Assume that \(p\) is prime and \(X^p - a\) is irreducible in \(\mathbb{Q}[X]\). Show that the Galois group of \(X^p - a\) over \(\mathbb{Q}\) is isomorphic to the group (with respect to composition) of all functions \(f : \mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}\) such that for some \(k, l \in \mathbb{Z}/p\mathbb{Z}\) with \(k \neq 0\), \[f(y) = ky + l \text{ for all } y \in \mathbb{Z}/p\mathbb{Z}.\]
Solution
Choose a root \(\alpha\in\mathbb C\) of \(X^p-a\) and a primitive \(p\)th root of unity \(\zeta\). Irreducibility implies \(a\ne0\), since \(X^p\) is reducible for \(p\geq2\).
1. The splitting field is \(K=\mathbb Q(\alpha,\zeta)\) and\([K:\mathbb Q]=p(p-1)\).
The roots are the \(p\) distinct numbers \(\alpha\zeta^j\) for \(j\in\mathbb Z/p\mathbb Z\). They lie in \(K\), and the field they generate contains \(\alpha\) and \(\zeta=(\alpha\zeta)/\alpha\). Thus \(K\) is exactly the splitting field.
Let \(F=\mathbb Q(\zeta)\). The polynomial \[ \Phi_p(X)=1+X+\cdots+X^{p-1} \] vanishes at \(\zeta\). Its translate is \[ \Phi_p(Y+1)=\frac{(Y+1)^p-1}{Y} =\sum_{j=1}^p\binom pj Y^{j-1}. \] Every coefficient other than the leading coefficient is divisible by \(p\), and the constant coefficient is \(p\), not divisible by \(p^2\). Eisenstein’s criterion [@DF04] and invertibility of translation show that \(\Phi_p\) is irreducible. Hence \([F:\mathbb Q]=p-1\).
Put \(d=[K:F]\). Since \(K=F(\alpha)\) and \(\alpha\) satisfies a degree-\(p\) polynomial over \(F\), we have \(1\leq d\leq p\). The original irreducibility gives \([\mathbb Q(\alpha):\mathbb Q]=p\). Applying the tower law through this subfield and through \(F\) yields \[ p\mid[K:\mathbb Q]=(p-1)d. \] As \(p\) is relatively prime to \(p-1\), it divides \(d\). The bounds on \(d\) force \(d=p\), giving the asserted total degree. For \(p=2\) the cyclotomic field is \(\mathbb Q\) and the same argument applies.
2. Every Galois automorphism acts affinely on the root labels.
Let \(\sigma\in\operatorname{Gal}(K/\mathbb Q)\). It preserves the multiplicative order of \(\zeta\), so \(\sigma(\zeta)=\zeta^k\) for a unique \(k\in(\mathbb Z/p\mathbb Z)^\times\). It sends \(\alpha\) to a root of \(X^p-a\), so \(\sigma(\alpha)=\alpha\zeta^l\) for a unique \(l\in\mathbb Z/p\mathbb Z\). Consequently \[ \sigma(\alpha\zeta^j)=\alpha\zeta^{kj+l}. \] Thus the permutation of the labels is \(j\mapsto kj+l\). The root action is a homomorphism into the permutation group, and is faithful because the roots generate \(K\).
3. Every affine permutation occurs, proving the isomorphism.
The affine maps form a group: composition and inversion are \[ f_{k,l}\circ f_{u,v}=f_{ku,kv+l},\qquad f_{k,l}^{-1}=f_{k^{-1},-k^{-1}l}. \] Their number is \(p(p-1)\), because \(l=f(0)\) and \(k=f(1)-f(0)\) determine the map uniquely, with \(p\) choices for \(l\) and \(p-1\) nonzero choices for \(k\).
1.The injective homomorphism in step 2 is therefore a bijectiononto this affine group, as required.