Exercise 12.1.16

Proof.

$\text{∙}$

Suppose that $\psi \in F(x_1,\dots ,x_n)$ is invariant under $S_n$.

Let $\phi =1$. Then $\phi$ is invariant under $S_n$, so $\psi$ is fixed by every permutation fixing $\phi$. By Theorem 12.1.6. $\psi$ is a rational function of $\phi$ with coefficients in $K=F({\sigma }_1,\dots ,{\sigma }_n)$, i.e., $\psi \in K(\phi )=K(1)=K$. So $\psi \in F({\sigma }_1,\dots ,{\sigma }_n)$.

$\text{∙}$

Suppose that $\psi \in F(x_1,\dots ,x_n)$ is invariant under $A_n$. Let $\phi =\sqrt{\mathrm{\Delta }}$. As the characteristic is not 2, by Proposition 2.4.1, $\sigma \cdot \sqrt{\mathrm{\Delta }}=\sqrt{\mathrm{\Delta }}$ if and only if $\sigma \in A_n$, so $H(\phi )=H(\sqrt{\mathrm{\Delta }})=A_n$. Thus $\psi$ is fixed by every permutation fixing $\phi$.

By Theorem 12.1.6. $\psi$ is a rational function of $\phi =\sqrt{\mathrm{\Delta }}$ with coefficients in $K=F({\sigma }_1,\dots ,{\sigma }_n)$, so $\psi \in K(\sqrt{\mathrm{\Delta }})$.

$\sqrt{\mathrm{\Delta }}\notin K$, because $\tau \cdot \sqrt{\mathrm{\Delta }}=-\sqrt{\mathrm{\Delta }}\ne \sqrt{\mathrm{\Delta }}$ for every transposition $\tau$. Therefore $K\subset K(\sqrt{\mathrm{\Delta }})$ is a quadratic extension, and $(1,\sqrt{\mathrm{\Delta }})$ is a basis of $K(\sqrt{\mathrm{\Delta }})$ over $K$. Therefore

$$\psi =A+B\sqrt{\mathrm{\Delta }},A,B\in K=F({\sigma }_1,\dots ,{\sigma }_n).$$

So Theorem 7.4.4 follows from Theorem 12.1.6.