Let $L\subset \mathbb C^3$ be the line defined by $x=y=0$, and $p\in L$ the point defined by $x=y=z=0$. Let's consider the blowup of $\mathbb C^3$ at $p$ and then blow up the strict transform of $L$ and denote the new space as $X$.

We can also consider blowup of $\mathbb C^3$ along $L$ and then blow up the preimage of $p$ and denote the new space as $Y$. The two spaces are not the same because the fibers over $p$ are not Edited: The fibers of $X$ and $Y$ over $p$ are the same: On one side, it is the blowup a point on $\mathbb P^2$, while on the other side, it can be viewed as a $\mathbb P^1$-bundle over $\mathbb P^1$.

Question 1: Are $X$ and $Y$ isomorphic?

According to Hartshorne II. Theorem 7.17, $$X\to \mathbb C^3$$ is a blowup of some ideal $I$ on $\mathbb C^3$, and the same for $Y$ and we denote the corresponding ideal as $J$.

Question 2: How to determine the ideals $I$ and $J$ explicitly?

It is elementary to check that both $I$ and $J$ are contained in the ideal $(x,y)$ of $L$, reduced on $L\setminus \{p\}$ and non-reduced at $p$ (otherwise, they would be isomorphic to $Bl_{L}\mathbb C^3$), but they have different non-reduced structure at $p$.

$(x^2,y^2,xy,xz,yz)$ as Youngsu suggested is probably the first ideal to consider that is supported on $L$ and non-reduced only at $p$, but currently, I can't determine it is $I$ or $J$, or perhaps neither of them.

How to find such $I$ and $J$ explicitly? Thanks in advance for any help.

  • $\begingroup$ If $E \subset X$ is the exceptional divisor and $\pi:X \to \mathbb C^3$ is the blowup map, IIRC we have the more general identity $I^{\otimes m} \cong \pi_* \mathcal O_X(-mE)$, but I don't have a reference off the top of my head. I believe it's discussed somewhere in the first chapter or two of Lazarsfeld's Positivity in Algebraic Geometry, somewhere near the discussion of Castelnuovo-Mumford regularity. $\endgroup$ Dec 21, 2020 at 23:40
  • $\begingroup$ @TabesBridges I worked that fact out here if you're looking for a quick link. $\endgroup$
    – KReiser
    Dec 22, 2020 at 0:47
  • $\begingroup$ @TabesBridges The identity $I=\pi_*\mathcal{O}_X(-E)$ requires both the variety and the blowup center to be smooth, but it is not the case here: the blowup center $\mathbb C[x,y,z]/I$ is not reduced at $p$. Actually, my computation shows that $\pi_*\mathcal{O}_X(-E)=(x,y)$, which is $I_{red}$ but not $I$, so the identity breaks down. $\endgroup$
    – AG learner
    Dec 22, 2020 at 1:33
  • $\begingroup$ Ah true, I couldn't remember if it was still valid for compositions of smooth blowups or not... $\endgroup$ Dec 22, 2020 at 1:55
  • $\begingroup$ Have you tried the product of the ideals $(x,y,z)$ and $(x,y)$? $\endgroup$
    – Youngsu
    Dec 22, 2020 at 2:15

1 Answer 1


Inspired by Sasha's answer here: https://mathoverflow.net/q/267573, I'm able to show that the two spaces arising from different orders of blowups are actually the same. Subsequently, the ideal of the blowup can be determined.

Let's first consider the following diagram. $\sigma_1$ blows up the point $p=[1:0:0:0]$ with exceptional divisor $E$, while $\sigma_2$ blows up the strict transform $\tilde{L}$ of the line $L$ through $p$, and the exceptional divisor is $D$. We work in projective space instead of affine space because we'd like to apply Mori's theorem. $\require{AMScd}$ \begin{CD}\label{1}\tag{1} \text{Bl}_{\tilde{L}}\text{Bl}_{p}\mathbb P^3 @>\sigma_2>> \text{Bl}_{p}\mathbb P^3\\ @VVfV @VV\sigma_1V\\ N @>g>> \mathbb P^3 \end{CD}

Then I claim that $X=\text{Bl}_{\tilde{L}}\text{Bl}_{p}\mathbb P^3$ can also be obtained by successive blowup along $L$ and then blow up along the preimage of $p$ (the other bridge of the diagram). This answers Question 1.

How to show that? Note that the strict transform $\tilde{E}$ of $E\cong\mathbb P^2$ is isomorphic to blowup of a point on $\mathbb P^2$, which is a ruled surface. We need Mori's cone theorem to show that we can contract $\tilde{E}$ in the ambient 3-fold!

First, the relative Picard group $\text{Pic}(X/\mathbb P^3)$ is generated by $\tilde{E}$ and $D$.

Second, as a ruled surface, $\tilde{E}$ has a fiber $F$ and a section $C$ at infinity. Note that $C$ is also in the ruling of $D$, one find the relative Mori cone $\overline{NE}(X/\mathbb P^3)$ is generated by $$C,F.$$

By computing the intersection number: $C\cdot \tilde{E}=0$, $C\cdot D=-1$, $F\cdot \tilde{E}=-1$, and $F\cdot D=1$, one concludes that the curve class $[F]\in \overline{NE}(X/\mathbb P^3)$ is an extremal ray. The fact that $$F\cdot K_X=F\cdot (2\tilde{E}+D)=-1$$ shows $F$ is $K_X$-negative. Mori's cone theorem [KM, Theorem 3.7] tells us:

Proposition: There is a unique birational morphism $f:X\to N$ which contracts $\tilde{E}$ to a curve $C'$. Moreover, $N$ is smooth and projective.

Similarly, one shows that one can contract the divisor $f(D)\cong D$ on $N$ to $L$ and obtain $\mathbb P^3$. This completes the other bridge in the commutative diagram $\eqref{1}$.

Finally, to answer Question 2, according to Sasha's explanation, the ideal sheaf of the birational morphism $X\to \mathbb P^3$ is the product of the ideal sheaf of the blowup of $\sigma_1$ and the ideal sheaf of the blowup $g$. So the ideal of $X\to \mathbb P^3$ (locally) is

$$(x,y,z)\cdot (x,y)=(x^2,y^2,xy,yz,xz),$$

which is indeed the same one suggested by Youngsu.

[KM]: J. Kollár and S. Mori. Birational geometry of algebraic varieties. Cambridge Tracts in Mathematics, 134. Cambridge University Press, Cambridge, 1998.


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