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:

- Blowup directly the entire scheme $Z$. In this way I get a singular variety $X$.
- 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.