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Let $X$ be a projective surface over $\mathbb{C}$, let $x\in X$ be the only singular point of $X$. Let $L$ be an ample line bundle on $X$. Consider the blow up $Y$ of $X$ along $x$, $f:Y\longrightarrow X$. Let $L'$ be the pull back of $L$ to $Y$ and let $E$ be the exceptional divisor.

We have the following short exact sequence on $Y$,

$0\longrightarrow L'-E\longrightarrow L'\longrightarrow L'|_E\longrightarrow 0$.

So there is an injection $|L'-E|\hookrightarrow |L'|$ which sends $C\mapsto C+E$.

1) Now what are the curves in $|L'-E|$? Are these curves that contain $E$ or curves that pass through a point in $E$?

2) What does the divisor $C+E$ mean? It means a union of the curves $C$ and $E$?

3) If so $C$ is a closed subscheme of $C+E$. Hence we have the surjection $\mathcal{O}_{C+E} \longrightarrow \mathcal{O}_C$. What is the ideal sheaf? It looks to me to be $\mathcal{O}_E$.

4) If $C$ is indeed a closed subscheme of $C+E$, and we start with a line bundle $A$ on $C+E$, and call the pullback to $C$ as $A'$, the degrees will be same I suppose. But what is the relationship between $h^0(C,A')$ and $h^0(C+E,A)$?

Sorry about the long post. These questions have been bothering me for a while. Thanks in advance!

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1) They may or may not contain $E$. 2) Union is a set theoretic notion. $C+E$ is a scheme theoretic notion.Locally, they are defined by the product of the equations defining $C$ and $E$. 3) No, the ideal sheaf is in fact $\mathcal{O}_E(-C)$. 4) The degree does not make sense for a possibly non-irreducible curve. You only have a multi degree, restricting to each component.In general, the relationship of the $h^0$ will depend, by the long exact sequence, on $h^i(\mathcal{O}_E(A-C)$.

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  • $\begingroup$ thank you! I still have the following doubts. (1) Then how does one describe curves in $|L'-E|$? Is $C+E$ a curve which avoids $E$? If we start with a curve $C'$ from $Y$ which avoids $x$, then how does one think of $C'-E$? $\endgroup$ – gradstudent Oct 30 '15 at 3:46
  • $\begingroup$ Do smooth curves in $|L'-E|$ have to intersect $E$? $\endgroup$ – gradstudent Oct 30 '15 at 3:53
  • $\begingroup$ I do not know how to answer (1), it will depend on a specific situation. $C+E$ clearly has $E$ in it, so in what sense does it avoid $E$? If $C'$ as you said, $C'-E$ is not an effective divisor, while it may be linearly equivalent to one. Smooth curves in $|L'-E|$ may or may not intersect $E$. It will be good for you to work out a bunch of simple examples to get an idea. $\endgroup$ – Mohan Oct 30 '15 at 16:10

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