Exercise 13.3.15

Question

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

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ewcommand{\ssi} {si et seulement si }

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ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

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ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $f=x^3-c_1x^2+c_2x-c_3 \in F[x]$ and let $D_f(y) \in F[y]$ be as in the previous exercise.
\be
\item[(a)] Show that $D_f(y)=y^2-(c_1c_2-3c_3)y+c_2^3+c_1^3c_3-6c_1c_2c_3+9c_3^2$.
\item[(b)] Assume in addition that f is separable and irreducible. Explain how $D_f(y)$ determines Galois group of f up to isomorphism.
\ee

\begin{proof}
\be
\item[(a)] By (2.26), we obtain
$$\Delta(f)=-4c_2^3-27c_3^2+c_1^2c_2^2-4c_1^3c_3+18c_1c_2c_3,$$
and by Ex.2.2.10,
 $$D+D'=\sum_{\sigma \in A_3}\sigma\cdot x_1^2x_2+(12)\sum_{\sigma \in A_3}\sigma\cdot x_1^2x_2=\sum_{\sigma \in S_3}\sigma\cdot x_1^2x_2=\sigma_1\sigma_2-3\sigma_3,$$ 
 so that
 $$D(f) + D'(f) = c_1c_2 - 3c_3.$$
(Note: If the characteristic is not 2, we could write
$$D(f) = \frac{1}{2}(D(f) + D'(f) + (D(f) - D'(f)) = \frac{1}{2}(c_1c_2 - 3c_3 \pm \sqrt{\Delta(f)}),$$
and compute $D(f)D'(f)$, but here we compute in all characteristic.)

But $DD'$ is symmetric, where
\begin{align*}
D D' &= \left(\sum_{\sigma \in A_3}\sigma\cdot x_1^2x_2ight)\left(\sum_{\sigma \in A_3}(	au\sigma)\cdot x_1^2x_2ight)\
&=(x_1^2x_2+x_2^2x_3+x_3^2x_1)(x_2^2x_1+x_1^2x_3+x_3^2x_2)\
&=x_{1}^{3} x_{2}^{3} + x_{1}^{4} x_{2} x_{3} + x_{1} x_{2}^{4} x_{3} + 3 x_{1}^{2} x_{2}^{2} x_{3}^{2} + x_{1}^{3} x_{3}^{3} + x_{2}^{3}x_{3}^{3} + x_{1} x_{2} x_{3}^{4} 
\end{align*}
With the usual Sage instructions, we obtain $D(f)D'(f)$
\begin{verbatim}
R.<x1,x2,x3,y1,y2,y3> = PolynomialRing(QQ, order = 'degrevlex')
elt = SymmetricFunctions(QQ).e()
e = [elt([i]).expand(3).subs(x0=x1, x1=x2, x2=x3) for i in range(4)]
J = R.ideal(e[1]-y1, e[2]-y2, e[3]-y3)
G = J.groebner_basis()
f = (x1^2*x2+x2^2*x3+x3^2*x1)*(x2^2*x1+x1^2*x3+x3^2*x2)
var('sigma_1,sigma_2,sigma_3')
var('c1,c2,c3')
g=f.reduce(G).subs(y1=c1,y2=c2,y3=c3)
g
\end{verbatim}
$$c_{1}^{3} c_{3} + c_{2}^{3} - 6 \, c_{1} c_{2} c_{3} + 9 \, c_{3}^{2}$$
Then 
\begin{align*}
 D(f)D'(f) &= c_2^3+c_1^3c_3-6c_1c_2c_3+9c_3^2
\end{align*}
and 
\begin{align*}
D_f(y) &=y^2-(D(f)+D'(f))y+D(f)D'(f) \
& =y^2-(c_1c_2-3c_3)y+c_2^3+c_1^3c_3-6c_1c_2c_3+9c_3^2
\end{align*}
\item[(b)] By Ex.6.2.6, $f$ being irreducible and separable, we know that $3 \mid |G_f|$. Therefore $G_f = A_3$ or $G_f = S_3$. Then Ex.14(d) proves that
\begin{align*}
G_f &= A_3 \iff D_f(y)	ext{ splits over } F,\
G_f &=S_3 \iff D_f(y) 	ext{ is irreducible over } F.
\end{align*}
\ee
\end{proof}

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}}

\subsection{Other Methods}

Exercise 13.3.15

Answers

0.1 Other Methods

Comments

Exercise 13.3.15

Answers

0.1 Other Methods

Comments

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