Exercise 2.2.10

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}}

Apply the proof of theorem 2.2.2 to express $\sum_3 x_1^2x_2$ in terms of $\sigma_1, \sigma_2, \sigma_3$.

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

Note that $x_1^2 x_2 = x_1^2 x_2^1 x_3^0$ is the leading term for the graded lexicographic order, so the following term in the sequence is $g = f - \sigma_1^{2-1} \sigma_2^{1-0} \sigma_3^0 = f - \sigma_1 \sigma_2$.

\begin{align*}
\sigma_1 \sigma_2 &= (x_1+x_2+x_3) (x_1 x_2+x_1x_3+x_2x_3)\
&= x_1^2x_2 + x_1^2 x_3+x_1x_2x_3+x_1x_2^2+x_2^2x_3+x_1x_2x_3+x_1x_3^2+x_2x_3^2+x_1x_2x_3\
&=f+ 3x_1x_2x_3,
\end{align*}
thus
$$f = \Sigma_3 x_1^2 x_2  = \sigma_1\sigma_2 - 3 \sigma_3.$$
\end{proof}

Exercise 2.2.10

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Exercise 2.2.10

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