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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 how do we describe curves in $|L'-E|$? Are these curves that contain $E$ or curves that pass through a point in $E$? Do they have to intersect $E$?

2) What does the divisor $C+E$ mean? The divisor $C+E$ comes from $X$ which passes through $x$?

3) Is $C$ is a closed subscheme of $C+E$. Then 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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    $\begingroup$ 1) Note that when you blow up a point $x$, the sheaf $\pi^*I_x$ is locally free - corresponding to the locally free sheaf $O(-mE)$ where $m$ is the multiplicity of the point. So sections of $O(L'-E)=\pi^*L\otimes O(-E)=\pi^*L\otimes \pi^*I_x$ correspond to sections of $L$ vanishing along $x$. In other words, the linear system $|L'-E|$ corresponds to (strict transforms of) curves of $|L|$ passing though $x$. $\endgroup$
    – Walter Neff
    Commented Oct 30, 2015 at 17:23
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    $\begingroup$ 2) $C-E$, as a divisor, is not effective (it has a negative coefficient). It might however still be linearly equivalent to one, say $D$, with non-negative coefficients. In that case $D\in |O(C-E)|$ corresponds, as above to a curve passing through $x$. 3),4) If $C$ is a curve (irreducible!) then $C-E$ is not a subscheme of $C$. The only way one can make sense of these statements is in the case when $C$ itself is a reducible divisor, and $E$ is one of the components. Then some $C-mE$ could mean the divisor corresponding to the other components of $C$. $\endgroup$
    – Walter Neff
    Commented Oct 30, 2015 at 17:23
  • $\begingroup$ @WalterNeff, I had one more doubt, the linear system $|L'-E|$ is basepoint free right even though the corresponding curves in $|L|$ have $x$ as a base point.? $\endgroup$
    – gradstudent
    Commented Nov 3, 2015 at 12:55

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In the following I suppose that by "curve" in a linear system you mean "effective divisor" vithout any claim about being irreducible and/or reduced.

1) Curves in |L'| are exactly the pull-back of curves in |L|. So "curves" in |L'-E| are obtained by pulling-back curves in L through x, and then remove $E$ with multiplicity 1. That's usually the strict transform, but may contain $E$ if the curve in $|L|$ passes through $x$ with high multiplicity.

2) $C+E$ is the reducible effective divisor union of $C$ and $E$ (of course) and equals the pull-back on $Y$ of the image of $C$ o $X$. In other words you get all pull-backs of curves in $|L|$ passing through $x$.

3) No, it is ${\mathcal O}_E(-C)$, its sections are supported on $E$ but vanish on the intersection with $C$.

4) No, the degree do not need to be the same. Assume for instance that $C$ is smooth irreducible intersecting transversally $E$ in $d$ points. Then every pair of line bundles on $C$ and $E$ may be "glued" to a line bundle on $C+E$.

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  • $\begingroup$ Thanks @Roberto Pignatelli! I had one more doubt, the linear system $|L'-E|$ is basepoint free right even though the corresponding curves in $|L|$ have $x$ as a base point.? $\endgroup$
    – gradstudent
    Commented Nov 3, 2015 at 12:54
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    $\begingroup$ Usually yes. But, as you are only supposing that $L$ be ample, it can be very small. I can construct examples where $|L'-E|$ is a single divisor, and then you get a lot of fixed points. $\endgroup$
    – Roberto Pignatelli
    Commented Nov 3, 2015 at 17:00
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    $\begingroup$ Only if $CE=1$. In this case, as you assume that $C$ does not contain $E$, $C$ intersects $E$ transversally in a smooth point of $C$, and then by transversality the curve $C'$ remains smooth in the image of that point. $\endgroup$
    – Roberto Pignatelli
    Commented Nov 8, 2015 at 17:45
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    $\begingroup$ If $CE \geq 2$, on the contrary, $C'$ will be singular. If for example $C$ and $E$ intersect in two or more points, all those points will map to the same point of $X$, which will be then singular for $C'$. But even if they intersect only in one point, as the intersection will not be transversal, $C'$ will have a cusp there. $\endgroup$
    – Roberto Pignatelli
    Commented Nov 8, 2015 at 17:48
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    $\begingroup$ No. Set $X$ the quadric cone in ${\mathbb P}^3$, $L={\mathcal O}_X(1)$. Then $Y$ is the Hirzebrich surface ${\mathbb F}_2$, $E$ is a rational curve with self intersection $(-2)$. You can pick as $C'$ an hyperplane section through the node $x$, two lines intersecting in $x$. Then $C$ is the union of two lines of the ruling of ${\mathbb F}_2$, each intersecting $E$ tranversally in a different point, so $CE=2$. The proper transform of $C'$ is in this case $C+E$. $\endgroup$
    – Roberto Pignatelli
    Commented Nov 8, 2015 at 18:26

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