The topology on $\mathbb A^2$ is not the product topology

Question

I'm trying to prove the Zariski topology on $\mathbb A^2$ is not the product topology on $\mathbb A^1\times \mathbb A^1$.

I'm looking for a counter-example based on the fact the closed subsets in $\mathbb A^1$ are the finite ones.

Thanks in advance

Answer 1

$(x,x)$ is closed in $\mathbb{A}^2$ being defined by the equation $y-x=0$.

The product topology gives finite sets of points and the horizontal and vertical lines.

Answer 2

@Rene has given you a great counterexample. Another way to arrive at an answer is to completely characterize closed sets in $\mathbb A^1 \times \mathbb A^1$ with the product topology, using the fact that closed subsets of $\mathbb A^1$ are finite.

Sets of the form $A \times B$ for $A,B$ sets in $\mathbb A^1$ with finite complements form a basis for the product topology. It's easy to verify that $A\times B = \mathbb A^1 \setminus \{L_1, \dots, L_k\}$ for $L_1, \dots, L_k$ "vertical" or "horizontal" lines (that is, one coordinate or the other is constant). Taking complements, closed subsets of $\mathbb A^1 \times \mathbb A^1$ with the product topology are intersections of finite unions of these lines. Can you show that not all closed sets of $\mathbb A^2$ are of this form?

Answer 3

A variation on the answer of Rene:

The diagonal $\Delta\subset \mathbb{A}^1\times\mathbb{A}^1$ in the Zariski topology is closed. If it were closed in the product topology, that would imply that $\mathbb{A}^1$ was Hausdorff (with the Zariski topology), which is obviously false. Thus the two topologies cannot coincide.