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}_ D1. 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
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