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Consider a foliation of $\mathbb P^2$ which is defined on $\mathbb A^2$ by a polynomial vector field

$$\mathscr F : M \frac \partial {\partial x} + N \frac \partial {\partial y}$$

Write the polynomials $M$ and $N$ as sums of their homogeneous parts:

$$
M = M_0 + \dots + M_{d+1}, \qquad \qquad \qquad
N = N_0 + \dots + N_{d+1}
$$

Without loss of generality, we may assume that the line at infinity $L_\infty$ does not contain any singularities of $\mathscr F$. In particular, $L_\infty$ is not $\mathscr F$-invariant, because every $\mathscr F$-invariant algebraic curve must pass through some singularity. To find the algebraic consequences of our assumption, we pass to the chart centered at the point at infinity on the $x$ axis, whose coordinates are

$$u = \frac 1x, \qquad \qquad \qquad \qquad v = \frac yx$$

Write $P_i = M_i(1, v)$ and $Q_i = N_i(1, v)$. Then define

$$
P = u^{d+1} P_0 + \dots + P_{d+1}, \qquad \qquad \qquad
Q = u^{d+1} Q_0 + \dots + Q_{d+1}
$$

In the new chart, the foliation is given by

$$\mathscr F : uP \frac \partial {\partial u} + (vP - Q) \frac \partial {\partial v}$$

Since $L_\infty$ is not $\mathscr F$-invariant, $u$ divides $vP - Q$. Hence there exists a polynomial $R(x,y)$ of degree $d$ such that

$$M_{d+1} = xR, \qquad \qquad \qquad N_{d+1} = yR$$

By construction, $P_{d+1} = R(1,v)$ has $d$ roots on $L_\infty$, possibly including the point at infinity.

Since $L_\infty$ does not contain any singularities of $\mathscr F$, the following system of equations has no solutions:

$$
\begin{cases}
P_{d+1} = 0 \\
vP_d - Q_d = 0
\end{cases}
$$

In other words, $u$ and $vP - Q$ are associates in the local ring $\mathcal O_p$, for any point $p \in M \cap N \cap L_\infty$. Hence,

$$I_p(P, Q) = I_p(P, Q - vP) = I_p(P, u) = \operatorname{ord}_p P _{d+1}$$

Therefore, the projective curves $M = 0$ and $N = 0$ meet at $d$ points on $L_\infty$, counted with multiplicity. By Bézout's theorem, these curves also meet at $d^2 + d + 1$ additional points on $\mathbb A^2$. Hence a foliation $\mathscr F$ of degree $d$ of the projective plane $\mathbb P^2$ has $d^2 + d + 1$ singularities, counted with multiplicity.

Tue, 09 Feb 2021 13:08 GMT