2
$\begingroup$

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.

$\endgroup$
  • 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
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.