Nonisomorphic quotient rings $k[x,y]/(x^2-y^{2n+1})$ for different values of $n$ I want to prove the following proposition:

For $n\neq m\in\mathbb{Z}_{\geq 0}$ the, $k$-algebras
  $R_n$ and $R_m$ are non-isomorphic, where $R_n=k[x,y]/(x^2-y^{2n+1})$, at least when $k=\mathbb{C}$.

In this case being the exponent of $y$ odd, I can see that all the $R_n$ are integral domains. But also as $k[x,y]$-modules they have the same length of free resolutions, thus the same projective dimension, etc., so I'm not sure what kind of invariant should I use to prove they're not isomorphic.
I think that for $n>0$ none of the rings are UFDs and that $R_0\cong k[t]$ so that would be one case.
 A: Your statement about UFDs is correct, so it suffices to treat the case when $0<m<n$. Here is a very low-tech solution involving picking a basis.
First, observe that our isomorphism must take $(x,y)_{R_n}$ isomorphically on to $(x,y)_{R_m}$: the preimage of a maximal ideal is maximal, and these are the only maximal ideals so that the localization of $R_n$ and $R_m$ at them are not regular local rings. So $\varphi(x)=ax+by+p$ and $\varphi(y)=cx+dy+q$ for our putative isomorphism $\varphi:R_m\to R_n$ where $p,q\in (x,y)^2$, $a,b,c,d\in k$ and we have $ad-bc\neq 0$. On the other hand, we must have $\varphi(x^2-y^{2m+1})=0$, so $\varphi(x)^2-\varphi(y)^{2m+1}=0$, or $(ax+by+p)^2-(cx+dy+q)^{2m+1}=0$.
We can pick a basis for $R_n$ as a $k$-vector space to be $x^iy^j$ with $i<2$. Expanding out in this basis and examining the quadratic terms, we get that $b=0$ since every term in $(cx+dy+q)^{2m+1}$ has total degree at least $3$ from our assumption that $m>0$. So $a\neq 0$ and in fact we may assume it to be $1$ by scaling $x$. One may also check that $2xp+p^2$ contributes no terms of degree $2m+1$ by counting degrees.
Now let's look at what happens when we expand out $(cx+dy+q)^{2m+1}$. In our chosen basis, the terms of smallest total degree are $(2m+1)cd^{2m}xy^{2m}+d^{2m+1}y^{2m+1}$. But the only way for this to be zero is if $d=0$, since we can't apply any rewriting trickery to this element. But if $d=0$, then $ad-bc=0$, contrary to our assumption.

For a more high-minded proof, one may note that if we let $C_n=\operatorname{Spec} R_n$, then the blowup of $C_n$ at the origin is $C_{n-1}$, so each $C_n$ will require $n$ blowups at the unique singular point to resolve its singularities and no smaller number will do. This can be formalized via an appropriate measure of the "badness" of the singularity, but I don't exactly recall the exact setup one should use at the moment. (It's been a while - probably the Hilbert-Samuel multiplicity is the right one to use. If anyone can set me straight, I'd welcome you responding with a comment or another answer!)
