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Let $D$ be an effective divisor on a smooth projective connected complex algebraic variety $X$. Suppose that $D\leq E$. Is it true that $$\chi(X,\mathcal{O}_X(D)) \leq \chi(X,\mathcal{O}_X(E))?$$

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This is clearly true for curves by classical Riemann-Roch thoerem which computes Euler characteristic of a line bundle in terms of it's degree.

I claim that this is already false for surfaces and will prove that using some basic theory. All of this is covered in Hartshorne.

Let $X$ be a non-singular, projective surface with trivial canonical sheaf, for example a generic surface in $\mathbb{P}^{3}$ of degree $4$ is of this type by adjunction. Choose some point $P \in X$ and let $\tilde{X}$ be the blow-up of $X$ in $P$ and $E \subseteq \tilde{X}$ be the exceptional divisor. By classical arguments, $E^{2} = -1$ and $K _{\tilde{X}} = E + K _{X} = E$. We can use this data to compute Euler characteristic of $\mathcal{O} _{\tilde{X}} (2E)$ using Riemann-Roch for surfaces, which tells us that

$\chi(\mathcal{O} _{\tilde{X}} (2E)) = \chi(\mathcal{O} _{\tilde{X}}) + \frac{1}{2} (2E).(2E - K _{\tilde{X}}) = \chi(\mathcal{O} _{\tilde{X}}) + \frac{1}{2} (2E).E = \chi(\mathcal{O} _{\tilde{X}}) - 1$.

So $\chi(\mathcal{O} _{\tilde{X}} (2E)) < \chi(\mathcal{O} _{\tilde{X}})$ even though $2E$ is an effective divisor.

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But the only effective divisor equivalent to $E$ is $E$ itself.. What happens if you add an effective divisor that does not introduce a new fixed part? –  Joachim Feb 10 '13 at 10:19
    
What do you mean by does not introduce a new fixed part? –  Piotr Pstrągowski Feb 10 '13 at 11:17
    
Well say we have a complete linear system of divisors, that is the set of all effective divisors linear equivalent to some given divisor D. This is denoted $|D|$. If a curve $C$ is contained in all divisors in the system $|D|$, one says that C belongs to the fixed part of the system. If E is the exceptional curve of a blowup of a point that is already smooth, the system $|D + E|$ will just be the the divisors in the system $|D|$ plus the divisor $E$ (proof needed) and hence $E$ is a fixed part. In particular this means $h^0(D+E) = h^0(D)$. –  Joachim Feb 11 '13 at 0:08
    
Elaboration on the last claim: for any divisor $D$, the system $|D|$ of effective divisors linearly equivalent to it, is naturally isomorphic to the projectification of the linear space $h^0(X, \mathcal{O}_X(D))$. So if $\dim(|D + E|) = \dim(|D|)$ which holds if $E$ is as above, we also know $h^0(X, O(D)) = h^0(X, O(D+E))$ –  Joachim Feb 11 '13 at 0:11
    
Now my question reads as, what if we work under the additional assumption that $h^0(X, O(D)) < h^0(X, O(D+E))$? (This is equivalent to $E$ not introducing a new fixed part, i hope i managed the slightly impossible task of explaining this in the few lines above) –  Joachim Feb 11 '13 at 0:13
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