Exercise 13.2.6

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}}

In this exercise, you will use Maple or Mathematica (or Sage!), to prove (13.23) and (13.24).
\be
\item[(a)] The first step is to enter (13.17) and call it, for example $u1$. Then use substitution commands and (13.19) to create $u2,\ldots u6$. For example, $u2$ is obtained by applying $(1\,2\,3)$ to $u1$. In Maple, this is done via the command

\begin{verbatim}
           u2 := subs({x1=x2,x2=x3,x3=x1},u1);
\end{verbatim}
whereas in Mathematica one uses
\begin{verbatim}
           u2 := u1 /. {x1->x2,x2->x3,x3->x1}
\end{verbatim}
\item[(b)] Now multiply out $\Theta(y) = (y-u1)\cdots(y-u6)$ and use the methods of section 2.3 to express the coefficients of $\Theta(y)$ in terms of the elementary symmetric polynomials.
\item[(c)] Show that your results imply (13.23) and (13.24).
\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}
Sage instructions :
\begin{verbatim}
R.<y,x,x1,x2,x3,x4,x5,y1,y2,y3,y4,y5> = PolynomialRing(QQ, order = 'degrevlex')
elt = SymmetricFunctions(QQ).e()
e = [elt([i]).expand(5).subs(x0=x1, x1=x2, x2=x3, x3 = x4, x4 = x5) 
               for i in range(6)]
J = R.ideal(e[1]-y1, e[2]-y2, e[3]-y3,e[4]-y4,e[5]-y5)
G = J.groebner_basis()
u1 =x1*x2 + x2*x3 + x3*x4 + x4*x5 + x5*x1 - x1*x3 - x3*x5 -x5*x2 - x2*x4 -x4*x1
u2 = u1.subs(x1 = x2, x2 = x3, x3 = x1)
u3 = u1.subs(x2 = x3, x3 = x4, x4 = x2)
u4 = u1.subs(x3 = x4, x4 = x5, x5 = x3)
u5 = u1.subs(x1 = x4, x4 = x5, x5 = x1)
u6 = u1.subs(x1 = x2, x2 = x5, x5 = x1)
f1 = (y-u1) * (y-u2) * (y-u3) * (y-u4) * (y-u5) * (y-u6) 
var('sigma_1,sigma_2,sigma_3,sigma_4,sigma_5')
g = f1.reduce(G).subs(y1=sigma_1, y2=sigma_2, y3=sigma_3, y4=sigma_4, y5= sigma_5)
h = g.collect(y);
\end{verbatim}

\bigskip

Now we can verify (13.23) and (13.24):

\begin{verbatim}
h.coefficient(y^5),h.coefficient(y^3)
\end{verbatim}
$$(0,0)$$
\begin{verbatim}
B2 = h.coefficient(y^4); B2
\end{verbatim}
$$	
-3 \, \sigma_{2}^{2} + 8 \, \sigma_{1} \sigma_{3} - 20 \, \sigma_{4}
$$

\begin{verbatim}
B4 = h.coefficient(y^2); B4
\end{verbatim}
\begin{align*}
&3 \, \sigma_{2}^{4} - 16 \, \sigma_{1} \sigma_{2}^{2} \sigma_{3} + 16 \, \sigma_{1}^{2} \sigma_{3}^{2} + 16 \, \sigma_{1}^{2} \sigma_{2} \sigma_{4} - 64 \, \sigma_{1}^{3} \sigma_{5} + 16 \, \sigma_{2} \sigma_{3}^{2} - 8 \, \sigma_{2}^{2} \sigma_{4} \
&- 112 \, \sigma_{1} \sigma_{3} \sigma_{4} + 240 \, \sigma_{1} \sigma_{2} \sigma_{5} + 240 \, \sigma_{4}^{2} - 400 \, \sigma_{3} \sigma_{5}
\end{align*}
\begin{verbatim}
B6 = h.subs(y = 0);B6
\end{verbatim}
\begin{align*}
&-\sigma_{2}^{6} + 8 \, \sigma_{1} \sigma_{2}^{4} \sigma_{3} - 16 \,
\sigma_{1}^{2} \sigma_{2}^{2} \sigma_{3}^{2} - 16 \, \sigma_{1}^{2}
\sigma_{2}^{3} \sigma_{4} + 64 \, \sigma_{1}^{3} \sigma_{2} \sigma_{3}
\sigma_{4}\
 &- 64 \, \sigma_{1}^{4} \sigma_{4}^{2} - 16 \, \sigma_{2}^{3}
\sigma_{3}^{2} + 64 \, \sigma_{1} \sigma_{2} \sigma_{3}^{3} + 28 \,
\sigma_{2}^{4} \sigma_{4} - 112 \, \sigma_{1} \sigma_{2}^{2} \sigma_{3}
\sigma_{4} - 128 \, \sigma_{1}^{2} \sigma_{3}^{2} \sigma_{4} + 224 \,
\sigma_{1}^{2} \sigma_{2} \sigma_{4}^{2} \
&+ 48 \, \sigma_{1}
\sigma_{2}^{3} \sigma_{5} - 192 \, \sigma_{1}^{2} \sigma_{2} \sigma_{3}
\sigma_{5} + 384 \, \sigma_{1}^{3} \sigma_{4} \sigma_{5} - 64 \, 
\sigma_{3}^{4} + 224 \, \sigma_{2} \sigma_{3}^{2} \sigma_{4} - 176 \,
\sigma_{2}^{2} \sigma_{4}^{2} - 64 \, \sigma_{1} \sigma_{3} \sigma_{4}^{2} \
&- 80 \, \sigma_{2}^{2} \sigma_{3} \sigma_{5} + 640 \,
\sigma_{1} \sigma_{3}^{2} \sigma_{5} - 640 \, \sigma_{1} \sigma_{2} 
\sigma_{4} \sigma_{5} - 1600 \, \sigma_{1}^{2} \sigma_{5}^{2} + 320 \,
\sigma_{4}^{3} - 1600 \, \sigma_{3} \sigma_{4} \sigma_{5} + 4000 \,
\sigma_{2} \sigma_{5}^{2}
\end{align*}

The coefficient $c_1$ of $y$ in $h = \Theta(y)$ is not symmetric in $x_1,\ldots,x_5$, but we verify that $c_1 = 2^5\sqrt{\Delta}\, y$, computing first $\sqrt{\Delta}$:
\begin{verbatim}
c1  = f1.coefficient(y)
x = [1,x1,x2,x3,x4,x5]
sqrtDelta = 1
for i in range(1,6):
    for j in range(i+1,6):
        sqrtDelta *= (x[i] -x[j])
sqrtDelta
c1 + 2^5 * sqrtDelta
\end{verbatim}
$$0$$

So (13.23) and (13.24) are verified.
\end{proof}

Exercise 13.2.6

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Exercise 13.2.6

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