This booklet is meant to function a textbook for a direction in algebraic topology before everything graduate point. the most issues coated are the category of compact 2-manifolds, the basic workforce, masking areas, singular homology thought, and singular cohomology idea. those themes are built systematically, warding off all unecessary definitions, terminology, and technical equipment. at any place attainable, the geometric motivation in the back of a few of the strategies is emphasised. The textual content involves fabric from the 1st 5 chapters of the author's previous ebook, ALGEBRAIC TOPOLOGY: AN advent (GTM 56), including just about all of the now out-of- print SINGULAR HOMOLOGY conception (GTM 70). the fabric from the sooner books has been conscientiously revised, corrected, and taken modern.
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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 ) .