Exercise 11.2.4

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 Theorem 11.1.4 we used slitting fields to show that a field of order $p^n$ exists for any prime $p$ and integer $n\geq 1$. When Galois and others considered this question in the nineteenth century, their approach was to prove the existence of an irreducible polynomial in $\F_p[x]$ of degree $n$. In other words, they needed to prove that $N_n>0$.
\be
\item[(a)] Prove that $N_n>0$ using Theorem 11.1.4.
\item[(b)] Suppose that we have proved Theorem 11.2.4 but not Theorem 11.1.4. Use this to prove that $N_n>0$.
\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)] By Theorem 11.1.4, there exists a finite field $\F_{p^n}$ with $p^n$ elements. We know that the group $\F_{p^n}^*$ is cyclic. If $\alpha$ is a generator of $\F_{p^n}^*$, then $\alpha^{p^n-1} = 1$, and
$$\F_{p^n} =\{0,1,\alpha,\alpha^2,\ldots,\alpha^{p^{n}-2}\}= \F_p(\alpha).$$ This proves the Primitive Element Theorem in the case where the field is finite. Let $f$ be the minimal polynomial of $\alpha$. Then $\deg(f) = [\F_{p^n}:\F_p] = n$ and $f$ is monic irreducible over $\F_p$, so $N_n>0$.

\item[(b)] By Theorem 11.2.4,
 $$N_n = \frac{1}{n} \sum_{d\mid n} \mu\left(\frac{n}{d}ight) p^{d}.$$
 Let $n = p_1^{\alpha_1}\cdots p_s^{\alpha_s}$ be the decomposition of $n$ in primes.The factors $d$ of $n$ such that $\mu\left(\frac{n}{d}ight) 
e 0$ are the integers $d = p_1^{\alpha_1-\beta_1} \cdots p_s^{\alpha_s-\beta_s}$ where $\beta_i = 0,1$. The minimum such factor is $d_{min} = p_1^{\alpha_1-1} \cdots p_s^{\alpha_s-1}$.
 
 If $N_n=0$ then $\sum_{d\mid n} \mu(\frac{n}{d}) p^{d} = 0$. Dividing by $p^{d_{min}}$,
 $$\sum_{d\mid n,\, d > d_{min}} \mu\left(\frac{n}{d}ight) p^{d-d_{min}}=-\mu\left(\frac{n}{d_{min}}ight)  =  \pm 1.$$
 As $p$ divides the left sum, $p\mid 1$. This is a contradiction, so $N_n >0$.
 
 (This gives another proof of Theorem 11.1.4.)
 \ee
 \end{proof}

Exercise 11.2.4

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Exercise 11.2.4

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