Prove that the cyclotomic extension \(\mathbb{Q}(\zeta_{43})/\mathbb{Q}\) has degree 42 and that its Galois group is cyclic of order 42: \[\operatorname{Gal}(\mathbb{Q}(\zeta_{43})/\mathbb{Q}) \cong (\mathbb{Z}/43\mathbb{Z})^\times \cong \mathbb{Z}/42\mathbb{Z}.\]
Solution
For the prime \(43\), \[ \Phi_{43}(x)=1+x+\cdots+x^{42}. \] The shifted polynomial \(\Phi_{43}(x+1)\) is Eisenstein at \(43\), so \(\Phi_{43}\) is irreducible over \(\mathbb Q\). Hence \[ [\mathbb Q(\zeta_{43}):\mathbb Q]=\deg\Phi_{43}=42. \]
The field \(\mathbb Q(\zeta_{43})\) is the splitting field of \(\Phi_{43}\), hence is Galois. Every automorphism is determined by \[ \zeta_{43}\longmapsto \zeta_{43}^a, \qquad a\in(\mathbb Z/43\mathbb Z)^\times, \] and every such \(a\) occurs. Therefore \[ \operatorname{Gal}(\mathbb Q(\zeta_{43})/\mathbb Q) \cong(\mathbb Z/43\mathbb Z)^\times. \] Since the multiplicative group of a finite field is cyclic, \[ (\mathbb Z/43\mathbb Z)^\times\cong C_{42}. \] Thus the extension has degree \(42\) and cyclic Galois group of order \(42\).