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

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Is there something missing in this question? What is the assumed topology on $\mathbb{A}$? –  Manos Sep 8 '12 at 19:00
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up vote 6 down vote accepted

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

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Note also that if $k$ is finite, the two topologies do coincide: both are the discrete topology on $k \times k$. –  Pete L. Clark May 21 '12 at 13:06
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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.

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