Exercise 5.3.3

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}}

Let $F$ be a field of characteristic $p$. The $n$th roots of unity are defined to be the roots of $x^n-1$ in the splitting field $F\subset L$ of $x^n-1$.
\begin{enumerate}
\item[(a)] If $p 
mid n$, show that there are $n$ distinct $n$th roots of unity in $L$.
\item[(b)] Show that there is only one $p$th root of unity, namely $1\in F$.
\end{enumerate}

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}
\begin{enumerate}
\item[(a)] Here $n \geq 1$.

As $f = x^n-1$, then $f'=nx^{n-1}$.
 
 If $p 
mid n$, then $n
eq 0$ in the field $F$ of characteristic $p$, thus $n$ is a unit in $F[x]$.
 
 $x(nx^{n-1}) - n(x^n-1) = n = xf'-nf$ is a B\'ezout's relation  between $f$ and $f'$, which proves that $f\wedge f'=1$. So $f$ is a separable polynomial, and the $n$ roots of $f$ in its splitting field, which are the $n$th roots of unity, are distinct.
 
 \item[(b)]
If the characteristic of $F$ is $p$, by Exercise 2,
 $$x^p-1 = (x-1)^p.$$
The only $p$th root of unity is thus 1.
 \end{enumerate}
\end{proof}

Exercise 5.3.3

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Exercise 5.3.3

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