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.