Introduction to Arithmetic Geometry Fall 2013 Lecture #25 12/05/2013

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1 Introducton to Arthmetc Geometry Fall 2013 Lecture #25 12/05/ Overvew of Mordell s theorem In the last lecture we proved that the torson subgroup of the ratonal ponts on an ellptc curve E/Q s fnte. In ths lecture we wll prove a specal case of Mordell s theorem, whch states that E(Q) s fntely generated. By the structure theorem for fntely generated abelan groups, ths mples E(Q) Z r T, where Z r s a free abelan group of rank r, and T s the (necessarly fnte) torson subgroup. 1 Thus Mordell s theorem provdes an alternatve proof that T s fnte, but unlke our earler proof, t does not provde an explct method for computng T. Indeed, Mordell s theorem s notably neffectve; t does not gve us a way to compute a set of generators for E(Q), or even to determne the rank r. It s a major open queston as to whether there exsts an algorthm to compute r; t s also not known whether r can be unformly bounded. 2 Mordell s theorem was generalzed to number felds (fnte extensons of Q) and to abelan varetes (recall that ellptc curves are abelan varetes of dmenson one) by André Wel and s often called the Mordell-Wel theorem. All known proofs of Mordell s theorem (and ts generalzatons) essentally amount to two provng two thngs: (a) E(Q)/2E(Q) s a fnte group. (b) For any fxed Q E(Q), the heght of 2P + Q s greater than the heght of P for all but fntely many P. We note that there s nothng specal about 2 here, any nteger n > 1 works. We wll explan what (b) means n a moment, but let us frst note that we really do need some sort of (b); t s not enough to just prove (a). To see why, consder the addtve abelan group Q. the quotent Q/2Q s certanly fnte (t s the trval group), but Q s not fntely generated. To see ths, note that for any fnte S Q, we can pck a prme p such that under the canoncal embeddng Q Q p we have S Z p, and therefore S Z p, but we never have Q Z p. The heght of a projectve pont P = (x : y : z) wth x, y, z Z sharng no common factor s defned as H(P ) := ( x, y, z ), where s the usual archmedean absolute value on Q. The heght H(P ) s a postve nteger that s ndependent of the representaton of the representaton of P, and for any bound B, the set {P E(Q) : H(P ) B} s fnte, snce t cannot possbly have more than (2B + 1) 3 elements. We wll actually use a slghtly more precse noton of heght, the canoncal heght, whch we wll defne later. Now let us suppose that we have proved (a) and (b), and see why ths mples that E(Q) s fntely generated. Snce E(Q)/2E(Q) s fnte, for any suffcently large B the fnte set S = {P E(Q) : H(P ) B} must contan a set of representatves for E(Q)/2E(Q), and 1 Any fntely generated abelan torson group must be fnte; ths does not hold for nonabelan groups. 2 Most number theorsts thnk not, but there are some notable dssenters. Andrew V. Sutherland

2 we can pck B so that (b) holds for all Q S and P S. If S does not generate E(Q), then there s a pont P 0 E(Q) S of mnmal heght H(P 0 ). Snce S contans a set of representatves for E(Q)/2E(Q), we can wrte P 0 n the form P 0 = 2P + Q, for some Q S and P E(Q). Snce P 0 S, we must have P S, but (b) mples H(P ) < H(P 0 ), contradctng the mnmalty of H(P 0 ). So the set E(Q) S must be empty and S s a fnte set of generators for E(Q). We should note that ths argument does not yeld an algorthm to compute S because we do not have an effectve bound on B (we know B exsts, but not how bg t s) Ellptc curves wth a ratonal pont of order 2 In order to smplfy the presentaton, we wll restrct our attenton to ellptc curves E/Q that have a ratonal pont of order 2 (to prove the general case one can work over a cubc extenson of Q for whch ths s true). In short Weerstrass form any pont of order 2 s an affne pont of the form (x 0, 0). After replacng x wth x + x 0 we obtan an equaton for E of the form E : y 2 = x(x 2 + ax + b), on whch P = (0, 0) s a pont of order two. Snce E s not sngular, the cubc on the RHS has no repeated roots, whch mples b 0, a 2 4b 0. The algebrac equatons for the group law on curves of ths form are slghtly dfferent than for curves n short Weerstrass form; the formula for the nverse of a pont s the same, we smply negate the y-coordnate, but the formulas for addton and doublng are slghtly dfferent. To add two affne ponts P 1 = (x 1, y 1 ) and P 2 = (x 2, y 2 ) wth x 1 x 2, as n Lecture 23 we consder the lne L through P 1 and P 2 wth equaton L: (y y 1 ) = λ(x x 1 ), where λ = (y 2 y 1 )/(x 2 x 1 ). Solvng for y and pluggng nto equaton for E, we have λ 2 x 2 = x(x 2 + ax + b) 0 = x 3 + (a λ 2 )x 2 + The x-coordnate x 3 of the thrd pont n the ntersecton L E s a root of the cubc on the RHS, as are x 1 and x 2, and the sum x 1 + x 2 + x 3 must be equal to the negaton of the quadratc coeffcent. Thus x 3 = λ 2 a x 1 x 2, y 3 = λ(x 1 x 3 ) y 1, where we computed y 3 by pluggng x 3 nto the equaton for L and negatng the result. The doublng formula for P 1 = P 2 s the same, except now λ = (3x 2 + 2ax + b)/(2y).

3 sogenes In order to prove that E(Q)/2E(Q) s fnte, we need to understand the mage of the multplcaton-by-2 map [2]. We could use the doublng formula derved above to do ths, but t turns out to be smpler to decompose [2] as a composton of two sogenes [2] = ˆϕ ϕ, where ϕ: E E and ˆϕ: E E for some ellptc curve E that we wll determne. The kernel of ϕ wll be {O, P }, where P = (0, 0) s our ratonal pont of order 2. Smlarly, the kernel of ˆϕ wll be {O, P }, where O s the dstngushed pont on E and P s a ratonal pont of order 2 on E. Recall from Lecture 24 that for any sogeny ϕ: E E we have an njectve map ker ϕ Aut(Q(E)/ϕ (Q(E ))) defned by P τp, where τ P s the translaton-by-p morphsm. In our present stuaton there s only one non-trval pont n the kernel of ϕ, the pont P = (0, 0), and t s ratonal, so we can work over Q. We can determne both E and the morphsm ϕ by computng ϕ (Q(E )) as the fxed feld of the automorphsm τp : Q(E) Q(E). Remark Ths strategy apples n general to any separable sogeny wth a cyclc kernel (a cyclc sogeny), all we need s a pont P that generates the kernel. For an affne pont Q = (x, y) not equal to P = (0, 0) the x-coordnate of τ P (Q) = P +Q s λ 2 a x, where λ = y/x s the slope of lne throught P and Q. Usng the curve equaton for E, we can smplfy ths to λ 2 a x = y2 ax 2 x 3 x 2 = bx x 2 = b x. The y-coordnate of τ P (Q) s then λ(0 b/x) 0 = by/x 2. Thus for Q {O, P } the map τ P s gven by (x, y) (b/x, by/x 2 ). To compute the fxed feld of τp, note that f we regard the slope λ = y/x as a functon n Q(E), then composton wth τ P merely changes ts sgn. Thus ( ) by/x τp (λ ( ) y 2 ) = = = λ 2. b/x x We also note that the pont Q + τ P (Q) s fxed by τ P, hence the sum of the y-coordnates of Q and τ P (Q) s fxed by τ P (when represented as affne ponts (x : y : 1)). Thus ( x τp (y by/x 2 ) = τp 2 ) y by x 2 = (b/x)2 ( by/x 2 ) b( by/x 2 ) (b/x) 2 = y by/x 2. Note that λ 2 = y 2 /x 2 = x(x 2 + ax + b)/x 2 = x + a + b/x, so let us defne X = x + a + b/x and Y = y(1 b/x 2 ) Then Q(X, Y ) s a subfeld of E(Q) = Q(x, y) fxed by τ P, hence a subfeld of ϕ (Q(E )), and we clam that t s a subfeld of ndex 2. To see ths, note that x = (X + Y X a)/2 and y = x X,

4 thus [Q(E) : Q(X, Y )] 2 and [Q(E) : Q(X, Y )] 1 because Q(E) contans x/y = X and Q(X, Y ) does not. We also know that [Q(E) : ϕ (Q(E ))] 2, snce ker ϕ Q(E) has order 2 and njects nto Aut(Q(E)/ϕ (Q(E))), therefore ϕ (Q(E )) = Q(X, Y ). Snce ϕ s a feld embeddng, we have Q(E ) Q(X, Y ). We now know the functon feld of E ; to compute an equaton for E we just need a relaton between X and Y. Y 2 = y 2 (1 b/x 2 ) 2 = x(x 2 + ax + b)(1 2b/x 2 b 2 /x 4 ) = X(x 2 2b + b 2 /x 2 ) = X ( (x + b/x) 2 4b ) = X ( (X a) 2 4b ) = X(X 2 2aX + a 2 4b). Let us now defne A = 2a and B = a 2 4b. Then the equaton Y 2 = X(X 2 + AX + B) has the same form as that of E, and snce B = a 2 4b 0 and A 2 4B = 16b 0, t defnes an ellptc curve E wth dstngushed pont O = (0 : 1 : 0), and the affne pont P = (0, 0) has order 2. The 2-sogeny ϕ: E E sends O to O and each affne pont (x, y) on E to (X, Y ) = (x + a + b/x, y(1 b/x 2 )) on E. Snce E has the same form has E, we can repeat the process above to compute the 2-sogeny ˆϕ: E E that sends O to O and (X, Y ) to (X + A + B/X, Y (1 B/X 2 )). One can then verfy that [2] = ˆϕ ϕ, by composng ˆϕ and ϕ and comparng the result to the doublng formula on E. But we can see ths more drectly by notng that ker( ˆϕ ϕ) = E[2] and deg( ˆϕ ϕ) = deg ˆϕ deg ϕ = 2 2 = 4 = #E[2] = # ker( ˆϕ ϕ). Thus the njectve homomorphsm E[2] Aut(Q(E)/( ˆϕ ϕ) (Q(E))) s an somorphsm, and the same holds for Aut(Q(E)/[2] Q(E)). Snce we are n characterstc zero, both extensons are separable, and t follows from Galos theory that there s a unque subfeld of Q(E) fxed by the automorphsm group {τp : P E[2]}. Thus the functon feld embeddngs ( ˆϕ ϕ) and [2] are equal, and the correspondng morphsms must be equal (by the functoral equvalence of smooth projectve curves and ther functon felds). Remark The constructon and argument above apples qute generally. Gven any fnte subgroup H of E( k) there s a unque ellptc curve E and separable sogeny E E wth H as ts kernel; see [2, Prop. III.4.12] The weak Mordell-Wel theorem We are now ready to prove that E(Q)/2E(Q) s fnte (n the case that E(Q) has a ratonal pont of order 2). Ths s a specal case of what s known as the weak Mordell-Wel theorem, whch says that E(k)/nE(k) s fnte, for any postve nteger n and any number feld k. Our strategy s to prove that E(Q)/ϕ(E(Q)) s fnte, where ϕ: E E s the 2-sogeny from the prevous secton. Ths wll also show that E (Q)/ ˆϕ(E(Q)) s fnte, and t wll follow that E/2E(Q) s fnte. We begn by characterzng the mage of ϕ n E (Q).

5 Lemma An affne pont (X, Y ) E (Q) les n the mage of ϕ f and only f ether X Q 2, or X = 0 and a 2 4b Q 2. Proof. Suppose (X, Y ) = ϕ(x, y). T If X 0 then X = (y/x) 2 Q 2. If X = 0 then x(x 2 +ax+b) = 0, and x 0 (snce ϕ(0, 0) = O ), so x 2 +ax+b = 0 has a ratonal soluton, whch mples a 2 4b Q 2. Conversely, f X Q 2 then x = (X + Y X a)/2 and y = x X gves a pont (x, y) E(Q) for whch ϕ(x, y) = (X, Y ), and f X = 0 and a 2 4b Q 2, then x 2 + ax + b has a nonzero ratonal root x for whch ϕ(x, 0) = (0, 0) = (X, Y ). Now let us defne the map π : E (Q) Q /Q 2 by { X f X 0, (X, Y ) a 2 4b f X = 0, and let π(o ) = 1. Lemma The map π : E (Q) Q /Q 2 s a group homomorphsm. Proof. By defnton, π(o ) = 1, so π preserves the dentty element and behaves correctly on sums nvolvng O. and snce π(p ) = π( P ) and the square classes of X and 1/X are the same, π preserves nverses. We just need to verfy π(p + Q) = π(p )π(q) for affne ponts P, Q that are not nverses. So let P and Q be affne ponts whose sum s an affne pont R, let Y = lx + m be the lne L contanng P and Q (the lne L s not vertcal because P + Q = R O ). Pluggng the equaton for Y gven by L nto the equaton for E gves (lx + m) 2 = X(X 2 + AX + B) 0 = X 3 + (A l 2 )x 2 + (B lm)x m 2. The X-coordnates X 1, X 2, X 3 of P, Q, R are all roots of the cubc on the RHS, hence ther product s equal to m 2, the negaton of the constant term. Thus X 1 X 2 X 3 s a square, whch means that π(p )π(q)π(p + Q) = 1, and therefore π(p )π(q) = 1/π(P + Q) = π(p + Q), snce π(p + Q) and 1/π(P + Q) are n the same square-class of Q. Lemma The mage of π : E (Q) Q /Q 2 s fnte. Proof. Let (X, Y ) be an affne pont n E (Q) wth X 0, and let r Z be a square-free nteger representatve of the square-class π(x, Y ). We wll show that r must dvde B, whch clearly mples that m π s fnte. The equaton Y 2 = X(X + ax + B) for E mples that X and X + ax + B le n the same square-class, thus X 2 + AX + B = rs 2 X = rt 2, for some s, t Q. Let us wrte t = l/m wth l, m Z relatvely prme. Pluggng X = rt 2 nto the frst equaton gves r 2 t 4 + Art 2 + B = rs 2 r 2 l 4 /m 4 + Arl 2 /m 2 + B = rs 2 r 2 l 4 + Arl 2 m 2 + Bm 4 = rm 4 s 2,

6 and snce the LHS s an nteger, so s the RHS. Let p be any prme dvdng r. Then p must dvde Bm 4, snce t dvdes every other term. If p dvdes m then p 3 must dvde r 2 l 4, snce t dvdes every other term, but then p dvdes l, snce r s squarefree, whch s mpossble because l and m are relatvely prme. So p does not dvde m and therefore must dvde B. Ths holds for every prme dvsor of the squarefree nteger r, so r dvdes B as clamed. Corollary E (Q)/ϕ(E(Q)) and E(Q)/ ˆϕ(E(Q)) are fnte. Proof. Lemma 25.3 mples that ker π = ϕ(e(q)), thus E (Q)/ϕ(E(Q)) m π s fnte, and ths remans true f we replace E wth E and ϕ wth ˆϕ. Corollary E(Q)/2E(Q) s fnte. Proof. The fact that [2] = ˆϕ ϕ mples that each ˆϕ(E (Q))-coset n E(Q) can be parttoned nto 2E(Q)-cosets. Two ponts P and Q n the same ˆϕ(E (Q))-coset le n the same 2E(Q)- coset f and only f (P Q) 2E(Q) = ( ˆϕ ϕ)(e(q)), equvalently, ˆϕ 1 (P Q) ϕ(e(q). Thus the number of 2E(Q)-cosets n each ˆϕ(E (Q))-coset s precsely E (Q)/ϕ(E(Q)), thus s fnte. #E(Q)/2E(Q) = #E(Q)/ ˆϕ(E (Q)) #E (Q)/ϕ(E(Q)) Remark The only place n our work above where we really used the fact that we are workng over Q, as opposed to a general number feld, s n the proof of Lemma Specfcally, we used the fact that the rng of ntegers Z of Q s a UFD, and that ts unt group Z s fnte. Nether s true of the rng of ntegers O k of a number feld k, n general, but there are analogous facts that one can use; specfcally, O k s a Dedeknd doman, hence deals can be unque factored nto prme deals, the class number of O k s fnte, and O k s fntely generated. We also assumed that E has a ratonal pont of order 2, but after a base extenson to a number feld we can assume ths wthout loss of generalty Heght functons Let k be any number feld. Recall from Lecture 6 that (up to equvalence) the absolute values of k consst of non-archmedean absolute values, one for each prme deal p of the rng of ntegers O k (these are the fnte places of k), and archmedean absolute values, one for each embeddng of k nto R and one for each conjugate par of embeddngs of k nto C (these are the nfnte places of k). Let P k denote the set of (fnte and nfnte) places of k. For each place p P k we want to normalze the assocated absolute value p so that (a) The product formula x p = 1 holds for all x k. (b) For any number feld k k and any place p of k we have q p x q = x p, where q p means that the restrcton of q to k s equvalent to p. Both requrements are satsfed by usng the standard normalzaton for Q, wth x p = p vp(x) for p < and x = x, and then for each q P k wth q p defnng x q = N kq/q p (x) 1/[k:Q] p,

7 where k q and Q p denote the completons of k at q and Q at p, respectvely. 3 Defnton The (absolute) heght of a projectve pont P = (x 0 : : x n ) P n (Q) s H(P ) := x p, where k = Q(x 0,..., x n ). For any λ Q, f we let k = Q(x 0,..., x n λ), then λx p = ( λ p x p ) = λ p p x = thus H(P ) s well defned (t does not depend on a partcular choce of x 0,..., x n ). For k = Q we can wrte P = (x 0 : : x n ) wth the x Z havng no common factor. Then x p = 1 for p < and H(P ) = x ; ths agrees wth the defnton we gave earler. Lemma For all P = (x 0 : : x n ) P n (Q) we have H(P ) 1. Proof. Pck a nonzero x j and let k = Q(x 0,..., x n ). Then H(P ) = x p x j p = 1. Defnton The logarthmc heght of P P n (Q) s the nonnegatve real number h(p ) := log H(P ). We now consder how the heght of a pont changes when we apply a morphsm to t. We wll show that there for any fxed morphsm φ: P m P n there are constants c and d (dependng on φ) such that for any pont P P m (Q) we have Ths can be wrtten more succnctly wrte as dh(p ) c h(φ(p )) dh(p ) + c. h(φ(p )) = dh(p ) + O(1), where the O(1) term ndcates a bounded real functon of P (the functon h(φ(p )) dh(p )). We frst prove the upper bound; ths s easy. Lemma Let k be a number feld and let φ: P n P m be a morphsm (φ 0 : : φ n ) defned by homogeneous polynomals φ k[x 0,..., x n ] of degree d. There s a constant c such that h(φ(p )) dh(p ) + c for all P P n ( k). 3 The correctness of ths defnton reles on some standard results from algebrac number theory that we wll not prove here; the detals are not mportant, all we need to know s that a normalzaton satsfyng both (a) and (b) exsts, see [1, p. 9] or [2, pp ] for a more detaled exposton. x,

8 Proof. Let c = N p j c j p, where c j ranges over coeffcents that appear n any φ, and N bounds the number of monomals appearng n any φ. If P = (a 0 :... : a n ) and k = Q(a 0,..., a n ), then H(φ(P )) = φ (P ) p,j c j a d p = ch(p ) d, by the multplcatvty of p and the trangle nequalty. The lemma follows. We now make a few remarks about the morphsm φ: P n P m appearng n the lemma. Morphsms wth doman P n are tghtly constraned, more so than projectve morphsms n general, because the deal of P n ( as a varety), s trval; ths means that the polynomals defnng φ are essentally unque up to scalng. Ths has several consequences. The polynomals φ defnng φ cannot have a common zero n P n ( k); otherwse there would be a pont at whch φ s not defned. Ths requrement s not explctly stated because t s mpled by the defnton of a morphsm as a regular map. The mage of φ n P m s ether a pont (n whch case d = 0), or a subvarety of dmenson n; f ths were not the case then the polynomals defnng φ would have a common zero n P n ( k). The fact that m φ s a varety follows from the fact that projectve varetes are complete (so every morphsm s a closed map). In partcular, f φ s non-constant then we must have m n. If φ s non-constant, then d = [k(p n ) : φ (k(m φ))] s equal to the degree of the φ. In partcular, f d = 1 then φ s a bjecton from P n to ts mage. Note that ths agrees wth out defnton of the degree of a morphsm of curves. Corollary It φ s any automorphsm of P n, then h(φ(p )) = h(p ) + O(1). (1) Proof. We must have d = 1, and we can apply Lemma to φ 1 as well. The corollary acheves our goal n the case d = 1 and m = n. If d = 1 and m > n, after applyng a sutable automorphsm to P m we can assume that m φ s the lnear subvarety of P m defned by x n+1 = x n+2 = = x m+1 = 0, and t s clear that the orthogonal projecton (x 0 : : x m ) (x 0 : : x n ) does not change the heght of any pont n ths subvarety. It follows that (2) holds whenever d = 1, whether m = n or not. We now prove the general case Theorem Let k be a number feld and let φ: P n P m be a morphsm (φ 0 : : φ n ) defned by homogeneous polynomals φ k[x 0,..., x n ] of degree d. Then h(φ(p )) = dh(p ) + O(1). (2) Proof. If d = 0 then φ s constant and the theorem holds trvally, so we assume d > 0. We wll decompose φ as the composton of four morphsms: a morphsm ψ : P n P N, an automorphsm of P N, an orthogonal projecton P N P n P m, and an automorphsm of P m. All but the morphsm ψ change the logarthmc heght of a pont P by at most an addtve constant that does not depend on P, and we wll show that h(ψ(p )) = dh(p ).

9 The map ψ = (ψ 0 : : ψ N ) s defned as follows. We let N = ( ) n+d d 1, and take ψ 0,..., ψ N to be the dstnct monomals of degree d n the varables x 0,..., x n, n some order. Clearly the ψ N have no common zero n P n (Q), so ψ defnes a morphsm P n P N. Let P = (a 0 : : a n ) be any pont n P n, and let k = Q(a 0,..., a n ). For each p P k, and t follows that ψ (P ) p = j H(ψ(P )) = a d j p = j ψ (P ) p = a j d p = ( a j p ) d, j ( j a j p ) d = H(P ) d. Thus h(ψ(p )) = dh(p ) as clamed. We now note that each φ s a lnear combnaton of the ψ j, thus φ nduces an automorphsm ˆφ: P N P N, and after applyng a second automorphsm of P N we may assume that the mage of ˆφ ψ n P N s the varety defned by x n+1 = = x N = 0. Takng the orthogonal projecton from P N to P n embedded n P m as the locus of x n+1 = = x m = 0 does not change the heght of any pont, and we may then apply an automorphsm of P m to map ths embedded copy of P n to m φ. Remark For an alternatve proof of Theorem usng the Nullstellensatz, see [2, VIII.5.6]. Lemma Let k/q be a fnte Galos extenson. Then h(p σ ) = h(p ) for all P P n (k) and σ Gal(k/Q). Proof. The acton of σ permutes P k, so f P = (x 0 : : x n ) wth x k, then H(P σ ) = x σ p = x σ p σ = x p = x p = H(P ). p σ P k p σ P k Remark Lemma also holds for k = Q. Theorem (Northcott). For any postve ntegers B, d, and n, the set s fnte. {P P n (k) : h(p ) B and [k : Q] d} Proof. Let P = (x 0 : : x n ) P n (k) wth [k : Q] d. We can vew each x as a pont P = (x : 1) n P 1 (k), and we have H(P ) = x p ( x p, 1) = H(P ). p P Thus t suffces to consder the case n = 1, and we may assume P = (x : 1) and k = Q(x). Wthout loss of generalty we may replace k by ts Galos closure, so let k/q be Galos wth Gal(k/Q) = {σ 1,... σ d }. The pont Q = (x σ 1 : : x σ d) P d 1 (k) s fxed by Gal(k/Q), hence by Gal(Q/Q), so Q P d 1 (Q). By Lemma 25.16, h(q) = h(p ), so we have reduced to the case k = Q, and by the argument above we can also assume n = 1. The set {P P 1 (Q) : h(p ) B} s clearly fnte; each P can be represented as a par of relatvely prme ntegers of whch only fntely many have absolute value at most e B.

10 25.6 Canoncal heght functons on ellptc curves Theorem (Tate). Let S be a set and let r > 1 a real number. Let φ: X X and h: X R be functons such that h φ = rh + O(1), and let 1 ĥ φ (x) := lm n r n h(φn (x)). Then ĥφ s the unque functon S R for whch () ĥφ = h + O(1); () ĥφ φ = rĥφ. Proof. Choose c so that 1 r h(φ(x)) h(x) c r for all x S. For all n > 1 we have 1 r n h(φn (x)) 1 r n 1 h(φn 1 (x)) = 1 1 r n 1 r h(φ(φn 1 (x)) h(φ n 1 (x)) c r n 1, thus for all x S the sequence 1 r h(φ n (x)) converges, so n ĥφ s well defned. For all x S we have ĥφ(x) h(x) 1 r n h(φn (x)) 1 r n 1 h(φn 1 (x)) n=1 n=1 c r n = c r 1, so () holds. Property () s clear, and for unqueness we note that f f = h + O(1) and f φ = rf then applyng the constructon above wth h replaced by f yelds ˆf φ = ĥφ, but t s also clear that ˆf φ = f, so f = ĥφ. We now want to apply Theorem to the set S = E(Q) wth φ = [2] the multplcatonby-2 map and r = 4, It mght seem natural to let h be the heght functon on the projectve plane P 2 contanng our ellptc curve E, but as E s a one-dmensonal varety, t s better to work wth P 1, so we wll use the mage of E under the projecton P 2 P 1 defned by (x : y : z) (x : z). To understand how [2] operates on π(e), we recall the formula to double an affne pont P = (x 1 : y 1 : 1) wth y 1 0 computes the x-coordnate of 2P = (x 3 : y 3 : 1) va x 3 = λ 2 2x 1, wth ( ) 3x λ = 1 + a 4 = 9x a 4x a2 4 2y 1 4y 2 = 9x a 4x a2 4 4x a, 4x 1 + 4a 6 where we have used the curve equaton y 2 = x 3 +a 4 x+a 6 to get a formula that only depends on x 1. We then have x 3 = 9x a 4x a2 4 4x a 4x 1 + a 6 2x 1 = x a 4x 2 1 8a 6x 1 + a 2 4 4x a 4x 1 + a 6. Puttng ths n projectve form, we now defne the map φ: P 1 P 1 by φ(x : z) = (x 4 + 2a 4 x 2 z 2 8a 6 xz 3 + a 2 4z 4 : 4x 3 z + 4a 4 xz 3 + a 6 z 4 ). The fact that 4a a2 6 0 ensures that the polynomals defnng φ have no common zero n P 1 (Q), thus φ: P 1 P 1 s a morphsm of degree 4, and Theorem mples that h(φ(p )) = 4h(P ) + O(1).

11 Defnton Let E be an ellptc curve over a number feld k. The canoncal heght ĥ: E( k) R s the functon ĥ = ĥφ π, where ĥφ s the functon gven by Theorem 25.19, wth φ: P 1 P 1 as above and h the absolute heght on P 1. It satsfes ĥ(2p ) = 4ĥ(P ) for all P E(Q). Theorem Let E be an ellptc curve over a number feld k. For any bound B the set {P E(k) : ĥ(p ) B} s fnte. Proof. Ths follows mmedately from Northcott s theorem and Theorem part (). Theorem (Parallelogram Law). Let ĥ be the canoncal heght functon of an ellptc curve E over a number feld k. Then for all P, Q E( k) we have ĥ(p + Q) + ĥ(p Q) = 2ĥ(P ) + 2ĥ(Q) Proof. Ths s a straght-forward but tedous calculaton that we omt; see [2, VIII.6.2] Proof of the Mordell s Theorem Wth all the peces n place we now complete the proof of Mordell s theorem for an ellptc curve E/Q wth a ratonal pont of order 2. Theorem Let E/Q be an ellptc curve wth a ratonal pont of order 2. Then E(Q) s fntely generated. Proof. By the weak Mordell-Wel theorem that we proved n 25.4 for ths case we know that E(Q)/2E(Q) s fnte. So let us choose a bound B such that the set S : = {P E(Q) : ĥ(p ) B} contans a set S 0 of representatves for E(Q)/2E(Q). We clam that S generates E(Q). Suppose for the sake of obtanng a contradcton that ths s not the case. Then there s a pont Q E(Q) S of mnmal heght ĥ(q); the fact that every set of bounded heght s fnte mples that ĥ takes on dscrete values, so such a Q exsts. There s then a pont P S 0 S such that Q = P + 2R for some R E(Q). Snce Q S, we must have R S, so ĥ(r) ĥ(q), by the mnmalty of ĥ(q). By the parallelogram law, 2ĥ(P ) = ĥ(q + P ) + ĥ(q P ) 2ĥ(Q) 0 + ĥ(2r) 2ĥ(Q) = 4ĥ(R) 2ĥ(Q) 2ĥ(Q) So ĥ(q) ĥ(p ) B and therefore Q S, a contradcton. References [1] J-P. Serre, Lectures on the Mordell-Wel theorem, 3rd edton, Sprnger Fachmeden Wesbaden, [2] J. H. Slverman, The arthmetc of ellptc curves, Sprnger, 2009.