THERE ARE NO POINTS OF ORDER 11 ON ELLIPTIC CURVES OVER Q.

Transcription

1 THERE ARE NO POINTS OF ORDER 11 ON ELLIPTIC CURVES OVER Q. IAN KIMING We shall prove the followng result from [2]: Theorem 1. (Bllng-Mahler, 1940, cf. [2]) An ellptc curve defned over Q does not have a ratonal torson pont of order 11. In other words, f E s an ellptc curve over Q then 11 #E(Q) tors. The proof we gve s a paraphrase of that of the artcle [2], augmented wth greater detal n a number of places. 1. Intal analyss. The proof of theorem 1 s by contradcton, so let us begn by assumng that we have an ellptc curve defned over Q wth a ratonal pont P of order 11 (much of the ntal analyss below goes through even f we assume that P has some smaller order, but ths s not mportant for us here). and: Defne: Then we have: P := P for Z. P P j j (11), ( ) P, P j, P k on a lne P + P j + P k = 0 + j + k 0 (11). Remember that the expresson on a lne has specal nterpretatons n degenerate case where two or more of the ponts P k concde. Consder now the followng smple lemma: Lemma 1. Let k be a feld and let (a, b, c), (α, β, γ) P 2 (k) be dstnct ponts. Then there s a unque lne through these ponts, and t s gven by the equaton: x y z a b c α β γ = 0. Two lnes gven by equatons ux + vy + wz = 0 and u x + v y + w z = 0 concde f and only f the ponts (u, v, w) and (u, v, w ) concde as ponts n P 2 (k). Two dstnct lnes n P 2 (k) ntersect n precsely 1 pont. Proof. Exercse n lnear algebra. 1

2 2 IAN KIMING We have P 0 = O = (0, 1, 0). Put: P1 = (a, b, c), P2 = (α, β, γ). By ( ) these 3 ponts are not on a lne. The lemma then mples that the 3 vectors (0, 1, 0), (a, b, c), and (α, β, γ) n Q 3 are lnearly ndependent. There s then an nvertble lnear map φ of Q 3 nto tself that maps these 3 ponts to the ponts P 0 := (0, 1, 0), P 1 := (1, 0, 0), and P 2 := (0, 0, 1), respectvely. We can consder the map φ n a natural way as a bjectve map of P 2 (Q) nto tself. As such t maps lnes to lnes, and consequently ( ) mples that: P, P j, P k on a lne + j + k 0 (11) f we put P := φ( P ) for Z. In partcular, we have that the ponts P 0, P 1, and P 3 are not on a lne. So, P 3 s not on the lne through the ponts P 0 and P 1; snce that lne s gven by the equaton z = 0 we see that w 0 f P 3 = (u, v, w). Smlarly we see that u 0 and that v 0 because P 3 s not on lne wth ether P 0 and P 2, or wth P 1 and P 2. Consequently, we may consder the nvertble lnear map ψ of Q 3 nto tself gven by x x/u, y y/v, z z/w. Agan, ψ gves a bjectve map of P 2 (Q) nto tself that maps lnes to lnes. Addtonally, we see that ψ fxes the ponts P 0, P 1, P 2. Puttng: P := ψ(p ) = ψ(φ( P )) for Z we have now that: ( ) P 0 = (0, 1, 0), P 1 = (1, 0, 0), P 2 = (0, 0, 1), P 3 = (1, 1, 1), ( ) P = P j j (11), and: ( ) P, P j, P k on a lne + j + k 0 (11). Let us put: P 4 = (x 1, x 2, x 3 ). Proposton 1. In the above settng we have: P 3 = (1, 0, 1), and the coordnates x 1, x 2, x 3 of the pont P 4 satsfy the equaton: x 2 1x 2 x 2 1x 3 + x 1 x 2 3 x 2 2x 3 = 0. Proof. Let us use ( ), ( ), and ( ) freely. So f j (11) there s a unque lne through the ponts P and P j ; we shall denote that lne by L,j. Lemma 1 tells us how to fnd an equaton for L,j f we know the coordnates of P and P j. Furthermore, f k,, j, m, n are ntegers such that k + + j k + m + n 0 (11) we can conclude that the pont P k s on the ntersecton L,j L m,n (n general these lnes may concde of course). Lemma 1 gves us equatons for certan lnes: (1) L 0,1 : z = 0, (2) L 0,2 : x = 0,

3 THERE ARE NO POINTS OF ORDER 11 ON ELLIPTIC CURVES OVER Q. 3 (3) L 0,3 : x z = 0, (4) L 1,2 : y = 0, (5) L 1,4 : x 3 y x 2 z = 0, (6) L 2,3 : x y = 0. Now, the lnes L 0,3 and L 1,2 are dstnct (ths s obvous, but n prncple we could also ether use lemma 1, or note that otherwse P 0,P 1,P 3 would be on a lne). Consequently, the pont P 3 s the unque pont of ntersecton between L 0,3 and L 1,2 ; combnng (3) and (4) we then fnd: (7) P 3 = (1, 0, 1). Then we fnd the equaton for L 3,4 : (8) L 3,4 : x 2 x + (x 1 x 3 )y + x 2 z = 0. The pont P 1 s the unque pont of ntersecton between L 0,1 and L 3,4, so combnng (1) and (8) we fnd: (9) P 1 = (x 1 x 3, x 2, 0). Then we fnd the equaton for L 1,3 : (10) L 1,3 : x 2 x (x 1 x 3 )y + (x 1 x 2 x 3 )z = 0. The pont P 2 s the unque pont of ntersecton between L 0,2 and L 1,3, so combnng (2) and (10) we fnd: (11) P 2 = (0, x 1 x 2 x 3, x 1 x 3 ). Combnng (7) and (11) we fnd an equaton for L 2, 3 : (12) L 2, 3 : (x 1 x 2 x 3 )x + (x 1 x 3 )y (x 1 x 2 x 3 )z = 0. Also, the pont P 5 s the unque pont of ntersecton between L 1,4 and L 2,3, so combnng (5) and (6) we fnd: (13) P 5 = (x 2, x 2, x 3 ). Then we fnd the equaton for L 0, 5 : (14) L 0, 5 : x 3 x x 2 z = 0. The pont P 5 s the unque pont of ntersecton between L 0, 5 and L 2, 3, so combnng (14) and (12) we fnd: (15) P 5 = ((x 1 x 3 )x 2, x 1 x 2 + x 1 x 3 + x 2 2 x 2 3, (x 1 x 3 )x 3 ) ; note that x 1 x 3 0 snce otherwse (11) would mply the contradcton P 2 = P 0 ; also, x 2 0 snce otherwse (13) would mply the contradcton P 5 = P 2 ; thus we have that P 5 n (15) certanly s a pont n the projectve plane; that ts coordnates satsfy (12) and (14) s mmedately seen. Now, snce (11) we know that the 3 ponts P 2, P 4, and P 5 are on a lne. Combnng lemma 1 wth P 2 = (0, 0, 1), P 4 = (x 1, x 2, x 3 ), and (15), we deduce that: x 1 x 2 x 3 (x 1 x 3 )x 2 x 1 x 2 + x 1 x 3 + x 2 2 x 2 3 (x 1 x 3 )x 3 = 0,

4 4 IAN KIMING whence: x 2 1x 2 x 2 1x 3 + x 1 x 2 3 x 2 2x 3 = 0. Corollary 1. If there exsts an ellptc curve defned over Q that has a ratonal pont of order 11, then the cubc curve C gven by the equaton: has more than 5 ratonal ponts. u 2 v u 2 w + uw 2 v 2 w = 0 Proof. The curve C clearly has the followng 5 ratonal ponts: P 0 = (0, 1, 0), P 1 = (1, 0, 0), P 2 = (0, 0, 1), P 3 = (1, 1, 1), P 3 = (1, 0, 1). Assumng the exstence of a ratonal pont of order 11 on some ellptc curve over Q, the above analyss and Proposton 1 revealed the exstence of a ratonal pont P 4 on C, dfferent from any of the ponts P 0, P 1, P 2, P 3, P 3 because of ( ). 2. The curve C and a specal ellptc curve. Theorem 1 s clearly mpled by corollary 1 and the followng proposton: Proposton 2. The cubc curve C gven by the equaton: u 2 v u 2 w + uw 2 v 2 w = 0 has exactly 5 ratonal ponts (namely the 5 ponts (0, 1, 0), (1, 0, 0), (0, 0, 1), (1, 1, 1), and (1, 0, 1)). Now, C s n fact an ellptc curve (.e., a smooth projectve curve of genus 1 wth a ratonal pont) although not n Weerstraß form. There s an algorthm due to T. Nagell, cf. [4], that puts the curve n Weerstraß form by means of a bratonal transformaton. The proof of the followng explct result s n fact a specal case of that algorthm. Ths result s all we need n order to prove proposton 2 (.e., we do not really need to know that C s an ellptc curve although ths follows from the result below). Proposton 3. Consder the cubc curve C gven by the equaton: u 2 v u 2 w + uw 2 v 2 w = 0, as well as the ellptc curve E gven by the Weerstraß equaton: The map f defned by: y 2 z = x 3 4x 2 z + 16z 3. f(u, v, w) := (4uv, 8v 2 4uw, uw) maps ponts (u, v, w) on C wth uv 0 to ponts (x, y, z) on E wth x(y + 4z) 0. Conversely, the map g defned by: g(x, y, z) := (2x 2, x(y + 4z), 4z(y + 4z)) maps ponts (x, y, z) on E wth x(y + 4z) 0 to ponts (u, v, w) on C wth uv 0, and we have: (f g)(x, y, z) = (x, y, z) whenever x(y + 4z) 0, (g f)(u, v, w) = (u, v, w) whenever uv 0.

5 THERE ARE NO POINTS OF ORDER 11 ON ELLIPTIC CURVES OVER Q. 5 Proof. Suppose that (u, v, w) s a pont on C wth uv 0. Then we also have w 0 (for snce (u, v, w) s on C we deduce u 2 v = 0 and hence uv = 0 f w = 0). Puttng then: V := v u, W := w u, t := V W, we fnd that t 2 W 3 W 2 + (1 t)w = 0, and so: snce W 0. Ths equaton mples f then Consequently, f we put so that t 2 W 2 W + (1 t) = 0, ± R = 2t 2 W 1 = 2 v2 uw 1 R := 1 4t 2 (1 t) = 4t 3 4t x := 4t = 4 v w, y := ±4 R = 8 v2 uw 4 y 2 = 4 2 R = 4 3 t t = x 3 4x , s a pont on the ellptc curve E. (x, y, 1) = (4uv, 8v 2 4uw, uw) If on the other hand (x, y, z) s a pont on E wth x(y + 4z) 0 then (u, v, w) := (2x 2, x(y + 4z), 4z(y + 4z)) s a pont n the projectve plane. It s n fact a pont on C, snce: u 2 (v w) + uw 2 v 2 w = 4x 2 (y + 4z)(x 2 (x 4z) + 8(y + 4z)z 2 (y + 4z) 2 z) = 4x 2 (y + 4z)(x 3 4x 2 z + 16z 3 y 2 z) = 0. The remanng clams of the proposton are now easly checked. Corollary 2. The cubc curve C has exactly 5 ratonal ponts f and only f the ellptc curve: E : y 2 z = x 3 4x 2 z + 16z 3 has exactly 5 ratonal ponts. Proof. Whenever the maps f and g from proposton 3 are defned they clearly map ratonal ponts on C to ratonal ponts on E, and ratonal ponts on E to ratonal ponts on C, respectvely. By the last statement of proposton 3 we can thus conclude that we have a bjecton between the sets and A := {(u, v, w) C(Q) uv 0} B := {(x, y, z) E(Q) x(y + 4z) 0}. Now, the ratonal ponts (u, v, w) on C wth uv = 0 are easly seen to be the followng 4 ponts: (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 0, 1).

6 6 IAN KIMING Also, the ratonal ponts (x, y, z) on E wth x(y + 4z) = 0 are found to be the 4 ponts: (0, 1, 0), (0, ±4, 1), (4, 4, 1). So: #C(Q) = 5 #A = 1 #B = 1 #E(Q) = The ellptc curve E. By corollary 2 the followng proposton mples proposton 2 and hence, as we noted above, theorem 1. Proposton 4. The ellptc curve: has exactly 5 ratonal ponts. E : y 2 = x 3 4x Proof. Usng Nagell-Lutz one determnes that E(Q) tors has order 5 (and s generated by the pont (0, 4)). Thus, the clam bols down to the clam that E(Q) has rank 0. To prove ths clam we assume a lttle background n algebrac number theory. The polynomal f(x) := x 3 4x s rreducble wth dscrmnant: Dsc(f) = (whch s also the dscrmnant of E of course). Let θ = θ 1, θ 2, θ 3 denote the roots of f. One of these s real and the other 2 are complex conjugates. Consder the cubc number feld: K := Q(θ). Usng for nstance PARI, see [5], or otherwse, one computes the followng data for K: The dscrmnant of K s: Dsc(K) = 44 = , the rng of ntegers s: O K = Z + Z 1 2 θ + Z 1 4 θ2, the unt rank of K s 1 (by Drchlet), and a fundamental unt s: η := θ so that the unts of O K are: The class number of K s: O K = 1 η. h K = 1. Now recall, see for nstance the notes [3], that we have a homomorphsm: µ : E(Q) K /(K ) 2 defned by: µ(o) = 1, µ(x, y) := (x θ) mod (K ) 2.

7 THERE ARE NO POINTS OF ORDER 11 ON ELLIPTIC CURVES OVER Q. 7 The homomorphsm µ has kernel 2E(Q) (see agan, for nstance, [3] for a proof). Snce we know that E(Q) tors = Z/Z5, we have: for some r Z 0. Consequently, E(Q) = Z/Z5 Z r Im(µ) = E(Q)/2E(Q) = (Z/Z2) r. What we have to prove, namely that r = 0, thus bols down to showng that µ has trval mage. Suppose to the contrary that (x, y) s a ratonal pont on E such that µ(x, y) s non-trval,.e., such that x θ s not a square n K. We know that x and y may be wrtten: x = r t 2, y = s t 3 wth ntegers r, s, t, such that gcd(r, t) = gcd(s, t) = 1. We have: so we know that µ(x, y) = (x θ) mod (K ) 2 = (r t 2 θ) mod (K ) 2 r t 2 θ (K ) 2. Consder the ntegral deal (r t 2 θ) of O K. Usng the analyss of [3] we fnd that we can wrte: ( ) ( ) (r t 2 θ) = A 2 where A s some ntegral deal, a {0, 1}, and the p are dstnct prme deals of O K such that each p dvdes the dscrmnant: (θ j θ k ) 2 = Dsc(f) = , j k.e., such that each p s a prme dvsor of ether 2 or 11. To see that the p dvde the dscrmnant, consder ther prme factorzaton n L := Q(θ 1, θ 2, θ 3 ). If p ramfes n L/K t s a dvsor of ether 2 or 11 snce these are the only ramfed prmes n L. If p does not ramfy,.e., s a product of dstnct prme deals of O L then the analyss of [3] shows that each of those prme dvsors must dvde some deal (θ j θ k ) and hence also the above dscrmnant. We now clam that all exponents a n the product pa n ( ) are 0. To see ths, let us frst note that the prme decompostons of 2 and 11 n K are as follows: p a (2) = p 3, (11) = q 2 q wth q q (as Dsc(K) = we already knew that 2 and 11 ramfy n K, and hence each has one of these two types of decompostons, but we actually need the precse decomposton for each). We have: N K/Q (p) = 2, N K/Q (q) = N K/Q (q ) = 11.

8 8 IAN KIMING The product pa can thus be wrtten: = p a1 q a2 (q ) a3 where a 1, a 2, a 3 {0, 1}, and we have: N K/Q (p ) a = 2 a1 11 a2+a3. p a On the other hand, N K/Q (p ) a N K/Q (A) 2 = N K/Q (r t 2 θ) so N K/Q(p ) a s a square. = ((r t 2 θ 1 )(r t 2 θ 2 )(r t 2 θ 3 )) = (t 6 (x θ 1 )(x θ 2 )(x θ 3 )) = (t 6 y 2 ) = (s) 2 We deduce a 1 = 0, and ether a 2 = a 3 = 0 or a 2 = a 3 = 1. Suppose that a 2 = a 3 = 1. Then by ( ) we would have qq (r t 2 θ); snce 11 = q 2 q we could then conclude that: 11 q 2 (q ) 2 (r t 2 θ) 2 = r 2 2rt 2 θ + t 4 θ 2 whence the number r 2 2rt 2 θ + t 4 θ 2 11 would be n O K = Z + Z 1 2 θ + Z 1 4 θ2, and so 11 would necessarly dvde both r and t, contradctng the fact that gcd(r, t) = 1. Hence a 1 = a 2 = a 3 = 0, and ( ) says smply that (r t 2 θ) = A 2 for some ntegral deal A. Snce K has class number 1, we have A = (α) for some α O K. Thus: r t 2 θ = u α 2 where u s a unt that s not a square n K (snce r t 2 θ s not a square n K). Adjustng α f necessary we may assume u { 1, η, η}. Now, N K/Q (u) N K/Q (α) 2 = N K/Q (r t 2 θ) = s 2 whence N K/Q (u) > 0. As N K/Q ( 1) = 1, and as we compute N K/Q (η) = 1, the only possblty left s that: r t 2 θ = η α 2 for some α O K. Puttng β := ηα and wrtng β = a + b 1 2 θ + c 1 4 θ2 wth a, b, c Z, we deduce the exstence of ntegers a, b, c such that: ( ) η (r t 2 θ) = (1 1 2 θ)(r t2 θ) = (a + b 1 2 θ + c 1 4 θ2 ) 2. Usng θ 3 = 4θ 2 16, θ 4 = 4θ 3 16θ = 16θ 2 16θ 64, we compute that ( ) means: r ( r 2 + t2 )θ + t2 2 θ2 = (a 2 4c 2 4bc) + (ab c 2 )θ + ( b2 4 + ac 2 + bc + c2 )θ 2,

9 THERE ARE NO POINTS OF ORDER 11 ON ELLIPTIC CURVES OVER Q. 9.e., the followng 3 equatons: () r = a 2 4c 2 4bc, () r 2t 2 = 2ab 2c 2, () 2t 2 = b 2 + 2ac + 4bc + 4c 2. Then () shows that r s even, and then () shows that a s even. Also, () shows that b s even. But snce a s even, the rght hand sde of () s then dvsble by 4, so 2t 2 s dvsble by 4, and t s even. Hence both r and t are even. Ths contradcts gcd(r, t) = 1. Remark: The bulk of the reasonng n the prevous proof can be generalzed to an arbtrary ellptc curve over Q n the specal Weerstraß form y 2 = x 3 + ax 2 + bx + c. Ths leads to the so-called Bllng upper bound on the rank of the curve. See [1]. References [1] G. Bllng: Beträge zur arthmetschen Theore der ebenen Kubschen Kurven vom Geschlecht ens, Nova Acta Regae Soc. Sc. Upsalenss 11 (1938), [2] G. Bllng, K. Mahler: On exceptonal ponts on cubc curves, J. London Math. Soc. (2) 15 (1940), [3] I. Kmng: Mordell s theorem: The rreducble case, lecture notes, [4] T. Nagell: Sur les proprétés arthmétques des cubques planes du premer genre, Acta Math. 52 (1928), [5] PARI/GP, Bordeaux, Department of Mathematcs, Unversty of Copenhagen, Unverstetsparken 5, DK Copenhagen Ø, Denmark. E-mal address: kmng@math.ku.dk