Exercise 13.3.5

Proof. (a) $G$ is the symmetry group of $\phi$, therefore $G\subset H$ is the isotropy subgroup of $\phi$ under the action of $H$ ( ${\mathrm{Stab}}_H(\phi )=H\cap {\mathrm{Stab}}_{S_n}(\phi )=H\cap G=G$), and $[H:G]=|H\cdot \phi |=l$ (cf. Theorem A.4.9). The degree of ${\Theta }_f^H(y)$ is equal to $l$, hence $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Theta }_f^H(y))=[H:G]$. (b) Let ${\tau }_1,...,{\tau }_l$ be representatives for the left cosets of $G$ in $H$. Then

Since $\phi \in F[x_1,...,x_n]$, for each $i,1\le i\le l$, $({\tau }_i\cdot \phi )({\alpha }_1,...,{\alpha }_n)=\phi ({\alpha }_{{\tau }_i(1)},\dots ,{\alpha }_{{\tau }_i(n)})\in F[{\alpha }_1,...,{\alpha }_n]$, hence the coefficients of ${\Theta }_f^H(y)$ are in $F[{\alpha }_1,...,{\alpha }_n]$.

Suppose $\sigma \in G_f$, then $\sigma \in H$ and $\sigma \cdot {\Theta }_f^H(y)={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^l(y-(\sigma {\tau }_i)\cdot \phi ({\alpha }_1,...,{\alpha }_n))$. But the set $\sigma {\tau }_1,...,\sigma {\tau }_l$ is also a set of left coset representatives of $G$ in $H$. Thus the action of $\sigma$ has merely permuted the roots of ${\Theta }_f^H(y)$ leaving the coefficients fixed. It means that the coefficients of ${\Theta }_f^H(y)$ are in $F$, i.e., ${\Theta }_f^H(y)\in F[y]$. (c) Assume that $G_f$ is congugate within $H$ to a subgroup $K$ of $G$, so that $G_f=\mathit{\tau K}{\tau }^{-1},\tau \in H$. Then ${\tau }^{-1}{G}_f\tau \subset G$.

Let $\tilde{\sigma }\in \mathrm{Gal}(L∕F)$ (where $L=F({\alpha }_1,\dots ,{\alpha }_n)$), and $\sigma \in G_f$ the associated permutation, so that $\tilde{\sigma }({\alpha }_i)={\alpha }_{\sigma (i)},i=1,\dots ,n$. Since ${\tau }^{-1}\mathit{\sigma \tau }\in G$ fixes $\phi$, $({\tau }^{-1}\mathit{\sigma \tau })\cdot \phi =\phi$, so that

Since this is true for any $\tilde{\sigma }\in \mathrm{Gal}(L∕F)$, this proves that $(\tau \cdot \phi )({\alpha }_1,\dots ,{\alpha }_n)\in F$.

Moreover $\tau \in H$, thus $\tau \cdot \phi$ is in the orbit $H\cdot \phi$, so $\beta =(\tau \cdot \phi )({\alpha }_1,\dots ,{\alpha }_n)$ is a root of ${\Theta }_{\phi }^H(y)=\underset{\psi \in H\cdot \phi }{\prod <!--\; FUNCTION\; APPLICATION\; -->}(y-\psi )$, and $\beta$ is in $F$. (d) We mimic the proof of Proposition 13.3.2, with an explicitation of the relabeling of the roots, as in part (c).

Let $\beta$ be a simple root of ${\Theta }_f^H(y)$ in $F$. By definition of ${\Theta }_f^H(y)$, there is $\tau \in H$ such that

We will prove that ${\tau }^{-1}{G}_f\tau \subset G$. If not, there is some $\sigma \in {\tau }^{-1}{G}_f\tau$ such that $\sigma \notin G$.

Then ${\sigma }^{\prime }=\mathit{\tau \sigma }{\tau }^{-1}\in G_f$ corresponds to some $\rho \in \mathrm{Gal}(L∕F)$, so that $\rho ({\alpha }_i)={\alpha }_{{\sigma }^{\prime }(i)},i=1,\dots ,n$.

Since $\beta \in F$, $\beta =\rho (\beta )$, so that

We conclude that

Since $\beta$ is a root of ${\Theta }_f^H$, $y-\beta =y-(\tau \cdot \phi )({\alpha }_1,\dots ,{\alpha }_n)$ is a factor of ${\Theta }_f^H(y)$.

But $\sigma \notin G$, thus $\phi \ne \sigma \cdot \phi$, therefore $\tau \cdot \phi \ne (\mathit{\tau \sigma })\cdot \phi$.

From $\tau \in H,G_f\subset H$, and $\sigma \in {\tau }^{-1}{G}_f\tau$, we know that $\mathit{\tau \sigma }\in H$, thus

is another factor of ${\Theta }_f^H(y)$, therefore the relative resolvent may be written

This is a contradiction, since by assumption $\beta$ is a simple root of ${\Theta }_f^H(y)$. This proves that ${\tau }^{-1}{G}_f\tau \subset G$. If $K={\tau }^{-1}{G}_f\tau$, $K$ is a subgroup of $G$, and $G_f=\mathit{\tau K}{\tau }^{-1},\tau \in H$, is conjugate within $H$ to a subgroup of $G$. □