The theorem $8.15$(p.177) from the Hartshorne's book "Algebraic Geometry" says:

" Let $X$ be an irreducible separated scheme of finite type over an algebraically closed field $k$. Then $\Omega_{X/k}$ is a locally free sheaf of rank $n= \dim \ X$ iff $X$ is nonsingular variety over $k$."

I can't understand where does we use the condition that $X$ is separated.

  • 2
    $\begingroup$ I doubt that this is needed. After all, we may reduce to the affine case, since both statements are local on $X$. Hartshorne's book is full with imprecise assumptions. Better consult other books (EGA, Liu, Görtz-Wedhorn, Bosch, ...). $\endgroup$ – Martin Brandenburg Jun 14 '14 at 19:14
  • $\begingroup$ @MartinBrandenburg: Being nonsingular is certainly local, but being a variety is not. $\endgroup$ – RghtHndSd Jun 16 '14 at 15:55
  • $\begingroup$ Ah, ok, but it becomes true when we just omit the word "variety". For details, see Matt's answer. $\endgroup$ – Martin Brandenburg Jun 16 '14 at 16:52

Hartshorne's definition of variety is an integral separated scheme over $k$. From the assumption on $\Omega_{X/k}$ he can deduce that $X$ is reduced, and since he assumes $X$ separable, he can then deduce that it is integral. But the assumption on $\Omega_{X/k}$ can't give that $X$ is separated. Thus he has to assume this in order to conclude that $X$ is a variety. (The fact that it is non-singular then follows from the assumption on $\Omega_{X/k}$.)

If he had worked with a more expansive definition of variety, one that allowed non-separated objects (e.g. what Mumford calls a pre-variety in the Red Book), then he wouldn't need the separated assumption.


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