Exercise 6.5.2

Proof. As in Example 6.5.1, where

let ${𝜃}_1(x)=x,{𝜃}_2(x)=-x,{𝜃}_3(x)=10x-x^3,{𝜃}_4(x)=-10x+x^3$, so the solutions of the equation are ${\alpha }_i={𝜃}_i(\alpha ),i=1,2,3,4$.

The roots of $f$ being polynomials in $\alpha$, the splitting field of $f$ is $F(\alpha )$ (See Exercise 1).

Moreover, as ${𝜃}_1=x,{𝜃}_2=-x,{𝜃}_4=-{𝜃}_3$ and ${𝜃}_3,{𝜃}_4$ are odd functions,

${𝜃}_1({𝜃}_i(\alpha ))={𝜃}_i(\alpha )={𝜃}_i({𝜃}_1(\alpha )),i=2,3,4$.

${𝜃}_2({𝜃}_i(\alpha ))=-{𝜃}_i(\alpha )={𝜃}_i(-\alpha )={𝜃}_i({𝜃}_2(\alpha )),i=3,4$.

${𝜃}_3({𝜃}_4(\alpha ))={𝜃}_3(-{𝜃}_3(\alpha ))=-{𝜃}_3^2(\alpha )=-{𝜃}_4^2(\alpha )={𝜃}_4(-{𝜃}_4(\alpha ))={𝜃}_4({𝜃}_3(\alpha ))$.

Thus ${𝜃}_i({𝜃}_j(\alpha ))={𝜃}_j({𝜃}_i(\alpha ))$, for $1\le i<j\le 4$, thus also for $1\le i,j\le 4$.