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Let $k$ be an algebraically closed field of characteristic $\neq 2$ and consider the affine varieties $X_1 := V(x^2-y^3)$ (the cusp) and $X_2 = V(y^2-x^2 (x+1))$ (the node). I would like to show that $X_1$ and $X_2$ are not isomorphic. This is easy if one knows some basic cohomology or singularity theory; but this exercise appears in an introductory script to AG where only absolute basics about affine varieties have been established. Thus, I'm looking for an elementary proof. But of course I would like to avoid any fiddly computation with polynomials (please don't post these as an answer).

Here is what I've done: There are bijective morphisms $\mathbb{A}^1 \to X_1, t \mapsto (t^3,t^2)$ and $\mathbb{A}^1 \to X_2, t \mapsto (t^2-1,t(t^2-1))$. They induce isomorphisms $\mathbb{A}^1 - \{0\} \cong X_1 - \{(0,0)\}$ (the inverse takes $(x,y) \mapsto x/y$) and $\mathbb{A}^1 - \{\pm 1\} \cong X_2 - \{(0,0)\}$ (the inverse takes $(x,y) \mapsto y/x$). Now any isomorphism $X_1 \cong X_2$ must preserve the origin. The reason is that it is the unique singular point (which can be formulated elementarily, see here). Thus, it would yield an isomorphism $\mathbb{A}^1 - \{0\} \cong \mathbb{A}^1 - \{\pm 1\}$. On coordinate rings, this is an isomorphism $k[t]_t \cong k[t]_{t^2-1}$ of $k$-algebras. It induces an isomorphism on the groups of units $k^* \cdot \langle t \rangle \cong k^* \cdot \langle t+1,t-1 \rangle$, preserving $k^*$, thus an isomorphism of groups $\mathbb{Z} \cong \mathbb{Z} \oplus \mathbb{Z}$, contradiction!

I hope that there is a shorter proof? As I've said, I'm only looking for proofs which are accessible to students which have just started to learn about affine varieties.

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By the way, your map to $X_2$ is not bijective. – Mariano Suárez-Alvarez Apr 7 '12 at 10:27
Sure. But this doesn't affect the proof, right? I only need this local isomorphism. – Martin Brandenburg Apr 7 '12 at 10:35
up vote 6 down vote accepted

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.

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Thanks. Isn't it a bit nasty to verify that no non-zero element of $n/n^2$ squares to zero? – Martin Brandenburg Apr 7 '12 at 9:44
Not really. That space has basi the classes $\mod n^2$ of the elements of $\{x,y\}$; take a linear combination and square: you want to know when it is in $n^3$. – Mariano Suárez-Alvarez Apr 7 '12 at 9:48
Yes, I know. This is what I meant by "nasty". But at least it's only linear algebra. I'll accept your answer :) – Martin Brandenburg Apr 7 '12 at 10:01
Hmm. That is not a nasty computation by any sensible standard! – Mariano Suárez-Alvarez Apr 7 '12 at 10:25

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