Exercise 2.2.4

Question

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\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

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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}}

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Let $f = x^3+bx^2+cx+d \in F[x]$ have roots $\alpha_1,\alpha_2,\alpha_3$ in the field $L$ containing $F$, and let $g$ be the polynomial defined in (2.17). Show carefully that
$$g(x) = x^3 + 2bx^2 + (b^2+c)x +bc-d.$$

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
\begin{align*}
f &= x^3+bx^2+cx+d\
&=( x-\alpha)(x-\beta)(x-\gamma)\
&= x^3 - \sigma_1(\alpha,\beta,\gamma)x^2 + \sigma_2(\alpha,\beta,\gamma)x - \sigma_3(\alpha,\beta,\gamma),\
\end{align*}
which gives
\begin{align*}
& \sigma_1(\alpha,\beta,\gamma) = -b,\
& \sigma_2(\alpha,\beta,\gamma)=+c,\
& \sigma_3(\alpha,\beta,\gamma)=-d.
\end{align*}
Let
\begin{align*}
G(x) &= (x -(x_1+x_2))(x-(x_1+x_3))(x-(x_2+x_3)).\
\end{align*}
Then $$g(x) = (x -(\alpha_1+\alpha_2))(x-(\alpha_1+\alpha_3))(x-(\alpha_2+\alpha_3))$$ is obtained from $G$ by the evaluation morphism which sends $x_1,x_2,x_3$ on $\alpha_1,\alpha_2,\alpha_3$.

Let  $p = (x+x_1)(x+x_2)(x+x_3) = x+ \sigma_1x^2+\sigma_2 x +\sigma_3$.

Then
\begin{align*}
G &= (x - \sigma_1 + x_3)(x-\sigma_1+x_2)(x - \sigma_1 + x_1)\
&=p(x-\sigma_1)\
&=(x-\sigma_1)^3 + \sigma_1 (x-\sigma_1)^2 + \sigma_2(x-\sigma_1) + \sigma_3\
&=(x^3 -3 \sigma_1 x^2 + 3 \sigma_1^2 x - \sigma_1^3) + (\sigma_1x^2 - 2 \sigma_1^2x + \sigma_1^3)+ (\sigma_2 x - \sigma_1 \sigma_2) + \sigma_3\
&=x^3 - 2 \sigma_1 x^2 + (\sigma_1^2+\sigma_2)x + (\sigma_3 - \sigma_1\sigma_2).
\end{align*}
The previous  evaluation morphism sends $\sigma_1$ on $\sigma_1(\alpha_1,\alpha_2,\alpha_3)=-b$, 
$\sigma_2$ on $\sigma_2(\alpha_1,\alpha_2,\alpha_3)=c$, $\sigma_3$ on $\sigma_3(\alpha_1,\alpha_2,\alpha_3)=-d$.

\begin{align*}
g(x) &= x^3 +2bx^2+(b^2+c)x+bc-d.
\end{align*}

In the example 2.2.6, 
$$f(x) = x^3+2x^2+x+7,$$
where
$$b=2,c=1,d=7,$$
$\alpha_1,\alpha_2,\alpha_3$ being the roots of $g$ in $\mathbb{C}$,
we obtain
\begin{align*}
g(x) &= (x -(\alpha_1+\alpha_2))(x-(\alpha_1+\alpha_3))(x-(\alpha_2+x\alpha_3))\
&=x^3 +2bx^2+(b^2+c)x+bc-d\
&=x^3 + 4x^2+5x-5.
\end{align*}
\end{proof}

Exercise 2.2.4

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Exercise 2.2.4

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