Exercise 11.2.6

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}}

This exercise is concerned with Example 11.2.8.
\be
\item[(a)] Show that $N_4 = 3$ when $p=2$. Then write down these three irreducible polynomials explicitly.

\item[(b)] Verify the factorization of $\Phi_{15}(x)$ given in the example.

\item[(c)] Show that the roots of $x^4+x^3+1$ and $x^4+x+1$ are the reciprocals of each other.
\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)] For $p=2$, by Exercise 4, $$N_4 = \frac{1}{4}(2^4-2^2) = 3.$$ With the Sage instructions \begin{verbatim} R. = GF(2)[] factor(x^16-x) \end{verbatim} we obtain $$x^{16} - x = x \cdot (x + 1) \cdot (x^{2} + x + 1) \cdot (x^{4} + x + 1) \cdot (x^{4} + x^{3} + 1) \cdot (x^{4} + x^{3} + x^{2} + x + 1).$$ So the irreducible polynomial of degree 4 are $$ x^{4} + x + 1 , \qquad x^{4} + x^{3} + 1 ,\qquad x^{4} + x^{3} + x^{2} + x + 1.$$ \item[(b)]As $x^{15} - 1 = \Phi_1(x) \Phi_3(x) \Phi_5(x) \Phi_{15}(x)$, \begin{align*} \Phi_{15}(x) &= \frac{(x^{15}-1)(x-1)}{(x^3-1)(x^5-1)}\ &= \frac{x^{10}+x^5+1}{x^2+x+1}\ &= x^8+x^7 +x^5 +x^4+x^3 + x+1 \end{align*} The Sage instructions \begin{verbatim} R.= GF(2)[] p = (x^10+x^5+1)/(x^2+x+1) factor(p) \end{verbatim} give the factorization in $\F_2[x]$ $$\Phi_{15}(x) = (x^{4} + x + 1) \cdot (x^{4} + x^{3} + 1).$$ \item[(c)] If $\alpha$ is a root of $x^4+x^3+1$, then $\alpha e 0$ and $\alpha^4 + \alpha^3+1 = 0$. Dividing by $\alpha^4$, $$1 + \frac{1}{\alpha} + \frac{1}{\alpha^4} = 0,$$ $\alpha^{-1}$ is a root of $x^4+x+1$. Similarly, if $\beta$ is a root of $x^4+x+1$, $\beta^{-1}$ is a root of $x^4+x^3+1$. (All these roots are the primitive $15$th roots of unity in $\F_{16}$.) \ee \end{proof}

Exercise 11.2.6

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Exercise 11.2.6

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