Very ample divisors on curvesPosted on December 11, 2018
All of the sources that I have found characterizing very ample divisors reference Hartshorne or otherwise use, in my opinion, “non-elementary” arguments. Here are some “elementary” proofs that I worked out with some fellow grad students.
Let be a projective non-singular irreducible curve over an algebraically closed field. Suppose that is a divisor on such that Let Here is the Riemann-Roch space of and is the dimension of . We define the map by
Lemma 1. If and are linearly equivalent divisors, then there is an automorphism of given by linear forms such that
Let be such that Then the map gives an isomorphism
If is defined by the basis , then By the isomorphism above, given any basis for the set is a basis for Thus there exists a change of basis on defined by linear forms such that for each Therefore also gives a projective linear map defined by Moreover, as desired.
Definition 2. A divisor is said to be very ample if the map gives an isomorphism to its image.
Theorem 3. A divisor with is very ample if and only if for every two points we have
Suppose that is very ample. Let Pick such that Observe that by applying a linear automorphism to (changing basis for ) we can assume that Let be defined by the basis Then and Since is effective, we must have and Moreover, and Thus similarly we have but So It remains to show that Note we have already shown above that Suppose that Then every that has a zero at must have a double zero at Note that must be a local parameter at on for some since generate But by our assumptions above it must be that This is a contradiction to being an isomorphism.
Conversely suppose that for all We will first show that is injective. Let and assume By our assumption we may pick and therefore and Thus
It remains to show that is nonsingular. Again suppose that Let us pick and We may then extend this to a basis and note that
Note that we may find a non-singular model for Therefore there exists a birational map between and and hence a is a birational map from to So is an isomorphism.
Observe that generates Moreover, for we have and Therefore for we have where and is a unit in So and thus has order at . Thus is a basis for and is nonsingular at Thus is nonsingular and defines an isomorphism.