Multiple blow-ups

algebraic-geometrybirational-geometryblowup

Suppose I consider, inside $\mathbb{P}^6$, the subvarieties $Y=V(x_2,\ldots,x_6)\simeq \mathbb{P}^1$ and $Z=V(x_0,\ldots,x_3)\simeq \mathbb{P}^2 $.

I want to understand what is the blow up of $\mathbb{P}^6$ along $Y,Z$. Following Harris' Algebraic Geometry book, I started from $Bl_Y(\mathbb{P}^6)$: the ideal of $Y$ is $I(Y)=(x_2,\ldots,x_6)$, and therefore I have a rational map

$$\phi:\mathbb{P}^6 \dashrightarrow \mathbb{P}^4, p\mapsto [p_2:\ldots:p_6]$$
and $Bl_Y(\mathbb{P}^6)\subset \mathbb{P}^6\times\mathbb{P}^4$ is the closed variety associated to the graph of $\phi$.

Now I'd like to blow up $\mathbb{P}^6$ along $Z$, thus I suppose I need to study

$$Bl_Z(Bl_Y(\mathbb{P}^6))$$

and here I have several questions:

  • $Bl_Z(Bl_Y(\mathbb{P}^6))\subset \mathbb{P}^6\times \mathbb{P}^4 \times \mathbb{P}^3$? I guess so since I'm repeating the above procedure, so I was wondering if a sequence of $n$-blow ups for example live in a cartesian product of $n+1$-projective spaces (altough I can obviously embed them via a Segre map);
  • Can I simultaneously blow up $\mathbb{P}^6$ along $Y,Z$? Do I obtain the same construction?
  • (Linked to the above question): what happens if $Y\cap Z \neq \emptyset$?

Best Answer

  1. Yes, you can do this - whenever you blow up something (locally) cut out by $d$ equations inside $X$, you can embed your blowup in the product $X\times\Bbb P^{d-1}$. This follows from the description of the blowup as a relative proj: $Bl_ZX=\underline{\operatorname{Proj}} \bigoplus \mathcal{I}_Z^n$, and locally on $X$ we get a surjection $\mathcal{O}_X[t_1,\cdots,t_d]\to \bigoplus \mathcal{I}_Z^n$ which (after gluing) corresponds to a closed immersion $Bl_ZX\hookrightarrow \Bbb P^{d-1}_X$.
  2. Yes, if $Y,Z$ are two disjoint closed subschemes then all of the blowups $Bl_Y(Bl_Z X)$, $Bl_Z(Bl_Y X)$ and $Bl_{Y\cup Z} X$ are the same. This can be verified locally, since blowups are isomorphisms away from the thing you blow up.
  3. It turns out that part of our conclusion from 2 holds once we generalize appropriately: $Bl_{\pi^{-1}(Y)}(Bl_Z X) \cong Bl_{\pi^{-1}(Z)}(Bl_Y X)$ (this is Lemma IV-41 in Eisenbud and Harris' The Geometry of Schemes). Note that this is the total transform, not the strict transform - strict transforms are sensitive to order. Whether or not this is equal to $Bl_{Y\cup Z}(X)$ should probably depend on how $Y$ and $Z$ intersect, but I don't have instructive examples handy right now, and basically all the general theory for resolution of singularities is developed assuming a blowup along a smooth center, which puts you back in the situation of 2.
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