By Torsten Ekedahl
Those lecture notes grew out of a one semester introductory direction on elliptic curves given to an viewers of machine technological know-how and arithmetic scholars, and think basically minimum historical past wisdom. After having lined uncomplicated analytic and algebraic points, placing specific emphasis on explaining the interaction among algebraic and analytic formulation, they move directly to a few extra really good themes. those contain the $j$-function from an algebraic and analytic point of view, a dialogue of elliptic curves over finite fields, derivation of recursion formulation for the department polynomials, the algebraic constitution of the torsion issues of an elliptic curve, advanced multiplication, and modular varieties. in order to inspire easy difficulties the ebook begins very slowly yet considers a few points equivalent to modular sorts of better point which aren't often taken care of. It offers greater than a hundred routines and a Mathematica TM laptop that treats a few calculations regarding elliptic curves. The booklet is aimed toward scholars of arithmetic with a common curiosity in elliptic curves but in addition at scholars of computing device technological know-how attracted to their cryptographic points. A e-book of the eu Mathematical Society (EMS). dispensed in the Americas through the yankee Mathematical Society.
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Extra info for One semester of elliptic curves
Sample text
This sum has the form that we intended for ℘ and thus is obviously periodic. Now ℘ (z + ω1 |ω1 , ω2 ) and ℘ (z|ω1 , ω2 ) are two functions with the same derivative and hence differ by a constant. By putting z = −ω1 /2 we get that this constant equals ℘ (ω1 /2) − ℘ (−ω1 /2). , ℘ (−z) = ℘ (z) as can be seen from the definition: ℘ (−z|ω1 , ω2 ) = 1 + z2 1 = 2+ z m,n∈Z 1 1 − 2 (−z − mω1 − nω2 ) (mω1 + nω2 )2 1 m,n∈Z (z − (−m)ω1 − (−n)ω2 )2 − 1 ((−m)ω1 + (−n)ω2 )2 = ℘ (z|ω1 , ω2 ). We thus see that ℘ (z) is an even elliptic function and ℘ , being the derivative of an even function, is odd.
Let U be an open subset of C. i) A function f : U → C is holomorphic if for every a ∈ U there is a disc D(a, r) ⊆ U , D(z, r) := {w ∈ C | |z − w| < r}, such that the restriction of f to D(a, r) has the form f (z) = an (z − a)n , n≥0 where the right-hand side is convergent in D(a, r). ii) A function f : U → C is meromorphic if for every a ∈ U there is a disc D(a, r) ⊆ U , D(z, r) := {w ∈ C | |z − w| < r}, such that the restriction of f to D(a, r) has the form f (z) = an (z − a)n , n≥−N where the right-hand side is convergent in D(a, r) \ {a} and f (a) = a0 if an = 0 for n < 0 and ∞ otherwise.
6. If u, v, w ∈ C \ with u + v + w = 0, then we have the following relation: ℘ (u) ℘ (u) 1 ℘ (v) ℘ (v) 1 = 0. ℘ (w) ℘ (w) 1 Proof. We solve directly for w; w = −u − v and consider the expression as a meromorphic function in u. We start by expanding ℘ (u) and ℘ (−u − v) as Taylor series: ℘ (u) = 1 + O(u2 ), u2 ℘ (−u − v) = ℘ (v) + ℘ (v)u + ℘ (v) u + O(u3 ). 2 Expanding8 the determinant gives ℘ (u) ℘ (u) 1 ℘ (v) ℘ (v) 1 ℘ (w) ℘ (w) 1 = O(1) and hence the determinant is holomorphic at u = 0. At u = −v we may introduce u = −u − v and then u = −u − v and the determinant becomes the same, but with top and bottom row interchanged (which of course only changes the sign).