Exercise 14.2.10

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 $\Phi_p(x)$ be the cyclotomic polynomial whose roots are the primitive $p$th roots of unity, where $p$ is prime. We know that $\Phi_p(x)$ is irreducible of degree $p-1$. In the quotation given in the Historical Notes, Galois asserts that $\Phi_p(x)$ is imprimitive.
\be
\item[(a)]
Prove Galois's claim for $p>3$ using Exercise 9.
\item[(b)] Explain why we need to assume that $p>3$ in part (a).
\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}
\item[(a)]
We know that the splitting field of $\Phi_p(x)$ over $\Q$ is $L = \Q(\zeta_p)$ (where $\zeta_p = e^{\frac{2i\pi}{p}}$), and that $\Gal(L/\Q) \simeq (\Z/p\Z)^*$, via the isomorphism 
$$
\left\{
\begin{array}{ccl}
\Gal(L/\Q) & 	o & (\Z/p\Z)^*\
\sigma & \mapsto & a : \sigma(\zeta_p) = \zeta_p^a.
\end{array}
ight.
$$
Therefore, $\Gal(L/\Q)$ is Abelian, and even cyclic with order $p-1$. Let $\Gal(L/\Q) \simeq G  \subset S_{p-1}$. If $p>3$, then $p-1$ is not prime, and Exercise 9 prove that  $G$ is imprimitive, so that $\Phi_p(x)$ is imprimitive.

\item[(b)] If $p = 3$, $p-1 = 2$ is prime, and $(\Z/3\Z)^* = \{1 ,-1 \}$ is cyclic of prime order. Moreover $\Phi_3(x) = x^2+x + 1$ is not imprimitive, as every polynomial of degree 2: by Definition 14.2.2, if $f$ is imprimitive, since $k>1,l>1$, $\deg(f) \geq |R_1|+|R_2| >2$.

If $p = 2$, $(\Z/2\Z)^* = \{1\}$, and $\Phi_2(x) = x+1$. Since $\deg(\Phi_2) = 1$, $\Phi_2(x)$  is not imprimitive.
 
\end{proof}

Exercise 14.2.10

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Exercise 14.2.10

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