Exercise 8.3.2

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}}

Assume that $F\subset L$ is a Galois extension and that $F$ has characteristic 0. Also, consider the extension $L \subset L(\zeta)$ obtained by adjoining a primitive $m$th root of unity. Prove that $F\subset L(\zeta)$ is Galois.

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}
Let $\zeta$ a primitive root of $f = x^m-1$, in other words a generator of $\mathbb{U}_m$.

As the characteristic of $F$ is 0, by Exercise 1, $f$ is a separable polynomial, and $f=(x-1)(x-\zeta)\ldots(x-\zeta^{m-1})$

$F(\zeta) = F(1,\zeta,\cdots,\zeta^{m-1})$ is the splitting field over $F$ of the separable polynomial $f$, so $F\subset F(\zeta)$ is a Galois extension. By hypothesis, $F\subset L$ is also a Galois extension. By the Theorem of the Primitive Element, there exists $\alpha\in L$ such that $L=F(\alpha)$. Then the compositum of $L=F(\alpha)$ and $F(\zeta)$  is $F(\alpha,\zeta) = L(\zeta)$. By Exercise 8.2.7, this is a Galois extension of $F$.
\begin{center}
$F \subset L(\zeta)$ is a Galois extension.
\end{center}
\end{proof}

Exercise 8.3.2

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Exercise 8.3.2

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