What is the Galois group of \(x^7 - 1\) over the rationals?
Solution
The roots of \(x^7-1\) are the seventh roots of unity, so its splitting field over \(\mathbb Q\) is \[ K=\mathbb Q(\zeta_7), \] where \(\zeta_7\) is primitive. Since \[ \Phi_7(x)=x^6+x^5+\cdots+x+1 \] is irreducible over \(\mathbb Q\), we have \([K:\mathbb Q]=6\).
Every \(\mathbb Q\)-automorphism of \(K\) is determined by \[ \zeta_7\longmapsto \zeta_7^a, \qquad a\in(\mathbb Z/7\mathbb Z)^\times, \] and every such choice gives an automorphism. Hence \[ \operatorname{Gal}(K/\mathbb Q) \cong(\mathbb Z/7\mathbb Z)^\times. \] The latter group is cyclic of order \(6\), so \[ \operatorname{Gal}(x^7-1/\mathbb Q)\cong C_6. \]