Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

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

share|cite|improve this answer
Thanks! I just realised that too - it was a silly question :) – Anna B May 22 '12 at 19:58

The exceptional divisor intersects trivially the total transform of a curve in the plane, not the strict transform.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.