Given a field $K$ (say, of characteristic zero) and $K[x_1,\dots,x_n]$ the ring of polynomials in $n$-variables, consider the action of the cyclic group $C_n=<\sigma>$ with $n$ elements by $\sigma(x_i)=x_{i+1}$ if $i<n$, $\sigma(x_n)=x_1$. The polynomials $p(x_1,\dots,x_n)$ such that $\sigma(p)=p$ are called cyclic polynomials.

Question: Can we describe the ring $CP_n$ of cyclic polynomials, giving generators and relations?

If $n=2$, then cyclic is the same than symmetric, hence the two elementary symmetric polynomials $s_1$ and $s_2$ generate and there are no relations. So $CP_2=K[s_1,s_2]$.

If $n=3$, then cyclic is the same than alternating, and it is known in this case that the elementary symmetric polynomials $s_1,s_2,s_3$ plus the polynomial $d:=(x_1-x_2)(x_2-x_3)(x_1-x_3)$ generate. But $d^2=\Delta$ is the discriminant, which can be express as a polynomial in $s_1,s_2,s_3$. Then $$CP_3=K[s_1,s_2,s_3,d]/(d^2-\Delta).$$

Can we give a description as above in the cases $n=4$ and $n=5$?


1 Answer 1


Not a full answer. Just describing what happens at the level of rational functions as an answer, because it does not fit into a comment. Invariant theory has a lot to say about this and related problems.

Let $E=K(x_1,x_2,\ldots,x_n)$ by the field of rational functions in $n$ independent indeterminates. Let $F=K(s_1,s_2,\ldots,s_n)$ be the subfield of symmetric rational functions. It is well known that $E/F$ is a Galois extension with Galois group $G\cong S_n$ acting by permuting the indeterminates.

Consider the element $$ u=\frac{x_1}{x_2}+\frac{x_2}{x_3}+\cdots+\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}. $$ It is easy to see that $\sigma(u)=u$ if and only if $\sigma$ is a power of the $n$-cycle $\alpha=(123\ldots n)$. By basic Galois theory this implies that $F(u)$ is the fixed field $E^H$ of $H=\langle\alpha\rangle$. Alternatively, you can show that the orbit of $u$ under $G$ has $(n-1)!$ elements.

We can bring all the fractions in $u$ together as follows $$ u=\frac{\sum_{i=1}^nx_i^2\prod_{j,j\neq i, j\not\equiv i+1\pmod n}x_j}{x_1x_2\ldots x_n}. $$ The denominator is in $E$. It may be tempting to conjecture that elementary symmetric polynomials together with the numerator might generate the ring of polynomial invariants. Unfortunately I don't have any intuition about this. It need not be that simple.

In the case of finite groups generated by reflections (= finite Coxeter groups), there is a beautiful description of the polynomial invariants due to Chevalley. See for example Chapter 3 of J.E. Humphreys, Reflection Groups and Coxeter Groups, Cambridge studies in advanced mathematics #29. I suspect the answer to the present question is also known, but I don't have the time to search for it.

  • $\begingroup$ Hmm. A more obvious candidate to replace $u$ would be $$\sum_{cyc}x_i^2x_{i+1}.$$ The analogue (if any survive) to Coxeter groups suggests that a more natural generating set exists (and may have only $n$ elements). $\endgroup$ Commented Aug 7, 2019 at 10:02
  • $\begingroup$ Actually somewhat tempted to delete this. Deserves a closer look. $\endgroup$ Commented Aug 7, 2019 at 10:04
  • 1
    $\begingroup$ Before you decide to delete it, the invariant theory (of this example) is also discussed in detail in this answer (mathoverflow.net/questions/14613/…). Summary: it's pretty messy/complicated. $\endgroup$
    – user687721
    Commented Aug 7, 2019 at 12:04

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