Exercise 10.2.6

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}}

Now suppose that $\zeta_n$ is constructible for some $n>2$. The goal of this exercise is to prove that if $p$ is an odd prime dividing $n$, then $p$ is a Fermat prime and $p^2 
mid n$. This and Exercise 5 will give a proof of Theorem 10.2.1 that doesn't require knowing that $\Phi_n(x)$ is irreducible for arbitrary $n$.
\be
\item[(a)] Let $p$ be an odd prime dividing $n$. Use Exercises 3 and 4 to show that $p$ is a Fermat prime.
\item[(b)] Now assume that $p$ is an odd prime and $p^2\mid n$. Use Exercise 4 to show that $\zeta_{p^2}$ is constructible. Then use Theorem 10.1.12 and Exercise 2 to obtain a contradiction.
\ee

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}
\be
\item[(a)] If $p$ is an odd prime factor of $n$, then by Exercise 4, $\zeta_p = (\zeta_n)^{n/p}$ is constructible, so $p$ is a Fermat prime by Exercise 3.
\item[(b)]Assume that $p$ is an odd prime and $p^2\mid n$.

By Exercise 4, $\zeta_{p ^2}= (\zeta_n)^{n/p^2}$ is then constructible. The splitting field of $\zeta_{p^2}$ over $\Q$ is $L = \Q(\zeta_{p^2})$. As $\Phi_{p^2}$ is irreducible, by Exercise 2,
 $$[L:\Q] = \deg(\Phi_{p^2}) = p(p-1).$$
 By Theorem 10.1.12, $\zeta_{p ^2}$ being constructible, $[L:\Q] = p(p-1)$ is a power or 2, so $p$ is a power of 2 , where $p$ is an odd prime. This is a contradiction.
 
 Conclusion: if $\zeta_n$ is constructible, then $n = 2^s p_1\cdots p_r$, where $p_1,\ldots,p_r$ are distinct Fermat primes.
\ee
\end{proof}

Exercise 10.2.6

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Exercise 10.2.6

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