Describe the Galois theory of \(f(x) = x^{2n} - 2\) over \(\mathbb{Q}\): irreducibility, splitting field, and Galois group structure.
Solution
Put \(m=2n\) and \(\alpha=2^{1/m}\). By Eisenstein at \(2\), \[ x^m-2 \] is irreducible over \(\mathbb Q\), so \([\mathbb Q(\alpha):\mathbb Q]=m\). Its roots are \[ \alpha\zeta_m^j,\qquad j\in\mathbb Z/m\mathbb Z, \] where \(\zeta_m\) is a primitive \(m\)th root of unity. Hence the splitting field is \[ K=\mathbb Q(\alpha,\zeta_m). \]
Let \[ E=\mathbb Q(\alpha),\qquad F=\mathbb Q(\zeta_m). \] Then \[ [E:\mathbb Q]=m,\qquad [F:\mathbb Q]=\varphi(m), \] and therefore \[ [K:\mathbb Q] =\frac{m\varphi(m)}{[E\cap F:\mathbb Q]}. \]
Every \(\sigma\in\operatorname{Gal}(K/\mathbb Q)\) is determined by \[ \sigma(\alpha)=\alpha\zeta_m^b,\qquad \sigma(\zeta_m)=\zeta_m^a, \] with \(b\in\mathbb Z/m\mathbb Z\) and \(a\in(\mathbb Z/m\mathbb Z)^\times\). Thus the Galois group embeds in the affine group \[ \mathbb Z/m\mathbb Z\rtimes(\mathbb Z/m\mathbb Z)^\times, \] where \((a,b)\) acts on the root index \(j\) by \(j\mapsto aj+b\). The affine group has order \(m\varphi(m)\), so the image has index \([E\cap F:\mathbb Q]\) in it.
Over \(F\), the extension \(K/F\) is a cyclic Kummer extension of degree dividing \(m\), generated by adjoining an \(m\)th root of \(2\). Hence \(\operatorname{Gal}(K/\mathbb Q)\) is solvable: its normal subgroup \(\operatorname{Gal}(K/F)\) is cyclic, with quotient the abelian group \[ \operatorname{Gal}(F/\mathbb Q)\cong(\mathbb Z/m\mathbb Z)^\times. \]
For example, \(m=2\) gives \(K=\mathbb Q(\sqrt2)\) with Galois group \(C_2\), while \(m=4\) gives \(K=\mathbb Q(\sqrt[4]2,i)\) with Galois group \(D_4\) of order \(8\).