Lemma 53.3.5. Let $k$ be a field. Let $X$ be a proper curve over $k$ with $H^0(X, \mathcal{O}_ X) = k$. If $X$ has a $\mathfrak g^ r_ d$, then $r \leq d$. If equality holds, then $H^1(X, \mathcal{O}_ X) = 0$, i.e., the genus of $X$ (Definition 53.8.1) is $0$.

Proof. Let $(\mathcal{L}, V)$ be a $\mathfrak g^ r_ d$. Since this will only increase $r$, we may assume $V = H^0(X, \mathcal{L})$. Choose a nonzero element $s \in V$. Then the zero scheme of $s$ is an effective Cartier divisor $D \subset X$, we have $\mathcal{L} = \mathcal{O}_ X(D)$, and we have a short exact sequence

$0 \to \mathcal{O}_ X \to \mathcal{L} \to \mathcal{L}|_ D \to 0$

see Divisors, Lemma 31.14.10 and Remark 31.14.11. By Varieties, Lemma 33.44.9 we have $\deg (D) = \deg (\mathcal{L}) = d$. Since $D$ is an Artinian scheme we have $\mathcal{L}|_ D \cong \mathcal{O}_ D$1. Thus

$\dim _ k H^0(D, \mathcal{L}|_ D) = \dim _ k H^0(D, \mathcal{O}_ D) = \deg (D) = d$

On the other hand, by assumption $\dim _ k H^0(X, \mathcal{O}_ X) = 1$ and $\dim H^0(X, \mathcal{L}) = r + 1$. We conclude that $r + 1 \leq 1 + d$, i.e., $r \leq d$ as in the lemma.

Assume equality holds. Then $H^0(X, \mathcal{L}) \to H^0(X, \mathcal{L}|_ D)$ is surjective. If we knew that $H^1(X, \mathcal{L})$ was zero, then we would conclude that $H^1(X, \mathcal{O}_ X)$ is zero by the long exact cohomology sequence and the proof would be complete. Our strategy will be to replace $\mathcal{L}$ by a large power which has vanishing. As $\mathcal{L}|_ D$ is the trivial invertible module (see above), we can find a section $t$ of $\mathcal{L}$ whose restriction of $D$ generates $\mathcal{L}|_ D$. Consider the multiplication map

$\mu : H^0(X, \mathcal{L}) \otimes _ k H^0(X, \mathcal{L}) \longrightarrow H^0(X, \mathcal{L}^{\otimes 2})$

and consider the short exact sequence

$0 \to \mathcal{L} \xrightarrow {s} \mathcal{L}^{\otimes 2} \to \mathcal{L}^{\otimes 2}|_ D \to 0$

Since $H^0(\mathcal{L}) \to H^0(\mathcal{L}|_ D)$ is surjective and since $t$ maps to a trivialization of $\mathcal{L}|_ D$ we see that $\mu (H^0(X, \mathcal{L}) \otimes t)$ gives a subspace of $H^0(X, \mathcal{L}^{\otimes 2})$ surjecting onto the global sections of $\mathcal{L}^{\otimes 2}|_ D$. Thus we see that

$\dim H^0(X, \mathcal{L}^{\otimes 2}) = r + 1 + d = 2r + 1 = \deg (\mathcal{L}^{\otimes 2}) + 1$

Ok, so $\mathcal{L}^{\otimes 2}$ has the same property as $\mathcal{L}$, i.e., that the dimension of the space of global sections is equal to the degree plus one. Since $\mathcal{L}$ is ample (Varieties, Lemma 33.44.14) there exists some $n_0$ such that $\mathcal{L}^{\otimes n}$ has vanishing $H^1$ for all $n \geq n_0$ (Cohomology of Schemes, Lemma 30.16.1). Thus applying the argument above to $\mathcal{L}^{\otimes n}$ with $n = 2^ m$ for some sufficiently large $m$ we conclude the lemma is true. $\square$

[1] In our case this follows from Divisors, Lemma 31.17.1 as $D \to \mathop{\mathrm{Spec}}(k)$ is finite.

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