Let $[x_1,x_2,x_3,x_4]$ be coordinates of $\mathbb{P}^3$ and $Z\subset \mathbb{P}^3$ the subscheme given by the ideal $$I_Z=(x_1,x_2,x_3^2) \subset \mathbb{C}[x_1,x_2,x_3,x_4]$$ i.e. $Z$ is a double point supported on the line $L:(x_1=x_2=0)$.

I want to consider the blowup of $\mathbb{P}^3$ along $Z$, i.e. $X=Bl_Z\mathbb{P}^3$. In order to do so I try two different approaches:

  1. Blowup directly the entire scheme $Z$. In this way I get a singular variety $X$.
  2. First blow up the reduced point $p=[0,0,0,1]$, with $p$ equal to the support of $Z$, and then blow up the point $q$ in $E_p\subset Bl_p\mathbb{P}^3$ corresponding to the direction of the line $L$ in the exceptional divisor $E_p$. In this way I obtain a smooth variety $$Y=Bl_q(Bl_p\mathbb{P}^3)$$ with the exceptional divisor $E_q \subset Y$ and the strict transform $E'_p$ of $E_p$ with $E'_p$ isomorphic to the blowup of $\mathbb{P}^2$ at one point.

My questions now are the following:

a) It seems that the output of procedure 1) is not the same as the output of 2), in fact $X$ is singular while $Y$ is smooth. But the resulting scheme that I'm blowing up it seems the same to me, with the only difference that in procedure 1) I'm blowing it up in a single step while in 2) I've done it in two subsequent steps. Why am I obtaining two different varieties?

b) I know that the exceptional divisor of a point in $\mathbb{P}^3$ is isomorphic to $\mathbb{P}^2$. Is this also the case in the second blowup when I'm considering the blowing up of $q \in E_p$? In other words is still $E_q$ isomorphic to $\mathbb{P}^2$? I've the feeling that the answer is no, essentially because the line $l=E'_p \cap E_q$ seems to have negative self intersection in $E_q$. So at this point what even is $E_q$?

Thanks in advance for eventual answers and observations/remarks.


1 Answer 1


These are definitely not the same. $Z$ is a complete intersection subscheme, so if you blow up $Z$, then the exceptional divisor (=set) is a single $\mathbb P^2$, while in the second case it's a union of a copy of $\mathbb P^2$ and a copy of a $\mathbb P^2$ blown up at a point (as you point out it is easy to see that it is not irreducible). I don't know how to answer your question of why you are obtaining different varieties, if not by a counter question: why would you get the same? In the second blow up you are blowing up something that does not come from the original variety. They are just not the same.

For your questions in b): If you blow up a non-singular point on any variety, then the exceptional set will be a projective space of one dimension less then what you are blowing up. I doubt that $l$ has negative self-intersection on $E_q$, because the latter is also a $\mathbb P^2$. (It has negative self-intersection on $E_p'$, but that makes sense.)


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.