Fix $n > 1$ and let $\zeta \in \mathbb{C}$ be a primitive $n$th root of unity. Let $G \subset \text{SL}_2(\mathbb{C})$ be a cyclic subgroup of order $n$ generated by the diagonal matrix $g = \text{diag}(\zeta, \zeta^{1})$. The group $G$ acts naturally on $\mathbb{C}[x, y]$. Let $\mathbb{C}[x, y]^G$ be the algebra of $G$invariant polynomials, namely $$\mathbb{C}[x, y]^G = \{f \in \mathbb{C}[x, y] : f(\zeta^{1}x, \zeta y) = f(x, y)\}.$$What is the description of the algebra $\mathbb{C}[x, y]^G$ in terms of generators and relations?

1$\begingroup$ That is, roughly, a cone over a rational normal curve. The generators are $u = xy$ and then $v_{l,m} = x^l y^m$ for nonnegative integers $l$, $m$ such that $l + m$ equals $n$. There are some obvious relations among the $v_{l,m}$, namely $v_{l,m}v_{l',m'} = v_{l'',m''}v_{l''',m'''}$ if $l+l'$ equals $l''+l'''$ and $m+m'$ equals $m''+m'''$. If $n$ is even, then there is a relation $u^{n/2} = v_{n/2,n/2}$. If $n$ is odd, then the relation is $v_{l,m}v_{m,l} = u^n$. $\endgroup$ – Jason Starr Dec 5 '15 at 14:41

1$\begingroup$ Francesco is correct: some of my relations were incorrect. I was confusing the $(\zeta^{1},\zeta)$action with the $(\zeta,\zeta)$action. $\endgroup$ – Jason Starr Dec 5 '15 at 15:02

1$\begingroup$ To correct what I wrote: among the $v_{l,m}$ I wrote down, only $v_{n,0}$ and $v_{0,n}$ are invariant. These satisfy the single relation $v_{n,0}v_{0,n} = u^n$. Sorry about the mistake. $\endgroup$ – Jason Starr Dec 5 '15 at 15:04

1$\begingroup$ @JasonStarr: I was confused, too, and my answer was wrong. I've edited it, and now it should be ok. $\endgroup$ – Francesco Polizzi Dec 5 '15 at 19:34
EDIT. This is a new version of the answer. Since at some point there was confusion between the $(\zeta, \, \zeta)$action and the $(\zeta^{1}, \, \zeta)$action, for the sake of clarity let me discuss both.
The $(\zeta^{1}, \, \zeta)$action (the OP case).
In this case there are three invariant monomials, namely $$T_0:= x^n, \quad T_1=xy, \quad T_2=y^n.$$ Then the invariant algebra $\mathbb{C}[x, \, y]^G$ is isomorphic to $$\mathbb{C}[T_0, \, T_1, \, T_2]/(T_0T_2T_1^n),$$ which is is a Rational Double Point (= Du Val singularity) of type $A_{n1}$, whose resolution graph is the Dynkin diagram with the same label, corresponding to a chain of $n1$ smooth rational curves with selfintersection $(2)$. This is a Gorenstein singularity for all $n \geq 2$, for instance because it is a hypersurface singularity, or because $G \subset \textrm{SL}_2(\mathbb{C})$.
The $(\zeta, \, \zeta)$action.
In this case the quotient $\mathbb{C}^2/G$ is isomorphic to the affine cone over a rational normal curve $C_n \subset \mathbb{P}^n$.
Therefore $\mathbb{C}[x, \, y]^G$ is the ring of regular functions of such a cone, namely the ring $\mathbb{C}[T_0, \ldots, T_n]/I$, where $I$ is the ideal generated by the $2 \times 2$ minors of the matrix $$\begin{equation} \begin{pmatrix} T_0 & T_1 & \ldots & T_{n1}\\ T_1 & T_2 & \ldots & T_n \end{pmatrix}. \end{equation}$$
Here the variables $T_i$ correspond to the invariant monomials for the action, namely $T_i:=x^{ni}y^i$. The singularity at the vertex of the cone is of type $\frac{1}{n}(1, \, 1)$, has embedding dimension $n+1$ and is not Gorenstein for $n \geq 3.$ The resolution is a unique smooth rational curve with selfintersection $(n)$.
Remark.
The two cases coincide for $n=2$, when the action is simply given by $(x, \, y) \to (x,\, y)$. The invariant algebra is generated by the three invariant polynomials $$T_0:=x^2, \quad T_1:=xy, \quad T_2:=y^2$$ subject to the unique relation $T_0T_2T_1^2=0,$ which is the affine cone over a smooth conic in $\mathbb{P}^2$.

2$\begingroup$ I'm totally confused; what are your generators $T_i$? Even if this presentation is right, it's extremely redundant: as Jason said, this is a hypersurface in $\mathbb{A}^3$ with the equation $uw=z^n$. $\endgroup$ – Ben Webster♦ Dec 5 '15 at 17:26

1$\begingroup$ @BenWebster: I think that was my fault. My comment confused the $(\zeta^{1},\zeta)$action and the $(\zeta,\zeta)$action. Once one person says something wrong, it is hard to realize that is not what the OP asked. $\endgroup$ – Jason Starr Dec 5 '15 at 18:03

1$\begingroup$ @BenWebster: you are right. I was confused, too. I will correct the answer. $\endgroup$ – Francesco Polizzi Dec 5 '15 at 19:11

1$\begingroup$ @JasonStarr: as I said before, the previous version of the answer was wrong. I hope that now all mistakes are corrected. $\endgroup$ – Francesco Polizzi Dec 5 '15 at 19:36