Exercise 6.1.7

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 $F \subset L$ be a finite extension, and let $\sigma : L 	o L$ be a ring homomorphism that is the identity on $F$. This exercise will show that $\sigma$ is an automorphism.
\begin{enumerate}
\item[(a)] Show that $\sigma$ is one-to-one.
\item[(b)] Show that $\sigma$ is onto.
\end{enumerate}

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}
\begin{enumerate}
\item[(a)]
Let $a \in L, a
eq 0$. Then $a$ has an inverse $b$ in the field $L$, so $ab=1$, $\sigma(a) \sigma(b) = \sigma(1) = 1 $, $\sigma(a) 
eq 0$. Therefore $\ker(\sigma) =\{0\}$, thus $\sigma$ is injective.

$\sigma : L 	o L$ is an injective field homomorphism.

\item[(b)]
As $K \subset L$ is a finite extension, $L$ is a finite dimensional vector space over $F$. As $\sigma$ is identity on $F$, $\sigma : L 	o L$ is an injective linear application on a finite dimensional vector space, thus $\sigma$ is also surjective :   $$\sigma \in \mathrm{Gal}(L/F).$$
\end{enumerate}
\end{proof}

Exercise 6.1.7

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Exercise 6.1.7

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