Exercise 2.3.3

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}}

Express $\Sigma_3 x_1^3 x_2^2$ in terms of elementary symmetric polynomials.

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}
Explicitly, 
$$f = \Sigma_3 x_1^3 x_2^2 = x_1^2x_2^3+x_1^2 x_3^3 + x_1^3 x_2^2+x_1^3 x_3^2 +x_2^2x_3^3+x_2^3 x_3^2.$$

Note that $x_1^3 x_2^2 = x_1^3 x_2^2 x_3^0$ is the leading term for the graded lexicographic order, so the following term in the sequence is $g = f - \sigma_1^{3-2} \sigma_2^{2-0} \sigma_3^0 = f - \sigma_1 \sigma_2^2$.

\begin{align*}
\sigma_1 \sigma_2^2 &= (x_1+x_2+x_3) (x_1 x_2+x_1x_3+x_2x_3)^2\
&= (x_1+x_2+x_3) (x_1^2 x_2^2+x_1^2x_3^2+x_2^2x_3^2+2x_1x_2x_3(x_1+x_2+x_3))\
&= x_1^2x_2^3+x_1^2 x_3^3 + x_1^3 x_2^2+x_1^3 x_3^2 +x_2^2x_3^3+x_2^3 x_3^2+x_1x_2x_3(x_1 x_2+x_1x_3+x_2x_3)+2\sigma_3 \sigma_1^2 \
&=f+ \sigma_3 \sigma_2+2\sigma_3 \sigma_1^2,
\end{align*}
thus
$$f =\sigma_1 \sigma_2^2- \sigma_3 \sigma_2-2\sigma_3 \sigma_1^2.$$
Verification with Sage:
\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(5)]
    J = R.ideal(e[1]-y1, e[2]-y2, e[3]-y3)
    G = J.groebner_basis()
    D = x1^3*x2^2;
    u = D + D.subs(x1=x2, x2=x1) + D.subs(x1=x3, x2=x2) + \
    	D.subs(x1=x2, x2=x3) + D.subs(x1=x3, x2=x1)+D.subs(x1=x1, x2=x3)
    var('sigma_1, sigma_2, sigma_3')
    u.reduce(G).subs(y1 = sigma_1, y2 = sigma_2,y3 = sigma_3)
\end{verbatim}
$$\sigma_1 \sigma_2^2-2\sigma_1^2\sigma_3 - \sigma_2 \sigma_3$$
\end{proof}

Exercise 2.3.3

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Exercise 2.3.3

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