REDUCTION MODULO p. We will prove the reduction modulo p theorem in the general form as given by exercise 4.12, p. 143, of [1].

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1 REDUCTION MODULO p. IAN KIMING We wll prove the reducton modulo p theorem n the general form as gven by exercse 4.12, p. 143, of [1]. We consder an ellptc curve E defned over Q and gven by a Weerstraß equaton: E : y 2 = x 3 + ax 2 + bx + c where a, b, c Z. To E we have attached the dscrmnant: D = 4a 3 c + a 2 b abc 4b 3 27c 2. Fx a prme number p, and recall that to p we have attached the p-adc valuaton ord p on Q as well as the subrng R p := fx Q j ord p (x) 0g. For x R we denote by x the reducton modulo p of x; thus, x s an element of F p. Defnton 1. The reducton modulo p of E s the cubc curve Ē defned over F p by the equaton: y 2 = x 3 + āx 2 + bx + c. When s Ē an ellptc curve over F p? By our crtera, ths s so precsely when:.e., f and only f: p 2 and D = 4ā 3 c + ā 2 b ā b c 4 b 3 27 c 2 0 p 2 and p D = 4a 3 c + a 2 b abc 4b 3 27c 2. So, there are only fntely many bad prmes p for whch the curve Ē s sngular. For p 2D we wll consder the reducton modulo p map E(Q) Ē(F p ) denoted by P P as follows: If P E(Q) we can wrte P = (x, y, z) wth x, y, z R p and at least 1 a unt n R p ; the vector (x, y, z) s then unquely determned up to multplcaton by a unt n R p, and hence the reducton: P = ( x, ȳ, z) s well-defned as a pont n P 2 (F p ). Clearly, P Ē(F p ). (Alternatvely, wrte P = (x, y, z) wth x, y, z Z and gcd(x, y, z) = 1. Wth ths requrement the vector (x, y, z) s unquely determned up to sgn). 1

2 2 IAN KIMING Recall our prevous notaton: For ν N: E(p ν ) = f(x, y) j ord p (x) 2ν, ord p (y) 3νg f0g = f(x, y) j ord p (x) 2νg f0g = f(x, y) j ord p (y) 3νg f0g, and that ord p (x) (resp. ord p (y)) must have shape 2ν 0 (resp. 3ν 0 ) f (x, y) E(Q) wth ord p (x) < 0 (resp., wth ord p (y) < 0). Theorem. Assume that p s such that the reduced curve Ē s an ellptc curve over F p,.e., assume that p 2D. The reducton modulo p map s a homomorphsm: wth kernel: E(Q) Ē(F p ) E(p) = f(x, y) j ord p (x) < 0g f0g = f(x, y) j ord p (y) < 0g f0g. Proof. Clearly 0 = 0 Ē(F p ). Let P = (x, y, 1) E(Q). If x R p then y R p, and so P = ( x, ȳ, 1) 0. On the other hand, f ord p (x) < 0 then ord p (x) and ord p (y) have shape: ord p (x) = 2ν, ord p (y) = 3ν for some ν N. But then P = (p 3ν x, p 3ν y, p 3ν ); snce p 3ν x, p 3ν R p p, but p 3ν y s a unt n R p we then deduce P = (0, 1, 0) = 0 Ē(F p ). So we certanly have for P E(Q) that: P = 0 P E(p). We now only have to show that the reducton modulo p map s a homomorphsm. To show that the reducton modulo p map s a homomorphsm we must show that f P 1, P 2, P 3 E(Q) wth P 1 + P 2 + P 3 = 0 then P 1 + P 2 + P 3 = 0. Assume then that P 1, P 2, P 3 E(Q) wth P 1 + P 2 + P 3 = 0. We easly see that ( P ) = P for P E(Q). From ths we see that we may well assume P 0 for = 1, 2, 3. Wrte then P = (x, y ) for = 1, 2, Suppose frst that 2 of the ponts P s n E(p), say P 1, P 2 E(p). Snce E(p) as prevously shown s a subgroup of E(Q) we can then conclude that P 3 = (P 1 + P 2 ) E(p). But then P = 0 for = 1, 2, 3, and thus trvally P 1 + P 2 + P 3 = Suppose then that exactly 1 of the ponts P s n E(p), say P 3 E(p). 2 a. If addtonally P 1 P 2 then the x-coordnate x 3 of the pont P 3 = (P 1 + P 2 ) s: ( ) 2 y2 y 1 x 3 = a x 1 x 2. x 2 x 1

3 REDUCTION MODULO p. 3 As ord p (x 3 ) < 0 and as now x 1, x 2, y 1, y 2 R p we must then necessarly have ord p (x 2 x 1 ) > 0,.e.: Now consder the dentty: x 1 x 2 (p). (y 2 + y 1 )(y 2 y 1 ) = y 2 2 y 2 1 = x 3 2 x a(x 2 2 x 2 1) + b(x 2 x 1 ) ; snce the rght hand sde has shape (x 2 x 1 ) A wth A R p, we can conclude that f we had y 2 y 1 (p) then we could deduce the mplcaton: ord p (x 2 x 1 ) s ord p (y 2 y 1 ) s for any s N; but ths would mply ord p ( y2 y 1 x 2 x 1 ) 0 and hence the contradcton ord p (x 3 ) 0. So n addton to the congruence x 1 x 2 (p) we have also y 2 y 1 (p). But then: P 2 = ( x 2, ȳ 2 ) = ( x 1, ȳ 1 ) = P 1, and P 1 + P 2 + P 3 = P 1 + P 2 = 0 as requred. 2 b. Suppose now nstead that P 1 = P 2. Snce P 3 0 the x-coordnate x 3 of P 3 = 2P 1 s now gven by: ( f ) 2 (x 1 ) x 3 = 2x 1 a, 2y 1 where f(x) := x 3 + ax 2 + bx + c. We conclude that ord p ( f (x 1) 2y 1 ) ord p (y 1 ) = ord p (2y 1 ) > 0 (here we used that p 2). But then P 1 = ( x 1, ȳ 1 ) = ( x 1, 0) s a pont of order 2; we conclude that P 1 + P 2 + P 3 = 2 P 1 = 0. < 0, hence 3. Suppose now fnally that P E(p) for = 1, 2, 3. So, the coordnates x, y are all n R p. Snce P 1 +P 2 +P 3 = 0 the ponts P 1, P 2, P 3 are precsely the ponts of ntersecton (counted wth multplctes) between E and some lne L. Let: αx + βy + γz = 0 be an equaton for L. We may assume that α, β, γ are n R p and at least 1 of them a unt n R p, say γ s a unt (as wll be seen, the other cases are analogous to deal wth). We can then assume an equaton for L of form: wth λ, ν R p. z = λx + νy Now let F (x, y, z) = 0 be a homogeneous equaton of degree 3 and coeffcents n R p that defnes E. E.g.: F (x, y, z) := x 3 + ax 2 z + bxz 2 + cz 3 y 2 z.

4 4 IAN KIMING As the ponts P = (x, y ), = 1, 2, 3, are the ponts of ntersecton between E and L, we must have: ( ) F (x, y, λx + νy) = ξ (xy 1 yx 1 )(xy 2 yx 2 )(xy 3 yx 3 ) for some non-zero, ratonal ξ. Suppose that we had ord p (ξ) < 0. Puttng then s := ord p (ξ) we would have s > 0 and the number p s ξ would be a unt n R p ; snce F has coeffcents n R p we could deduce that the polynomal: p s ξ (xy yx ) = p s F (x, y, λx + νy) has all coeffcents n R p ; but then: p s ξ (xȳ y x ) = 0 whence (snce p s ξ were a unt n R p ) (xȳ y x ) = 0 for at least 1 ; but ths means x = ȳ = 0 for that whch s mpossble, as 1 = λx + νy. So we have ξ R p. If we had ξ = 0 we would deduce from ( ) that: F (x, y, λx + νy) = 0 where F denotes the (coeffcentwse) reducton of F ; but that would mean that Ē would contan the whole lne z = λx + νy, somethng that we know s not possble. So we have n fact that ξ s a unt n R p. Then: F (x, y, λx + νy) = ξ (xȳ y x ) wth ξ 0. Snce clearly F = 0 s an equaton defnng Ē over F p, we can now conclude that the ponts P = ( x, ȳ ) are the ponts (counted wth multplctes) of ntersecton between Ē and the lne gven over F p by the equaton z = λx + νy. We conclude that P 1 + P 2 + P 3 = 0. Corollary 1. For p a prme such that p 2D the reducton modulo p map gves an njectve homomorphsm: E(Q) tors Ē(F p ). Proof. The statement follows from the theorem snce the subgroup E(p) s torson free. Corollary 2. If p s a prme such that p 2D then #E(Q) tors j #Ē(F p ). Proof. E(Q) tors s somorphc to a subgroup of the fnte abelan group Ē(F p ). By the last corollary, f we have prmes p 1,..., p t, none of whch dvdes 2D, then #E(Q) tors j gcd(#ē(f p1 ),..., #Ē(F pt )).

5 REDUCTION MODULO p. 5 In practce, ths often gves a very effcent method of determnng #E(Q) tors and n fact E(Q) tors tself. Let us look at a couple of examples. If p 2D one can compute Ē(F p ) wth the program PARI. The relevant commands are: gp : Start PARI E = ellnt([a 1, a 2, a 3, a 4, a 5 ]) : Intalze the curve E : y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6. ellap(e, p) : Compute the number a p := p + 1 #Ē(F p ) (for p 2D). Example 1: If E s the curve y 2 = x then D = One computes a 7 = 5, a 11 = 0, so that #Ē(F 7 ) = 8 ( 5) = 13 and #Ē(F 11 ) = 12 0 = 12. Snce gcd(12, 13) = 1 we deduce that E(Q) tors = 0. Example 2: Consder the curve E gven by y 2 = x 3 + x. D = 4. E has the obvous torson pont (0, 0) of order 2. The dscrmnant s One computes: #Ē(F 3 ) = 4, #Ē(F 5 ) = 4, #Ē(F 7 ) = 8,.... In fact, one can show that #Ē(F p ) s always dvsble by 4 whenever p s an odd prme. But we clam that E(Q) tors = f0, (0, 0)g and that ths can stll be deduced from the above theorems: For let us for p = 3, 5 look not only at the order of the group #Ē(F p ) but also at ts structure: We fnd: and hence that Ē(F 3 ) = Z/4. Also: and consequently Ē(F 5 ) = Z/2 Z/2. Ē(F 3 ) = f0, (0, 0), (2, 1), (2, 2)g Ē(F 5 ) = f0, (0, 0), (2, 0), (3, 0)g Snce E(Q) tors embeds nto both Ē(F 3 ) and Ē(F 5 ) we deduce that the fnte abelan group E(Q) tors s both cyclc and has exponent a dvsor of 2. As E(Q) tors contans n any case a subgroup of order 2, namely the subgroup generated by (0, 0), the clam follows. References [1] J. H. Slverman, J. Tate: Ratonal ponts on ellptc curves. Undergraduate Texts n Mathematcs, Sprnger-Verlag Department of Mathematcs, Unversty of Copenhagen, Unverstetsparken 5, DK Copenhagen Ø, Denmark. E-mal address: kmng@math.ku.dk