Very ample divisors on curves

Posted 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 C be a projective non-singular irreducible curve over an algebraically closed field. Suppose that D is a divisor on C such that l(D)=n>0. Let      L(D)=\langle f_1,\ldots,f_n\rangle. Here L(D) is the Riemann-Roch space of D and l(D) is the dimension of L(D). We define the map \phi_D:C\to\mathbb{P}^{n-1} by      \phi_D(P)=(f_1(P):\ldots:f_n(P)).

Lemma 1. If D and D&39; are linearly equivalent divisors, then there is an automorphism \psi of \mathbb{P}^{n-1} given by linear forms such that \phi_{D&39;}(C)=\psi(\phi_D(C)).

Let g\in k(C)^* be such that D+\text{div}(g)=D&39;. Then the map f\mapsto gf gives an isomorphism L(D)\to L(D&39;).

If \phi_{D&39;} is defined by the basis \{ gf_1,\ldots,gf_n\}, then  \begin{aligned}[t]     \phi_{D&39;}(P) &= (g(P)f_1(P),\ldots,g(P)f_n(P)) \\                  &= (f_1(P),\ldots,f_n(P))         \\                  &= \phi_D(P). \end{aligned} By the isomorphism above, given any basis \{h_1,\ldots,h_n\} for L(D&39;) the set \{g^{-1}h_1,\ldots,g^{-1}h_n\} is a basis for L(D). Thus there exists a change of basis on L(D) defined by linear forms T_i\in k[X_1,\ldots,X_n]_1 such that T_i(f_1,\ldots,f_n)=g^{-1}h_i for each i. Therefore T also gives a projective linear map \psi defined by      (x_1,\ldots,x_n)\mapsto (T_1(x_1,\ldots,x_n),\ldots,T_n(x_1,\ldots,x_n)). Moreover, \phi_{D&39;}(P)=\psi(\phi_{D}(P)) as desired.

Definition 2. A divisor D is said to be very ample if the map \phi_D gives an isomorphism to its image.

Theorem 3. A divisor D with l(D)>0 is very ample if and only if for every two points P,Q\in C we have l(D-P-Q)=l(D)-2.

Suppose that D is very ample. Let P\in C. Pick Q\in C such that Q\ne P. Observe that by applying a linear automorphism to \mathbb{P}^{n-1} (changing basis for L(D)) we can assume that      \phi_D(P)=(1:0:\ldots:0), \ \phi_D(Q)=(0:1:0:\ldots:0). Let \phi_D be defined by the basis f_1,\ldots,f_n\in L(D). Then f_1(P)=1 and f_1(Q)=0. Since D is effective, we must have f_1\in L(D) and f_1\not\in L(D-P). Moreover, f_2(Q)=1 and f_2(P)=0. Thus similarly we have f_2\in L(D-P) but f_2\not\in L(D-P-Q). So l(D-P-Q)=l(D)-2. It remains to show that l(D-2P)=l(D)-2. Note we have already shown above that l(D-P)=l(D)-1. Suppose that L(D-P)=L(D-2P). Then every f\in L(D) that has a zero at P must have a double zero at P. Note that x_i/x_1 must be a local parameter at \phi_D(P) on \phi_D(C) for some i\in\{2,\ldots,n\} since \{ x_i \, : \, 2\le i\le n\} generate \mathfrak{m}_{\phi_D(P)}. But by our assumptions above it must be that      \text{ord}_P(\phi_D^*(x_i/x_1))=\text{ord}_P(f_i/f_1)\ge 2. This is a contradiction to \phi_D being an isomorphism.

Conversely suppose that l(D-P-Q)=l(D)-2 for all P,Q\in C. We will first show that \phi_D is injective. Let P,Q\in D and assume P,Q\not\in C. By our assumption we may pick f_1\in L(D-P)\setminus L(D-P-Q) and therefore f_1(P)=0 and f_1(Q)\ne 0. Thus \phi_D(P)\ne\phi_D(Q).

It remains to show that \phi_D(C) is nonsingular. Again suppose that P=(1:0:\ldots:0). Let us pick f_1\in L(D)\setminus L(D-P) and f_2\in L(D-P)\setminus L(D-2P). We may then extend this to a basis \{f_1,\ldots, f_n\} and note that f_3,\ldots,f_n\in L(D-2P).

Note that we may find a non-singular model X for \phi_D(C). Therefore there exists a birational map between C and X and hence a \phi_D is a birational map from C to \phi_D(C). So \phi_D^*:k(\phi_D(C))\to k(C) is an isomorphism.

Observe that \{ x_i/x_1 \, : \, 2\le i\le n\} generates \mathfrak{m}_{\phi_D(P)}. Moreover, for i\ge 3 we have      \text{ord}_P(\phi_D^*(x_i/x_1))=\text{ord}_P(f_i/f_1)>1 and      \text{ord}_P(\phi_D^*(x_2/x_1))=\text{ord}_P(f_2/f_1)=1. Therefore for i\ge 3 we have      \phi_D^*(x_i/x_1)=\phi_D^*(u)\phi_D^*(x_2/x_1)^m where m>1 and u is a unit in \mathcal{O}_{\phi_D(P)}. So x_i/x_1=u(x_2/x_1)^m and thus x_i/x_1 has order m at \phi_D(P). Thus \{x_2/x_1\} is a basis for \mathfrak{m}_{\phi_D(P)}/\mathfrak{m}_{\phi_D(P)}^2 and \phi_D(C) is nonsingular at \phi_D(P). Thus \phi_D(C) is nonsingular and \phi_D defines an isomorphism.