# Blow-up of 3-dimensional affine space through the line $x_1=x_2=0$.

I'm trying to understand blow-ups. In the Gathmann's notes there is an exercise:

Let $$\widetilde{\mathbb A^3}$$ be the blow-up of $$\mathbb A^3$$ at the line $$V(x_1,x_2)\equiv \mathbb A^1$$. When the stric transforms of two lines in $$\mathbb A^3$$ through $$V(x_1,x_2)$$ intersect in the blow-up? What is therefore the geometric meaning of the points in the exceptional set?

I proved that $$\widetilde{\mathbb A^3}=\{(x_1,x_2,x_3)\times[a:b]\in \mathbb A^3\times \mathbb P^1\ ;\ x_1b=x_2a\}$$.

A line in $$\mathbb A^3$$ through $$V(x_1,x_2)$$ is given by $$\ell_{z_0\times[a:b:c]}: \frac{x_1}{a}=\frac{x_2}{b}=\frac{x_3-z_0}{c}$$. So, its strict transform in $$\widetilde{\mathbb A^3}$$ is $$\widetilde{\ell_{z_0\times[a:b:c]}}=\{(x_1,x_2,x_3)\times[y_1:y_2]\in\widetilde{\mathbb A^3}\ / \ \ [a:b]=[y_1:y_2],\ cx_2=b(x_3-z_0)\}$$.

So, if $$\ell'_{z_1\times[a':b':c']}: \frac{x_1}{a'}=\frac{x_2}{b'}=\frac{x_3-z_1}{c'}$$ is other line in $$\mathbb A^3$$ trough $$V(x_1,x_2)$$. And its strict transform in $$\widetilde{\mathbb A^3}$$ is $$\widetilde{\ell'_{z_1\times[a':b':c']}}=\{(x_1,x_2,x_3)\times[y_1:y_2]\in\widetilde{\mathbb A^3}\ / \ \ [a':b']=[y_1:y_2],\ c'x_2=b'(x_3-z_1)\}$$.

The intersection is

$$\widetilde{\ell'_{z_1\times[a':b':c']}}\cap \widetilde{\ell_{z_0\times[a:b:c]}}=\{(x_1,x_2,x_3)\times[y_1:y_2]\in\widetilde{\mathbb A^3}\ / \ \ [a:b]=[y_1:y_2],\ x_3=\frac{c'bz_0-cb'z_1}{c'b-cb'}\}$$.

Is this correct?

In the case of the blow-up of $$\mathbb A^2$$ at the origen $$(0,0)$$ for every direction $$[a:b]$$, line through the origen, we have a line in the blow-up of $$\mathbb A^2$$ through the point $$[a:b]$$ in the execptional divisor. But in the case of the blow-up of $$\mathbb A^3$$ through the line $$V(x_1,x_2)$$ I don't understand the geometric meaning of the points in the exceptional set, $$\mathbb A^1 \times \mathbb P^1$$. I will appreciated some explanation about this fact, also.

• Dear @quasi-coherent, could you please add a small explanation regarding why a line in $\mathbb{A}^3$ through $V(x_1,x_2)$ is of that form? I really don't understand how you computed them: what is $z_0$, and why you consider $[a:b:c]$ (which belong to $\mathbb{P}^2$ and doesn't seem to be involved in your reasoning)... Thanks in advance! Jun 19, 2022 at 18:08
• Yes, I'm sorry. I forgot to say that $(a,b,c)$ is the direction of the line and $(0,0,z_0)\in V(x_1,x_2)$ is a point in the line. The equation follows from the line's equation $(x_1,x_2,x_3)=t(a,b,c)+(0,0,z_0)$ for every $t$ in a field $K$. Well, I'm not considering the cases where some $a,b,c$ is $0$ because for my question this is sufficiente, I think. Maybe is not necessary consider the notation $[a:b:c]$ and just $(a,b,c)$, but I consider $[a:b:c]$ because it represent a line in $\mathbb A^3$ through the origin, and the line's equation is true for every representant of $[a:b:c]$. Jun 19, 2022 at 21:16

The points $$(p,[a:b]) \in \mathbb A^1 \times \mathbb P^1$$ in the exceptional divisor parametrize pairs of a point $$p$$ on the center of the blow up (the line $$Z = V(x_1,x_2)$$) and a normal direction to $$Z$$ at $$p$$, i.e. a point of $$\mathbb P(T_p\mathbb A^3/T_p Z)\cong \mathbb P^1$$.
The connection to the case of blowing up a point in a variety $$p \in X$$ is that "the tangent space to a point" is trivial, so the normal space at $$p$$ is just the tangent space $$T_p X$$ without modding out by any subspaces.
In general (at least for blowing up a smooth center $$Z$$ in a smooth variety $$X$$), the exceptional divisor is isomorphic to the projectivization of the normal bundle $$N_{Z\subset X}$$.