cusp and node are not isomorphic

Solution 1:

A cusp has a double tangent at the singularity, which a node has simple tangents at its singular point. This is the reason why the two curves are not isomorphic.

The way to see this is to study the structure of the completion of the coordinate rings at the singular points (before doing that, students must get to the point that they can show that the node and the cusp are not affinely equivalent, though, to appreciate what all this means...), but we can trim this down to the following:

Let $A=k[x,y]/(x^2-y^3)$ and let $\newcommand\m{\mathfrak{m}}m=(x,y)$ the maximal ideal at the singular point. Then we can construct a graded ring $\tilde A=A/\m\oplus\m/\m^2\oplus\m^2/\m^3$ in the obvious way. One can check easily that there is a one dimensional subspace of $\m/\m^2$ whose elements square to zero.

On the other hand, if $B=k[x,y]/(y^2-x^3-x^2)$ and $\newcommand\n{\mathfrak{n}}\n=(x,y)$, in the graded ring $\tilde B=B/\n\oplus\n/\n^2\oplus\n^2/\n^3$ there are no non-zero elemens of $\n/\n^2$ which square to zero.

Since any isomorphism $A\cong B$ induces an isomorphism of graded rings $\tilde A\cong\tilde B$ (because you can already show that it must map $\m$ to $\n$), this is enough to conclude.

Of course, this is just doing things in a big-enough quotient of the completions of $A$ and $B$ at $\m$ and $\n$ that the fact that they are non-isomorphic shows up.