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 ndimensional complex projective space.

$\begingroup$ It is not true. The singularities are where the gradiant is zero and this is way more than 6 if you write the equations down.!! Also if you blow up a surface then you wont have a minimal surface(?)!! $\endgroup$– user13559Mar 10 '11 at 19:47

$\begingroup$ I don't understand the comment. What is not true? $\endgroup$– Karl SchwedeMar 11 '11 at 1:39

2$\begingroup$ @unknown I think that your calculation for the singular locus is wrong. The singular locus is made of only six isolated points. There is no line of singularities. To avoid missing points, you might actually want to work with the projective coordinates, it is also more easy.For example, you are missing all the singular points with $S=0$. $\endgroup$– JMEMar 11 '11 at 11:05

1$\begingroup$ @unknown I notice that your space is actually an elliptic surface. Why are you interested in it? This can help to identify the kind of resolutions you want. $\endgroup$– JMEMar 11 '11 at 11:45

11$\begingroup$ The deleted comment was mine. I deleted it because I was afraid of all those exclamation marks.. $\endgroup$– J.C. OttemMar 11 '11 at 12:06
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 $$X^3+Y^3+Z^33(1+\lambda^2)XYZ=0.$$ 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 $S^23XY=0$. 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 $I_6$.

$\begingroup$ @Francesco Polizzi: I would like to thank you for your very nice answer. I donts know if I can vote your answer but it is 100. Thanks $\endgroup$ Mar 16 '11 at 3:05

$\begingroup$ Dear Francesco...Once again thanks for your nice answer. I was thinking how do you know that the reducible fibres are of type I_6? Thanks $\endgroup$ Mar 23 '11 at 4:27

$\begingroup$ @unknown: The six singular points of $X$ are the vertices of two "triangles" forming two singular fibres of the elliptic fibration. In the resolution, you are blowingup these vertices, hence you are adding three more smooth rational curves in each of the two fibres. Consequently, you obtain two reducible fibres made of six (3+3) smooth rational curves forming an "exagon", that is two reducible fibres of type $I_6$. $\endgroup$ Mar 23 '11 at 10:02
The deleted comment was mine  I just stated that the singular locus consisted of six isolated singular points. Here is a Macaulay2 session to back up this claim:
Resolution of the surface S^2(X^3+Y^3+Z^3)3(S^2+T^2)XYZ=0 in P2xP1:
Macaulay2, version 1.4
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases,
PrimaryDecomposition, ReesAlgebra, TangentCone
i1 : R:=QQ[x,y,z,s,t]
o1 : PolynomialRing
i2 : I=ideal(s^2*(x^3+y^3+z^3)3*(s^2+t^2)*x*y*z)
3 2 3 2 2 3 2 2
o2 = ideal(x s + y s  3x*y*z*s + z s  3x*y*z*t )
o2 : Ideal of QQ[x, y, z, s, t]
i3 : hyper=R/I
i4 :
P5=QQ[v_0..v_5]  making a map to P5 using the Segre embedding
o4 : PolynomialRing
i5 : segre=map(hyper,P5,matrix{{x*s,y*s,z*s,x*t,y*t,z*t}});
o5 : RingMap hyper < P5
i6 : J=ker segre
2 2 2
o6 = ideal (v v  v v , v v  v v , v v  v v , v v + v v  3v v v + v v
2 4 1 5 2 3 0 5 1 3 0 4 0 3 1 4 0 1 5 2 5

3 3 3
 3v v v , v + v  3v v v + v  3v v v )
3 4 5 0 1 0 1 2 2 0 4 5
o6 : Ideal of P5
i7 :
V=variety(J)  this is the surface in P5
o7 = V
o7 : ProjectiveVariety
i8 : dim V
o8 = 2
i9 : I=ideal singularLocus V  this is the singular locus, it has dimension 0
o9 = ideal (v v , v v , v v , v v , v v , v v , v v , v v , v v , v v , v v ,
4 5 3 5 2 5 1 5 0 5 3 4 2 4 1 4 0 4 2 3 1 3

2 2 2
v v , v  v v , v v  v , v  v v )
0 3 1 0 2 0 1 2 0 1 2
o9 : Ideal of P5
i10 : decompose I  Primary decomposition
o10 = {ideal (v , v , v , v , v ), ideal (v , v , v , v , v ), ideal (v , v ,
2 1 0 4 3 2 1 0 5 3 2 1

2 2
v , v , v ), ideal (v + v + v , v , v , v , v + v v + v ), ideal (
0 5 4 0 1 2 5 4 3 1 1 2 2

v + v , v  v , v , v , v )}
1 2 0 2 5 4 3
o10 : List
i11 : W=variety(I)
o11 : ProjectiveVariety
i12 : dim W
o12 = 0
i13 : degree I  and the sigular locus consists of 6 points, each with multiplicity 1
o13 = 6

$\begingroup$ Is it possible to transform this code to the minimal resolution graphs for the normal surface singularities of the form $\{ x^p + y^q + z^r = 0 \}$. $\endgroup$– user150450Feb 11 '20 at 13:14
A simple computation shows that the equation
$$ u(x^3+y^3+z^3)3vxyz=0 $$
defines a nonsingular 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 $\zeta=\dfrac{vu}u$, $\xi=\left(\dfrac{x+y+z}{3xyz}\right)\cdot(x+\omega y+\omega^2 z)$, 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 $x^3+y^3+z^33xyz=(x+y+z)(x+\omega y+\omega^2 z)(x+\omega y+\omega^2 z)$. 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 $\zeta=\dfrac{vu}u$, 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 twotoone 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.

$\begingroup$ Maybe I am missing something, but it seems to me that you are considering a different surface than the one in the question. Shouldn't your base change be $[s:t]\mapsto [s^2:s^2+t^2]$? Otherwise you end up with the surface The surface $s^2 (x^3+y^3+z^3)3 t^2 x y z=0$ which has only 3 singular points and over $[1:0]$, the fiber is just $x^3+y^3+z^3=0$ which is a smooth elliptic fiber. $\endgroup$– JMEMar 11 '11 at 20:51

1$\begingroup$ Dear JME, you are right, in my original answer there was a $+t^2$ missing from the definition of the base change. There was and still is not anything wrong with the argument. I included the expression of the $\xi,\eta,\zeta$ as requested. $\endgroup$ Mar 12 '11 at 1:44

3$\begingroup$ I wonder why it is generally the first assumption of MO users that there is something wrong with the argument and not the more obvious (like in this case) that the person writing the answer committed a typo? Sometimes it feels like some people are eager to find an error instead of working together towards a correct solution. $\endgroup$ Mar 12 '11 at 1:47

1$\begingroup$ Uh, and to my stalker: I get it that you have a problem with me, but perhaps next time you could make a comment why you are downvoting my answer. Not that it matters too much, but it's kind of creepy that the last few weeks any time I answered a question I got a downvote in a few minutes. If this is not personal, then I would really like to know what's wrong with my answers lately. Thanks! $\endgroup$ Mar 12 '11 at 1:50

5$\begingroup$ Sándor, if you think there is some sort of weirdness going on, you should contact the moderators. I don't think it's right that those of us who use our real names can get penalized for it. $\endgroup$ Mar 12 '11 at 2:27
Francesco explained beautifully the resolution. Since I had prepared a geometric description of the resolution, I thought I will still post it.
The singular surface $$ E: S^2 (X^3+Y^3+Z^3)3 (S^2+T^2) X Y Z=0 $$ is an hypersurface of bidegree $(2,3)$ in $\mathbb{P}^1\times \mathbb{P}^2$. The rational curve $\mathbb{P}^1$ is parametrized by the projective coordinates $[S:T]$ and $[X:Y:Z]$ are projective coordinates of $\mathbb{P}^2$. For every point of $\mathbb{P}^1$, the equation defines a cubic in $\mathbb{P}^2$ which is in the form of Hesse pencil: $$ H: s (X^3+Y^3+Z^3)+ t XYZ=0, \quad [s:t]\in \mathbb{P}^1. $$ Hesse pencil is famous in number theory, in cryptography and also shows up examples of mirror symmetry in physics. It is related to the Hesse configuration of 9 points and 12 lines in $\mathbb{P}^2$. There is a nice review by Artebani and Dolgachev. Hesse pencil can be seen as an elliptic surface with base $\mathbb{P}^1$. It admits singular fibers of Kodaira type $I_3$ (three lines forming a triangle).
The fibration considered in the question is obtained from Hesse pencil with the following map: $$ [s:t]\mapsto [s^2:3(s^2+t^2)]. $$ This map is twotoone eveywhere except at $s=0$ and at $t=0$ where it is onetoone. This is related to the $\mathbb{Z}_2$ singularities described by Francesco in his answer. The six singular points of the elliptic surface $E$ are the intersection points of the three lines that form the fibers $I_3$ above $[S:T]=[1:0]$ and $[S:T]=[0:1]$. After the resolution, the singular points are replaced by $(2)$curves. The resolution describes a topological transition where two singular fibers of type $I_3$ are replaced by fibers of Kodaira type $I_6$. The transition is realized by replacing on each $I_3$ fiber, each of the 3 intersections points of the three lines by a $\mathbb{P}^1$.

$\begingroup$ Dear JME, I would like to thank you for your nice interpretation!! $\endgroup$ Mar 16 '11 at 5:07