Foliations on $$\mathbb {CP}^2$$ with a unique singular point without invariant algebraic curves

Foliations on $\mathbb{P}^2$ with only one singular point

“…We point out that these result was already obtained in [2] when the singularity is a saddle-node. Anyway, for the sake of completeness we also include a proof of this case.…”

“…Consider the 1-form ω = x + A 2 (x, y) + xA2 3 (x, y) + xφ(x, y) dy + tx 2 + B 3 (x, y) − yφ(x, y) dx, withA 2 (x, y) = αx 2 + wxy + ry 2 A 2 3 (x, y) = tw(4rt − 3s) s x 2 + βxy + rαy (x, y) = tw(3s − 2rt) s x 2 y + sxy 2 φ(x, y) = t(2rt − s) 2 (rt − s) x 3 + 2t(rt − s)w 2 s x 2 y + (rt − s)(s + 2rt)w s xy 2 + r(rt − s)y 3 , (rt − s) (s 4 − rs 3 t − r 2 s 2 t 2 + s 2 tw 3 − 2rst 2 w 3 + r 2 t 3 w 3 ),where r, s, t ∈ C * , w ∈ C, rt = s. The 1-form ω define a foliation on P 2 with a unique singularity which is a saddle-node. If we set w = 0 the 1-form ω adopt the form x + ry 2 + βx 2 y + xφ(x, y) dy + tx 2 + sxy 2 − yφ(x, y) dx, whereφ(x, y) = t(2rt − s) 2 (rt − s) x 3 + r(rt − s)y 3 , β = 1 rt − s (s 2 − rst − r 2 t 2 )and r, s, t ∈ C * , rt = s. This family is equivalent to that given by Alcántara and Pantaleón-Mondragón in[2]. Also note that if w = 0, r = t = 1 and s = b we recover Example 5.…”

Plane polynomials and Hamiltonian vector fields determined by their singular points

“…Case 1 in (10). For P with δpdq " 1 4 pd 2 `3d ´2q points, the interpolation matrix φ has 1 2 pd 2 `3dq odd columns and 1 2 pd 2 `3d ´2q even rows, for example for pd `1q " 3, 6, 7. Moreover, (number of columns of φ)-1 = (number of rows of φ).…”

“…The dimension of the kernel of φ is at least one, thus dim K pL d pPqq ě 0. There are 1 2 pd 2 `3dq minors A j from the matrix φpx 1 , y 1 , . .…”

“…Case 2 in (10). The dimension of Krx, ys 0 ďn is even and we assume 1 4 pd 2 `3dq P N points in P. The interpolation matrix φ is square of even size, and there are 1 2 pd 2 `3dq columns and rows; for example when d " 4, 5. If we assume P such that tdetpφpx 1 , y 1 , .…”