I am trying to compute the exceptional divisor of the affine singularity $x^2 + y^2 + z^4 = 0$.

First, I take the chart $U_1$ with coordinates $x,y_1,z_1$ verifying $ y = y_1x, z = z_1x$. The equation is $x^2(1 + y_1^2 + z_1^4x) = 0$, so the strict transform is smooth and the exceptional divisor $E_1$ has equation $ \{ (0,\pm i, t) : t \in \Bbb C \}$.

Now, let's take the chart $U_2$ with coordinates $z, x = x_2z, y = y_2z$. The equation is $z^2(x_2^2 + y_2^2 + z^2) = 0$, the strict transform is singular and the exceptional divisor is $E_2 = \{ (t, \pm it, 0) : t \in \Bbb C\}$. We have $E_1 = E_2$ and it is corresponds to two copies of $P^1$ intersecting transversally.

We do a second blow-up in the chart $U_2$, with respect to the coordinate $x_2$, i.e we have a third chart $U_3$ with new coordinates are $x_2, y_3,z_3$ where $y_2 = x_2y_3, z_2 = z_3x_2$. The equation is $ x_2^2(1 + y_3^2 + z_3^2) = 0$. This is smooth so we don't need to blow-up anymore. The exceptional divisor $E_3$ is given by $x_2 = 0, 1 + y_3^2 + z_3^2 = 0$. It is a plane conic, i.e $E_3 \cong P^1$. I want to compute $E_3 \cap E_2$.

Coordinates $U_2$ and $U_3$ are related by $y_3 = y_2x_2^{-1}, z_3 = x_2z, x_2 = x_2$. In particular, $U_2 \cap U_3 = \{(x_2,zy_2,z) : x_2 \neq 0)\}$. Since $E_3$ was defined as $x_2 = 0, 1 + y_3^2 + z_3^2 = 0$ we have $E_3 \cap E_2 = \emptyset$ in this chart.

Similary I got that $E_3 \cap E_2 = \emptyset$ in the other charts (I can add details if necessary). So I get that the exceptional divisor is $E_2 \sqcup E_3$. But by toric geometry I I should get $3$ copies of $P^1$, with one intersecting the two other transversally, which is clearly not what I did obtain. If someone could check my computations and tell me where is my mistake, I would be very happy !

  • $\begingroup$ First, $(0, \pm i,t)$ is not irreducible, so why do you call it $E_1$ and pretend there is only one exceptional curve? $\endgroup$
    – Mohan
    May 23, 2017 at 21:09
  • $\begingroup$ @Mohan : I agree with whay you said. I called it $E_1$ because it was intersection of the exceptional divisor with the chart $U_1$. I agree it is not irreducible since $E_1$ is send on $E_2$ which is the union of two lines meeting at a point. Maybe it was not very clever notation. I found that there were $3$ exceptional curves, but they don't intersect like they should do : I find that two $P^1$ intersect at a point (this is the reducible variety I called $E_1$) and there is another $P^1$ but it should intersect something else, but according to my computations it is disjoint from $E_1$. $\endgroup$
    – user378546
    May 23, 2017 at 22:08

1 Answer 1


Your $U_2$ is a open chart in a blowup of $\mathbb A^3$. Your $U_3$ is an open chart in a blowup of a blowup of $\mathbb A^3$. So $U_2$ and $U_3$ are open charts on different varieties, and it makes no sense to talk about their "intersection" $U_2 \cap U_3$.

Instead, we could proceed as follows. After the first blowup, the exceptional divisor consists of a pair of rational curves intersecting at a point. In the $U_2$ chart, they are given by the equations: $$ C_1: \ \ \ y_2 = ix_2 ,\ \ \ z = 0, \\ C_2 : \ \ \ y_2 = - ix_2, \ \ \ z=0. $$

Let us now perform the second blowup, and let us find the strict transforms of $C_1$ and $C_2$ after the second blowup. In the $U_3$ chart, these are given by: $$ \hat C_1 : \ \ \ y_3 = i, \ \ \ z_3 = 0, \\ \hat C_2: \ \ \ y_3 = -i, \ \ \ z_3 = 0.$$ Notice that $\hat C_1$ and $\hat C_2$ are also rational curves, but they do not intersect. This should not surprise us. The second blowup was performed at $(x_2, y_2, z) = (0,0,0)$, which is the intersection point of $C_1$ and $C_2$. Since $C_1$ and $C_2$ intersect transversally, it is absolutely right that they should be separated by the second blowup.

However, in doing the second blowup, the exceptional divisor acquires a third irreducible component, which is the preimage of $(x_2, y_2, z) = (0,0,0)$. This is your plane quadric: $$ Q: \ \ \ 1 + y_3^2 + z_3^2 = 0, \ \ \ \ x_2 = 0.$$ The full exceptional divisor is then the union of the three rational curves, $\hat C_1$, $\hat C_2$ and $Q$.

It remains to examine how $Q$ intersects $\hat C_1$ and $\hat C_2$. The intersection points are: $$ Q \cap \hat C_1 = \{(x_2, y_3, z_3) = (0, i , 0) \}, \\ Q \cap \hat C_2 = \{ (x_2, y_3, z_3) = (0, -i, 0) \},$$ and each of these intersections is transversal. This is precisely the intersection pattern that you were looking for.

  • $\begingroup$ Thank you, I will read carefully your answer. $\endgroup$
    – user378546
    May 24, 2017 at 14:43
  • $\begingroup$ Everything is clear, thanks again ! $\endgroup$
    – user378546
    May 24, 2017 at 14:52
  • 1
    $\begingroup$ @rain Cool - glad it helped! $\endgroup$
    – Kenny Wong
    May 24, 2017 at 14:53

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