Exercise 12.1.9

Proof.

(a)

Here $K=F({\sigma }_1,\dots ,{\sigma }_n),L=F(x_1,\dots ,x_n)$, where $F$ has characteristic $0$.

We know (Theorem 6.4.1) that $K\subset L$ is a Galois extension, and that

$$\psi :\{\begin{array}{ccc}\hfill S_n\hfill & \hfill \to \hfill & \hfill \mathrm{Gal}(L∕K)\hfill \\ \hfill \tau \hfill & \hfill \mapsto \hfill & \hfill \tilde{\tau }\{\begin{array}{ccc}\hfill L\hfill & \hfill \to \hfill & \hfill L\hfill \\ \hfill f\hfill & \hfill \mapsto \hfill & \hfill \tau \cdot f\hfill \end{array}\hfill \end{array}$$

(where $\tau \cdot f(x_1,\dots ,x_n)=f(x_{\tau (1)},x_{\tau (2)},\dots ,x_{\tau (n)})$)

is an isomorphism from $S_n$ to $\mathrm{Gal}(L∕K)$.

Write $\tilde{H}=\psi (H)$ the subgroup of $\mathrm{Gal}(L∕K)$ corresponding to $H\subset S_n$, and $L_{\tilde{H}}$ its fixed field (we can write $L_H=L_{\tilde{H}}$).

$K\subset L$ is a finite extension, and $K\subset L_H\subset L$, so $K\subset L_H$ is a finite extension. Since the characteristic of $F$ is $0$, the Theorem of the Primitive Element (Corollary 5.4.2 (b)) applies to give $\phi \in L_H$ such that $L_H=K(\phi )$.

Since $K\subset L$ is a Galois extension, the Galois correspondence (Theorem 7.3.1) gives

$$\tilde{H}=\mathrm{Gal}(L∕L_{\tilde{H}})=\mathrm{Gal}(L∕K(\phi )).$$

We show that $H=H(\phi )$:

$\text{∙}$

If $\tau \in H$, then $\tilde{\tau }=\psi (\tau )\in \tilde{H}=\mathrm{Gal}(L∕K(\phi ))$. Since $\phi \in K(\phi )$, $\tau \cdot \phi =\tilde{\tau }(\phi )=\phi$, so $\tau \in H(\phi )$.

$\text{∙}$

If $\tau \in H(\phi )$, then $\tau \cdot \phi =\phi$. If $u(x_1,\dots ,x_n)\in K(\phi )$, then $u(x_1,\dots ,x_n)=f(\phi (x_1,\dots ,x_n))$, where $f\in K(x)$. Therefore $$\tau \cdot u(x_1,\dots ,x_n)=f(\phi (x_{\tau (1)},\dots ,x_{\tau (n)})=f(\phi (x_1,\dots ,x_n))=u(x_1,\dots ,x_n),$$

so $\tilde{\tau }(u)=\tau \cdot u=u$ for all $u\in K(\phi )$, thus $\tilde{\tau }\in \mathrm{Gal}(L∕K(\phi ))=\tilde{H}$, and so $\tau \in H$.

Conclusion: if $H$ is a subgroup of $S_n$, there is $\phi \in L$ such that $H=H(\phi )$.

(b)

Let $\phi ={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{\sigma \in H}\sigma \cdot m$, where $m=x_1^{a_1}\cdots x_n^{a_n}$ with distinct exponents $a_1,\dots ,a_n$.

$\text{∙}$

If $\tau \in H$, by (6.7), $$\tau \cdot \phi =\underset{\sigma \in H}{\sum }(\mathit{\tau \sigma })\cdot m=\underset{{\sigma }^{\prime }\in H}{\sum }{\sigma }^{\prime }\cdot m=\phi ({\sigma }^{\prime }=\mathit{\tau \sigma }).$$

Therefore $\tau \in H(\phi )$.

$\text{∙}$

If $\tau \in H(\phi )$, $\tau \cdot \phi =\phi$, where $\phi ={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{\sigma \in H}\sigma \cdot m$, so $$\underset{\sigma \in H}{\sum }(\mathit{\tau \sigma })\cdot m=\underset{\chi \in H}{\sum }\chi \cdot m,$$

$$\underset{\sigma \in H}{\sum }{x}_{(\mathit{\tau \sigma })(1)}^{a_1}\cdots x_{(\mathit{\tau \sigma })(n)}^{a_n}=\underset{\chi \in H}{\sum }{x}_{\chi (1)}^{a_1}\cdots x_{\chi (n)}^{a_n}.$$

Moreover,

$$\underset{i=1}{\overset{n}{\prod }}{x}_{\chi (i)}^{a_i}=\underset{j=1}{\overset{n}{\prod }}{x}_j^{a_{{\chi }^{-1}(j)}},(j=\chi (i)),$$

so

$$\underset{\sigma \in H}{\sum }{x}_1^{a_{{(\mathit{\tau \sigma })}^{-1}(1)}}\cdots x_n^{a_{{(\mathit{\tau \sigma })}^{-1}(n)}}=\underset{\chi \in H}{\sum }{x}_1^{a_{{\chi }^{-1}(1)}}\cdots x_n^{a_{{\chi }^{-1}(n)}}$$

Since the exponents $a_1,\dots ,a_n$ are distinct, the $k$ terms of ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{\chi \in H}\chi \cdot m$, where $k=|H|$, are distinct, so there exists exactly one term in the right member which is the same as the term $x_1^{a_{{\tau }^{-1}(1)}}\cdots x_n^{a_{{\tau }^{-1}(n)}}$ of the left member corresponding to $\sigma =e$, so there exists $\chi \in H$ such that

$$x_1^{a_{{\tau }^{-1}(1)}}\cdots x_n^{a_{{\tau }^{-1}(n)}}=x_1^{a_{{\chi }^{-1}(1)}}\cdots x_n^{a_{{\chi }^{-1}(n)}}.$$

This implies $a_{{\tau }^{-1}(i)}=a_{{\chi }^{-1}(i)},1\le i\le n$. Since the exponents are distinct, $a_k=a_l$ implies $k=l$, so we obtain ${\tau }^{-1}(i)={\chi }^{-1}(i)$ for all $i$, therefore ${\tau }^{-1}={\chi }^{-1}$ and $\tau =\chi \in H$.

We have proved $H=H(\phi )$.