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