Zariski Topology question

Question

Could you please give a hint how to show that the zariski topology on $\mathbb{A}^2$ is not the product topology on $\mathbb{A}^1\times\mathbb{A}^1$

Answer 1

Assuming $k$ is infinite, consider the topological properties of the line $x=y$.

Answer 2

Just to add to Prof Magidin's example: it's a common exercise in point set topology to prove that a topological space $X$ is Hausdorff if and only if the diagonal $\Delta = \{(x, x) : x \in X\}$ is closed in the product topology on $X \times X$. And when $k$ is infinite [in particular, when $k$ is algebraically closed] you know that $\mathbf A^1$ is not Hausdorff: any two non-empty open sets intersect.

By the way, this consideration of the diagonal motivates the correct analogue of the Hausdorff property in algebraic geometry: the notion of a separated morphism of schemes.