Welcome to mathimatikoi.org forum; Enjoy your visit here.
Galois group of $x^42x^2+9$
 Grigorios Kostakos
 Founder
 Articles: 0
 Posts: 460
 Joined: Mon Nov 09, 2015 1:36 am
 Location: Ioannina, Greece
Galois group of $x^42x^2+9$
Find the Galois group of polynomial $x^42x^2+9$ over $\mathbb{Q}$.
Grigorios Kostakos

 Former Team Member
 Articles: 0
 Posts: 77
 Joined: Mon Nov 09, 2015 11:52 am
 Location: Limassol/Pyla Cyprus
 Contact:
Re: Galois group of $x^42x^2+9$
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 4group. 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 nontrivial 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 \(V_4\).
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 4group. 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 nontrivial 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 \(V_4\).