find the singularities of an algebraic curve
singularities(f, x, y)
a polynomial specifying an algebraic curve
Let f be a squarefree polynomial in x and y. Then f defines an algebraic curve in the plane C^2, and also in the projective plane P^2 by making f homogeneous. This procedure computes the singular points of the curve in the projective plane. The points are given by homogeneous co-ordinates [X,Y,Z].
For each singularity this procedure also computes the multiplicity m, the delta invariant delta, and the number of local branches r. An ordinary double point is characterized by m=2,δ=1,r=2. For a cusp one has m=2,δ=1,r=1. In general r≤m and m⁢m−12≤δ, and both of these are equalities when the singularity is an ordinary m-multiple point. The Milnor number equals 2⁢δ−r+1.
The output of this procedure is a set consisting of lists of the following form point,m,δ,r.
This procedure computes all singularities up to conjugation. So if a singularity RootOf⁡_Z2−2,1,1 is given in the output, and if RootOf⁡_Z2−2 does not appear in the input, then −RootOf⁡_Z2−2,1,1 is a singular point as well but will not be given in the output.
The genus of a curve is the number (d-1)*(d-2)/2 - Sum(delta invariants) where d is the degree of the curve. Note that if we apply this formula to compute the genus, then for each singularity we must multiply the delta invariant by the degree of the algebraic extension over which the singularity is defined, because only one singularity of each conjugacy class is given in the output.
f ≔ −8⁢y5−207⁢y⁢x4+180⁢x5−35⁢y4−128⁢y3⁢x+621⁢y⁢x3−450⁢x4+82⁢y3−521⁢y⁢x2+369⁢x3−19⁢y2+135⁢y⁢x−100⁢x2−28⁢y−7⁢x+8
Note that the conjugate (replace RootOf⁡_Z2+1 by −RootOf⁡_Z2+1 is also a singularity. So the genus is (5-1)*(5-2)/2-1-1-1-2*1=1
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