Let $X$ be a smooth complete complex (algebraic) 3-fold, $D$ an effective divisor on $X$, and $C$ a smooth integral curve contained in the support of $D$. Let $X'$ be the blowup of $X$ along $C$, and denote by $D'$ the strict transform of $D$, $E$ the exceptional divisor, and $C'=D' \cap E$ the (set-theoretic) intersection.

Then is it true that $C \cong C'$, $D' \cong D$, and the normal bundles $N_{C/D} \cong N_{C'/D'}$ under the identification of $C \cong C'$ ? Are the degrees of the normal bundles the same?

  • 4
    $\begingroup$ I don't think $D' \simeq D$. If $D$ has two components $D_1,D_2$ with transverse intersection $C=D_1 \cap D_2$, then the strict transform under the blow up is disconnected. $\endgroup$ – J.C. Ottem May 20 '11 at 13:45
  • 3
    $\begingroup$ J.C. Ottem is correct; on the other hand I think $D^\prime \simeq D$ should be true under the assumption that $D$ is smooth along $C$. $\endgroup$ – user5117 May 20 '11 at 13:53
  • $\begingroup$ Yes, I see now. Thank you for your comments. $\endgroup$ – Parsa May 21 '11 at 16:21

Let $X=\mathbb P^3$, $D$ a quadric cone and $C$ a line on $D$ that goes through the vertex of $D$. Then $D'$ is isomorphic to $D$ blown up at its vertex, in particular it is smooth, $E\simeq C\times \mathbb P^1$ and $C'$ is isomorphic to the union of one member of each of the rulings on $E$, that is $C\cup \mathbb P^1$ intersecting in a single point transversally.


In general $D' \not\cong D$. In fact, $D'$ is the blow-up of $D$ in (the sheaf of ideals of) $C$ and $C'$ is the exceptional divisor of this blowup. So, if $C$ is a Cartier divisor on $D$ (locally given by 1 equation) then $D' \cong D$, $C' \cong C$. This is true if $D$ has multiplicity 1 along $C$.

  • $\begingroup$ Aha, while Sandor's example is excellent and answers the question, this is really answers the question I meant to ask. Thank you. $\endgroup$ – Parsa May 21 '11 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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