Exercise 11.1.7

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 $f \in \F_p[x]$ be irreducible of degree $
u$. Use (7.1) and Theorem 11.1.7 to prove Galois's observation that if $i$ is one root of $f$ in a splitting field, then the other roots are given by $i^p,i^{p^2},\ldots,i^{p^{
u-1}}$.

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}
 Since $f$ is irreducible, $L = \F_p[x] / \langle f angle$ is a field, and $[L : \F_p] = \deg(f) = 
u$, so $\vert L \vert = q= p^
u$, and $L = \F_{q}$. 
 
 By Theorem 11.1.7, $ \F_p \subset L = \F_{q}$ is a Galois extension, therefore $L$ contains all roots of $f$, so $L$ is the splitting field of $f$ over $\F_p$.

Write $F = \mathrm{Frob}_p$ the Frobenius automorphism of $\F_q=  \F_{p^
u}$. By Theorem 11.1.7,  $\Gal(\F_q/\F_p) = \langle F angle$.

Since $\F_p \subset F_q$ is a Galois extension (Theorem 11.1.7(a)), the irreducible polynomial $f \in \F_p[x]$ is separable, so $f$ has $
u$ distinct roots.

If $i$ is a root of $f$ in $L = \F_q$, then $F^k(i),\ 0 \leq k < 
u$ is also a root of $f$. If $j$ is any root of $f$, since $f$ is irreducible, $\Gal(\F_q/\F_p) $ acts transitively on the set of the roots of $f$, so  there exists $\sigma \in \Gal(\F_q/\F_p)$ such that $\sigma(i) = j$. Since $\Gal(\F_q/\F_p) = \langle F angle$ and since the order of $F$ is $
u$, there exists $k, \ 0\leq k < 
u$, such that $F^k(i) = j$. Thus the set of the roots of $f$ is
$$\{i,F(i)=F^2(i), \ldots, F^{
u -1}(i)\} = \{i,i^p,i^{p^2},\ldots,i^{p^{
u - 1}}\}.$$
Since $f$ is separable, $f$ has $
u$ distinct roots, so these elements are distinct. Therefore
$$f(x) = (x-i)(x-i^p)(x - i^{p^2})\cdots(x- i^{p^{
u - 1}}).$$
(This is an example of (7.1).)
\end{proof}

Exercise 11.1.7

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Exercise 11.1.7

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