TITLE: Showing that $\mathbb{C}[x,y]^{\mu_n}$ and $\mathbb{C}[x,y,z]/(xy-z^n)$ are isomorphic as rings
QUESTION [7 upvotes]: The problem: Let $\mu_n$ act on $\mathbb{C}[u,v]$ with weights $(1,-1)$. I would like to show that the rings $\mathbb{C}[u,v]^{\mu_n}$ and $\mathbb{C}[x,y,z]/(xy-z^n)$ are isomorphic.
Explanation of terminology: $\mu_n \subset \mathbb{C}$ is the cyclic group of nth roots of unity, with multiplication as the group action. Pick $\epsilon \in \mathbb{C}$ to be a primitive nth root of unity, so that $\mu_n = \langle \epsilon \rangle$. To say that $\mu_n$ acts on $\mathbb{C}[u,v]$ with weights $(1,-1)$ means that the action of $\mu_n$ on $\mathbb{C}[u,v]$ is defined by setting $\epsilon \cdot u = \epsilon u$, $\epsilon\cdot v = \epsilon^{-1}v$ and extending polynomially. Define $\mathbb{C}[u,v]^{\mu_n}$ to be the subring of $\mathbb{C}[u,v]$ consisting of polynomials that are invariant under the given action of $\mu_n$ on $\mathbb{C}[u,v]$.
I want to show that there is an isomorphism of rings: $\mathbb{C}[u,v]^{\mu_n} \cong \mathbb{C}[x,y,z]/(xy-z^n)$.
My partial solution: $\mathbb{C}[u,v]^{\mu_n}$ can be identified as the subring $\mathbb{C}[u^n,v^n,uv] \subset \mathbb{C}[u,v]$. So the problem is to show that $\mathbb{C}[u^n,v^n,uv] \cong \mathbb{C}[x,y,z]/(xy-z^n)$. There is a surjective ring homomoprhism $\varphi: \mathbb{C}[x,y,z] \to \mathbb{C}[u^n,v^n,uv]$ defined by $x \mapsto u^n, y\mapsto v^n, z \mapsto uv$. A direct check shows that $(xy-z^n) \subset \ker{\varphi}$.
The difficulty: I am unable to show the reverse inclusion, $\ker{\varphi} \subset (xy-z^n)$. Any help will be appreciated. Is there another way to see that $\ker{\varphi} = (xy-z^n)$, without chasing inclusions?
Thank you in advance.
EDIT: Based on Alistair's answer below, here is how I understand one should proceed: suppose $\alpha$ (as defined below) lies in $\ker{\varphi}$, so that $\sum{a_{i,j,k}u^{ni+k}v^{nj+k}} = 0$ --- (1).
I want to show that $b_{i,j,k} \in \mathbb{C}$ can be chosen s.t. $(\sum{b_{i,j,k}x^iy^jz^k})(xy-z^n) = \alpha$ i.e. $\sum(b_{i-1,j-1,k} - b_{i,j,k-n})x^iy^jz^k = \alpha$. Apply $\varphi$ to this last equation and use linear independence of $u^av^b$ in $\mathbb{C}[u,v]$ and equation (1) to get linear equations of the form:
(expression in $b$s) = (expression in $a$s).
Using these I can make a choice of $b$s, so that the equation $(\sum{b_{i,j,k}x^iy^jz^k})(xy-z^n) = \alpha$ is satisfied, thus proving the reverse inclusion.

REPLY [5 votes]: You can prove this by "brute force".  Suppose that an element $\alpha = \sum_{i,j,k} a_{i,j,k} x^i y^j x^k \in \mathbb{C}[x,y,z]$ is mapped to zero.  Then $\sum_{i,j,k} a_{i,j,k} u^{ni} v^{nj} (uv)^k = \sum_{i,j,k} a_{i,j,k} u^{ni+k} v^{nj+k} = 0$.  Now, using the fact that the elements $u^av^b$ are linearly independent in $\mathbb{C}[u,v]$, show that $\alpha$ is a $\mathbb{C}[x,y,z]$-multiple of $xy-z^n$.