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PostPosted: 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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PostPosted: Mon Jan 18, 2016 3:42 am 

Joined: Mon Nov 09, 2015 11:52 am
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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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