## Aven Bross

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### Very ample divisors on curves

Dec 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

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

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

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

If $\phi_{D'}$ is defined by the basis $\{ gf_1,\ldots,gf_n\}$, then

By the isomorphism above, given any basis $\{h_1,\ldots,h_n\}$ for $L(D')$ 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

Moreover, $\phi_{D'}(P)=\psi(\phi_{D}(P))$ 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

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

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

It remains to show that

Note we have already shown above that

Suppose that

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

This is a contradiction to $\phi_D$ being an isomorphism.

Conversely suppose that

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

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

and

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

and

Therefore for $i\ge 3$ we have

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.