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mrw
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It is just a consequence of the Riemann-Roch theorem. But another way to see it, is using the fact that if $L$ is any line bundle then $L=A\otimes B^{-1}$ where $A$ and $B$ are very ample line bundles. Thus, if $Z$ is the zero set of a general section of $B$ you have a sequence

$0\to F\otimes A^n\otimes B^{-(i+1)}\to F\otimes A^n\otimes B^i\to F\otimes A^n\otimes B^i\otimes O_Z\to 0$$0\to F\otimes A^n\otimes B^{-(i+1)}\to F\otimes A^n\otimes B^{-i}\to F\otimes A^n\otimes B^{-i}\otimes \mathcal O_Z\to 0$

Now just sum over $i$ and proceed by induction on the dimension.

It is just a consequence of the Riemann-Roch theorem. But another way to see it, is using the fact that if $L$ is any line bundle then $L=A\otimes B^{-1}$ where $A$ and $B$ are very ample line bundles. Thus, if $Z$ is the zero set of a general section of $B$ you have a sequence

$0\to F\otimes A^n\otimes B^{-(i+1)}\to F\otimes A^n\otimes B^i\to F\otimes A^n\otimes B^i\otimes O_Z\to 0$

Now just sum over $i$ and proceed by induction on the dimension.

It is just a consequence of the Riemann-Roch theorem. But another way to see it, is using the fact that if $L$ is any line bundle then $L=A\otimes B^{-1}$ where $A$ and $B$ are very ample line bundles. Thus, if $Z$ is the zero set of a general section of $B$ you have a sequence

$0\to F\otimes A^n\otimes B^{-(i+1)}\to F\otimes A^n\otimes B^{-i}\to F\otimes A^n\otimes B^{-i}\otimes \mathcal O_Z\to 0$

Now just sum over $i$ and proceed by induction on the dimension.

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mrw
  • 438
  • 2
  • 7

It is just a consequence of the Riemann-Roch theorem. But another way to see it, is using the fact that if $L$ is any line bundle then $L=A\otimes B^{-1}$ where $A$ and $B$ are very ample line bundles. Thus, if $Z$ is the zero set of a general section of $B$ you have a sequence

$0\to F\otimes A^n\otimes B^{-(i+1)}\to F\otimes A^n\otimes B^i\to F\otimes A^n\otimes B^i\otimes O_Z\to 0$

Now just sum over $i$ and proceed by induction on the dimension.