Exercise 7.22

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

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{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

(continuation) Set $\mathrm{tr}(\alpha) = \alpha+\alpha^p+\cdots+\alpha^{p^{n-1}}$. Prove that
\begin{enumerate}
\item[(a)] $\mathrm{tr}(\alpha) + \mathrm{tr}(\beta) = \mathrm{tr}(\alpha+\beta)$.
\item[(b)] $\mathrm{tr}(a\alpha) = a\, \mathrm{tr}(\alpha)$ for $a \in \Z/p\Z$.
\item[(c)] There is an $\alpha \in F$ such that $\mathrm{tr}(\alpha) 
e 0$.
\end{enumerate}

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

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{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Let $F$ the Frobenius automorphism of $\F_q$ introduced in Ex.7.21.

(a),(b) : If $x,y \in \F_q$, and $a \in \F_p$, then $a^p = a$, so $F(x+y) =(x+y)^p = x^p+y^p = F(x)+F(y)$, and $F(ax) = a^p x^p = a x^p =a F(x)$, so $F$ is $\F_p$-linear, and also $tr = I + F + F^2+\cdots+F^{n-1}$.

(c) The polynomial $p(x) = x+x^p+x^{p^2}+\cdots+x^{p^{n-1}}$ has degree $p^{n-1}$, so $p(x)$ has at most $p^{n-1}$ roots in $\F_q$, and $|\F_q| =p^n >\deg(p) = p^{n-1}$. Therefore there exists in $\F_q$ some element $\alpha$ which is not a root of $p(x)$, and so  $\mathrm{tr}(\alpha) = p(\alpha) 
e 0$.
\end{proof}

Exercise 7.22

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Exercise 7.22

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