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for the embedding defined by very ample divisors, if they are lin. eq, then the embeddings are the "same" (up to a linear transformation). What do we know if given that the divisors are algebraically equivalent? doe this imply something? (e.g. maybe the embeddings they defined appear in an algebraic family of embeddings?)

[BTW, I want to ask the same question for numerically equivalent divisors, any comments are welcome.]

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A little comment. A divisior D defines an embedding to the projective space of dimenison $h^0(O(D))-1$. Now for two algebraically euivalent divisors D and D' this dimension can be different. For example you can consider a curve of genus g. Then for a canoncial divisor you get an embedding to $P^{g-1}$, while for a generic divisor you get $P^{g-2}$. If you a happy to call two embeddings to spaces of diffent dimesnion algebraically equivalent, than the answer to your quesiton is yes. Overwise it is no.

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Let me see, your question depends strongly on the variety your line bundle is over. The equivalence classes, algebraic or numeric, of line bundles are measuring some facts about your variety. the former is a cohomology group whereas the latter may have information on the rational curves that your variety may have. That is to say, we've got to say what is the base variety like. Supposing the variety $X$ over $k=\overline{k}$ is a projective scheme which is regular in codimension one, then two effective divisors algebraically equivalent give rise to the same line bundle $\mathcal{O}(D)$ (invertible sheaf) which eventually gives rise to morphism $\phi:X\rightarrow \mathbb{P}(H^0(X,mD))=\mathbb{P}^r$ (up to an automorphism of $\mathbb{P}^r$). Meaning we are getting a map $\phi$ out of the line bundle $\mathcal{O}(mD)$ which is going to be an embedding provided the linear system $|\mathcal{O}(mD)|$ separate points and separate tangent vectors. If the former property is not true, we may get only a rational map $\phi:X\rightarrow \mathbb{P}^r$, whereas the latter property is taking care of the singularities of the image $\phi(X)\subset \mathbb{P}^r$. The map $\phi$ though depends on the numerical class of the divisor we started with, then that's why I asked for an effective divisor at the beginning.

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Thanks for your explanation, I wonder when you say two alg equiv effective divisors give rise to the same line bundle $\mathcal{O}(D)$, doesn't that mean they are actually linear equivelant? – Ying Zhang Oct 5 '10 at 3:19

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