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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.