Exercise 11.2.7

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}}

As in the discussion of Berlekamp's algorithm, let $R = \F_p[x]/\langle f angle$ and consider the $p$th power map $T : R 	o R$. Prove that $T$ is a linear map when $R$ is regarded as a vector space over $\F_p$.

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 $\overline{g} = g+\langle f angle, \overline{h} = h+\langle f angle \in T$, and $\lambda \in \F_p$. Since the characteristic is $p$, $(g+h)^p = g^p+h^p$, so
\begin{align*}
T(\overline{g} + \overline{f}) 
&=(g+h)^p + \langle f angle\
&=g^p+h^p+\langle f angle\
&= (g^p+\langle f angle)+ (h^p+\langle f angle)\
&= T(\overline{g}) + T(\overline{h}).
\end{align*}
Since $\lambda \in \F_p$, $\lambda^p=\lambda$, and $\lambda \langle f angle = \langle \lambda f angle$, $\lambda(g +\langle f angle) = \lambda g + \langle f angle$, so
\begin{align*}
T(\lambda \overline{g}) &= T(\lambda g + \langle f angle)\
&=(\lambda g)^p + \langle f angle\
&=\lambda g^p +\langle f angle\
&=\lambda(g^p + \langle f angle)\
&=\lambda T(\overline{g}).
\end{align*}
Thus $T:R	o R$ is linear.
\end{proof}

Exercise 11.2.7

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Exercise 11.2.7

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