If we identify \mathbb A^2 with \mathbb A^1\times\mathbb A^1 in the natural way, show that the Zariski topology on \mathbb A^2 is not the product topology of the Zariski topologies on the two copies of \mathbb A^1.
This is clearly much more a problem in topology than in algebra, and in particular we can reduce it to a fairly well-known statement in general topology. We note that the diagonal Z(y-x) in \mathbb A^2 is an algebraic set and thus closed. Recall that a topological space is Hausdorff if and only if its diagonal \Delta =\{(x,x):x\in X\} is closed in the product \mathbb A^1\times\mathbb A^1. Thus, if \mathbb A^1 is not a Hausdorff space, then we have found a distinction.
To do this, we will use the fact that the Zariski topology on \mathbb A^1 is just the finite complement topology. More generally, we will show that the finite complement topology on any infinite set is not Hausdorff. Consider any such infinite space X endowed with the finite complement topology and suppose x,y\in X are separated by neighborhoods x\in U,y\in V. Then (U\cap V)^c = X since U\cap V=\emptyset. By De Morgan, U^c\cup V^c = X, but U^c,V^c are both finite and X is infinite so this is a contradiction. Hence this topology is not Hausdorff as desired.
No comments:
Post a Comment