Download Algebraic Curves and One-Dimensional Fields by Fedor Bogomolov PDF

By Fedor Bogomolov

Algebraic curves have many certain homes that make their research really worthwhile. therefore, curves offer a average creation to algebraic geometry. during this booklet, the authors additionally deliver out features of curves which are specific to them and emphasize connections with algebra. this article covers the fundamental subject matters within the geometry of algebraic curves, similar to line bundles and vector bundles, the Riemann-Roch Theorem, divisors, coherent sheaves, and zeroth and primary cohomology teams. The authors make some extent of utilizing concrete examples and specific tips on how to make sure that the fashion is apparent and comprehensible. numerous chapters improve the connections among the geometry of algebraic curves and the algebra of one-dimensional fields. this is often an engaging subject that's infrequently present in introductory texts on algebraic geometry. This e-book makes an exceptional textual content for a primary path for graduate scholars.

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Example text

2 2 Notice that we have added 1 in order to account for the point at infinity. 13. Suppose that E/Q is an elliptic curve that has bad reduction at a prime p. How many points does the singular curve E have over Fp ? 1. 14. Let E/Q be the elliptic curve y 2 = x3 + 3. Its minimal discriminant is ΔE = −3888 = −24 · 35 . Thus, the only primes of bad reduction are 2 and 3 and E/Fp is smooth for all p ≥ 5. 6. Elliptic curves over finite fields 41 where all the coordinates should be regarded as congruences modulo 5.

7. The rank and the free part of E(Q) 45 (2) For all P ∈ E(Q) and m ∈ Z, h(mP ) = m2 · h(P ). ) (3) Let P ∈ E(Q). Then h(P ) ≥ 0, and h(P ) = 0 if and only if P is a torsion point. For the proofs of these properties, see [Sil86], Ch. VIII, Thm. 3, or [Mil06], Ch. IV, Prop. 5 and Thm. 7. As we mentioned at the beginning of this section, we can calculate upper bounds on the rank of a given elliptic curve (see [Sil86], p. 2). 4 ([Loz08], Prop. 1). Let E/Q be an elliptic curve given by a Weierstrass equation of the form E : y 2 = x3 + Ax2 + Bx, with A, B ∈ Z.

Y02 = (x0 − e1 )(x0 − e2 )(x0 − e3 ). Thus, each term (x0 − ei ) must be almost a square, and we can make this precise by writing (x0 − e1 ) = au2 , (x0 − e2 ) = bv 2 , (x0 − e3 ) = cw2 , y02 = abc(uvw)2 , where a, b, c, u, v, w ∈ Q, the numbers a, b, c ∈ Q are square-free, and abc is a square (in Q). 1. Let E : y 2 = x3 − 556x + 3120 = (x − 6)(x − 20)(x + 26) so that e1 = 6, e2 = 20 and e3 = −26. The point (x0 , y0 ) = 66469980 ( 164184 289 , 4913 ) is rational and on E. We can write x0 − e1 = 164184 −6=2· 289 285 17 2 293 2 2 and, similarly, x0 − e2 = ( 398 17 ) and x0 − e3 = 2 · ( 17 ) .

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