Exercise 6.1.1

Question

\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}}

Let $L = F(\alpha_1,\ldots,\alpha_n)$, and let $p_i \in F[x]$ be a nonzero polynomial vanishing at $\alpha_i$. Explain why the proof of Corollary 6.1.5 implies that $|\Gal(L/F)| \leq \deg(p_1)\cdots\deg(p_n)$.

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}
$L=F(\alpha_1,\cdots,\alpha_n)$, where $\alpha_i$ is algebraic over $F$. Then $\alpha_i$ is the root of a polynomial $p_i \in F(x)$.

By Proposition 6.1.4, every $\sigma \in \mathrm{Gal}(L/F)$ is uniquely determined by the images of $\alpha_i,\  i=1,\cdots,n$. $\alpha_i$ being a root of $p_i \in F[x]$, $\sigma(\alpha_i)$ is also a root of $p_i$. So there exist only $\deg(p_i)$ possibilities for the choice of $\sigma(\alpha_i)$.

More formally, write $R_i$ the set of the roots of  $p_i$ in $L$, then $\sigma(\alpha_i) \in R_i$, with $\vert R_i \vert \leq \deg(p_i)$, and the map
$$
\left\{
\begin{array}{ccc}
 \mathrm{Gal}(L/F) & 	o  &  R_1	imes \cdots 	imes R_n\
\sigma  & \mapsto  &   (\sigma(\alpha_1), \ldots,\sigma(\alpha_n))
\end{array}
ight.
$$
is injective (one-to-one), since $\sigma \in \mathrm{Gal}(L/F)$ is uniquely determined by the images of $\alpha_i,\ i=1,\cdots,n$.

Therefore $$\vert \mathrm{Gal}(L/F) \vert \leq \vert R_1 \vert 	imes \cdots  	imes \vert R_n\vert \leq \deg(p_1)\cdots \deg(p_n). $$
\end{proof}

Exercise 6.1.1

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Exercise 6.1.1

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