Exercise 12.1.12

Question

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The discussion following Theorem 12.1.10 shows that if we are going to use Lagrange's strategy when $n\geq 5$, then we need to begin with $\varphi = \sqrt{\Delta}$, which has isotropy subgroup $A_n$. Suppose that $\psi \in L$ is our next choice, and let $	heta(x)$ be the resolvent of $\psi$. Since we regard $K(\sqrt{\Delta})$ as known, we may assume that $\psi 
ot \in K(\sqrt{\Delta})$. The idea is to factor $	heta(x)$ over $K(\sqrt{\Delta})$, say $	heta = R_1\cdots R_s$, where $R_i \in K(\sqrt{\Delta})[x]$ is irreducible. This is similar to how (12.13) factors the resolvent of $t_1$ over $K(y_1)$. Suppose that $\psi$ enables us to continue Lagrange's inductive strategy. This means that some factor of $	heta$, say $R_j$, has degree $<n$. Your goal is to prove that this implies the existence of a proper subgroup of $A_n$ of index $<n$.
\be
\item[(a)] Prove that $\deg(R_j) \geq 2$.
\item[(b)] Since $	heta$ splits completely over $L$, the same is true for $R_j$. Let $\psi_j \in L$ be a root of $R_j$ and consider the fields
$$K \subset K(\sqrt{\Delta}) \subset M = K(\sqrt{\Delta},\psi_j) \subset L.$$
Let $H_j \subset S_n$ be the subgroup corresponding to $\Gal(L/M) \subset \Gal(L/K)$ under (12.1). Prove that $H_j \subset A_n$ and that $[A_n:H_j]$ is the degree of $R_j$.
\item[(c)] Conclude that $\deg(R_j)<n$ implies that $H_j$ is a proper subgroup of $A_n$ of index $<n$. With more work, one can show that $\deg(R_i) = [A_n:A_n\cap H(\psi)]$ for all $i$ and that
$$s = \frac{2}{[H(\psi):A_n \cap H(\psi)]}.$$
It follows that $s = 1$ or $2$.
\ee

Richard Ganaye · Accepted Answer

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\begin{proof}
\be
\item[(a)]
Here $K = F(\sigma_1,\ldots,\sigma_n)$ and $L = F(x_1,\ldots,x_n)$.

The roots of the resolvent $	heta$ are all the distinct $\sigma \cdot \psi$, where $\sigma \in S_n$. If $\deg(R_j)=1$, then $R_j(x) = x - \sigma \cdot \psi$ for some $\sigma \in S_n$. Since $R_j \in K(\sqrt{\Delta})[x]$, then $\sigma \cdot \psi \in K(\sqrt{\Delta})$. If $\sigma \in A_n$ then $\sigma^{-1} \in A_n$ fixes $\sqrt{\Delta}$,  and so $\psi = \sigma^{-1} \cdot (\sigma \cdot \psi) \in K(\sqrt{\Delta})$, which contradicts our assumption, therefore $\sigma \in S_n \setminus A_n$ and $\sigma \cdot \sqrt{\Delta} = - \sqrt{\Delta}$.

As $\sigma \cdot \psi \in K(\sqrt{\Delta})$, $\sigma \cdot \psi = A + B \sqrt{\Delta}, \ A,B \in K = F(\sigma_1,\ldots,\sigma_n)$. Therefore $\psi = \sigma^{-1}\cdot (A+B \sqrt{\Delta}) = A - B \sqrt{\Delta} \in K(\sqrt{\Delta})$: this is a contradiction.

Thus $\deg(R_j) \geq 2$.

\item[(b)] Since $K \subset K(\sqrt{\Delta}) \subset M$, the Galois correspondence being order reversing,
$$\Gal(L/M) \subset \Gal(L/K(\sqrt{\Delta})) \subset \Gal(L/K).$$
The same inclusions are true for the corresponding subgroups of $S_n$:
$$H_j \subset A_n \subset S_n.$$
By the fundamental Theorem (Theorem 7.3.1), since $K \subset L$, a fortiori $K(\sqrt{\Delta}) \subset L$ are Galois extensions,  the index  $(A_n:H_j) = (\Gal(L/K(\sqrt{\Delta}) : \Gal(L/M))$ is equal to $[M : K(\sqrt{\Delta})] = [K(\sqrt{\Delta},\psi_j ): K(\sqrt{\Delta})]$. The minimal polynomial of $\psi_j$ over $K(\sqrt{\Delta})$ being $R_j$, $[K(\sqrt{\Delta},\psi_j):K(\sqrt{\Delta})] = \deg(R_j)$, so
$$(A_n:H_j) = \deg(R_j).$$

\item[(c)] If $H_j = A_n$, then by the Galois correspondence $K(\sqrt{\Delta},\psi_j) = K(\sqrt{\Delta})$, and then $\psi_j \in K(\sqrt{\Delta})$. But this implies that $R_j = x - \psi_j$ has degree 1, which is impossible by part (a). So $H_j$ is a proper subgroup of $A_n$.  If $\deg(R_j) < n$, then $H_j$ is a proper subgroup of $A_n$ such that $(A_n:H_j) <n$. By Theorem 12.1.10(b), this is impossible for all $n\geq 5$.
\ee
\end{proof}

Exercise 12.1.12

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Exercise 12.1.12

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