Exercise 4.3.4

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}}

Suppose that $F\subset L$ is a finite extension with $[L:F]$ prime.
\begin{enumerate}
\item[(a)] Show that the only subfields of $L$ containing $F$ are $F$ and $L$.
\item[(b)] Show that $L = F(\alpha)$ for any $\alpha \in L \setminus F$.
\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)]
If a subfield  $K$ of $L$ satisfies  $F \subset K \subset L$, then
$$[L:F] = [L:K][K:F],$$ 
so $[K:F]$ divides $p = [L:F]$, where $p$ is a prime.

If $[K:F] = 1$, then $K = F$, and if $[K:F]=p$, then $[L:K]= 1$, thus $K = L$.

Conclusion: If  $[L:F]$ is a prime number, the only intermediate subfields of the extension $F \subset L$  are $L$ and $F$.

\item[(b)] Since $\alpha \in L$, $F \subset F(\alpha) \subset L$.
If $\alpha 
ot \in F$, then $F(\alpha) 
eq F$, thus by (a), $F(\alpha) = L$.

\end{enumerate}
\end{proof}

Exercise 4.3.4

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Exercise 4.3.4

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