Exercise 9.2.5 (Extension solvable iff Galois group is solvable)

Question

Prove the $\Longleftarrow$ direction of Lemma 9.2.4.

Hassaan Naeem · Accepted Answer

\textbf{Solution:}
If $Gal(M:K)$ is solvable then we have:
$$\{1\} = Gal(M:L_r) \unlhd Gal(M:L_{r-1}) \unlhd ... \unlhd Gal(M:L_0) = Gal(M:K) $$
where $Gal(M:L_{i})$ is normal in $Gal(M:L_{i-1})$  and $Gal(M:L_{i-1})/Gal(M:L_{i})$ is an abelian group for $i=r, r-1, ..., 1$.
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Since $M:K$ is finite, normal, and seperable, by \textbf{Lemma 7.2.16} we have that $M:L_i$ are seperable and by \textbf{Corollary 7.1.6}
we have that $M:L_i$ are finite and normal, for each $i \in \{ 1, ..., r\}$.
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Since $M:K$ is finite, normal, and seperable, by the fundmental theorem we can conclude that,
$L_1$ is a normal extension of $K=L_0$ since $Gal(M:L_1)$ is a normal subgroup of $Gal(M:K)$.
Similarly, $M:L_i$ is finite, normal, and seperable, then by the fundamental theorem $L_i$ is a normal extension of $L_{i-1}$ since $Gal(M:L_i)$ is a normal subgroup of $Gal(M:L_{i-1})$,
for each $i \in \{1,..., r\}$
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In addition, again by the fundamental theorem for each $i \in \{1,...,r\}$:
$$ \frac{Gal(M:L_{i-1})}{Gal(M:L_{i})} \cong Gal(L_i:L_{i-1})$$
where we have shown earlier that the left hand side is an abelian group, hence the right side must also
be abelian.
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Hence we have $K=L_0 \subseteq L_1 \subseteq ... \subseteq L_r = M$ where $r \ge 0$ and 
where we have proved that for each $i \in \{1,...,r\}, \ L_i:L_{i-1}$ is normal and $Gal(L_i:L_{i-1})$ is abelian.
Hence $M:K$ is solvable.

Richard Ganaye · Answer

ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} ewcommand{\F}{\mathbb{F}} ewcommand{\U}{\mathbb{U}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\im}{\,\mathrm{Im}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{\Gal}{\mathrm{Gal}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} \begin{proof} Here we assume that $M:K$ is finite, normal and separable. We assume also that $G = \Gal(M:K)$ is solvable. By definition, there exist $r \geq 0$ and subgroups $$\{e\} = G_r \unlhd G_{r-1} \unlhd \cdots \unlhd G_1 \unlhd G_0 = G$$ such that $G_{i-1}/G_i$ is abelian for $i \in \{1,\ldots,r\}$. Consider the intermediate fields $L_i = \mathrm{Fix}(G_i)$, so that we obtain a Galois correspondance \medskip \begin{center} \begin{tikzpicture} ode(H0) at (4,5) {$G_r = \{1\}$} ; ode (H1) at (4,4) {$\vdots$} ; ode (H2) at (4,3) {$G_i$} ; ode (H3) at (4,2) {$G_{i-1}$} ; ode (H4) at (4,1) {$\vdots$} ; ode (H5) at (4,0) {$G_0 = G$} ; \draw [<-] (H5) edge (H4) ; \draw [<-] (H4) edge (H3) ; \draw [<-] (H3) edge (H2) ; \draw [<-] (H2) edge (H1) ; \draw [<-] (H1) edge (H0) ; ode(L0) at (0,5) {$L_r = M$} ; ode (L1) at (0,4) {$\vdots$} ; ode (L2) at (0,3) {$L_i$} ; ode (L3) at (0,2) {$L_{i-1}$} ; ode (L4) at (0,1) {$\vdots$} ; ode (L5) at (0,0) {$L_0 = K$} ; \draw [->] (L5) edge (L4) ; \draw [->] (L4) edge (L3) ; \draw [->] (L3) edge (L2) ; \draw [->] (L2) edge (L1) ; \draw [->] (L1) edge (L0) ; \end{tikzpicture} \end{center} Consider the chain $L_i \subseteq L_{i+1} \subseteq M$. The extension $M : L_i$ is finite, normal and separable, so the Fundamental Theorem applies here. Since $$G_i = \Gal(M:L_i) \unlhd G_{i-1} = \Gal(M : L_{i-1}),$$ the extension $L_i : L_{i-1}$ is normal, and $$\Gal(L_i : L_{i-1}) \simeq \Gal(M : L_{i-1})/ \Gal(M:L_i) = G_{i-1}/G_i$$ is abelian for $i \in \{1,\ldots,s\}$. This shows that $M : K$ is solvable. \end{proof}

Exercise 9.2.5 (Extension solvable iff Galois group is solvable)

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Exercise 9.2.5 (Extension solvable iff Galois group is solvable)

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