Exercise 7.4.3

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}}

This exercise will study part (b) of Theorem 7.4.4 when $f$ is a polynomial in $x_1,\ldots,x_n$ that is invariant under $A_n$. The theorem implies that $f = A+B\sqrt{\Delta}$ for some $A,B \in F(\sigma_1,\ldots,\sigma_n)$. You will prove that $A$ and $B$ are polynomials in the $\sigma_i$. Recall that $F$ is a field of characteristic $
e 2$.
\be
\item[(a)] Show that $f + (1\,  2)\cdot f = 2A$.
\item[(b)] In part (a), the left-hand side is a polynomial while the right-hand side is a symmetric rational function. Use theorem 2.2.2 to conclude that $A$ is a polynomial in the $\sigma_i$.
\item[(c)] Let $P$ denote the product of $f-A$ and $(1\,2)\cdot (f-A)$. Show that $P = -B^2\Delta$.
\item[(d)] Let $B = u/v$, where $u,v \in F[\sigma_1,\ldots, \sigma_n]$ are relatively prime (recall that $F[\sigma_1,\ldots,\sigma_n]$ is a UFD). In Exercise 8 of section 2.4 you showed that $\Delta$ is irreducible in $F[\sigma_1,\ldots,\sigma_n]$. Use this and the equation $v^2P =-u^2\Delta$ to show that $v$ must be constant. This will prove that $B\in F[\sigma_1,\ldots,\sigma_n]$.
\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}
Let  $f = f(x_1,\cdots,x_n) \in F[x_1,\cdots,x_n]$ that is invariant under $A_n$.
By Theorem 7.4.4, $f=A + B\sqrt{\Delta}, \ A,B \in F(\sigma_1,\cdots,\sigma_n)$.
\begin{enumerate}
\item[(a)] Let $	au = (1\, 2)$.

By (7.16),  $	au \cdot\sqrt{\Delta} = \mathrm{sgn}(	au) \sqrt{\Delta} = - \sqrt{\Delta}$.

As $	au$  fixes $A,B \in F(\sigma_1,\cdots,\sigma_n)$, $	au \cdot f = A - B \sqrt{\Delta}$, thus
$$f + 	au \cdot f = 2 A.$$

\item[(b)] The polynomial $A =\frac{1}{2} (  f + 	au \cdot f) \in F[x_1,\cdots,x_n]$ satisfies $\sigma \cdot A = A$. By Theorem 2.2.2, $A = h(\sigma_1, \cdots,\sigma_n)$, where $h$ is a polynomial.

\item[(c)] Let $P = (f-A)(	au\cdot(f-A))$.

Then $P = (B\sqrt{\Delta})(-B \sqrt{\Delta}) = -B^2 \Delta$.

\item[(d)] Let $B = u/v, \ u,v \in F[\sigma_1,\cdots,\sigma_n]$, where $u,v$ are relatively prime. Then $$v^2 P = -u^2 \Delta.$$

As $	au \cdot P = (	au\cdot(f-A))(	au \cdot(	au\cdot(f-A))) = (	au\cdot(f-A))(f-A) = P$, $P$ is invariant under $A_n$ and also invariant under $	au$, thus is invariant under $S_n$, and $P$ is a polynomial in $x_1,\cdots,x_n$, since $f, A \in F[x_1,\ldots,x_n]$. Therefore there exists a polynomial $g$ such that $P = g(\sigma_1, \cdots, \sigma_n)$, and 
$v^2 g = -u^2 \Delta$ is an equality in $F[\sigma_1, \cdots \sigma_n]$: $u,v,g,\Delta \in F[\sigma_1,\cdots,\sigma_n]$.

By Exercise 2.4.8, $\Delta$ is irreducible in $F[\sigma_1,\cdots, \sigma_n]$. Moreover $v^2$ divides $u^2 \Delta$ and is relatively prime with $u^2$, thus $v^2$ divides $\Delta$, where $\Delta$ is irreducible. This is impossible, unless $v$ is a constant $\lambda \in F^*$. Therefore $B = \lambda^{-1} u$ is a polynomial in $\sigma_1,\cdots,\sigma_n$.

Conclusion: if  $f = f(x_1,\cdots,x_n) \in F[x_1,\cdots,x_n]$ is invariant under $A_n$, where the characteristic of $F$ is not 2,
then  $$f=A + B\sqrt{\Delta}, \ A,B \in F[\sigma_1,\cdots,\sigma_n].$$

\end{enumerate}
\end{proof}

Exercise 7.4.3

Answers

Comments

Exercise 7.4.3

Answers

Comments

Add answer