Number Theory

## Elementary Methods in the Analytic Theory of Numbers by A. O Gelfond, Yu. V. Linnik, L. J. Mordell

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By A. O Gelfond, Yu. V. Linnik, L. J. Mordell

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Extra info for Elementary Methods in the Analytic Theory of Numbers

Sample 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.