Projective spaces with Zariski topology

Question

Why $\mathbb{P}^1\times\mathbb{P}^1\not\cong\mathbb{P}^2$ where the projective spaces have the Zariski topology?

Answer 1

One way to see this is to look at the Picard groups: that of $\mathbb{P}^2$ is $\mathbb{Z}$ (this is true for any projective space over a field), while that of $\mathbb{P}^1$ is $\mathbb{Z} \times \mathbb{Z}$ (in fact, the Picard group of a projective bundle over a noetherian scheme $S$ is always $\mathrm{Pic} S \times \mathbb{Z}$). The case of interest here (namely, $\mathbb{P}^1 \times \mathbb{P}^1$) can be proved by explicitly calculating with Weil divisors.

Here is another approach. In fact, $\mathbb{P}^2$ is not the product of any two varieties. To see this, recall that any two varieties of dimension one in $\mathbb{P}^2$ (i.e. curves) intersect (and the intersection number is given by Bezout's theorem). But if we had $\mathbb{P}^2 = A \times B$, then $A, B$ are necessarily both curves, and we could take $A \times b_1, A \times b_2$ as two disjoint projective plane curves. (It follows that $\mathbb{P}^2$ is not even topologically isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$ (the latter having the Zariski topology, not the product!), because a "curve" is just a closed subset of dimension one...)

Answer 2

Suppose that they are homeomorphic under Zariski topology.

Let $C$ be an irreducible, irrational algebraic curve on $\mathbb{P}^2$ (e.g. Fermat curve), then it is an infinite closed set under Zariski topology. This means that the preimage of $C$ in $\mathbb{P} \times \mathbb{P}$ is also a infinite closed set (it just looks like a grid with finite rows and colones, because closed sets on $\mathbb{P}$ are either finite sets or $\mathbb{P}$).

Since $C$ is irreducible, we know that the preimage of $C$ is something like $\{x\}\times \mathbb{P}$ (because the image of any $\{x\}\times \mathbb{P}$ is an algebraic curve contained in $C$). So we deduce that $\mathbb{P} \simeq C$, which yields a contradiction because $C$ is supposed irrational.

Yuan's argument is correct, but only for usual topology.

Answer 3

Working with the Zariski topology makes even affine product spaces nonisomorphic. Since $\mathbb{P}^n$ can be charted by open sets isomorphic to affine spaces $\mathbb{A}^n$, it is enough to prove it for the affine case. The closed subsets of $\mathbb{A}^1\times\mathbb{A}^1$ in the product Zariski topology are the intersections of subsets $Z_1\times Z_2\,$, product of a closed subset $Z_1\subseteq \mathbb{A}^1$ (i.e., a finite set of points of $\mathbb{A}^1$ or the whole $\mathbb{A}^1$) and another closed subset $Z_2\subseteq \mathbb{A}^1$. Therefore, the diagonal $\{(x,\,x)\vert x\in\mathbb{A}^1\}$ is not closed in $\mathbb{A}^1\times\mathbb{A}^1$ using this topology since it is neither a finite cartesian product of points of $\mathbb{A}^1$ with with other points (i.e. a finite number of points in the space $\mathbb{A}^1\times\mathbb{A}^1$), nor a finite product of points of $\mathbb{A}^1$ with lines $\mathbb{A}^1$ (i.e. a finite number of straight lines parallel to axis $x=0$ or $y=0$ in $\mathbb{A}^1\times\mathbb{A}^1$). However, this diagonal is in fact a closed subset in the Zariski topology of $\mathbb{A}^2$ because it is given by the zero set $Z(x-y)$.