3.6 The Galois group of a polynomial

If a polynomial $f\in F[X]$ is separable, then its splitting field $F_{f}$ is Galois over $F$, and we call $\operatorname{Gal}(F_{f}/F)$ the Galois group $G_{f}$ of $f.$

Let $f(X)=\prod_{i=1}^{n}(X-\alpha_{i})$ in a splitting field $F_{f}$. We know that the elements of $\operatorname{Gal}(F_{f}/F)$ map roots of $f$ to roots of $f$, i.e., they map the set $\{\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\}$ into itself. Being automorphisms, they act as permutations on $\{\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\}$. As the $\alpha_{i}$ generate $F_{f}$ over $F$, an element of $\operatorname{Gal}(F_{f}/F)$ is uniquely determined by the permutation it defines. Thus $G_{f}$ can be identified with a subset of $\operatorname{Sym}(\{\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\})\approx S_{n}$ (symmetric group on $n$ symbols). In fact, $G_{f}$ consists exactly of the permutations $\sigma$ of $\{\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\}$ such that, for $P\in F[X_{1},\ldots,X_{n}]$,

To see this, note that the kernel of the map

consists of the polynomials $P(X_{1},\ldots,X_{n})$ such that $P(\alpha_{1},\ldots,\alpha_{n})=0$. Let $\sigma$ be a permutation of the $\alpha_{i}$ satisfying the condition (8). Then the map

factors through the map (9), and defines an $F$-isomorphism $F_{f}\rightarrow F_{f}$, i.e., an element of the Galois group. This shows that every permutation satisfying the condition (8) extends uniquely to an element of $G_{f}$, and it is obvious that every element of $G_{f}$ arises in this way.

This gives a description of $G_{f}$ not mentioning fields or abstract groups, neither of which were available to Galois. Note that it shows again that $(G_{f}\colon 1)$, hence $[F_{f}\colon F]$, divides $\deg(f)!.$