TITLE: minimal resolution of singularities
QUESTION [7 upvotes]: What is the minimal resolution of singularities of the surface
$S^2(X^3+Y^3+Z^3)-3(S^2+T^2)XYZ=0$ which is a subset of $\mathbb{P}^1\times\mathbb{P}^2$
 Please note that in this equation $[S:T]\in{\mathbb{P^1}}$ and $[X:Y:Z]\in{\mathbb {P^2}}$ and by $\mathbb{P^n}$ we mean n-dimensional complex projective space.

REPLY [7 votes]: A simple computation shows that the equation
$$
u(x^3+y^3+z^3)-3vxyz=0
$$
defines a non-singular surface $F\subset\mathbb P^1\times \mathbb P^2$. The projection to $\mathbb P^1$ gives an elliptic fibration $\sigma:F\to \mathbb P^1$. This has exactly two singular fibers, over $[0:1]$ and $[1:1]$, each consisting of three lines not going through a common point. The local equation for the projection at the singular points of the fibers is $$(\xi,\eta)\mapsto \zeta=\xi\eta.$$

Details (to satisfy popular demand): 
a) Near the point $[0:1]\times[0:0:1]$, let $\zeta=\dfrac uv$, $\xi=\left(\dfrac{3vz}{x^3+y^3+z^3}\right)\cdot x$, and $\eta=y$. Notice that $\dfrac{3vz}{x^3+y^3+z^3}$ is a unit near that point. Near the other singular points of the fiber over $[0:1]$ permute the variables accordingly.
b) Near the point $[1:1]\times [1:1:1]$, let $\zeta=\dfrac{v-u}u$, $\xi=\left(\dfrac{x+y+z}{3xyz}\right)\cdot(x+\omega y+\omega^2 z)$, and $\eta=x+\omega y+\omega^2 z$ where $\omega\neq 1$ is a $3$rd root of unity. In particular $1+\omega+\omega^2=0$. Notice that $x^3+y^3+z^3-3xyz=(x+y+z)(x+\omega y+\omega^2 z)(x+\omega y+\omega^2 z)$. Permute the three linear factors accordingly for the other two singular points of the fiber.

Note: Actually one can conclude the stated local condition without doing this explicit calculation. The point is this: we know that the singular fiber is three lines in the plane intersecting in three separate points. Therefore, locally each of the singularities of the fiber is defined by $\zeta=\xi\eta$. Since the nearby fibers are smooth, the family, locally, is a smoothing of a node. The versal deformation space of a node is one dimensional (it's exactly what the displayed equation claims) and hence this smoothing has to be locally isomorphic to that.  


Now consider a base change of $\sigma$ by taking square roots $\mathbb P^1\to \mathbb P^1$, $[s:t]\mapsto [s^2:s^2+t^2]$. The new surface $G=F\times_{\mathbb P^1}\mathbb P^1$ is the surface in the question. This will acquire singularities over the points where $\sigma$ was not a smooth morphism. We saw above that the local equation of the map at those points is given by 
$$(\xi,\eta)\mapsto \zeta=\xi\eta.$$
The base change replaces $\zeta$ by $\zeta^2$, so the local equation of the surface becomes $$\zeta^2=\xi\eta.$$

Details:
a) near $[0:1]$ we had above $\zeta=\dfrac uv$, so the base change makes it $\zeta=\dfrac{s^2}{s^2+t^2}=\left(\dfrac 1{1+\tau^2}\right)\rho^2$. Replace $\zeta$ with $\rho$ and $\xi$ with $\xi\cdot(1+\tau^2)$.
b) near $[1:1]$ we had $\zeta=\dfrac{v-u}u$, so the base change makes it $\zeta=\dfrac{t^2}{s^2}=\rho^2$. Replace $\zeta$ with $\rho$.

Note: Again, this can be done without the explicit computation. Any two-to-one map $\mathbb P^1\to \mathbb P^1$ is simply taking roots of the local coordinates defining the points where that map is branched. Therefore if $\zeta$ is the local equation of the branch point, then the cover replaces $\zeta$ with $\zeta^2$.


In other words, the surface has exactly $6$ singular points, each locally analytically isomorphic to the vertex of a quadratic cone, and hence blowing up these points (once) yields the minimal resolution.

Edit history:
  1) Thanks to JME for pointing out the typo in the definition of the base change map.
2) Edit 1: added the local calculation for the description of the map near the singular points.
3) Edit 2: added the theoretical argument (which in my mind actually preceded the calculation) that implies the same result as the calculation.