Two way of computing blowup -- which one is correct? I came across reading Corollary 7.15 in Hartshorne's book. A special case of this corollary is the following statement

If $Y,C\subset X$ are subvarieties, $\widetilde X\to X$ is the blowup of $X$ along $C$ and $\widetilde Y\to Y$ is the blowup along $Y\cap C$, then there is an inclusion $\widetilde Y\subset \widetilde X$. In other words, where exists a commutative diagram
$\require{AMScd}$
\begin{CD}
\widetilde{Y}=Bl_{C\cap Y}Y @>>> \widetilde{X}=Bl_C X\\
@V  V V @VV V\\
Y @>>> X.
\end{CD}

Now assume that $X=\mathbb C^n$ and thus that $Y\subset \mathbb C^n$ is an affine variety. Let $Z\subset Y$ be a fixed subvariety. My confusion arises when comparing the following two blowups:

*

*Take as center $C$ a variety such that $C\cap Y=Z$ and compute the blowup.

*Take as center $C$ the variety $Z$ itself.

The point is that $C$ in 1. can be defined by fewer equations than the $C$ in 2. -- that is because the intersection with $Y$ can give additional equations (coming from the equations of $Y$). So it seems that there are two ways of computing the blowup of of $Y$ along $Z$ and they result in different results. Which one is the correct one and why?
Example:
Take $X=\mathbb C^4$ with coordinates $x_1,,\dots, x_4$, $Y=\{x_1^2-x_2^2=0\}$, $Z=\{x_1=x_2=x_3=0\}$.
For 1. we can take $C=\{x_2=x_3=0\}$ because $C\cap Y=\{x_1^2-x_2^2=x_2=x_3=0\}=Z$. Then, $$\widetilde X=\{a_0x_2=a_1x_3\}\subset \mathbb C^4\times \mathbb P^1$$ and the strict transform of $Y$ can be computed in the corresponding charts to be

*

*chart 1 ($a_0\neq 0$): $\widetilde Y\cap \{a_0\neq 0\}=\{x_1^2-a_1^2x_3^2=x_2-a_1x_3\}\subset \mathbb C^5$,

*chart 2 ($a_1\neq 0$): $\widetilde Y\cap \{a_1\neq 0\}=\{x_1^2-x_2^2=x_3-a_0x_2=0\}\subset \mathbb C^5$.

On the other hand, we can also blowup $X$ along $Z$ and obtain
$$\widetilde X=\{a_0x_2-a_1x_1=a_0x_3-a_2x_1=a_1x_3-a_2x_2=0\}\subset \mathbb C^4\times \mathbb P^2$$

*

*chart 1 ($a_0\neq 0$): $\widetilde Y\cap \{a_0\neq 0\}=\{1-a_1^2=x_2-a_1x_1=x_3-a_2x_1=0\}\subset \mathbb C^6$,

*chart 2 ($a_1\neq 0$): $\widetilde Y\cap \{a_1\neq 0\}=\{a_0^2-1=x_1-a_0x_2=x_3-a_2x_2=0\}\subset \mathbb C^6$,

*chart 3 ($a_2\neq 0$): $\widetilde Y\cap \{a_2\neq 0\}=\{a_0^2-a_1^2=x_1-a_0x_3=x_2-a_3x_3=0\}\subset \mathbb C^6$.

In these charts we had to remove factors corresponding to the exceptional divisor. So it seems that $\widetilde Y=\{a_0^2-a_1^2=0\}\subset \widetilde X$.
Chart 3 of the second and chart 2 of the first are isomorphic. But I don't see any isomorphism between the remaining charts and there is also one more chart for the second than for the first.
Moreover, using the first method, the inverse image of $Z$ is $Z\times \mathbb P^1$, whilst for the second method it is $Z\times \mathbb P^2$ and from the equations there is nothing like an inclusion or so that I can see.
So what is the point here? With both methods I calculated $Bl_Z Y$ but the results seem to be very different from each other.
 A: I think what is happening is that you need to take the scheme-theoretic intersection, so that $Y\cap C=\{x_1^2=x_2=x_3=0\}$, which is then different from $Z=\{x_1=x_2=x_3=0\}$.
For notational convenience I will slightly rewrite what you have done.
In the first case you take $C=\{x_2=x_3=0\}$ to get the blow up $\tilde X\subset\mathbb C^4\times\mathbb P^1$ given by $\{a_3x_2=a_2x_3\}$ in coordinates $(x_1,x_2,x_3,x_4),(a_2:a_3)$.
The pullback gives $\tilde Y$, which has two charts
$$ \tilde Y_2 = \{x_1^2=x_2^2,\ x_3=a_3x_2,\ a_2=1\} \quad\textrm{and}\quad
\tilde Y_3 = \{x_1^2=x_2^2,\ x_2=a_2x_3,\ a_3=1\}. $$
In the second case you take $Z=\{x_1=x_2=x_3=0\}$ to get the blow up $\tilde{\tilde X}\subset \mathbb C^4\times\mathbb P^2$ given by $\{a_2x_1=a_1x_2,\ a_3x_1=a_1x_3,\ a_3x_2=a_2x_3\}$ in coordinates $(x_1,x_2,x_3,x_4),(a_1:a_2:a_3)$.
The pullback is now $\tilde{\tilde Y}$ which has three charts, the first two of which actually coincide (using that $(1:a_2:a_3)=(a_2:1:a_2a_3)$ when $a_2^2=1$). Thus we need just two charts
$$ \tilde{\tilde Y}_2 = \{x_1=a_1x_2,\ x_3=a_3x_2,\ a_1^2=1,\ a_2=1\} \quad\textrm{and}\quad
\tilde{\tilde Y}_3 = \{x_1=a_1x_3,\ x_2=a_2x_3,\ a_1^2=a_2^2,\ a_3=1\}. $$
It is now clear that there is a natural surjective map $\tilde{\tilde Y}\to\tilde Y$ induced from the map $\mathbb P^2\to\mathbb P^1$, $(a_1:a_2:a_3)\mapsto (a_2:a_3)$ (away from the point $(1:0:0)$).
Disclaimer: I am no expert.
