Let $X$ be an integer curve of (arithmetic) genus $g=0$. (the arithmetic genus $g$ is defined by $g:= 1 -\chi_k(\mathcal{O}_X)$ where $\mathcal{O}_X$ is the structure sheaf of $X$ and $\chi_k(\mathcal{O}_X) := \sum _{i \ge 0} (-1)^i dim_k H^i(X, \mathcal{O}_X)$ the Euler-Poincare characteristic of $\mathcal{O}_X$.

Let assump that there exist a $k$-valued point $a \in X(k)$. I wan't to show that the induced Cartier Divisor $\mathcal{O}_X(a)$ is globally generated, therefore for every $x \in X$ there exist a global section $s \in \Gamma(X, \mathcal{O}_X(a)= H^0(X,\mathcal{O}_X(a)) $ such that the stalk $s_x$ at $x$ is not zero:

My attempts:

I firstly considered the exact sequence

$0 \to \mathcal{O}_X \to \mathcal{O}_X(a) \to \mathcal{F} \to 0$

where $\mathcal{F}$ is cokernel of the sequence.

This provides a long exact sequence of $\mathcal{O}_X(X)=H^0(X, \mathcal{O}_X)$-modules

$0 \to H^0(X, \mathcal{O}_X) \to H^0(X, \mathcal{O}_X(a)) \to H^0(X, \mathcal{F}) \to H^1(X, \mathcal{O}_X) \to ...$

Because $g=0$ and $H^i(X, \mathcal{O}_X)=0$ for $i \ge 2$ we conclude $H^1(X, \mathcal{O}_X)=0$.

Futhermore I suppose that $H^0(X, \mathcal{O}_X) =k$ (why?)

So I can simplify my sequence to

$0 \to k \to H^0(X, \mathcal{O}_X(a)) \to \to H^0(X, \mathcal{F}) \to 0$

My goal will be to show that $1$ is mapped in the left module morphism to an invertible element in $H^0(X, \mathcal{O}_X(a))$, but I don't know how to conclude it

  • $\begingroup$ You have not used the fact, that you actually know how the cokernel looks like. It is a skyscraper sheaf supported at $a$ with stalk $k$. $\endgroup$
    – MooS
    Nov 30, 2017 at 9:59

1 Answer 1


Let $i: \{a\} \to X$ be the inclusion of your point. Note that $\{a\} = \operatorname{Spec} k$ by the assumption that $a$ is $k$-valued.

Note that your exact sequence comes from the exact sequence

$$0 \to \mathcal O_X(-a) \to \mathcal O_X \to i_* \mathcal O_a \to 0$$

We have $\mathcal O_a = k$, hence the co-kernel is just a skyscraper sheaf supported at $a$ with stalk $k$.

Also note that the first map of the sequence is an isomorphism away from $a$. Now we tensor with $\mathcal O_X(a)$ to obtain

$$0 \to \mathcal O_X \to \mathcal O_X(a) \to k \to 0.$$

Note that the co-kernel does not change, because $\mathcal O_X(a)$ is free of rank $1$ at $a$, i.e. tensoring with it does not change anything locally at $a$.

Of course the first map is still an isomorphism away from $a$. In particular the image of $1 \in H^0(X,\mathcal O_X)$ will generate $\mathcal O_X(a)$ at all points but $a$. To get a global section which generates at the point $a$, we take the long exact sequence and as you have figured out yourself, the genus assumption yields $H^1(X,\mathcal O_X)=0$, i.e. we get

$$0 \to k \to H^0(X,\mathcal O_X(a)) \to k \to 0$$

In particular $\dim H^0(X,\mathcal O_X(a))=2$ and the "other global section besides $1$", i.e. a pre-image of $1$ along the surjective map $H^0(X,\mathcal O_X(a)) \to k$ generates $\mathcal O_X(a)$ at $a$.

  • $\begingroup$ Hi. Thank you for your enlightening answer. Indeed your argument pointed precisely out why the preimage of $1 \in k$ in the map $H^0(X,\mathcal O_X(a)) \to k$ doesn't vanishes at stalk $\mathcal O_X(a)_a$. The only still unclear argument is your phrasing that "the map above is still an isomorphism away from $a$". Do you mean this in sense of restrictions to all stalks at $x \in X$ with $x \neq a$ and under using the fact that $\mathcal O_X$ as structure scheaf is always (?) globally generated? $\endgroup$
    – user267839
    Dec 1, 2017 at 12:05
  • $\begingroup$ If you have an isomorphism at stalks (away from $a$), this does not change after tensoring, because tensoring commutes with taking stalks. I do not need any property of the structure sheaf or global generation or whatever for this. $\endgroup$
    – MooS
    Dec 1, 2017 at 12:09

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