Very ample divisors on curves

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.