Is the affine algebraic set defined by the ideal $\langle xy\rangle$ isomorphic to the affine line?

Question

I was working on an exercise and got stuck on a question that asks:

Consider the ideal $I = \langle xy \rangle \in \mathbb{C}[x, y]$ and let $V = \mathbb{V}(I) \subseteq \mathbb{A}^2_{\mathbb{C}}$. Is $V$ isomorphic to $\mathbb{A}^1_{\mathbb{C}}$?

I've tried showing that $V$ is defined by the union of the two sets $\{x=0\}$ and $\{y=0\}$, which are two straight lines that intersect at the origin, thus removing the point $(0,0)$ would create $4$ separate sets of points whereas removing any point on the affine line only creates two disconnected sets, and thus cannot be isomorphic. Is this explanation valid?

Answer 1

I am assuming that you work in the language of schemes. The line and $V(xy)$ are not isomorphic. Here are some options:

• You can show that $V(xy)$ has two irreducible components, while $\mathbb A^1$ has just one.

• You can show that the ring $\mathbb C[x,y]/(xy)$ associated to $V(xy)$ has zero divisors and is not an integral domain (this is related to the first point).

• You can show that $V(xy)$ has a 2-dimensional tangent space (at the origin), while the tangent spaces of $\mathbb A^1$ are all $1$-dimensional.

As some people in the comments mentioned, the real picture of $V(xy)$ is misleading. If you remove the origin, then the resulting space will not be disconnected in the analytic topology, because you work over the complex numbers. A second problem with that line of argument is, that the Zariski topology is way too coarse for it to work well. You should think of the closed sets in the Zariski topology as of (finite unions) of lines, points, etc. and not as of analytically closed subsets. A typical open set of a scheme $X$ will be something like $X$ without a line, a surface and two points, and not sth. like a half space or an open interval.