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If you read Serre's FAC you will notice the definition of algebraic variety X (section 34 page 40), http://achinger.impan.pl/fac/fac.pdf

axiom 2: $\Delta= \{(x,x) \mid x \in X\} \subset X \times X$ is closed in the induced topology of X $\times$ X.

In the product topology (which looks like the induced topology... but I may be wrong here) $\Delta$ closed iff X Hausdorff

He later uses the Zariski topology on X. But we know the Zariski topology is not Hausdorff.

What is wrong with my interpretation of the induced topology?

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This is because the topology on $X \times X$ is not the product topology when we use the Zariski topology. You can think of $\mathbb A^2 = \mathbb A^1 \times \mathbb A^1$ for convince yourself of this fact.

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  • $\begingroup$ I was thinking the same thing... but I'm not convinced about the topology. Can you give an example of what the open sets look like in $\mathbb{A}^2$? $\endgroup$ – user352102 Mar 13 '17 at 17:11
  • $\begingroup$ have a look here: math.stackexchange.com/questions/128102/… $\endgroup$ – user352102 Mar 13 '17 at 17:15
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    $\begingroup$ The answers the users did contains all the informations you need. The open set in $\mathbb A^2$ are complement of finite union of curves and points. You should think a bit more about the curve defined by $(x,x)$ in $\mathbb A^2$ or a parabola for example. $\endgroup$ – user171326 Mar 13 '17 at 17:40
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    $\begingroup$ No. We are in the category of algebraic varieties so we always take the Zariski topology. Maybe you should read more about algebraic geometry before reading FAC ? $\endgroup$ – user171326 Mar 13 '17 at 18:11
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    $\begingroup$ I believe Serre is recalling some stuff about algebraic varieties before giving this "axiom 2", and he's writing all the necessary precisions. In particular there is a remark that the Zariski topology on product spaces. If everything is still not clear I would suggest to study a bit more algebraic geometry before reading this article. $\endgroup$ – user171326 Mar 13 '17 at 18:45

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