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

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 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,

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 2 + 4 + 5 0 (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: 0 0 1 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 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.

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 2 + 1. x := 4t = 4 v w, y := ±4 R = 8 v2 uw 4 y 2 = 4 2 R = 4 3 t 3 4 4 2 t 2 + 16 = x 3 4x 2 + 16, 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 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) = 5. 3. 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 2 + 16 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 2 + 16 s rreducble wth dscrmnant: Dsc(f) = 2 8 11 (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 = 2 2 11, 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 := 1 1 2 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.

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) = 2 8 11, 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) = 2 2 11 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 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,

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), 1-165. [2] G. Bllng, K. Mahler: On exceptonal ponts on cubc curves, J. London Math. Soc. (2) 15 (1940), 32 43. [3] I. Kmng: Mordell s theorem: The rreducble case, lecture notes, http://www.math.ku.dk/~kmng/lecture_notes/2003-ellptc_curves/mordell.pdf [4] T. Nagell: Sur les proprétés arthmétques des cubques planes du premer genre, Acta Math. 52 (1928), 93 126. [5] PARI/GP, Bordeaux, http://par.math.u-bordeaux.fr/. Department of Mathematcs, Unversty of Copenhagen, Unverstetsparken 5, DK- 2100 Copenhagen Ø, Denmark. E-mal address: kmng@math.ku.dk