Exercise 13.1.8

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 Example 10.3.10, we showed that the roots of $f=7m^4 - 16 m^3 -21 m^2 + 8m+4 \in \Q[m]$ can be constructed using origami. Show that the splitting field of $f$ is an extension of $\Q$ of degree 24. By the results of Section 10.1, it follows that the roots of $f$ are not constructible with straightedge and compass, since $24$ is not a power of $2$.

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} The discriminant of $g = \frac{1}{7} f$ is $$\Delta(g) = \frac{174446784}{117649} = 2^6\cdot 3^6\cdot 3739\cdot 7^{-6},$$ so $\Delta(g)$ is not a square in $\Q$. The Ferrari resolvent is $$ heta_f(y) = y^{3} + 3 y^{2} - \frac{240}{49} y - \frac{3824}{343}.$$ and $$7^3 heta_f(y) = 343 y^{3} + 1029 y^{2} - 1680 y - 3824$$ has no root in $\Q$, so is irreducible over $\Q$. By theorem 13.1.1, $G = S_4$. Therefore the splitting field $L$ of $f$ has degree $$[L:\Q] = |G| = 24.$$ \bigskip Sage instructions : \begin{verbatim} var('m') R. = QQ[m] f= 7*m^4-16*m^3-21*m^2+8*m+4 g=f/7 d=g.discriminant() d.factor() \end{verbatim} $$2^{6} \cdot 3^{6} \cdot 7^{-6} \cdot 3739$$ \begin{verbatim} R. = QQ[] l = f.coefficients(sparse=False); c1 = -l[3]/l[4]; c2 = l[2]/l[4];c3 = -l[1]/l[4]; c4 = l[0]/l[4]; theta_f = y^3 -c2*y^2 +(c1*c3-4*c4)*y - c3^2-c1^2*c4 + 4*c2*c4; \end{verbatim} $$y^{3} + 3 y^{2} - \frac{240}{49} y - \frac{3824}{343}$$ \begin{verbatim} theta_f.is_irreducible() \end{verbatim} $$ ext{True}$$ \end{proof}

Exercise 13.1.8

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Exercise 13.1.8

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