Let $f:C\longrightarrow C'$ be a finite morphism of curves. Let $A'$ be a line bundle on $C'$ and let $A$ be it's pullback to $C$. If $A'$ is globally generated, is it true that $A$ is globally generated too?

Since $A'$ is globally generated, we have a surjection $H^0(C',A')\otimes \mathcal{O}_{C'}\longrightarrow A'\longrightarrow 0$.

Pullback operation is right exact, therefore we get $H^0(C',A')\otimes\mathcal{O}_C \longrightarrow A \longrightarrow 0$.

But since $f$ is a finite morphism, $h^0(C,A)$ could be greater than $h^0(C',A')$. Therefore we could compose with projection to get the required surjection. Is that right?

But $H^0(C',A')\otimes\mathcal{O}_C \longrightarrow A \longrightarrow 0$ --- this surjection shows that $A$ can be generated by fewer global sections isn't it? Thank you!


Given a section $\sigma'\in H^0(C',A')$ and its pullback $\sigma=f^*(\sigma') \in H^0(C,A)$, we have for every $c\in C$ the formula: $$\sigma(c)=\sigma'(f(c))\in A[c]=A'[f(c)]$$ So if $\sigma'(f(c))\neq 0$, which can always be achieved for a suitable choice of $\sigma '$ by the hypothesis of global generation of $A'$, we will have $\sigma(c)\neq0$, and this proves that $A$ is globally generated too.

Nota Bene
The above is valid for arbitrary morphisms (finite or not) between arbitrary varieties of arbitrary (maybe different) dimensions.

  • $\begingroup$ Thanks Georges! But doesn't it mean that $A$ can be generated by a subspace $H^0(C',A')$ of $H^0(C,A)$? That is the stalks of $A$ are generated by fewer sections. But will this not mean that the complement say $\mathcal{O}^r$ is in the kernel of the surjection $H^0(C,A)\otimes\mathcal{O}_C\longrightarrow A\longrightarrow 0$? $\endgroup$ – gradstudent Sep 9 '15 at 11:42
  • $\begingroup$ Yes, sections coming from $H^0(C',A')$ suffice to generate $A$. $\endgroup$ – Georges Elencwajg Sep 9 '15 at 12:20
  • $\begingroup$ But the kernel $K$ of $H^0(C,A)\otimes \mathcal{O}_C\longrightarrow A$ has no global sections right? Then how does it contain copies of $\mathcal{O}$ $\endgroup$ – gradstudent Sep 9 '15 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.