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Let $X$ be an abelian surface. $C$ be a curve in $X$. Consider the projective bundle $\pi:\mathbb{P}^1_C\longrightarrow C$. This is a projective morphism. I have two questions :

1) Can we find an effective divisor $D\neq 0$ on $C$ such that the line bundle corresponding to it pulls back to the trivial line bundle, that is $\pi^*\mathcal{O}_C(D)=\mathcal{O}_{\mathbb{P}^1_C}$? I suppose that if $D$ is a divisor linearly equivalent to the zero divisor, then this happens. But is it true that every degree 0 line bundle, that is an element of $Pic^O(C)$ pulls back to the trivial line bundle?

2) While I was trying this I was wondering if the degree of the line bundle is preserved by pullback. And what the degree of the morphism $\pi$ is.

Thanks in advance!

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There are several things I don't understand here. What role does the abelian surface play? (None that I can see.) What is the projective bundle? Is it just $\mathbf P^1 \times C$? (You didn't specify any rank 2 vector bundle on $C$ to be projectivised).

Now to your questions:

  1. For a surjective projective morphism, a nontrivial bundle never pulls back to the trivial bundle. This is an immediate consequence of the projection formula. So the answer to both questions here is no.

  2. There is no such thing as degree of a morphism that isn't finite, nor degree of a line bundle on a variety other than a curve.

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  • $\begingroup$ Thank you! By $\mathbb{P}^1_C$, I mean $\mathbb{P}^1\times C$ only, is that not $\mathbb{P}(\mathcal{O}^{\oplus 2})$? $\endgroup$ – gradstudent Aug 18 '15 at 16:08
  • $\begingroup$ @poorna: yes, that's right.It was just confusing to me that you denoted it $\mathbf P^1_C$ and used the term "projective bundle" rather than just calling it the product. $\endgroup$ – Schemer Aug 19 '15 at 8:12

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