# Arithmetic Cohen-Macaulayness of curves/surfaces defined by weighted power sums in 3 variables

Pick $p,q,r$ complex numbers (I am most interested in the case when they are positive integers). Define the function

$P_i = px^i + qy^i + rz^i$

where $x,y,z$ are coordinates. I have a few related questions:

Has the algebra generated by $P_1, P_2, P_3, P_4, ...$ been considered before in the literature? For example, is it known how many $P_i$ are needed to generate the rest (this is a ring of invariants under a finite group for a finitely generated ring so there is always a bound for fixed $p,q,r$).

Has the Cohen-Macaulay property been considered for such algebras?

I am also interested in the related question where we impose the relation that $P_1 = 0$, i.e., $px+qy+rz=0$.

More generally, are there known criteria (say, on singularities) for a projective curve/surface that guarantee that it is arithmetically Cohen-Macaulay? (Probably there is nothing for surfaces, but I might hope it is possible for curves). Cohomological criteria aren't so useful in this situation, I think.

(This comes up in a research project related to subspace arrangements -- when $p,q,r$ are positive integers, these algebras are invariant rings under a symmetric group of their coordinate rings -- look at the locus in $C^{p+q+r}$ where there is a group of $p$ equal coordinates, $q$ equal coordinates, and $r$ equal coordinates).

EDIT: I know that the algebra can fail to be Cohen-Macaulay in some cases, for example when $(p,q,r)=(3,2,1)$, so I am interested in determining which parameters give which behavior.

Update: I thought more about it. The good news is two questions are equivalent, but the calculation seems a little more complicated then I thought.

So let $A$ be the algebra with condition $P_1=0$. Then $A=R/(P_1)$, so $A$ is CM iff $R$ is.

Let's focus on $A$ then. We still have $A[x,y]=k[x,y]$. But $x,y$ satisfies degree 6 (?) polynomials over $A$, so I am not sure if the trick will apply. There might be some symmetries I failed to see though. (end update)

Here is a strategy that works pretty well for the 2 variables case (see below), which should be similar to your case when you impose $P_1=0$.

Let $R$ be your algebra, it is trivial that $R[x,y] = k[x,y,z]$. We can use the following:

Lemma: Let $R$ be a domain and $a\in R$. Then $S= R[a^{1/n}]$ is CM iff $R$ is.

The point is $S\cong R[t]/(t^n-a)$ and you can pass CMness between the operations of adjoining a variable and killing a regular element.

By solving for $x,y$ one can probably write $k[x,y,z]$ as successive extensions of $R$ as in the Lemma.

Here is a back-of-napkin example calculation in the case of 2 variables, which should be similar to the case when $P_1=0$. Assume $pq(p-q)(p+q)\neq 0$ (the degenerate cases are simple), then we have $(x-y)^2= \frac{(p+q)P_2-P_1^2}{pq}$.

So let $x-y=c$, clearly $k[x,y] = R[c] = R[t]/(t^2-a)$ where $a= (p+q)P_2-P_1^2$. So $R$ is CM.

• I should have mentioned that the algebra isn't always CM: for example, when $(p,q,r)=(3,2,1)$ and with $P_1=0$, you can show that it is a finite module over $P_2,P_3$ and the numerator of its Hilbert series has negative coefficients. Jun 30 '14 at 12:05

I do not know if it is related to your problems, but there are two papers of R. Dvornicich and U. Zannier studying the field of fraction generated by subsets of the P_i in the case p,q and r are equal 1. See

R. Dvornicich, U. Zannier, Solution of a problem about symmetric functions, Rocky Mountain J. Math. 33 (2003) 1279–1287.

R. Dvornicich, U. Zannier, Newton functions generating symmetric fields and irreducibility of Schur polynomials. Adv. Math. 222 (2009), no. 6, 1982--2003.