Two affine varieties which are isomorphic, but their projectivisations are not I am learning Algebraic Geometry and came across the following question:

Show $V(y-x^3) \cong V(y-x^2)$ as affine varieties in $\mathbb{A}^2$. Prove that their projectivisations in $\mathbb{P}^2$ are not isomorphic by computing the local ring at the point $p=[0:1:0]$.

I know how to do the first part (I used the inverse maps $[x:y] \to [x,x^3]$ and $[x,y] \to [x,x^2]$). However, I am having trouble computing the local ring at $p$. Intuitively, the projectivisation of $V(y-x^3)$ contains a singularity on the open cover $y\neq 0$ where we get $z^2=x^3$, whereas this is not the case for $V(y-x^2)$. However, I don't know how to compute the local ring, and I would find it very useful to go through this computation. Any help is much appreciated!
 A: First we homogenize the equations to $yz^2 - x^3$ and $yz - x^2$, respectively, then "dehomogenize" in a neighborhood of $[0:1:0]$ to $z^2 - x^3$ and $z - x^2$. Now the local rings are
$$
k[x, z]/(z^2 - x^3)_{(x, z)}
$$
and
$$
k[x, z]/(z - x^2)_{(x, z)}.
$$
and we want to prove these are not isomorphic. The bottom ring is  isomorphic to $k[t]_{(t)}$ via the map $x\mapsto t$, $z \mapsto t^2$.
The top ring cannot be isomorphic to $k[t]_{(t)}$. As you said, the idea of the proof is that the self-intersection point of the curve $z^2 - x^3$ is not smooth. We can formalize this in a lot of ways, but they all come down to proving that this ring is not a DVR. The easiest way to show this is to show that it's not integrally closed in its fraction field. Indeed, we have an embedding
$$
k[x, z]/(z^2 - x^3)_{(x, z)} \hookrightarrow k[t]_{(t)}.
$$
mapping $x \mapsto t^3$, $z \mapsto t^2$. Now $t$ is not in the image of this embedding, but is integral over the image (which is precisely $k[t^2, t^3]_{(t^2, t^3)}.$
