Number Theory

## Birational Geometry of Foliations by Marco Brunella

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By Marco Brunella

The textual content offers the birational category of holomorphic foliations of surfaces.  It discusses at size the speculation built via L.G. Mendes, M. McQuillan and the writer to review foliations of surfaces  within the spirit of the type of complicated algebraic surfaces.

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Additional info for Birational Geometry of Foliations

Example text

Fn are holomorphic in a neighbourhood of w and that the derivative D = d/dz maps the ring K[f1 , . . , fn ] into itself. Assume that fi (w) ∈ K for 1 ≤ i ≤ n. Then there exists a number C1 having the following property. Let P (x1 , . . , xn ) be a polynomial with coeﬃcients in K and of degree deg(P ) ≤ r. If f = P (f1 , . . C1k+r . Moreover, the denominator of Dk f (w) is bounded by d(P )C1k+r , where d(P ) is the denominator of the coeﬃcients of P . Proof. There exist polynomials Pi (x1 , .

We begin by observing that ∞ 1 1−z zn = n=0 for |z| < 1. Upon diﬀerentiating both sides, we ﬁnd that ∞ 1 . (1 − z)2 nz n−1 = n=0 Thus, ∞ (1 − z)−2 − 1 = (n + 1)z n , n=1 a fact we will use below. Let r = min{|w| : w ∈ L }. Then, for 0 < |z| < r, we can write ∞ 1 1 − 2 = ω −2 [(1 − z/ω)−2 − 1] = (n + 1)z n /ω n+2 . 2 (z − ω) ω n=1 Summing both sides of this expression over ω ∈ L , we obtain ℘(z) − 1 = z2 ∞ (n + 1)z n /ω n+2 . ω∈L n=1 Interchanging the summations on the right-hand side and noting that for odd n ≥ 1, the sum ω −n−2 = 0 ω∈L (because both ω and −ω are in L ), we obtain the ﬁrst assertion of the theorem.

Here C is an absolute constant that depends only on K. Proof. Let t be the degree of the number ﬁeld K. We write each of the numbers αij in terms of an integral basis: t αij = aijk ωk , k=1 aijk ∈ Z. Siegel’s Lemma 25 From these equations, we also see that by inverting the t × t matrix ( ) (ωk ), we can solve for the aijk . Thus we see that |aijk | ≤ C0 A where C0 is a constant depending only on K (more precisely, on the integral basis ωi ’s and the degree t). We write each xj as xj = yj ω so that the system becomes n t t aijk yj ωk ω = 0 j=1 k=1 =1 with yj to be solved in Z.