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Surely a silly question, but anyway. Suppose I blow up a point P in the plane. Then the exceptional divisor E should have zero intersection with (the strict transform of) any curve in the plane, since they are all linearly equivalent to a line not passing through P. But the strict transform of a line L through P seems to be a line which intersects E transversely (if I've computed correctly) which would mean L.E=1. What have I done wrong?

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This just means that the pull-back process doesn't preserve linear equivalence of divisors. This would work if we had a flat morphism.

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Thanks! I just realised that too - it was a silly question :) –  Anna B May 22 '12 at 19:58
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The exceptional divisor intersects trivially the total transform of a curve in the plane, not the strict transform.

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True. This is because pulling-back invertible sheaves is well defined for any morphism. The strict transform is the pull-back of cycles, the total transform is the pull-back of invertible sheaves. –  user18119 May 22 '12 at 22:18
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