Exercise 13.1.10

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}}

As in Example 13.1.4, let $f=x^4+bx^2+d \in F[x]$, where $d 
ot \in F^2$. Compute $\Delta(f)$ and $	heta_f(y)$.

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 $f$ is
$$\Delta(f) =  16 b^{4} d - 128 b^{2} d^{2} + 256 d^{3} = 16\  d\,  ( b^{2} - 4 d)^{2}.$$
The Ferrari resolvent is
$$	heta_f(y) = y^3-by^2-4dy+4bd = (y-b)(y^2-4d).$$

Sage instructions:
\begin{verbatim}
R.<x,b,d> = QQ[]
f=x^4+b*x^2+d
c1 = 0; c2 = b; c3 = 0; c4 = d;
theta_f = x^3 - c2*x^2 + (c1*c3-4*c4)*x - c3^2-c1^2*c4 + 4*c2*c4;
factor(theta_f)
\end{verbatim}
$$(- x + b) \cdot (- x^{2} + 4 d)$$
\begin{verbatim}
Delta = theta_f.discriminant(x)
factor(Delta)
\end{verbatim}
$$	
\left(16ight) \cdot d \cdot (- b^{2} + 4 d)^{2}
$$
Thus $	heta_f(y) = (y-b)(y - 2\sqrt{d})(y+2\sqrt{d})$ has a unique root in $F$ if $d 
ot \in F^2$, and the discriminant is not a square in $F^2$.
\end{proof}

Exercise 13.1.10

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Exercise 13.1.10

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