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 Post subject: Galois group of $x^4-2x^2+9$Posted: Mon Jan 18, 2016 3:41 am
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Find the Galois group of polynomial $x^4-2x^2+9$ over $\mathbb{Q}$.

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Grigorios Kostakos

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 Post subject: Re: Galois group of $x^4-2x^2+9$Posted: Mon Jan 18, 2016 3:42 am

Joined: Mon Nov 09, 2015 11:52 am
Posts: 76
Location: Limassol/Pyla Cyprus
Let us denote the polynomial by $f$. It is not too difficult to see that the roots of $f$ are $\pm \sqrt{2} \pm i$. It follows that the splitting field $K$ of $f$ is contained in $\mathbb{Q}(i,\sqrt{2})$ and in fact it must be equal to it. Indeed since $\sqrt{2} = \frac{1}{2}((\sqrt{2} + i) + (\sqrt{2}-i))$ then $\sqrt{2}$ must belong to $K$ and a similar calculation shows that $i \in K$ as well.

We have $[K: \mathbb{Q}] = [K: \mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}): \mathbb{Q}] = 2 \cdot 2 = 4$. So the Galois group of $f$ must have order $4$ and so it must be either $C_4$ or $V_4$, the Klein 4-group. Because the extension has at least three intermediate fields, namely $\mathbb{Q}(\sqrt{2}),\mathbb{Q}(i)$ and $\mathbb{Q}(i\sqrt{2})$ by the fundamental theorem of Galois theory it follows that the Galois group must have at least 3 non-trivial subgroups and thus it cannot be $C_4$ which has only one such subgroup. So the Galois group must be $V_4$.

Alternatively, without using the fundamental theorem, we can observe that the Galois group must contain the automorphisms $\sigma$ and $\tau$ generated by $\sigma(i) = i,\sigma(\sqrt{2}) = -\sqrt{2}$ and $\tau(i) = -i,\tau(\sqrt{2}) = \sqrt{2}$ which generate $V_4$.

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