Blow up of $V(x^2 - y^2 + x^3) \subset \operatorname{Spec}k[x,y]$

Question

This is Example E from Mumford's red book Chapter III Section 3. Let $f= x^2 - y^2 + x^3$ and $X = V(f) \subset \operatorname{Spec}k[x,y]$. Then the blow up at the origin $B(X)$ will be a curve in $B_2$ (blow up $\operatorname{Spec}k[x,y]$ at the origin) entirely contained in $U = \operatorname{Spec}k[x,y/x]$, where $B_2$ is covered by $U$ and $V = \operatorname{Spec}k[y,x/y]$. I'm not seeing how $B(X)$ is contained in $U$. Any explanation is appreciated. Thank you.

Answer 1

If $p: B_2 \rightarrow \mathbb{A}^2_k$ is the blowup map, then $p^{-1}(X) \cap V$ is defined by the polynomial $$ y^2\left( \frac{x}{y} \right)^2 - y^2 + y^3\left( \frac{x}{y} \right)^3 = y^2\left( y\left( \frac{x}{y} \right)^3 + \left( \frac{x}{y} \right)^2 - 1\right). $$ So, $p^{-1}(X)$ is (set theoretically) the union of $X_1 = Z(y)$ and $X_2 = Z(y(x/y)^3 + (x/y)^2 - 1)$. Note also that $p^{-1}(0) \cap V = Z(x,y) = Z(y) = X_1$ (since $y = 0\implies x = y(x/y) = 0$ on $V$). Therefore, $$ (p^{-1}(X) \cap V) - (p^{-1}(0) \cap V) = X_2 - X_1. $$ The curves $X_1$ and $X_2$ in $V$ meet along $Z(y, (x/y)^2 - 1)$, which is just the pair of points $(x/y, y) = (1,0)$, $(x/y,y) = (-1,0)$. Therefore, $B(X) \cap V$ is the closure of $X_2 - \{(1,0), (-1,0)\}$ in $V$, which is exactly $X_2$.

We can now check that $B(X) \cap V = X_2$ is entirely contained in $U \cap V = D(x/y) \subset V$. Indeed, $(B(X) \cap V) \cap Z(x/y)$ is defined by the ideal $(y(x/y)^3 + (x/y)^2 - 1, x/y)$, which is the unit ideal. So, the intersection is empty.