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Let $F$ be a field and $f(x_1,\ldots,x_n), g(x_1,\ldots,x_n) \in F[x_1,\ldots, x_n]$. Suppose that $h(x_1,\ldots,x_n):=f(x_1,\ldots,x_n)/g(x_1,\ldots, x_n)$ is symmetric, in the sense that for every $\sigma \in S_n$, $\Phi(\sigma)(h) = h$, where $\Phi(\sigma)$ is the unique automorphism of $F(x_1,\ldots,x_n)$ which extends the automorphism $\phi(\sigma)$ of $F[x_1,\ldots,x_n]$ which fixes $F$ and sends $x_i$ to $x_{\sigma(i)}$. Suppose further that $f$ and $g$ have no common factor.

How to prove that both $f$ and $g$ are symmetric? I tried hard on this one but no ideas come to mind.

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HINT:If $f/g$ is in lowest terms and $\frac{\sigma f}{\sigma g} = \frac{f}{g}$ for all $\sigma$ then $\sigma g$ divisible by $g$ for all $\sigma$. Taking $\sigma \mapsto \sigma^{-1}$ we get that $\sigma g = \epsilon_{\sigma} g$ for all $\sigma$, where $\sigma \mapsto \epsilon_{\sigma}$ is a ($1$-) character of $S_n$. If $\epsilon_{\sigma} \equiv 1$ then $g$ symmetric, done. Otherwise, $\epsilon_{\sigma}$ is the signature. But then $g$ is skew-symmetric, and then, necessarily, $f$ is also skew symmetric ( since their quotient is symmetric). But every skew-symmetric polynomial is divisible by $\delta=\prod_{i< j}(x_i - x_j)$, and thus $f$, $g$ are not relatively prime, contradiction.

Obs: It may be convenient to produce explicit symmetric polynomial as a quotient $\frac{f}{\delta}$, with $f$ skew-symmetric, $\delta$ as above, ( see Schur functions), although the fraction is clearly not in lowest terms.

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  • $\begingroup$ I don't see why $\sigma g$ is divisible by $\sigma$ $\endgroup$ – Cauchy Aug 20 '17 at 10:18
  • $\begingroup$ @Cauchy: It's a property of fractions in lowest terms.. You just said that $f$, $g$ have no common factor... $\endgroup$ – Orest Bucicovschi Aug 20 '17 at 10:21
  • $\begingroup$ That's a property of fractions of stuff in Bezout domains. $F[x_1,\ldots,x_n]$ is not a Bezout domain. $\endgroup$ – Cauchy Aug 20 '17 at 10:23
  • $\begingroup$ @Cauchy: that's also a property of fractions for factorial domains, you can check that. So, if $a | b c $ and $a$, $b$ relatively prime, then $a| c$. $\endgroup$ – Orest Bucicovschi Aug 20 '17 at 10:25
  • $\begingroup$ Oh! Of course, we can do that, we don't need Bezout. Sorry. Thank you for your answer. Will take a look at it later (must learn what a character is and why every skew-symmetric polynomial is divisible by $\delta$) $\endgroup$ – Cauchy Aug 20 '17 at 10:30

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