Exercise 11.1.11

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 $f \in \F_p[x]$ be an irreducible polynomial of degree $n$. Prove that $f$ splits completely in $\F_{p^n}$.

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} 
Let $F = \F_p[x]/\langle f angle$. Since $f$ is irreducible over $\F_p$, $F$ is a field, and since $\deg(f) = n$, $|F | = p^n$, so $F$ is a field with $p^n$ elements.

Moreover $\overline{x} = x + \langle f angle$ is a root of $f$ in $F$.

By Theorem 11.1.2, $F$ is a splitting field over $\F_p$ of the separable polynomial $x^{p^n} -x$, therefore $F$ is a Galois extension of $\F_p$. As the irreducible polynomial $f$ has one root in $F$, it splits completely in $F$.

If $\F_{p^n}$ is a field with $p^n$ elements, there exists an isomorphism $\varphi : F 	o \F_{p^n}$, which sends the roots of $f$ in $F$ on roots of $f$ in $\F_{p^n}$, so $f$ splits completely in $\F_{p^n}$, and this doesn't depend of the choice of the field with $p^n$ elements that we name $\F_{p^n}$.
\end{proof}
\bigskip

Note: let $\alpha$ be a root of $f$ in $\F_{p^n}$, and $\alpha_1=\alpha,\ldots,\alpha_n$ the roots of $f$ in $\F_{p^n}$. Then $[\F_{p^n} : \F_p] = n =[\F_p(\alpha) : \F_p]$, hence $\F_p(\alpha) = \F_{p^n}$. Since $\F_{p^n} = \F_{p}(\alpha) \subset \F_{p}(\alpha_1,\ldots,\alpha_n) \subset \F_{p^n}$, we conclude that
$$\F_{p^n} = \F_p(\alpha_1,\ldots,\alpha_n)$$
is a splitting field of $f$ over $\F_p$.

Since $\F_{p^n}$ is a splitting field of $f$ over $\F_p$, by Exercise 7, 
$$f(x) = a (x -\alpha)(x - \alpha^p)(x-\alpha^{p^2})\cdots(x- \alpha^{p^{n-1}}),\ a \in \F_p.$$

Exercise 11.1.11

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Exercise 11.1.11

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