Let $X$ be a smooth projective variety over $\mathbb{C}$. Consider the blow up of $X$ about a closed subvariety $Z$. Let $X'=Bl_Z(X)$. Let $Y$ be a smooth irreducible divisor of $X$ properly containing $Z$. Let $Y'=Bl_Z(Y)$. I have the following doubts.

1) I believe that $Y'$ is the strict transform of $Y$ under the blow up $\pi:X'\rightarrow X$. Is that correct?

So $Y'$ is a divisor in $X'$.

2) What is the relation between $O_X(Y)$ and $O_{X'}(Y')$? I am getting that $\pi^*O_X(Y)\otimes O_{X'}(-E)= O_{X'}(Y')$. Here $E$ is the exceptional divisor. Is that right?

Help will be appreciated.

  • 1
    $\begingroup$ Certainly (1) is true. This ought to be in most books, e.g., Hartshorne II.8. $\endgroup$
    – Hoot
    Jun 13, 2016 at 16:20
  • $\begingroup$ @Hoot for the record, it's II.7 $\endgroup$
    – rollover
    Apr 26, 2017 at 10:51

1 Answer 1


Point (1) is correctly addressed in the comment by Hoot. As for point (2), your intuition is on the right track. On the other hand, you should keep track of the multiplicities of the loci involved. As a matter of example, let $X$ be the projective plane, $Y$ the cuspidal rational curve, and $Z$ the singular point of $Y$. $Z$ is a regular subvariety of $X$, so the exceptional divisor is just a copy of $\mathbb{P}^1$ (in general, if you blow up something singular, the exceptional locus might be pretty ugly though). The strict transform of $Y$ (i.e. $Y'$ in your notation) is going to be a smooth rational curve tangent to $E$. This reflects that $Y$ has multiplicity 2 along $Z$. This gives you that $\pi^*Y= Y'+2E$. As you see, the ingredients are exactly the ones you expected, but, in this case, they are weighted with coefficients depending on the singularities of $Y$ along $Z$.

Edit I am reading your answer more carefully now. If both $Y$ and $Z$ are smooth, then claim (2) is fine as well.

Addendum You comment is right. The blow up is an isomorphism over $X \setminus Z$. In particular, if $\widehat{Y}$ is disjoint from $Z$, its strict transform $\widehat{Y}'$ coincides with the pullback $\pi^*(\widehat{Y})$ , and it is isomorphic to $\widehat{Y}$. Now, if $Y$ and $\widehat{Y}$ are linearly equivalent, so are their pullbacks (just because the isomorphism between $\mathcal{O}_X(Y)$ and $\mathcal{O}_X(\widehat{Y})$ induces an isomorphism between their pullbacks). On the other hand, this is telling you that the strict transforms of linear equivalent divisors are not linearly equivalent if just one of the two goes through $Z$.

Let me be more explicit. Blow up a point $P$ in $\mathbb{P}^2$. Let $L_1$ be a line through $P$, and $L_2$ a line not containing $P$. Denote by $M_1$ and $M_2$ the respective strict transforms. Then, by what above said, we have $\pi^*L_1=M_1+E$, and $\pi^*L_2=M_2$. By Bezout theorem we know that the intersection products $L_1 \cdot L_2=(L_1)^2=(L_2)^2=1$. In particular $L_1$ and $L_2$ meet properly at one point, say $Q$. Now, since $L_2$ does not go through $P$, we have that the pullbacks $M_1+E$ and $M_2$ meet properly at one point (the only premiage of $Q$). Given that these divisors are also linearly equivalent to each other, we get $1=(M_2)^2=M_2 \cdot (M_1+E)=(M_1+E)^2$. In particular, we get $1=(M_1+E)^2=M_1^2+2M_1\cdot E+ E^2$. Since $M_1$ and $E$ meet properly at one point, we know $M_1 \cdot E=1$. Then, we know that $E^2=\mathrm{deg}\mathcal{O}_{X'}(E)_{|E}$. By the description in sections 7 and 8 in chapter 2 of Hartshorne, we know that this is the relative $\mathcal{O}(1)$ bundle, that $E=\mathbb{P}^1$; these things together tell us that $\mathcal{O}_{X'}(E)_{|E}\cong \mathcal{O}_{\mathbb{P}^1}(-1)$. Thus, that degree is $-1$, so $E^2=-1$. This is negative self intersection is phrased as $E$ does not deform: there is no other effective divisor equivalent to $E$. Now, putting this in our previous equation, we get that $(M_1)^2=0$. As you see, $(M_1)^2 \neq (M_2)^2$; in particular, they can not be linearly equivalent.

  • $\begingroup$ @Stephano, thank you! But for (2), we could have a $Y_1$ linearly equivalent to $Y$ and misses $Z$, in that case the strict transform $Y_1'$ is isomorphic to $Y_1$, and the $\pi^*O(Y_1)=O(Y_1')$ right? I am confused. $\endgroup$ Jun 13, 2016 at 18:51
  • $\begingroup$ @gradstudent I have edited the answer to address your comment. $\endgroup$
    – Stefano
    Jun 13, 2016 at 19:13
  • $\begingroup$ @Stephano, thanks for the great answer. So indeed we have that the line bundle $O_{X'}(Y')\simeq\pi^* O_X(Y)\otimes O_{X'}(-E)$, isn't it? $\endgroup$ Jun 13, 2016 at 19:24
  • $\begingroup$ @gradstudent Yes, it is what we have. $\endgroup$
    – Stefano
    Jun 13, 2016 at 19:30
  • 2
    $\begingroup$ One should probably prove that identity. I think you just want to imitate what Hartshorne does in the case of a surface in V.3 or somewhere around there. $\endgroup$
    – Hoot
    Jun 13, 2016 at 20:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .