Lemma 53.3.4 (0CCR)—The Stacks project

Lemma 53.3.4. Let $k$ be a field. Let $X$ be a nonsingular proper curve over $k$ . Let $(\mathcal{L}, V)$ be a $\mathfrak g^1_ d$ on $X$ . Then the morphism $\varphi : X \to \mathbf{P}^1_ k$ of Lemma 53.3.2 either

is nonconstant and has degree $\leq d$ , or
factors through a closed point of $\mathbf{P}^1_ k$ and in this case $H^0(X, \mathcal{O}_ X) \not= k$ .

Proof. By Lemma 53.3.2 we see that $\mathcal{L}' = \varphi ^*\mathcal{O}_{\mathbf{P}^1_ k}(1)$ has a nonzero map $\mathcal{L}' \to \mathcal{L}$ . Hence by Varieties, Lemma 33.44.12 we see that $0 \leq \deg (\mathcal{L}') \leq d$ . If $\deg (\mathcal{L}') = 0$ , then the same lemma tells us $\mathcal{L}' \cong \mathcal{O}_ X$ and since we have two linearly independent sections we find we are in case (2). If $\deg (\mathcal{L}') > 0$ then $\varphi$ is nonconstant (since the pullback of an invertible module by a constant morphism is trivial). Hence

$\deg (\mathcal{L}') = \deg (X/\mathbf{P}^1_ k) \deg (\mathcal{O}_{\mathbf{P}^1_ k}(1))$

by Varieties, Lemma 33.44.11. This finishes the proof as the degree of $\mathcal{O}_{\mathbf{P}^1_ k}(1)$ is $1$ . $\square$

Comments (2)

Comment #5100 by Tongmu He on May 12, 2020 at 05:17

It seems that the morphism $\varphi : X \to \mathbf{P}^1_k$ in 53.3.4 might be constant. Since $k$ might not be algebraically closed, even if the global sections $s_0, s_1$ of $\mathcal{L}$ are linearly independent over $k$ , they might not be algebraically independent over $k$ . A possible consequence is that the morphism $\varphi : X \to \mathbf{P}^1_k$ induced by these two sections might factor through a $k'$ -rational point of $\mathbf{P}^1_k$ where $k'/k$ is a nontrivial field extension.

Comment #5309 by Johan on June 17, 2020 at 09:39

Good catch! Thanks very much and fixed here.

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