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In most case, the definition of a variety over a field $k$ at least requires that being "of finite type" and being "separated". It has no question for me that being of finite type, since we always like finite.

I donot know the reason why we require being "separated" to a variety?

There is a reason since a scheme over a field $k$ being separated will have the property that the intersection of two affine open sets is still affine open set.

Are there any other acceptant reasons?

Thanks a lot.

2 Answers 2

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Separatedness is the analogue in the Zariski topology of Hausdorfness in the cohmology of a manifold or complex analytic space. Without it, strange behaviour can occur. E.g. if $Y$ is not separated, then two morphisms $f,g: X \to Y$ could coicide on a dense open subset of $X$ while not coinciding on all of $X$. Hence "analytic continuation'' is not valid for morphisms into $Y$.

Many arguments in geometry proceed by analytic continuation/Zariski density arguments, so it is natural to place oneself in a context where those arguments can be applied without reservation.

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Besides what Matt E has explained, if you include the separatedness in the definition of scheme i.e. as a separated (having closed diagonal) prescheme, the analogy will be more clear in this respect that a variety is in fact, a separated prevariety where prevariety by this definition is an irreducible, reduced prescheme of finite type over an algebraically closed field $k.$ This approach is closer to the way, Mumford in Red book of varieties and schemes has adopted to define the notion of variety.

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    @Pete L.Clark: I admit that, prescheme is old fashioned these days, but I just wanted to note that it fits nicely to the hierarchy of AG definitions.2012-05-12