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I'm trying to understand the proof at the top of page 3 of

http://math.stanford.edu/~vakil/02-245/sclass6A.pdf

http://math.stanford.edu/~vakil/02-245/sclass6B.pdf

Why can $D$ and $D'$ be moved in their linear equivalence class in such a way that they intersect transversely AND such that their intersection lies in $U$? Intuitively this makes sense, of course, but what are the formal arguments?

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    The file doesn't seem to work for me. Could you provide some more details for me please?2012-03-11
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    Does this one work? http://math.stanford.edu/~vakil/02-245/sclass6B.pdf2012-03-11
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    Or this one? http://math.stanford.edu/~vakil/02-245/sclass6.ps2012-03-11
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    I took a look, but I'm also a bit dazzled. I know that one can move $D$ and $D^\prime$ such that they intersect properly. Transversal intersection would mean that they intersect with multiplicity $1$. Maybe this follows from the fact that they are hyperplane sections? To see that you move the intersection into $U$ you can probably use the following. Let $S$ be a subset of codimension $2$ of $S$ (e.g. the set $D\cap D^\prime\cap (X-U)$). Then $D$ is lin eq to a divisor disjoint from $S$. So the philosophy is that you can move away from some finite set of closed points. Might be wrong.2012-03-11
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    This should follow from Bertini's type argument as in the answer of Matt. E. to a previous question of yours.2012-03-14
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    Thank you for answering. However I don't see how we can get a transversal intersection...2012-03-14

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