Exercise 11.1.3

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}}

Give a proof of Corollary 11.1.8 that uses neither Theorem 11.1.7 nor the Galois correspondence.

\medskip

{\bf Corollary 11.1.8} Let $\F_{p^m}$ and $\F_{p^n}$ be finite fields. Then $\F_{p^m}$ is isomorphic to a subfield of $\F_{p^n}$ if and only if $m\mid n$.

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} 
As in the text, if $\F_{p^n}$ has a subfield with $p^m$ elements, written $\F_{p^m}$, then
$$\F_p \subset \F_{p^m} \subset \F_{p^n},$$
therefore
$$n = [\F_{p^n} : \F_p] = [\F_{p^n} : \F_{p^m}]  [\F_{p^m} : \F_{p}] =  [\F_{p^n} : \F_{p^m}] 	imes m,$$
so $m \mid n$.

Conversely, suppose that $m \mid n$.

The elements of $\F_{p^n}$ are the $p^n$ distinct roots of $x^{p^n} - x$.

Since $m\mid n$, then $n = k m, k \in \N$, therefore $p^n - 1 = p^{km} - 1 = (p^m-1)\sum\limits_{i=0}^{k-1} p^{mi}$.

So $p^m-1$ divides $p^n - 1$, thus $p^n-1 = l (p^m-1), l \in \N$. Consequently
$$x^{p^n-1} - 1 = x^{l(p^m-1)}- 1 = (x^{p^m-1} - 1)\sum_{j=0}^{l-1} x^{j(p^m-1)},$$
therefore $ x^{p^m-1} - 1 \mid x^{p^n-1} - 1$, and also $x^{p^m} - x \mid x^{p^n}- x$.

Therefore the polynomial $x^{p^m} -x$ splits completely over $\F_{p^n}$. By Exercise 1, the set of its roots is a subfield of $\F_{p^n}$ with $p^m$ elements. By Corollary 11.1.3, it is isomorphic to $\F_{p^m}$.
\end{proof}

Exercise 11.1.3

Answers

Comments

Exercise 11.1.3

Answers

Comments

Add answer