Showing that $mathbb{C}[x,y]^{mu_n}$ and $mathbb{C}[x,y,z]/(xy-z^n)$ are isomorphic as rings

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 $epsilon cdot u = epsilon u$, $epsiloncdot v = epsilon^{-1}v$ 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 $x mapsto u^n, ymapsto v^n, z mapsto uv$. 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.