Intersection of the blowing up and the exceptional curve for a concrete example

Question

Let $f(x,y)=x^2y+xy^2 - x^4-y^4=0$ be an affine curve.

Consider its blowing up at the origin, namely $x^2y+xy^2 = x^4+y^4$ and $xu=ty$ in $\mathbb{A}^2 \times \mathbb{P}^1$.

If $t \neq 0$ then we can set $y=xu$ and substituting this to the equation of curve, we obatin $x^3(2u-x(1+u^4))=0$, which implies the intersection of the blowing up and the exceptional curve is the point $u=0,x=0$. Because $f(x,y)=f(y,x)$, we get the same result for the case $ u \neq 0$ with $x=ty$, namely, we obtain $t=0,y=0$. Thus intersection of the blowing up and the exceptional curve is constructed by these two points. Is it correct?

---

Receive a comment.

I was wrong.

Case I.

$u=1$ and $x^2y+xy^2 = x^4+y^4$ and $xu=ty$ in $\mathbb{A}^2 \times \mathbb{P}^1$. Then, $x^3\{u+u^2-x(1+u^4)\}=0$ and thus the intersection of exceptional curve $\{x=0\}$ and the blowing up is given by the equation $ u+u^2-0\times(1+u^4)=0$. Thus $u=0$ and $u=-1$ are the intersection.

Case II.

$t=1$ and $x^2y+xy^2 = x^4+y^4$ and $xu=ty$ in $\mathbb{A}^2 \times \mathbb{P}^1$. Then, $y^3\{t+t^2-y(1+t^4)\}=0$ and thus the intersection of exceptional curve $\{y=0\}$ and the blowing up is given by the equation $ t+t^2-0\times(1+t^4)=0$. Thus $t=0$ and $t=-1$ are the intersection.

Because $(t:u)=(1:-1)=(-1:1)$, $t=-1$ in case II and $u=-1$ in case I are same points, thus the intersection of the blowing up and the exceptional curve consist three distinct points.

Answer 1

No, your answer is not correct. After substituting $y=xu$ you simplified incorrectly: your $2u$ should be $-u-u^2$.

In general, points of the exceptional curve correspond to tangent directions at the origin, and points of the intersection are tangent directions to your original curve. You can find these by just taking the lowest-degree homogeneous part of your equation and factorising it. If you do this for your example, what do you get?

For additional fun, you can use your favourite graphing software to make a picture. The free online tool Desmos produces the following picture in your example:

Of course this "method" comes with a warning: the picture only shows real points of the curve, hence only the real tangent directions are visible. Nevertheless, in this case it shows that something has gone wrong in your original calculations.

Answer 2

As Lazzaro Campoetti says, you've made a mistake in your calculations. After substituting $y = xz$ and simplifying, I find the equation $$ 0 = x^3(z(1+z) - x(1+z^4)) \, . $$ The factor $x^3$ corresponds to the exceptional locus. However, it is not a point, but a line: since $y = xz$, then $x = 0$ implies $y = 0$ and leaves $z$ unconstrained. Thus the exceptional locus is the $z$-axis, i.e., the vertical line $x = 0, y = 0$.

The other factor $z(1+z) - x(1+z^4)$ is the strict transform of the curve. Since $y = xz$, setting $z(1+z) - x(1+z^4) = 0$ we obtain the parametric equation $$ (x,y,z) = \left(\frac{z(1+z)}{1+z^4}, \frac{z^2(1+z)}{1+z^4}, z \right) \, . $$ To find the intersection with the exceptional locus, we must find the points where $x = 0, y = 0$. Since $y = xz$, then it suffices to solve $0 = x = \frac{z(1+z)}{1+z^4}$. Thus we find that the strict transform intersects the exceptional locus when $z = 0$ and $z = -1$. However, there is also one more intersection at infinity, since $$ x = \frac{z(1+z)}{1+z^4} \to 0 \quad \text{as $z\to \infty$} \, . $$ (You could also show this by setting $z = Z_1/Z_0$, clearing denominators, and then setting $Z_0 = 0$.) You can see these features in the plot in this SageMath cell.

Moreover, based on considerations of the tangent directions at the singular point $(0,0)$ of the original curve, you should expect there to be $3$ intersections. The coordinate $z = y/x$ keeps track of the tangent slopes, and as I mentioned in this comment, the tangent lines at the origin and $x = 0$, $y = 0$, and $y = -x$, which have slopes $\infty$, $0$, and $-1$.