Exercise 9.2.2 (Solvability of extensions)

Question

Let $N:M:K$ be extensions, with $N:M,\ M:K$ and $N:K$
all finite, normal and separable. Show that if $N:M$ and $M:K$
are solvable then so is $N:K$.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
This is straightforward from the 	extbf{Definition 9.2.1}.
We have $K \subseteq M \subseteq N$. We know that $M:K$ and $N:M$ are finite and normal.
We also know that $Gal(M:K)$ and $Gal(N:M)$ are abelian since they are solvable.
Then since $N:K$ is a finite normal seperable extension, we have that $N:K$ is solvable, where $r = 2$.

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}}

\begin{proof} 
Since $M : K$ is solvable, there exist $r\geq 0$ and intermediate fields
\begin{eqnarray}
&K = L_0\subseteq L_1 \subseteq \cdots \subseteq L_r = M
\end{eqnarray}
such that $L_i : L_{i-1}$ is normal and $\Gal(L_i : L_{i-1})$ is abelian for each $i \in \{1,\ldots,r\}$.
Similarly, since $N:M$ is solvable, there exist $s\geq 0$ and intermediates fields
\begin{eqnarray}
 &M=  L'_0\subseteq L'_1 \subseteq \cdots \subseteq L'_s =N
 \end{eqnarray}
such that $L'_j : L'_{j-1}$ is normal and $\Gal(L'_j : L'_{j-1})$ is abelian for each $j \in \{1,\ldots,s\}$.
Define $L_i = L'_{i-r}$ for $i \in \{r+1,\ldots,r+s\}$, so that $L_{r+1} = L'_1,\ldots, L_{r+s} = L'_s$. Then
\begin{eqnarray}
K = L_0\subseteq L_1 \subseteq \cdots \subseteq L_r = M \subseteq L_{r+1} \subseteq L_{r+2} \subseteq \cdots \subseteq L_{r+s} = N.
\end{eqnarray}

Then $L_i : L_{i-1}$ is normal and $\Gal(L_i : L_{i-1})$ is abelian for each $i \in \{1,\ldots,r\}$ by (1), and $L_i : L_{i-1} = L'_{i-r} : L'_{i-r-1}$ is normal and $\Gal(L_i : L_{i-1}) = \Gal(L'_{i-r} : L'_{i-r-1})$ is abelian for each $i \in \{r+1,\ldots,r+s\}$ by (2) (for $i = r+1$, $\Gal(L_{r+1} : L_r) = \Gal(L_{r+1} : M) = \Gal(L'_1:L'_0)$).

Therefore $N:K$, which is  finite, normal and separable, is also solvable.
\end{proof}

Exercise 9.2.2 (Solvability of extensions)

Answers

Comments

Comments

Exercise 9.2.2 (Solvability of extensions)

Answers

Comments

Comments

Add answer