Simple normal crossings divisor and blow up

Question

I have to prove some things about simple normal crossings divisors and the blow up. The definition of a simple normal crossings divisor we use is a finite union $V = \cup_i V_i$ of irreducible quasi-projective curves on a nonsingular quasi-projective surface $S$ is a simple normal crossings divisor if each $V_i$ is nonsingular, no three different different $V_i$ intersect and if two curves $V_i \neq V_j$ intersect, this intersection is a single point and the two tangent lines of $V_i$ and $V_j$ in this point are distinct.

Let $\pi: \text{Bl}_0(\mathbb{A}^2) \rightarrow \mathbb{A}^2$ be the blow up at the origin. I have to prove the following.

1. If $Z=\mathbb{V}(x_0x_1(x_0+x_1)) \subseteq \mathbb{A}^2$, then $\pi^{-1}(Z) \subseteq \text{Bl}_0(\mathbb{A}^2)$ is a simple normal crossings divisor.
2. If $Z=\mathbb{V}(x_0^2-x_1^3) \subseteq \mathbb{A}^2$, then $\pi^{-1}(Z) \subseteq \text{Bl}_0(\mathbb{A}^2)$ is not a simple normal crossings divisor.
3. Repeatedly blow up $\mathbb{A}^2$ and $Z = \mathbb{V}(x_0^2-x_1^3)$ until the preimage of $Z$ is a simple normal crossings divisor.

For the first item I observed that $\pi^{-1}(Z) = \mathbb{V}(x_0) \cup \mathbb{V}(x_1) \cup \mathbb{V}(x_0+x_1)$. Set $V_1 = \mathbb{V}(x_0)$, $V_2 = \mathbb{V}(x_1)$ and $V_3 = \mathbb{V}(x_0+x_1)$. Then I computed the intersection with the charts using an example in our notes \begin{align*} V_1 \cap (\mathbb{A}^2 \times \mathbb{A}_{x_0}^1) & = \emptyset\\ V_1 \cap (\mathbb{A}^2 \times \mathbb{A}_{x_1}^1) & \cong \{(x_0,x_1,v)| x_0=vx_1,v=0\}\\ V_2 \cap (\mathbb{A}^2 \times \mathbb{A}_{x_0}^1) & \cong \{(x_0,x_1,u)| x_1=ux_0,u=0\}\\ V_2 \cap (\mathbb{A}^2 \times \mathbb{A}_{x_1}^1) & = \emptyset\\ V_3 \cap (\mathbb{A}^2 \times \mathbb{A}_{x_0}^1) & \cong \{(x_0,x_1,u)| x_1=ux_0,1+u=0\}\\ V_3 \cap (\mathbb{A}^2 \times \mathbb{A}_{x_1}^1) & \cong \{(x_0,x_1,v)| x_0=vx_1,1+v=0\}. \end{align*} Then it is clear that this a is a simple normal crossings divisor, since these intersections have no intersection and are nonsingular and irreducible.

Now I tried to do the same for the second item with $V=\pi^{-1}(Z)$ and found the following intersection with the charts \begin{align*} V \cap (\mathbb{A}^2 \times \mathbb{A}_{x_0}^1) & \cong \{(x_0,x_1,u)| x_1=ux_0,1-u^3x_0=0\}\\ V \cap (\mathbb{A}^2 \times \mathbb{A}_{x_1}^1) & \cong \{(x_0,x_1,v)| x_0=vx_1,v^2-x_1=0\}. \end{align*} I think these intersections are nonsingular and irreducible, so in order for this to not be a simple normal crossings divisor we need to find a contradiction with another property. I cannot prove this however.

I was thinking that there might be a problem in the computations in the first item, since I do not understand how the blow up behaves around the origin $(0,0)$.

Any help is welcome, involving these computations or just a general explanation about simple normal crossings divisor or blow up.

Answer 1

You've made a key mistake: you've forgotten to include the exceptional divisor in the preimage $\pi^{-1}(Z)$. In question 1, using your notation, on $\Bbb A^2\times\Bbb A^1_{x_0}$ the exceptional divisor is $V(x_1)$, and on $\Bbb A^2\times\Bbb A^1_{x_1}$ the exceptional divisor is $V(x_0)$. Each of $V_1,V_2,V_3$ intersects the exceptional divisor, but they all do so transversely and in one point, and by your work no three components meet at any point (as to do this, two of $V_1,V_2,V_3$ would need to meet).

For question 2, let me show you a slightly different way to do the calculations which will make things a little easier. Computing, on $\Bbb A^2\times \Bbb A^1_{x_0}$ (which has coordinate algebra isomorphic to $k[x_0,u]$) we have $\pi^{-1}(Z)$ is cut out by $x_0^2-(ux_0)^3=x_0^2(1-u^3x_0)$ - the $x_0^2$ term corresponds to the exceptional divisor, and the $1-u^3x_0$ term corresponds to the strict transform. It's easy to see that $V(x_0,1-u^3x_0)=\emptyset$, so there are no intersections in the first patch. On $\Bbb A^2\times \Bbb A^1_{x_1}$ (which has coordinate algebra isomorphic to $k[x_1,v]$) we have $\pi^{-1}(Z)$ is cut out by $(x_1v)^2-x_1^3=x_1^2(v^2-x_1)$ - the $x_1^2$ term corresponds to the exceptional divisor, and the $v^2-x_1$ term corresponds to the strict transform. These meet non-transversely at $v=0,x_1=0$, so you have more blowups to do. (You should need two more.)

Answer 2

Here I use a more algebraic description a divisor with normal crossings. Let $X$ be a smooth scheme over a field $k$, then we say that an effective Cartier divisor $D$ has normal crossings if irreducible components of $D$ are smooth and at each point $x \in X$, $D$ is cut out by $x_1^{n_1}\cdots x_r^{n_r}$ where $(x_1,...,x_d)$ is a system of regular parameters at $x$ and $r \leq d$.

Let's assume that the ground field is $k$, using KReiser's computations, in the second question, in the first chart $$V \simeq \operatorname{Spec}\left(\frac{k[x_0,u]}{x_0^2(1 -u^3x_0)} \right).$$ Now look at local ring at $(0,0)$, which is $k[x_0,u]_{(x_0,u)}$, $(x_0,1-u^3x_0)$ is the whole local ring, so it cannot be a regular system of parameters. In the second chart, the behavior at $(0,0)$ is also singular. Indeed, $$V \simeq \operatorname{Spec}\left(\frac{k[x_1,v]}{x_1^2(x_1-v^2)} \right)$$ so that two curves $x_1^2=0$ and $x_1-v^2=0$ intersect non-transversally at $(0,0)$. Algebraically, $(x_1-v^2,x_1)$ is not a regular system of parameters, there is a strict inclusion $(x_1-v^2,x_1)=(x_1,v^2) \subset (x_1,v)$.

So if you want a divisor with normal crossings, you should blow up the origins in the both charts.