Galois group of $x^7-3$

Let \[ \alpha=3^{1/7},\qquad \zeta=\zeta_7. \] The polynomial \(x^7-3\) is Eisenstein at \(3\), so \[ [\mathbb Q(\alpha):\mathbb Q]=7. \] Its splitting field is \[ K=\mathbb Q(\alpha,\zeta). \] Since \[ [\mathbb Q(\zeta):\mathbb Q]=6 \] and the two degrees are coprime, the intersection of \(\mathbb Q(\alpha)\) and \(\mathbb Q(\zeta)\) is \(\mathbb Q\). Hence \[ [K:\mathbb Q]=42. \]

Over \(\mathbb Q(\zeta)\), define \[ \sigma(\alpha)=\zeta\alpha, \qquad \sigma(\zeta)=\zeta; \] then \(\sigma\) has order \(7\). Choose \(3\) as a generator of \((\mathbb Z/7\mathbb Z)^\times\) and define \[ \tau(\alpha)=\alpha, \qquad \tau(\zeta)=\zeta^3; \] then \(\tau\) has order \(6\). They satisfy \[ \tau\sigma\tau^{-1}=\sigma^3. \] Thus \[ \operatorname{Gal}(K/\mathbb Q) \cong C_7\rtimes C_6, \] where \(C_6\cong(\mathbb Z/7\mathbb Z)^\times\) acts faithfully on \(C_7\). Equivalently, \[ \boxed{\operatorname{Gal}(K/\mathbb Q)\cong\operatorname{AGL}_1(\mathbb F_7)}, \] the Frobenius group of order \(42\).