TITLE: Projective spaces with Zariski topology
QUESTION [6 upvotes]: Why $\mathbb{P}^1\times\mathbb{P}^1\not\cong\mathbb{P}^2$ where the projective spaces have the Zariski topology?

REPLY [2 votes]: 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 $Z(x-y)$.