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.

  • $\begingroup$ 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! $\endgroup$
    – konoa
    Jun 19, 2022 at 18:08
  • 1
    $\begingroup$ 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 $[a:b:c]$. $\endgroup$ Jun 19, 2022 at 21:16

1 Answer 1


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 $N_{Z\subset X}$.


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