Exercise 6.3.5

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

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ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

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}\,}

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ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $f \in F[x]$ be separable, where $f = g_1\cdots g_s$ for $g_i \in F[x]$ of degree $d_i>0$, and let $L$ be the splitting field of $f$ over $F$. Show that $\Gal(L/F)$ is isomorphic to a subgroup of the product group $S_{d_1} 	imes \cdots 	imes S_{d_s}$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

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}
We show the proposition for $s=2$ to have lighter notations.

Suppose that $f = gh\in F[x]$ is separable, with $g,h \in F[x]$, $\deg(g)=r,\deg(h)=s$. Then $g,h$ are separable.

Write $\alpha_1,\cdots,\alpha_{r}$ the roots of $g$ in $M$, and $\beta_1,\cdots,\beta_{s}$ the roots of $h$ in $N$.

Let $M$ be a splitting field of $g$ over $F$, and $N$ be a splitting field of $h$ over $F$.
Then $M=F(\alpha_1,\cdots,\alpha_r), N = F(\beta_1,\cdots,\beta_{s})$, and  $L=F(\alpha_1,\cdots,\alpha_{r}, \beta_1,\cdots,\beta_{s})$. As $f$ is separable, the $d=r+s$ roots of $f$, $\alpha_1, \cdots,\alpha_{r},\beta_1,\cdots,\beta_{s}$ are distinct.

Write $A$ the set of the roots of $g$ in $L$, $B$ the set of roots of $h$ in $L$ : $\vert A \vert = r, \vert B \vert = s$, and write $S(A)$ the set of bijections of $A$ (and the same for $B$) : $S(A)\simeq S_r, S(B) \simeq S_s$.

Let $\sigma \in \Gal(L/F)$. As $g,h \in F[x]$, $\sigma$ induces a permutation of the roots of $g$ and of the roots of $h$, so the maps
\begin{center}
$\sigma_1 : 
\left\{
\begin{array}{ccc}
A  &	o   &  A \
  \alpha&  \mapsto  &   \sigma(\alpha) 
\end{array}
ight.
$ and
$\sigma_2 : 
\left\{
\begin{array}{ccc}
B  &	o   &  B \
  \beta&  \mapsto  &   \sigma(\beta) 
\end{array}
ight.
$
\end{center}
restrictions  of $\sigma$ on $A,B$, satisfy $\sigma_1 \in S(A), \sigma_2 \in S(B)$.

The map
\begin{center}
$\varphi : 
\left\{
\begin{array}{ccc}
\Gal(L/F) &	o   &  S(A)	imes S(B) \
  \sigma&  \mapsto  &  (\sigma_1,\sigma_2)
\end{array}
ight.
$
\end{center}
is a group homomorphism: if $\varphi(\sigma) = (\sigma_1,\sigma_2)$ and $\varphi(	au) = (	au_1,	au_2)$ (with $\sigma,	au \in \Gal(L/F)$), and also $\eta = \sigma \circ 	au, \varphi(\eta) = (\eta_1,\eta_2)$, then for all $\alpha$ in $A$ and $\beta \in B$, 
\begin{align*}
\eta_1(\alpha) = \eta(\alpha) = (\sigma  	au)(\alpha) = (\sigma_1  	au_1)(\alpha) , \eta_2(\beta) = \eta(\beta) = (\sigma  	au)(\beta) =(\sigma_2  	au_2)(\beta),
\end{align*}
thus $\eta_1 = \sigma_1	au_1, \eta_2 = \sigma_2	au_2$. Consequently

$\varphi(\sigma \circ 	au) = \varphi(\eta) = (\eta_1,\eta_2) = (\sigma_1  	au_1, \sigma_2  	au_2) = (\sigma_1, \sigma_2) (	au_1,	au_2) =  \varphi(\sigma)\varphi(	au)$.

$\varphi$ is injective : if $\varphi(\sigma) = (\sigma_1,\sigma_2) = (1_A,1_B)$, then $$\sigma(\alpha_i) = \alpha_i,\ i=1,\cdots r \ \mathrm{and}\ \sigma(\beta_j) = \beta_j,\ j=1,\cdots s.$$

As $L=F(\alpha_1,\cdots,\alpha_{r}, \beta_1,\cdots,\beta_{s})$, $\sigma = 1_L$.

$\Gal(L/F)$ is isomorphic to a subgroup of $S(A) 	imes S(B)$, and as $S(A)	imes S(B) \simeq S_r	imes S_s$,  $\Gal(L/F)$ is isomorphic to a subgroup of $S_r	imes S_s$.

We can generalize to $s$ polynomials similarly, or by induction.
\end{proof}

Exercise 6.3.5

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Exercise 6.3.5

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