Exercise 11.2.17

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}}

In section 4.2 we use the Sch"onemann-Eisenstein criterion to prove that $\Phi_p(x) = x^{p-1}+\cdots+x+1$ is irreducible over $\Q$, where $p$ is prime. Here is a very different proof. We know that primitive roots modulo $p$ exist. By Dirichlet's theorem on primes in arithmetic progression, it follows that there is a prime $l$ such that $[l] \in (\Z/p\Z)^*$ has order $p-1$. Prove that $\Phi_p(x)$ is irreducible modulo $l$ and conclude that it is irreducible over $\Q$. This argument is due to Sch"onemann in 1845 (see [5]).

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 $[g]$ a generator of $(\Z/p\Z)^*$. By Dirichlet's Theorem, since $\gcd(g,p)=1$, there is some integer $k$ such that $l = g+kp$ is prime, and  $[l] = [g]$ has order $p-1$.

Let $\overline{\Phi}_p(x)$ the reduction modulo $l$ of $\Phi_p$. By Theorem 11.2.7, $\overline{\Phi}_p(x)$ is the product of $\phi(p)/m = (p-1)/m$  irreducible polynomials in $\F_l[x]$ of degree $m$, where $m$ is the order of $[l] \in (\Z/p\Z)^*$, which is $p-1$. So $m = p-1$, and $\overline{\Phi}_p(x)$ is irreducible over $\F_l$.

Since $\Phi_p$ is monic, any decomposition $\Phi_p(x) = u(x) v(x)$ with $u,v \in \Z[x], \deg(u)<p-1, \deg(v)<p-1$ would give a decomposition $\overline{\Phi}_p(x)= \overline{u}(x) \overline{v}(x)$ in $\F_l[x]$, where $\deg(\overline{\Phi}_p(x)) = p-1, \deg(\overline{u}(x))<p-1,\deg( \overline{v}(x)) < p-1$. Since $\overline{\Phi}_p(x)$ is irreducible over $\F_l$, this is impossible, so $\Phi_p(x)$ is irreducible over $\Q$.
\end{proof}

Exercise 11.2.17

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Exercise 11.2.17

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