the ring of integers of K is the usual ring of integers Z and the ring of S-integers of Kis{±2 i 3 j n : i, j, n ∈ Z}.In this case the group of units of K is {±1} and the group of S-units of K is{±2 i 3 j : i, j ∈ Z},which is free abelian of rank 2.We now define the S-regulator and the S-norm of a fracti<strong>on</strong>al ideal, as in [13].Let {ε 1 , . . . ε s−1 } be a system of fundamental S-units in K. C<strong>on</strong>sider the matrix (log|ε i | υj )where j runs over s − 1 places of K. The absolute value of the determinant of the matrixis called the S-regulator of K. When the set S is the set of infinite places of Kthis coincides with the usual definiti<strong>on</strong> of regulator. The product formula implies that thedefiniti<strong>on</strong> is independent of the choice of places υ 1 , . . . , υ s−1 , for the (s − 1) × s matrixB = (log|ε i | υj ) where j runs over all the places of S has the property that its columnsadd up to 0. Hence the absolute value of the determinant of the matrix obtained bydeleting <strong>on</strong>e of the columns is independent of the choice of column. The definiti<strong>on</strong> isalso independent of the system of fundamental units {ε 1 , . . . ε s−1 } as the S-regulatoris, up to a c<strong>on</strong>stant, the volume of a fundamental domain of the lattice spanned by thevectors L(ε i ) = (log|ε i | υj ) ∈ R s c<strong>on</strong>tained in the s-dimensi<strong>on</strong>al subspace of vectorswhose entries add up to 0, and this lattice is the same for every choice of a system offundamental S-units.If α ∈ K is a n<strong>on</strong>zero algebraic integer, the fracti<strong>on</strong>al ideal generated by α canbe uniquely written in the form a 1 · a 2 , where a 1 is composed of primes outside S anda 2 is composed of primes c<strong>on</strong>tained in S. We define the S-norm of α as N(a 1 ) and wedenote it by N S (α). Note that N S (α) is a positive integer for every α ∈ O S \{0}.Lemma 3.2.1. Let K be a number field. For any n<strong>on</strong>zero algebraic number α ∈ K, wehave h(α −1 ) = h(α). For algebraic numbers α 1 , . . . , α n ∈ K, we haveh(α 1 α 2 · · · α n ) ≤ h(α 1 )+· · ·+h(α n ), h(α 1 +· · ·+α n ) ≤ log n+h(α 1 )+· · ·+h(α n ),25
and for any place υ ∈ M Klog‖α‖ υ ≤ [K : Q] h(α).Proof. The first equality is an immediate c<strong>on</strong>sequence of the definiti<strong>on</strong> of absolute logarithmicheight and the product formula. For the sec<strong>on</strong>d estimate note that for n<strong>on</strong>negativex 1 , . . . , x nmax{1, x 1 x 2 · · · x n } ≤ max{1, x 1 } · · · max{1, x n }.We also havemax{1, x 1 + · · · + x n } ≤ n max{1, x 1 } · · · max{1, x n },which implies the third estimate. Finally, the last inequality is also an immediate c<strong>on</strong>sequenceof the definiti<strong>on</strong> of absolute logarithmic height.Lemma 3.2.2. Let K be a number field of degree d and S a finite set of places of Kincluding the infinite places. Denote by s the cardinality of S. Let ε ∈ K ∗ be a S-unit.Let η ∈ M K be a place of K making ‖ε‖ η minimal. Thenh(ε) ≤ − s d log‖ε‖ η.Proof. Note ‖ε‖ υ = 1 for all υ /∈ S so we can choose η ∈ S. Then 0 < ‖ε‖ η ≤ 1. Now,h(ε) = h(ε −1 ) = 1 ∑max{log‖ε −1 ‖ υ , 0}dυ∈M K≤ 1 ∑log‖ε −1 ‖ η = − s dd log‖ε‖ η,υ∈Sand we have now proved the lemma.Lemma 3.2.3. Let K be a number field of degree d and S a finite set of places of Kincluding the infinite places. Let ε be a S-unit. Then∑|log‖ε‖ υ | = 2d h(ε).υ∈S26
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- Page 6 and 7: AcknowledgmentsTo the glory of my G
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- Page 10 and 11: Chapter 1IntroductionConsider the h
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- Page 16 and 17: Chapter 2The Mordell-Weil groupThe
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- Page 36 and 37: Proof. Recall that ‖ε‖ υ = 1
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- Page 46 and 47: Choose nowδ = 2c 9 (n, d)H h(µ 1
- Page 48 and 49: c ∗ 14 = 2H∗ d s 1+s 2 −2 P d
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- Page 66 and 67: Lemma 4.2.4 (I-3). Let P, Q ∈ R[X
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- Page 70 and 71: andU(x) = [Q, R](x) = (b + b ′
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- Page 74 and 75: Lemma 4.2.10 (III-3-a). We have z 1
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Since I n converges, T n converges
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and a sum of integral</stro
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2 = a 1a ′ 1 − b 1b ′ 1 + √
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• Else do- PutandIH i = θ √⎧
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and⎧⎪⎨ −1, (a n,i = b and x
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algorithm used for the multiplicati
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interested in. The roots of f are m
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assuming that the polynomial f defi
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the g × 2g matrix⎛∫∫∫ ⎞A
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curves with the point at infinity a
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We are now interested in the numeri
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conformed ourselves with estimates
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Let P be an integral</stron
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We will now find an upper bound for
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for the divisors D 1 = (−1, 1)
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Proof. Since c 16 > 0, then X(P ) i
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now want to estimate the height of
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and for all x ≤ −c 16max{ ∫ x
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andφ(2D i )2= (d i,1 , d i,2 ) ∈
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Again, by our assumption, the right
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The reader can find the MAGMA progr
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[8] Nils Bruin, Chabauty methods us
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[30] Serge Lang, Algebraic number t
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[51] Michel Waldschmidt, Diophantin