Ample line bundle

On a quasi-compact scheme $X$, a line bundle $L$ is ample if each point has an affine neighborhood of the formEquivalently, the affine nonvanishing loci of global sections of positive tensor powers form a basis of a topology. This makes ampleness a condition that global sections recover enough affine local geometry.^[ref]

For a proper scheme over a field, this is equivalent to some positive tensor power being very ample: $L^{\otimes m}$ is the pullback of $\mathcal{O}(1)$ by a closed immersion into projective space. A positive tensor power may be needed. On an elliptic curve, a degree-one line bundle is ample but has only one independent global section, so it cannot itself embed the curve. Its third power has degree three and gives a plane cubic embedding. On an affine scheme, every line bundle is ample under the scheme definition, even when the scheme is not proper.

On a projective scheme, a line bundle is ample exactly when its restriction to every reduced irreducible component is ample. The disjoint union of those components maps finitely and surjectively to the scheme, so finite-surjective descent of the finite pullback of an ample line bundle proves this. The same argument applies to restrictions to any closed subscheme. Thus nilpotent structure does not affect ampleness.

For a finite morphism $f:X\to Y$, the pullback of an ample line bundle is ample. If $f$ is also surjective, ampleness is equivalent in both directions. These results hold for proper schemes over a field, including nonreduced schemes. Cohomological ampleness and the exact finite pushforward prove the forward direction; coherent-sheaf devissage gives finite-surjective descent. In particular, a finite map to projective space makes the pulled-back hyperplane bundle ample.

If $A$ is an ample line bundle on a projective variety and $L$ is any line bundle, then $A^{\otimes m}\otimes L$ is very ample for all sufficiently large integers $m$. Taking $L$ to be trivial also makes $A^{\otimes m}$ very ample.