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Find the Galois group of $f(X) = X^4 + 2X^2+4$ over $\mathbb{Q}$.

Let $L$ be the splitting field of $f$ over $\mathbb{Q}$. Finding the roots of this polynomial, I got $$X^2 = \frac{-2\pm \sqrt{4-16}}{2} = -1 \pm \sqrt{3}i$$ so the roots are $\alpha_1 = \sqrt{-1+\sqrt{3}i}, \alpha_2 =\sqrt{-1-\sqrt{3}}i, \alpha_3 = -\sqrt{-1+\sqrt{3}i}$ and $ \alpha_4 -\sqrt{-1-\sqrt{3}}i$. Now $L = \mathbb{Q}(\alpha_1, \alpha_2)$, and since $$\alpha_1 \alpha_2 = \sqrt{(-1 + \sqrt{3}i)(-1-\sqrt{3}i)} = \sqrt{1+3} = 2$$ we see that $L = \mathbb{Q}(\alpha_1) = \mathbb{Q}(\alpha_2)$. It's not difficult to see that $[\mathbb{Q}(\alpha_1) : \mathbb{Q}] = 4$, so the Galois group of $L/\mathbb{Q}$ over $\mathbb{Q}$ is is either $\mathbb{Z}/4\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

My guess is that this Galois group is cyclic, since I think that $\mathbb{Q}(\sqrt{3}i)$ (a cyclotomic extension by a cubic root of unity) is the only intermediate subfield of $L/\mathbb{Q}$. How can I know for sure?

D_S
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  • The question has no sense ! It should be "find the Galois group $Gal(E/\Bbb Q)$ where $E$ is the splitting field of $f(X)\in\Bbb Q[X]$" – idm Jan 07 '15 at 00:48
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    The Galois group of a polynomial is a well-defined notion, which is exactly the one that you suggest by the way. I am not expert in notation but it seems that Galois theory was made first for polynomials and then for field extensions. http://en.wikipedia.org/wiki/Galois_theory – Jérémy Blanc Jan 07 '15 at 00:56
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    @idm That's the definition of "Galois group of $f(X)$" – Marco Flores Jan 07 '15 at 00:56
  • Related: https://math.stackexchange.com/questions/204709 – Watson Dec 26 '16 at 13:03

1 Answers1

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You can also write the solutions in the following way: $$\frac{1}{2}\left(\pm \sqrt{2}\pm \sqrt{6}i\right)$$ and then find that $L=\mathbb{Q}[\sqrt{2},\sqrt{3}i]$. The Galois group is then $(\mathbb{Z}/2\mathbb{Z})^2$.

One generator corresponds to the conjugation $z\mapsto \bar{z}$ and the second is $\alpha\mapsto -\alpha$, where $\alpha$ is one of your roots; this is an automorphism since $L=\mathbb{Q}[\alpha]$ and because $-\alpha$ is also a root.

Jérémy Blanc
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  • Could you explain the first step please? About how to write the roots in this new form? – Sigurd Apr 19 '20 at 18:37
  • Does this work in more generality? I am trying to do a similar trick for $\sqrt{1+i\sqrt{3}}$. – Sigurd Apr 19 '20 at 18:56
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    @Sigurd You can simply check that the solutions are like this by writing $\prod_{i=1}^4(X-a_i)$, where $a_1,a_2,a_3,a_4$ are the four roots I suggested, and develop it. – Jérémy Blanc Apr 22 '20 at 14:40
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    with $\sqrt{1+i\sqrt{3}}$ it works similarly with $\frac{1}{2}(\pm \sqrt{6}\pm \sqrt{2}i)$ – Jérémy Blanc Apr 22 '20 at 14:48
  • From this, I get the system of equations $$ \begin{cases} -s_1 = -a_1-a_2-a_3-a_4 = 0, \ s_2 = a_1a_2 + a_1a_3 + a_1a_4 + a_2a_3 + a_2a_4 + a_3a_4 = 2, \ -s_3 = -a_1a_2a_3 - a_1a_2a_4 - a_2a_3a_4 = 0, \ s_4 = a_1a_2a_3a_4 = 4. \end{cases} $$ I still don't see how to extract the formulas for the roots from this, maybe I'm thinking in the wrong direction and overlooking the easy solution? – Sigurd Apr 22 '20 at 15:00
  • I can see the roots you suggested are indeed correct, but I wonder what is the method for finding them? – Sigurd Apr 22 '20 at 15:05
  • Btw, the polynomial that I am working with in this case is $f = X^4 -2X^2+4$. – Sigurd Apr 22 '20 at 15:31