I need help to understand blowups of points in curves in $\mathbb A^2$

Question

I'm trying to understand how to blowup curves which I'm finding very difficult.

Example $V=V\bigg(y^2-x^2(x+1)\bigg)$

Blowup map $\pi$:

$$B=\{(x,l)\in \mathbb A^2\times \mathbb P^1|x\in l\}\to \mathbb A^2$$ $$(x,l)\mapsto x$$

The projective space $\mathbb P^1$ can be covered by its standart charts: $\mathbb P^1=\mathbb A^1_x\cup\mathbb A^1_y$, where $\{\mathbb A_x=(a,b)|a\neq0\}$ and $\{\mathbb A_y=(a,b)|b\neq0\}$.

So we can rewrite

$B\subset \mathbb A^2\times \mathbb P^1\cong (\mathbb A^2\times\mathbb A_x^1)\cup(\mathbb A^2\times \mathbb A_y^1)\to \mathbb A^2$

And now? what we have to do? restrict to some chart? if yes, how can we proceed?:

$V=\pi^{-1}\bigg(V\big(y^2-x^2(x+1))\cap(\mathbb A^2\times\mathbb A_y^1)\big)\bigg)=?$

I really need help to understand this example.

If someone could help me I would be very grateful.

Thanks a lot

Answer 1

You're correct in that we should consider the two charts. Concretely, the blowup is described as the following two equations $xs-yt=0$ and $y^2-x^2(x+1)=0$ where we demand that the equations should be homogeneous in $s,t$ (which they are).

Now, over the chart where $t \neq 0$, we can set $t=1$. Thus the equations become $y=xs$ and $y^2=x^2(x+1)$. But then we get $$ x^2s^2=x^2(x+1)$$

Since $s \neq 0$, it follows that the blowup is the union of $\{ x = 0 \}$ and $\{s^2=x+1\}$. The first is the exceptional divisor. The second is the proper transform of $C$ (the curve), where it has been transformed to a smooth quadratic.

Now, lets look over the chart $s \neq 0$. We set $s=1$. Similarly as above, we get the equation $x=yt$ and thus: $$ y^2 = y^2t^2(yt+1)$$

Thus, the blowup is the union of $\{y = 0\}$ and $\{1=t^2(yt+1)\}$, the latter a smooth quadric, again.