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?


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.

  • $\begingroup$ 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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