By N. S. Gopalakrishnan

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**Extra info for Commutative algebra**

**Sample text**

Let RI be an unbounded simply connected open region in the complex plane which does not contain the roots e l , e z , e 3 of the cubic 4x 3 - g2 X - g3' 28 I. 11 For uER 1, defme a function z = g(u) by where a fixed branch of the square root is chosen as t varies in R 1 • Note that the integral converges and is independent of the path in Rl from u to 00, since Rl is simply connected. The function z = g(u) can be analytically continued by letting R z be a simply connected region in IC - {e 1 , ez , e 3 } which overlaps with R 1 • If uER z , then choose uIERlnRz, and set z=g(u)=g(ul)+S~'(4t3_gzt g3)-I/zdt.

If yZ = x 3 - 112 x gives an elliptic curve over IFp. More generally, if yZ = f(x) is an elliptic curve E defined over an algebraic number field, and if p is a prime ideal of the number field which does not divide the denominators of the coefficients of f(x) or the discriminant of f(x) , then by reduction modulo p we obtain an elliptic curve defined over the (finite) residue field of p. At first glance, it may seem that the elliptic curves over finite fieldswhich lead only to finite abelian groups-are not a serious business, and that reduction modulo p is a frivolous game that will not help us in our original objective of studying i1J-points on yZ = x 3 - I1zX.

Y2' j\Z2) = (0, )li5'2' )l2 Zl) = (0,)11' Zl) = PI (where we have used the fact that p divides Y l Z2 - Y2Zl)' (ii)p does not divide Xl' Then P2 = (XlX2,Xl)l2,XlZ2) = (X l X2,X2)1l' X2Z1) = (Xl' )11' Zl) = Pl' Conversely, suppose that PI = P2 . , if p divides X 1Y2 - X2Yl and X1Z2 - X2Z1. Finally, we must show thatp divides Yl Z2 - Y2 Zl' If both Yl and Z1 are divisible by p, then this is trivial. Otherwise, the conclusion will follow by repeating the above argument with Xl' x 2 replaced by Yl' Y2 or by Zl' Z2' This concludes the proof of the lemma.

### Commutative algebra by N. S. Gopalakrishnan

by Paul

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