This question is about the proof of Theorem V.2.17 in Hartshorne's Algebraic Geometry. Here everything is defined over some algebraically closed field $k$. Define $\mathcal O = \mathcal O_{\mathbf P^1}$. Let $e \ge 0$ be an integer. Then the following is what I do not understand:

The proof says that when $n \ge e$, we can have a surjective map $\mathcal O \oplus \mathcal O(-e) \to \mathcal O (n-e) \to 0$ by arguing that we can take morphisms $\mathcal O \to \mathcal O(n-e)$ and $\mathcal O(-e) \to \mathcal O(n-e)$ corresponding to effective divisors of degrees $n-e$ and $n$ on $\mathbf P^1$ which do not meet.

To be more precise, what I do not understand is:

(1) how to obtain morphisms $\mathcal O \to \mathcal O(n-e)$ and $\mathcal O(-e) \to \mathcal O(n-e)$ from effective divisors of degree $n-e$ and $n$ on $\mathbf P^1$. I think these two morphisms are surjective.

(2) why we require the two effective divisors mentioned above do not meet in order to obtain a surjective morphism $\mathcal O \oplus \mathcal O(-e) \to \mathcal O (n-e) \to 0$

Thank you for anyone who gives some help.


1 Answer 1


(1) The morphisms are not surjective, as for example the map $\mathcal O \to \mathcal O(1)$ given by $1\mapsto x$ fails to be surjective at $(0,1)$ -- the section $x$ lies in the maximal ideal there. Similarly, {\em e.g.}, the map $1\mapsto x^2-xy$ gives a map $\mathcal O\to \mathcal O(2)$; tensoring this map by $\mathcal O(-1)$ gives a map $\mathcal O(-1)\to \mathcal O(1)$.

(2) If the divisors have a point in common, then the images of both maps will lie in the maximal ideal at that point, kind of as in (1) -- say you take the sum of the two maps above; then both of them land inside the maximal ideal at $(0,1)$ and hence so does image of the direct sum, and the map fails to be surjective. Take the second map to be $1\mapsto x^2-y^2$, and now one of the sections $x$ or $x^2-y^2$ is a generator at each point, so the map is onto (because it's surjective at the stalk level).

  • $\begingroup$ Thank you for your reply. I realized right after I posted the question is that to see maps in (1) are not surjective can also come from the fact that for any effective divisor $D$, $div(f)+D \ge 0$ does not imply $f \in \Gamma(X, \mathcal O) =k$. Here $X$ is a projective nonsingular curve over an algebraic closed field. But your (1) and (2) give more geometric insight to me. Thank you. $\endgroup$
    – user41541
    Commented Jun 27, 2015 at 4:07

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