Using blowups to desingularise curves - understanding a simple example

Question

I'm learning about blowups in Algebraic Geometry, and am having trouble understanding how to apply them to desingularise varieties.

To illustrate my confusion, I will use the first example from these notes. While I have linked the notes, I will develop everything we need here, in my own words.

Definition

Let $\mathbb{A}^n$ be an affine space. Define $$ B_0\mathbb{A}^n = \{(x, l)\in \mathbb{A}^n\times \mathbb{P}^{n-1} : x \in l\}, $$ and let $\pi :B_0\mathbb{A}^n\to \mathbb{A}^n$ be the projection onto $\mathbb{A}^n$. For an affine variety $X \subseteq \mathbb{A}^n$, let $B_0X = \overline{\pi^{-1}(X\setminus\{0\})}$, where the closure is in the topological space $\mathbb{A}^n \times \mathbb{P}^{n-1}$, defined as a subspace of $\mathbb{P}^n \times \mathbb{P}^{n-1}$ via the Segre embedding.

Example

This is the first example from the notes linked above. I will restate their argument, and then explain my confusion.

Consider the affine variety $X = \mathcal{V}(x^2 - y^2) \subseteq \mathbb{A}^2$. We would like to find $B_0X$. Firstly, we have $$ \pi^{-1}(X \setminus \{0\}) = \{((x, y), [a:b]) \in \mathbb{A}^2\times\mathbb{P}^1: ay = bx, x^2 = y^2, (x,y) \neq(0,0)\}. $$

Let $\mathbb{A}_a$ be the affine chart with $a \neq 0$. In particular, $$ \mathbb{A}_a = \{((x,y),[a:b]) \in \mathbb{A}^2 \times \mathbb{P}^1 : a \neq 0\}. $$ We may assume that $a = 1$ so that $$ \mathbb{A}_a = \{((x,y),[1:b]) \in \mathbb{A}^2 \times \mathbb{P}^1\}. $$

Then it is easy to see that $$ \pi^{-1}(X\setminus \{0\})\cap \mathbb{A}_a = \{((x,bx),[1:b]) : b^2=1, x \neq 0\}, $$ and this is isomorphic$^1$ to $$ \{(x, x, 1) : x \in k^*\} \cup \{(x, -x, -1): x \in k^*\} \subseteq \mathbb{A}^3. $$ Clearly the closure in $\mathbb{A}^3$ is $$ \mathcal{V}(y - x, z-1) \cup \mathcal{V}(y + x, z + 1), $$ so we have$^2$ $$ B_0X = \mathcal{V}(y - x, z-1) \cup \mathcal{V}(y + x, z + 1). $$

My confusion(s)

There are two steps in this argument that seem unjustified to me. I will list them here, corresponding to the citations above.

1. Why can we make the identification $(x, y, t) \mapsto ((x,y), [1:t])$? Just because they have the same underlying set, $k^3$, doesn't necessarily mean they will have the same Zariski topology.

2. All we have shown is that the closure of $\pi^{-1}(X\setminus \{0\}) \cap \mathbb{A}_a$ in $\mathbb{A}_a$ is isomorphic to $\mathcal{V}(y - x, z-1) \cup \mathcal{V}(y + x, z + 1)$. Why does this tell us the closure of $\pi^{-1}(X\setminus \{0\})$ in $\mathbb{A}^2 \times \mathbb{P}^1$? I am able to show in this case that indeed the closure does lie in $\mathbb{A}_a$, but I don't see why this should be the case in general.

Any help in understanding either of these would be very much appreciated.

Answer 1

1. The identification here is that $D(x_0)\subset \Bbb P^1_R$ is isomorphic to a copy of $\Bbb A^1_R$. This works everywhere, all the time, and it's mentioned in one of the other lectures in that series (page 2, "affine charts") if you're particularly attached to them - otherwise most textbooks dealing with varieties at this level should include a proof (see for instance Hartshorne proposition I.2.2; or if you upgrade to schemes this is basically part of the definition). The gist of the argument is that if you have a closed set which can be written as $V(F)\cap D(x_i)$ for some homogeneous ideal $F$, then the same set viewed as a subset of $\Bbb A^1$ by just setting $x_i=1$ in all of the equations. We can get $F$ back by rehomogenizing with respect to $x_i$ appropriately.

Next, we have that $\Bbb A^n\times\Bbb A^m \cong \Bbb A^{n+m}$ for any $n,m$, where we mean a product of varieties, not a product of topological spaces. While you would be correct that the product of topological spaces $\Bbb A^n\times\Bbb A^m$ is not isomorphic as a topological space to $\Bbb A^{n+m}$, the product operation in varieties is a little more subtle: it does make this isomorphism work as varieties, and that's what's being used here.

2. You've verified that all the points of your blowup do indeed lie in $\Bbb A_a$, as you call it, so that's sufficient here. You're right to be a little concerned, because it's not sufficient in general: think about the case $xy=0$, where if you use the same blowup charts as here you'll end up having to consider both. As you get more practice with blowups you'll get a better sense of when you'll have to check all the charts versus when you can leave some out. (You can always check all the charts and then verify that the ones you thought you could skip are indeed skippable.)