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.



minimal resolution of singularities
Let us start by writing down the computation of the singular points in the chart $S=1$. 


Writing $\lambda:=T/S$, in the chart $S=1$ we can rewrite the equation of the surface as
$$X^3+Y^3+Z^3-3(1+\lambda^2)XYZ=0.$$
This is an elliptic fibration over $\mathbb{C}$ (with coordinate $\lambda$), whose fibres are the curves of the Hesse pencil of cubics in $\mathbb{P}^2$.


Taking derivatives with respect to $X, Y, Z, \lambda$ we obtain the equations:
$$X^2-(1+\lambda^2)YZ=0,$$ $$Y^2-(1+\lambda^2)XZ=0,$$ $$Z^2-(1+\lambda^2)XY=0,$$ $$\lambda XYZ=0.$$
The only possibility is $\lambda=0$, so the singularities are the three points
$$[1:1:1], \; [1: a :a^2], \; [1:a^2:a], \quad a:=e^{2 \pi i /3}$$
in the fibre over $\lambda=0$. In fact, the fibre over $\lambda=0$ degenerates as the union of three distinct lines, which form a triangle whose vertices are the three points above.


An easy local computation shows that all these points are of type $A\_1$, so the minimal resolution for each of them is given by a $(-2)$-curve. In other words, the fibre of the resolved surface in $\lambda=0$, i.e over $[S:T]=[1:0]$, is of type $I\_6$ according to Kodaira classification. 


Now let us consider the chart $T=1$. The equation of the surface becomes
$$S^2(X^3+Y^3+Z^3)-3(S^2+1)XYZ=0.$$
We are interested only on the singularities lying over $S=0$, and a straightforward computation gives the three points 
$$[1:0:0], \; [0:1:0], \; [0:0:1].$$
In fact, the fibre over $[S:T]=[0:1]$ degenerates to $XYZ=0$, i.e. the union of the three coordinate lines.


In the chart $Z=1$ the equation becomes
$$S^2(X^3+Y^3+1)-3(S^2+1)XY=0,$$
so the tangent cone in $(X,\,Y)=(0,\,0)$ is the irreducible quadric $S^2-3XY=0$. In the other charts the situation is the same, so again we have three points of type $A\_1$.


Summing up, the surface has three points of type $A\_1$ over $[S:T]=[1:0]$, three points of type $A\_1$ over $[S:T]=[0:1]$ and no other singularities. 


The minimal resolution is an elliptic fibration over $\mathbb{P}^1$ with two reducible fibres of type $I\_6$.


