

Angina Seng
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  • then how can I do part b? If the extension isn't normal, its not Galois? – user558377 May 11 '18 at 17:23
  • Then how about finding the Galois group of the normal closure? @user558377 – Angina Seng May 11 '18 at 17:29
  • So I should find the splitting field $E$ of $f = x^8 -10x^4 +1$ and then find $Gal(E/\mathbb{Q})?$ Is that still the same question? – user558377 May 11 '18 at 17:35
  • So I know that the roots of $f= x^8 -10x^4 + 1$ are $\pm \sqrt{\pm \sqrt{5 \pm 2\sqrt6}}$ So would the normal closure be $\mathbb{Q}(a, i)$ or am I missing something that takes care of the root $\sqrt {\sqrt{5 - 2\sqrt6}}$? – user558377 May 11 '18 at 17:44
  • There's a relation between the square root of $\sqrt3+\sqrt2$ and the square root of $\sqrt3-\sqrt2$. @user558377 – Angina Seng May 11 '18 at 17:59
  • In other words, would the normal closure be $\mathbb{Q}(\sqrt[4]{5 + 2\sqrt6}, \sqrt[4]{5 - 2\sqrt6}, i)$ or $\mathbb{Q}(\sqrt[4]{5 + 2\sqrt6}, i)$? – user558377 May 11 '18 at 17:59
  • Sorry I think we were writing our comments at the same time. So the answer is $E = \mathbb{Q}(a = \sqrt[4]{5 + 2\sqrt6}, i)$, which would mean $\operatorname{Gal}(E/\mathbb{Q}) = [E:\mathbb{Q}] = [\mathbb{Q}(a, i): \mathbb{Q}(a)][\mathbb{Q}(a):\mathbb{Q}] = 2 \times 8 = 16$ so the Galois group is isomorphic to some subgroup of $S_8$ of order 16? – user558377 May 11 '18 at 18:05