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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 $L \cdot E=1$. 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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  • $\begingroup$ Thanks! I just realised that too - it was a silly question :) $\endgroup$
    – Anna B
    May 22, 2012 at 19:58
  • $\begingroup$ Maybe "strict transform" is a better word than pull-back. The canonical map $\operatorname{Pic}(X) \to \operatorname{Pic}(\widetilde X)$, which "pulls-back" the line bundle, certainly preserves linear equivalence, if we identify line bundles with divisors modulo linear equivalence. $\endgroup$ Sep 21, 2019 at 14:57
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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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    $\begingroup$ 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. $\endgroup$
    – user18119
    May 22, 2012 at 22:18

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