When I was reading a paper, I came across a statement like "Since ince $M$ is affine, the trivial bundle is ample and ..." I think that line bundle $L$ on a variety $M$ is ample if it the global sections of $L^{\otimes n}$ give an embedding $M$ to some projective space. Is this clear for the trivial line bundle of affine variety?


1 Answer 1


Quoting Hartshorne, Example II.7.4.2:

If [the scheme] $X$ is affine, then any invertible sheaf on $X$ is ample.

Hartshorne's definition of ampleness (equivalent to yours; see Theorem II.7.6) is that $L$ is ample if, for any coherent sheaf $F$ on $X$, there exists an integer $n_0$ such that $F \otimes L^n$ is generated by globals for all $n > n_0$. You get the example I quote above upon recalling (II.5.16.2) that coherent sheaves on an affine scheme are determined by their global sections, and you get the result you quote because the trivial bundle is invertible.

  • $\begingroup$ I see. Hartshorne's definition of ampleness is different from what I initially thought. As you said, they are the same in case $X$ is affine as global section defines the scheme. But he defines it via global generation... Thank you very much. $\endgroup$
    – M. K.
    Commented Sep 13, 2012 at 8:45
  • $\begingroup$ @M.K., Hartshorne's definition (=that of EGA) is the correct one. For an algebraic variety it is equivalent to the embedding property. $\endgroup$
    – user18119
    Commented Sep 13, 2012 at 9:49
  • $\begingroup$ Really? I will think about it. Thanks, QiL. $\endgroup$
    – M. K.
    Commented Sep 13, 2012 at 18:14

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