# algebraic geometry – Show that the Zariski topology on \$A^2\$ is not the product topology on \$A^1 times A^1\$. (Hint: consider the diagonal.)

As already pointed out in the comments one elegant way to show this is by using that \$mathbb{A^1}\$ is not Hausdorff, and thus the diagonal is not closed in the product topology. Another, more image-like proof is the following:

The Zariski topology on \$mathbb{A}^1\$ is just the cofinite topology. Hence a basis of the product topology is given by the products \$U times V\$ where \$U, V subseteq mathbb{A}^1\$ are both only missing finitely many points (if you try to visualize \$mathbb{A}^2\$ as \$mathbb{R}^2\$ then you can imagine these basis elements as the plane \$mathbb{R}^2\$ where we remove finitely many lines parellel to the axes.)

Now take a look at the diagonal \$Delta = {(x,x) mid x in mathbb{A}^1}\$. Using the above visualization we can already “see” that the complement of \$Delta\$ cannot be written as the union of the basis elements described above.

More formally let \$U, V subseteq mathbb{A}^1\$ be open. If \$U neq emptyset\$ or \$V neq emptyset\$, i.e. if \$U times V neq emptyset\$, then there exist \$x in U cap V\$ because we are dealing with the cofinite topology and \$mathbb{A}^1\$ is infinite. Thus \$(x,x) in U times V\$ and \$(U times V) cap Delta neq emptyset\$. Therefore we cannot write \$Delta^C\$, which is non-empty, as the union of basis elements \$U times V\$ with \$U,V subseteq mathbb{A}^1\$ open.

So \$Delta\$ is not closed in the product topology. But \$Delta\$ is closed in the Zariski topology of \$mathbb{A}^2\$, because it is the zero locus of \$P(x,y) = x-y\$.

(Notice however, that the product topology is coarser then the Zariski topology.)

PS: For visualization I also find it very useful to think about the Zariski topology on \$mathbb{R}^2\$ and the product of the Zariski topology on \$mathbb{R} times mathbb{R}\$. Instead of the diagonal an even more intuitive subset is then the unit circle \$S^1 subseteq mathbb{R}^2\$. This is certainly closed in the Zariski topology, as it is given as the zero locus of \$x^2+y^2-1\$. If \$S^1\$ were closed in the product topology on \$mathbb{R} times mathbb{R}\$, then we could write \$S^1\$ by taking a finite number of lines, all parallel to the axes, and then intersecting this by some other lines, all of which are also parallel to the axes. This seems really absurd.