# Projective spaces with Zariski topology

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

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If you're just looking for intuition, over $\mathbb{R}$ we have a torus on the LHS and $\mathbb{RP}^2$ on the RHS, so one is orientable and the other isn't. – Qiaochu Yuan Jun 11 '11 at 21:58
When you write $\mathbb P^1 \times \mathbb P^1$, do you mean the fibre product as varieties (or schemes), or do you mean the literal topological product. If you mean the former, than Akhil Mathew's answer is what you are looking for. If you mean the latter, then Hsueh-Yung Lin's answer is what you want. Regards, – Matt E Jun 12 '11 at 0:04

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...)

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Dear Akhil, I'm not sure whether the OP means $\mathbb P^1 \times \mathbb P^1$ to be understood in the sense of varieties and/or schemes (in which case your answer is perfectly correct), or in the sense of topological spaces, in which case Hsueh-Yung Lin's answer is correct. I've posted a comment on the question asking for clarification. Best wishes, – Matt E Jun 12 '11 at 0:06
@Matt: Dear Matt, Indeed, and thanks for pointing this out: I've added a clarification that, in the last sentence, I had assumed $\mathbb{P}^1 \times \mathbb{P}^1$ had the usual Zariski topology on it. – Akhil Mathew Jun 12 '11 at 0:41

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 $Z(x-y)$.

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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.

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