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Oct 6, 2013 at 23:23 comment added N B Sasha, of course you are right, by proceeding in this way for every fixed degree I didn't mean that $X$ stays unchanged topologically. So that for different degrees we have topologically different $X$. So the construction is not universal in sense of degree, but in each specified case it works.
Oct 6, 2013 at 20:33 comment added Sasha @N B: Not quite. For example, if $C = P^1$ then $X = P_{P^1}(O \oplus O(k))$ is a Hirzebruch surface, and as far as I understand those for even $k$ and for odd $k$ are not homeomorhpic.
Oct 6, 2013 at 17:27 comment added N B By the way, Sasha, I think your argument gives the answer for the divisors of the same degree, this is when you vary the algebraic structure of $X$, while topologically $X$ is the same. One can do it for every fixed degree in this way, of course.
Oct 6, 2013 at 17:15 history edited N B CC BY-SA 3.0
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Oct 6, 2013 at 17:12 comment added N B Artie, thank you. I apologize for not formulating it clearly. Actually, Sasha answered the version of the question stated in parentheses which allows to vary $X$ (he chose $X$ to be geometrically ruled). When one fixes $X$, it is still not answered here.
Oct 6, 2013 at 16:58 comment added user5117 N B and @Sasha: I thought the problem was for fixed $C$ and $X$? Those projective bundles certainly won't be isomorphic for all $L$...
Oct 6, 2013 at 14:25 history edited N B CC BY-SA 3.0
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Oct 6, 2013 at 14:23 comment added N B Sasha, thank you. This is elementary, of course, but somehow I was not able to get it at the moment of writing my comment.
Oct 6, 2013 at 5:28 comment added Sasha For any line bundle $L$ on $C$ you can embed $C$ into $P_C(O \oplus L)$ as a section in such a way that the normal bundle is $L$, so the normal bundle can be arbitrary.
Oct 6, 2013 at 3:00 history asked N B CC BY-SA 3.0