Let \(f(x) = x^4 - 4x^2 + 2 \in \QQ[x]\).
Find the splitting field \(K\) of \(f\), and compute \([K: \QQ]\).
Find the Galois group \(G\) of \(f\), both as an explicit group of automorphisms, and as a familiar abstract group to which it is isomorphic.
Exhibit explicitly the correspondence between subgroups of \(G\) and intermediate fields between \(\QQ\) and \(k\).
Not the nicest proof! Would be better to replace the ad-hoc computations at the end.
Solution
(of a)
Note that \(g(x) = x^2 - 4x + 2\) has roots \(\beta = 2 \pm \sqrt{2}\), and so \(f\) has roots \[ \alpha_1 &= \sqrt{2 + \sqrt 2} \\ \alpha_2 &= \sqrt{2 - \sqrt 2} \\ \alpha_3 &= -\alpha_1 \\ \alpha_4 &= -\alpha_2 .\]
and splitting field \(K = \QQ(\theset{\alpha_i})\).
(of b)
\(K\) is the splitting field of a separable polynomial and thus Galois over \(\QQ\). Moreover, Since \(f\) is irreducible by Eisenstein with \(p=2\), the Galois group is a transitive subgroup of \(S^4\), so the possibilities are:
\(S_4\)
\(A_4\)
\(D_4\)
\(\ZZ/(2) \cross \ZZ/(2)\)
\(\ZZ/(4)\)
We can note that \(g\) splits over \(L \definedas \QQ(\sqrt 2)\), an extension of degree 2.
We can now note that \(\min(\alpha, L)\) is given by \(p(x) = x^2 - (2 + \sqrt 2)\), and so \([K: L] = 2\).
We then have \[ [K: \QQ] = [K: L] [L : \QQ] = (2)(2) = 4 .\]
This \(\abs{\Gal(K/\QQ)} = 4\), which leaves only two possibilities:
\(\ZZ/(2) \cross \ZZ/(2)\)
\(\ZZ/(4)\)
We can next check orders of elements. Take \[ \sigma &\in \Gal(K/\QQ) \\ \alpha_1 &\mapsto \alpha_2 .\]
Computations show that
\(\alpha_1^2 \alpha_2^2 = 2\), so \(\alpha_1 \alpha_2 = \sqrt 2\)
\(\alpha_1^2 = 2 + \sqrt 2 \implies \sqrt 2 = \alpha_1^2 - 2\)
and thus \[ \sigma^2(\alpha_1) &= \sigma(\alpha_2) \\ &= \sigma\left(\frac{\sqrt 2}{\alpha_1}\right) \\ &= \frac{\sigma(\sqrt 2)}{\sigma(\alpha_1)} \\ &= \frac{\sigma(\alpha_1^2 - 2)}{\alpha_2} \\ &= \frac{\alpha_2^2 - 2}{\alpha_2} \\ &= \alpha_2 -2\inverseof{\alpha_2} \\ &= \alpha_2 - \frac{2\alpha_1}{\sqrt 2} \\ &= \alpha_2 -\alpha_1 \sqrt 2 \\ &\neq \alpha_1 ,\]
and so the order of \(\sigma\) is strictly greater than 2, and thus 4, and thus \(\Gal(K/\QQ) = \theset{\sigma^k \suchthat 1\leq k \leq 4} \cong \ZZ/(4)\).
(of c)
By part (b), \(G = \Gal(K/\QQ) \cong \ZZ/4\) is cyclic of order \(4\), generated by \(\sigma\) (with \(\sigma(\alpha_1) = \alpha_2\)). A cyclic group of order \(4\) has exactly three subgroups, forming a chain: \[ \{1\} \subset \generators{\sigma^2} \subset G, \] where \(\generators{\sigma^2}\) has order \(2\) (index \(2\) in \(G\)). By the fundamental theorem of Galois theory, the intermediate fields between \(\QQ\) and \(K\) form the reverse chain: \[ K \supset K^{\generators{\sigma^2}} \supset \QQ, \] with \([K : K^{\generators{\sigma^2}}] = |\generators{\sigma^2}| = 2\) and \([K^{\generators{\sigma^2}} : \QQ] = 2\).
We already know one intermediate field of degree \(2\) over \(\QQ\), namely \(\QQ(\sqrt 2)\) (from part (a), \(g(x) = x^2 - 4x + 2\) splits over \(\QQ(\sqrt 2)\), and \([K : \QQ(\sqrt 2)] = 2\)). Since the subgroup lattice is a chain, there is exactly one intermediate field of each degree, so \(K^{\generators{\sigma^2}} = \QQ(\sqrt 2)\).
Thus the full correspondence is: \[ \{1\} \leftrightarrow K, \qquad \generators{\sigma^2} \leftrightarrow \QQ(\sqrt 2), \qquad G \leftrightarrow \QQ. \]