Example 5

Let $F = K(X_1, ..., X_n)$ be the field of rational fractions in $n$ indeterminates $X_1, ..., X_n$ with coefficients in $K$. Put

for $1 \leq k \leq n$ and $E = K(s_1, ..., s_n)$; we denote by $f(T)$ the polynomial We have $f(T) = \prod_{i=1}^n (T - X_i)$, so that $F$ is a splitting field of the separable polynomial $f(T) \in E[T]$. Further, for every permutation $a \in \mathfrak{S}_n$ there exists a unique $K$-automorphism $h_a$ of $F$ such that $h_a(X_i) = X_{a(i)}$ for $1 \leq i \leq n$; we have $h_a(s_k) = s_k$ for $1 \leq k \leq n$, hence $h_a$ is an $E$-automorphism of $F$. In other words, $F$ is a Galois extension of $E$ and the restriction to the set of roots $\{ X_1, ..., X_n \}$ of $f(T)$ defines an isomorphism of $\mathrm{Gal}(F/E)$ onto the group $\mathfrak{S}_n$. In particular, $E$ consists of rational fractions $f$ such that for every $\sigma \in \mathfrak{S}_n$ (cf. IV, p. 67, Cor.).