Exercise 2.4.11

Question

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Exercise 8 of section 2.2 showed that if $\varphi \in F(x_1,\ldots,x_n)$ is symmetric, then $\varphi \in F(\sigma_1,\ldots,\sigma_n)$. In this exercise, you will refine this result as follows. Suppose that $\varphi \in F(x_1,\ldots,x_n)$ is symmetric, and write $\varphi = A/B$, where $A,B \in F[x_1,\ldots,x_n]$ are relatively prime. The claim is that A,B are themselves symmetric and hence lie in $F[\sigma_1,\ldots,\sigma_n]$. We can assume that $A$ and $B$ are nonzero.
\begin{enumerate}
\item[(a)] Use the previous exercise and Exercise 8 of section 2.2 to show that $\varphi = C/D$ where $C,D \in F[\sigma_1,\ldots,\sigma_n]$ are relatively prime in $F[x_1,\ldots,x_n]$.
\item[(b)] Show that $AD = BC$ and then use unique factorization in $F[x_1,\ldots,x_n]$ to show that $A$ and $B$ are constant multiples of $C$ and $D$ respectively.
\item[(c)] Conclude that $A,B \in F[\sigma_1,\ldots,\sigma_n]$ as claimed.
\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}
\begin{enumerate}
\item[(a)]
As $\varphi \in F(\sigma_1,\cdots,\sigma_n)$, by Exercise 2.2.8,
$$\varphi = C/D, \ C,D \in F[\sigma_1,\cdots,\sigma_n].$$
Reducing this fraction, we can suppose that $C,D$ are relatively prime in $F[\sigma_1,\cdots,\sigma_n]$, thus relatively prime in  $F[x_1,\cdots, x_n]$ by Exercise 2.4.10.

\item[(b)]
$\varphi = A/B = C/D$, so $AD = BC$, where $C,D$ are symmetric and relatively prime in  $F[x_1,\cdots,x_n]$ , and also $A,B$  relatively prime in  $F[x_1,\cdots,x_n]$.

As $F[x_1,\ldots,x_n]$ is a unique factorisation domain, as $A \mid BC$ and $A,B$ are relatively prime, $A \mid C$. Similarly, $C \mid AD$, and $C,D$ are relatively prime,  so $C \mid A$ : $A$ and $C$ are associate, therefore
$$A = k C, B = k D, k\in F^*.$$

\item[(c)]

Since $C,D$ are symmetric, $A,B$ are also symmetric.

Conclusion : if $\varphi=A/B$ is symmetric, where $A,B \in F[x_1,\ldots,x_n]$ are relatively prime, then $A,B$ are symmetric.
\end{enumerate}
\end{proof}

Exercise 2.4.11

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Exercise 2.4.11

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