Structure of the ring of cyclic polynomials

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 $$i, $$\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 $$n=5$$?

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.

• 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). – Jyrki Lahtonen Aug 7 at 10:02
• Actually somewhat tempted to delete this. Deserves a closer look. – Jyrki Lahtonen Aug 7 at 10:04
• 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. – user687721 Aug 7 at 12:04