The topology on $\mathbb A^2$ is not the product topology 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
 A: $(x,x)$ 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.
A: 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.
A: @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?
