Let \(f(x) \in \mathbb{Z}[x]\) be an irreducible polynomial of degree \(n \ge 2\), and let \(K/\mathbb{Q}\) be its splitting field with \(\operatorname{Gal}(K/\mathbb{Q}) \cong S_n\). What algebraic, geometric, and field-theoretic properties must the roots \(\alpha_1, \alpha_2, \dots, \alpha_n \in \mathbb{C}\) of \(f(x)\) satisfy?
Solution
Let \(K\) be the splitting field and identify \(G=\operatorname{Gal}(K/\mathbb Q)\) with the full permutation group \(S_n\) on the roots.
Since the characteristic is \(0\), the roots \(\alpha_1,\dots,\alpha_n\) are distinct. Irreducibility gives transitivity, hence each root has degree \[ [\mathbb Q(\alpha_i):\mathbb Q]=n. \] More generally, after ordering the roots, the pointwise stabilizer of \(\alpha_1,\dots,\alpha_k\) is naturally \(S_{n-k}\). By Galois correspondence, \[ [\mathbb Q(\alpha_1,\dots,\alpha_k):\mathbb Q] =[S_n:S_{n-k}]=\frac{n!}{(n-k)!} \] for \(0\le k\le n-1\); adjoining the last root does not enlarge the field. In particular \([K:\mathbb Q]=n!\).
Every polynomial relation over \(\mathbb Q\) among the ordered roots is transported to another valid relation by every permutation in \(S_n\): if \(P(\alpha_1,\dots,\alpha_n)=0\), then \[ P(\alpha_{\sigma(1)},\dots,\alpha_{\sigma(n)})=0 \quad(\sigma\in S_n). \] Thus the ideal of polynomials \(P\in\mathbb Q[x_1,\dots,x_n]\) with \(P(\alpha_1,\dots,\alpha_n)=0\) is stable under permuting the variables.
The discriminant \(\Delta(f)\) is not a square in \(\mathbb Q\), because otherwise the Galois group would lie in \(A_n\). For \(n\ge3\), the unique index-two subgroup of \(S_n\) is \(A_n\), so the unique quadratic intermediate field is \[ \mathbb Q(\sqrt{\Delta(f)}). \] For \(n=2\), \(K=\mathbb Q(\sqrt{\Delta(f)})\).
Finally, complex conjugation is an element of \(G\); as a permutation of the roots it fixes the real roots and swaps each nonreal conjugate pair. Thus its cycle type records exactly the number of real and nonreal roots.