Exercise 2.2.19

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}}

Suppose that $F$ is a field of characteristic 0.
\begin{enumerate}
\item[(a)] Use the Newton identities (2.22) and Theorem 2.2.2 to prove that every symmetric polynomial in $F[x_1,\ldots,x_n]$ can be expressed as a polynomial in $s_1,\ldots,s_n$.
\item[(b)] Show how to express $\sigma_4 \in F[x_1,x_2,x_3,x_4]$ as a polynomial in $s_1,s_2,s_3,s_4$.
\end{enumerate}

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}
For all $r, 1\leq r \leq n$,
 $$s_r = \sigma_1 s_{r-1}-\sigma_2 s_{r-2}+\cdots+(-1)^r\sigma_{r-1}s_1 + (-1)^{r-1} r \sigma_r,$$
 and $\sigma_1 = s_1$.
 
 If we suppose that $\sigma_1,\sigma_2,\cdots,\sigma_{r-1}$ are polynomials in $s_1,s_2,\cdots,s_n$, the characteristic of the field $F$ being 0 (this allows the division by $r$), then
 
 $\sigma_r =\frac{ (-1)^{r-1}}{r}(s_r - \sigma_1 s_{r-1}+\sigma_2 s_{r-2}+\cdots+(-1)^{r-1} \sigma_{r-1}s_1)$
is a polynomial in $s_1,\cdots,s_n$.
 
 Conclusion : for all $r,1\leq r \leq n$, $\sigma_r$ can be expressed as a polynomial in $s_1,\cdots,s_n$.
 
By Ex. 2.2.17, we obtain
 \begin{align*}
\sigma_1 &= s_1,\
\
 \sigma_2 &= -\frac{1}{2}(s_2 - \sigma_1 s_1 )\
 &=\frac{1}{2}(s_1^2-s_2),\
 \
 \sigma_3 &= \frac{1}{3}(s_3-\sigma_1s_2+\sigma_2s_1)\
 &=\frac{1}{3}\left[s_3-s_1s_2+\frac{1}{2}s_1(s_1^2-s_2) ight]\
 &=\frac{1}{6}(2s_3+s_1^3-3s_1s_2),\
 \
 \sigma_4 &=-\frac{1}{4}(s_4-\sigma_1s_3+\sigma_2s_2-\sigma_3s_1)\
 &=-\frac{1}{4} \left[  s_4 - s_1s_3 +\frac{1}{2} s_2(s_1^2-s_2) -\frac{s_1}{6}(2s_3-3s_1s_2+s_1^3) ight]\
 &=-\frac{1}{24}\left[   6s_4 - 6 s_1s_3 +3 s_2(s_1^2-s_2)-s_1(2s_3-3s_1s_2+s_1^3)ight]\
 &=\frac{1}{24}(-6s_4+8s_1s_3-6s_1^2s_2+3s_2^2+s_1^4).
 \end{align*}
\end{proof}

Exercise 2.2.19

Answers

Comments

Exercise 2.2.19

Answers

Comments

Add answer