Exercise 12.2.12

Question

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In the situation of Theorem 12.2.5, let $G \subset S_n$ correspond to $\Gal(L/F)$, and $H \subset S_n$ correspond to $\Gal(KL/K)$. By Exercise 8, we know that $H \subset G$. Also let $A$ be the set of arrangements studied in Exercise 11. Then a left coset $\sigma H \subset G$ gives a subset $\sigma H\cdot \alpha_1\cdots \alpha_n \subset A$, and since the map $\sigma \cdot \alpha_1\cdots \alpha_n$ is one-to-one and onto, the sets $\sigma H\cdot \alpha_1\cdots\alpha_n$ partition $A$ into disjoint subsets. We claim that these are the "groups" that appear in $\it 1^{\circ}$ and $\it 2^{\circ}$ of Galois Proposition II.
\be
\item[(a)] Given any two such "groups" $\sigma H \cdot \alpha_1\cdots\alpha_n$ and $	au H \cdot \alpha_1\cdots\alpha_n$, prove that there is $\gamma \in G$ such that (as Galois says in $\it 2^{\circ}$) one passes from one to the other by applying $\gamma$ to all arrangements in the first.

\item[(b)] So far, it seems like Galois describing cosets. However, as pointed out in [12], Galois thought of these "groups" differently. This is seen by explaining how they relate to $\it 1^{\circ}$ of Galois' proposition. Let $M'$ be the field used in Exercise 10, and let $	au \in \Gal(M'/F)$. Then $K' = 	au(K)$ is a conjugate of $K$. Let $\sigma \in G$ be the permutation corresponding to $	au|_L \in \Gal(L/F)$. Show that $\sigma H \sigma^{-1}$ is a subgroup of $S_n$ corresponding to $\Gal(K'L/K')$.

\item[(c)] Using the setup of part (b), consider the "group" $\sigma H\cdot \alpha_1\cdots\alpha_n \subset A$. Prove that $\sigma H \sigma^{-1} \subset S_n$ is the set of all permutations of $S_n$ that map the first element of this "group", namely $\sigma\cdot \alpha_1\cdots \alpha_n$, to another element of the "group". (Remember that this is the process for turning a "group" of arrangements into a subgroup of $S_n$.)
\ee
Combining parts (b) and (c), we see that what Galois says in $\it 1^{\circ}$ of Proposition II is fully consistent with what we did in Exercise 10.

Richard Ganaye · Accepted Answer

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\begin{proof}
\be
\item[(a)] Let $\gamma = 	au \sigma^{-1}$. Since $G$ is a subgroup of $S_n$, $\gamma \in G$, and, for all $h \in H$,
\begin{align*}
\gamma \cdot (\sigma h \cdot \alpha_1\cdots \alpha_n) &= \gamma \sigma h \cdot \alpha_1\cdots \alpha_n\
&= 	au h \cdot \alpha_1\cdots \alpha_n,
\end{align*}
so 
$$ \gamma \cdot (\sigma H \cdot \alpha_1\cdots \alpha_n) =	au H \cdot \alpha_1\cdots \alpha_n.$$
There exists $\gamma \in G$ such that one passes from one to the other by applying $\gamma$ to all arrangements in the first.

\item[(b)] By Exercise 10(d), we know that
$$\Gal(L/K' \cap L) = (	au \vert_L) \Gal(L/K\cap L) (	au \vert _L)^{-1}.$$
The map $\Gal(L/F) 	o S_n$ given by the action of the Galois group on the roots is a morphism. As $\sigma$ is the image of $	au\vert_ L$ by this morphism, we obtain
$$H' = \sigma H \sigma^{-1},$$
where $H$ is the subgroup of $S_n$ corresponding to ${\Gal(L/K\cap L) }$, and $H'$ to ${\Gal(L/K' \cap L)}$.

These two subgroups of $S_n$ are the images of $\Gal(KL/K)$ and $\Gal(K'L/K')$ under the injective homomorphism (12.29), which is compatible with (12.28) and (12.29) by Exercise 8, i.e. $	au$ and $	au\vert_L$ map to the same element of $S_n$.

Conclusion: if $H\subset S_n$ is corresponding to $\Gal(KL/K)$, then $\sigma H \sigma^{-1}$ is a subgroup of $S_n$ corresponding to $\Gal(K'L/K')$ (where $K' = 	au K$, and $\sigma \in G$ is the permutation corresponding to $	au\vert L \in \Gal(L/F)$).

\item[(c)] 
Let $\gamma \in S_n$. Then $\gamma$ maps $\sigma\cdot \alpha_1\cdots \alpha_n$ on $\sigma H\cdot \alpha_1\cdots\alpha_n $ if and only if, there exists $h \in H$ such that
$$\gamma \cdot(\sigma\cdot \alpha_1\cdots \alpha_n) = (\sigma h) \cdot \alpha_1\cdots\alpha_n.$$
This is equivalent to  $(\gamma \sigma)\cdot \alpha_1\cdots \alpha_n = (\sigma h) \cdot \alpha_1\cdots\alpha_n,\ h\in H$.

Since $\sigma \mapsto \sigma\cdot \alpha_1\cdots \alpha_n$ is bijective (Exercise 11(b)), this is equivalent to
$$\gamma \sigma = \sigma h,\  h \in H,$$
or equivalent to $\gamma \in \sigma H \sigma^{-1}.$
$$\gamma \cdot(\sigma\cdot \alpha_1\cdots \alpha_n) \in \sigma H\cdot \alpha_1\cdots\alpha_n  \iff \gamma \in \sigma H \sigma^{-1}.$$
\ee
\end{proof}

Exercise 12.2.12

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Exercise 12.2.12

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