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Let $(X,\mathcal O_X)$ be an algebraic prevariety, by definition, it is an algebraic variety iff the diagonal $\Delta(X)$ is closed in the product $X\times X$.

The above property is equivalent to the Hausdorff separation axiom, so $(X,\mathcal O_X)$ is a variety iff $X$ is Hausdorff. But there is a problem because $\mathbb A^n_k$ with Zariski topology is not Hausdorff, so every prevariety immersed in $\mathbb A^n_k$ cannot be a variety.

Certainly this reasoning is wrong, but I can't find the mistake.

thanks for help.

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It is not right that $(X,\mathcal{O}_X)$ is a variety iff $X$ is Hausdorff. In fact, almost no varieties are Hausdorff (because the Zariski topology is so sparse). –  Fredrik Meyer Jun 16 '12 at 18:54
Look at the definition 2.7 at page 120: books.google.it/… Perrin's algebraic varieties are in reality algebraic prevarieties –  Dubious Jun 16 '12 at 18:57
The diagonal is closed, but $X \times X$ does not have the product topology, so this does not imply that $X$ is Hausdorff. –  Dylan Moreland Jun 16 '12 at 19:10
and which is the topology on $X\times X$? –  Dubious Jun 16 '12 at 19:25
The set is not the cartesian product either, if you think of a variety as a scheme. It has a Zariski topology, like any other variety/scheme. –  Zhen Lin Jun 16 '12 at 19:48

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