How to calculate the rank of elliptic curve?

Question

Consider Mordell curve $E:y^2=x^3+2089$,prove that $E(\mathbb{Q})\cong \mathbb{Z}^4$and $P_1=(-12,19),P_2=(-10,33),P_3=(-4,45),P_4=(3,46)$ generated.

We can prove that $E(\mathbb{Q})=\{\mathcal{O}\}$. If $P\in E(\mathbb{Q})$ is a torsion, then $P$ is an integral point and $y(P)=0$ or $y(P)^2|\Delta_E$ where $\Delta_E=4A^3+27B^2=117825867=3^3\times 2089^2$,follow $y(P)^2\equiv 0\ (\text{mod}\ 3^3\times 2089^2)$.It's obviously impossible.

My questions:

(1)How to prove linear independence between $P_1=(-12,19),P_2=(-10,33),P_3=(-4,45),P_4=(3,46)$？

(2)How to prove that any $P\in E(\mathbb{Q})$ can be written in the form of $P=[a]P_1+[b]P_2+[c]P_3+[d]P_4$ where $a,b,c,d\in \mathbb{Z}$ ? Or is there any other method to calculate the rank of $E(\mathbb{Q})$? Thank you!

Answer 1

Consider $e$ to be a (symbolic) root of the equation $X^3 + 2089=0$. It is useful to work over $L:=L_e=\Bbb Q(e):=\Bbb Q[X]/(X^3 + 2089)$, since we have one torsion point over $L$, and one root of the polynomial $X^3 + 2089$ in $L$, and the following map $\phi=\phi_e$ has codomain related to $L$. This map is a map between abelian groups: $$ \phi=\phi_e\ : \ (E(\Bbb Q),+)\ \longrightarrow\ (\underbrace{L^\times\text{ modulo squares}}_{:=\Gamma_e}\ , \ \cdot\ )\ , \\ P=(a,b)\longrightarrow (a-e) \text{ modulo squares .} $$ Do the same for the other two (symbolic) roots $e'$, $e''$ of the equation $X^3+2089$. We obtain fields $L'=L_{e'}$, $L''=L_{e''}$, corresponding maps $\phi',\phi''$ from $(E(\Bbb Q),+)$ to corresponding multiplicative groups modulo squares $\Gamma',\Gamma''$. Finally define the global Kummer map $$ \Phi=(\phi,\phi',\phi''):(E(\Bbb Q),+)\longrightarrow \Gamma\times\Gamma'\times\Gamma''\ . $$ Let $D$ be the discriminant of the polynomial $X^3+2089$, and let $\Delta$ be the elliptic curve discriminant. We have: $$ D=(e-e')^2(e-e'')^2(e'-e'')^2=-3^3\cdot 2089^2=\Delta/16\ . $$ (Often the roots appear in notation as $(e_1,e_2,e_3)$, then $\Phi$ is $(\phi_1, \phi_2,\phi_3)$. Sometimes, $\delta =(\delta_1,\delta_2,\delta_3)$ is used instead.)

For the convenience of the reader, i am reproducing in the above notation the Proposition 3.2.1 from

• [1] the Elliptic curve handbook, Jan Connell, 1999 (draft), 327 pages,

the field $K$ from loc. cit. is in our case $\Bbb Q$, an $F2$-Krull domain:

(a) $\phi$, $\phi'$, $\phi''$ are group homomorhism, so $\Phi$ is a group homomorphism, too.

(b) The kernel of $\Phi$ is $2 E(\Bbb Q)$, the image of the multiplication-by-two map $[2]:E(\Bbb Q)\to E(\Bbb Q)$. Moreover, since $e,e',e''\not\in\Bbb Q$ the kernel of each component $\phi$, $\phi'$, $\phi''$ of $\Phi$ is also $2E(\Bbb Q)$.

(c) The image of $\Phi$ is finite.

Hence the group $E(\Bbb Q)/2E(\Bbb Q)$ is finite.

• See also [2] Computing the Rank of Elliptic Curves over Number Fields, Denis Simon, London Mathematical Society, received March 2000, published April 2002. In section 1.2 the image of $\phi$ (in our case of an irreducible polynomial $X^3+2089$) is shown to be inside an explicit set $L(S,2)$ of $\Gamma$, which is the set of all classes $\bar \lambda \in \Gamma$ coming from a $\lambda\in L$ which has even valuation for all places outside the set $S$ of places. And $S$ is a set of "bad places" $\mathfrak P$ of $L$ (over $2,3,2089,\infty$), see loc. cit. for details.

We have (Proposition 1.3 from [2]): $$ L(S,2)\cong \frac {U_{L,S}}{U_{L,S}^2}\times \operatorname{Cl}_S(L)[2] \ . $$ Notations are as in [2]. In our case, the unit group of $L$ has the structure $\Bbb Z/2\times \Bbb Z$, the two-torsion part is generated by $-1$, the free part is generated by some unit, computer aid...

sage: L.<a> = NumberField(x^3 + 2089)
sage: for u in LU.gens():
....:     print(f'Order {u.order()} for {L(u)}')
....: 
Order 2 for -1
Order +Infinity for 55567439267004*a^2 + 2115473969379288*a + 17962374576937249

The class number of $L$ is $12$. With an empty set as $S$, sage delivers...

sage: L.selmer_generators([], 2)
[-1,
 55567439267004*a^2 + 2115473969379288*a + 17962374576937249,
 205018041800/3*a^2 - 2620826236715/3*a + 33503052232613/3,
 -a^2 + 25*a + 483]

Adding as first argument some prime ideals in the factorization of some further relevant primes, like $(2)$ and $(3)$,

sage: factor(L.ideal(2))
(Fractional ideal (2, 1/3*a^2 - 1/3*a + 1/3)) * (Fractional ideal (2, a + 1))
sage: factor(L.ideal(3))
(Fractional ideal (3, 1/3*a^2 - 1/3*a + 1/3))^2 * (Fractional ideal (3, 1/3*a^2 - 1/3*a - 2/3))

lead to further Selmer generators, for instance

sage: L.selmer_generators( [L.ideal(3, 1/3*a^2 - 1/3*a - 2/3), L.ideal(2, a+1)], 2)
[277*a^2 - 3541*a + 45266,
 -53356/3*a^2 - 826367/3*a - 1844596/3,
 -1,
 55567439267004*a^2 + 2115473969379288*a + 17962374576937249,
 -2230617685/3*a^2 + 28514862895/3*a - 364516703776/3]

Such generators appear below in a full list for $L(S,2)$.

Also, $L(S,2)$ is in the kernel of the norm map of the extension $L:\Bbb Q$ (modulo squares).

This all is restricting the explicit structure of $E(\Bbb Q)/2E(\Bbb Q)$, and of the corresponding representatives, the search for them is algorithmic.

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With the above information, the rank computation of the given elliptic curve $E$ over $\Bbb Q$ has the following steps:

• compute the "bad places" $S$,
• compute the units of the field $L=\Bbb Q(e)$, in our case, we have two generators to consider, consider their classes in $L(S,2)$,
• compute non-unit elements of $L$, so that adding their classes we have a complete set of representatives for $L(S,2)$,
• restrict to the elements that locally lift,
• search for global lifts, this may fail if the search is exploding through the given practical limits,
• decide which is the two-torsion of $E(\Bbb Q)$,
• decide which is the rank of the two-Selmer-group,
• decide which is the rank of the two-Sha-group, thus knowing the rank of $E$
• use the global lifts, and get better (reduced) generators (having smaller heights).

---

It is time now to compare the above with the verbose protocol during a Simon-$2$-descent for the given curve (output was manually rearrange to fit in the width of the MSE page):

sage: E = EllipticCurve([0, 2089])
sage: E.simon_two_descent(verbose=3)
ellQ_ellrank([0, 0, 0, 0, 2089], [[-12, 19], [-10, 33], [-4, 45], [3, 46], [8, 51]]);
   starting ellQ_ellrank
 Elliptic curve: Y^2 = x^3 + 2089
 E[2] = [[0]]
   bnfinit(x^3 + 2089) done
  A = 0
  B = 0
  C = 2089

  Computing L(S,2)
   Constructing the field Selmer group: L(S,2)
   #LS2gen = 6
  L(S,2) = [Mod(-1, x^3 + 2089), 
            Mod(55567439267004*x^2 + 2115473969379288*x + 17962374576937249, x^3 + 2089), 
            Mod(277*x^2 - 3541*x + 45266, x^3 + 2089), 
            Mod(34528/3*x^2 - 441385/3*x + 5642398/3, x^3 + 2089), 
            Mod(3, x^3 + 2089), 
            Mod(894849*x^2 + 22995150*x + 147724168, x^3 + 2089)]

  Computing the Selmer group
   Computing the kernel of the norm map
   selmer = [0, 0, 0, 0; 1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 1, 0; 0, 0, 1, 0; 0, 0, 0, 1]
   p-adic points
  badprimes = [2, 3]~
  selmer = [0, 0, 0, 0; 1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 0, 1; 0, 0, 0, 1; 0, 0, 1, 0]
  Selmer rank = 4

  Search for trivial points on the curve
   trivial solution: lead(pol) is a square
   point found by ratpoint = [3, 46]
  Trivial points on the curve = [[-12, 19], [-10, 33], [-4, 45], [3, 46],
                                 [8, 51], [0], [3, 46]]

  Run through the Selmer group

  zc = Mod(55567439267004*x^2 + 2115473969379288*x + 17962374576937249, x^3 + 2089)
  does not come from a trivial point
  quartic: 1*Y^2 = 16*x^4 - 136*x^3 + 456*x^2 - 800*x + 617
   reduction of the quartic 16*x^4 - 136*x^3 + 456*x^2 - 800*x + 617
   reduced quartic = 16*x^4 - 8*x^3 + 24*x^2 - 96*x + 9
  reduced: 1*Y^2 = 16*x^4 - 8*x^3 + 24*x^2 - 96*x + 9
   trivial solution: lead(pol) is a square
  point on the reduced quartic = [1, 4, 0]
  point on the quartic = [1, 0, 4]~
   reconstruction of the point on the curve
 point on the curve = [-15/4, 361/8]

  zc = Mod(277*x^2 - 3541*x + 45266, x^3 + 2089)
  does not come from a trivial point
  quartic: 1*Y^2 = 9*x^4 - 20*x^3 - 138*x^2 - 204*x - 63
   reduction of the quartic 9*x^4 - 20*x^3 - 138*x^2 - 204*x - 63
   reduced quartic = 9*x^4 - 56*x^3 - 24*x^2 - 24*x + 32
  reduced: 1*Y^2 = 9*x^4 - 56*x^3 - 24*x^2 - 24*x + 32
   trivial solution: lead(pol) is a square
  point on the reduced quartic = [1, 3, 0]
  point on the quartic = [1, 0, 3]~
   reconstruction of the point on the curve
 point on the curve = [232/9, 3743/27]

  zc = Mod(894849*x^2 + 22995150*x + 147724168, x^3 + 2089)
  does not come from a trivial point
  quartic: 1*Y^2 = 212*x^4 - 504*x^3 + 516*x^2 - 272*x + 57
   reduction of the quartic 212*x^4 - 504*x^3 + 516*x^2 - 272*x + 57
   reduced quartic = 9*x^4 - 24*x^3 - 84*x^2 - 104*x + 4
  reduced: 1*Y^2 = 9*x^4 - 24*x^3 - 84*x^2 - 104*x + 4
   trivial solution: lead(pol) is a square
  point on the reduced quartic = [1, 3, 0]
  point on the quartic = [-1, -1, 3]~
   reconstruction of the point on the curve
 point on the curve = [18, 89]

  zc = Mod(34528*x^2 - 441385*x + 5642398, x^3 + 2089)
  comes from the trivial point [-10, 33]

  rank of found points     = 4
  rank of the Selmer group = 4
#E(Q)[2]      = 1
#S(E/Q)[2]    = 16
#E(Q)/2E(Q)   = 16
#III(E/Q)[2]  = 1
rank(E/Q)     = 4
listpoints = [[-15/4, 361/8], [232/9, 3743/27], [18, 89], [-10, 33]]
   Reduction of the generators [[-10, 33], [-15/4, 361/8], [18, 89], [232/9, 3743/27]]
   infinite order points = [[-10, 33], [-15/4, 361/8], [18, 89], [232/9, 3743/27]]
   infinite order points = [[-4, -45], [-10, 33], [8, -51], [-12, -19]]
   computing the 2-torsion
   E[2] = [1, [], []]
   infinite order points = [[-4, -45], [-10, 33], [8, -51], [-12, -19]]
  reduced generators = [[-4, 45], [-10, 33], [8, 51], [-12, 19]]
   end of ellQ_ellrank
v = [4, 4, [[-4, 45], [-10, 33], [8, 51], [-12, 19]]]
Rank determined successfully, saturating...
(4, 4, [(-4 : 45 : 1), (-10 : 33 : 1), (8 : 51 : 1), (-12 : 19 : 1)])

(Please scroll down.)

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Conclusion:

This answer is trying to go into details, thus making the hands "dirty" (and showing how the computational part looks like).

(1) To check the independence is enough to show that $\phi(P_1)=(-12-e)$, $\phi(P_2)=(-10-e)$, $\phi(P_3)=(-4-e)$, $\phi(P_4)=(3-e)$ are independent in the group $\Gamma$ of elements in $L^\times$ modulo squares. For this, one has to see the elements in $L(S,2)$ first, then make explicit the isomorphism to (units modulo squares) times (class group modulo two-torsion).

Alternatively, compute the height pairings for the given points, we obtain a $4\times 4$ matrix, if there would be a linear relation between them, it would lead to a degenerated matrix. In our case:

sage: P1, P2, P3, P4 = E((-12, 19)), E((-10, 33)), E((-4, 45)), E((3, 46))

sage: E.height_pairing_matrix([P1, P2, P3, P4])
[  2.62754246798053 -0.430822033802014  0.584794842483018  0.148063740654735]
[-0.430822033802014   2.05889278037727 -0.591083786458631  -1.50201151893930]
[ 0.584794842483018 -0.591083786458631   2.01143112262342  0.430246915129471]
[ 0.148063740654735  -1.50201151893930  0.430246915129471   3.01975805440168]

sage: E.height_pairing_matrix([P1, P2, P3, P4]).det()
17.5653394265919

With given precision in the computation of the height pairs, we are far away from a zero determinant.

(2) To see that the given points generate, a good way is to follow the above two-descent protocol. (Using specific isogenies is also a possibility.) Sooner or later it is hard to avoid solving local lifting problems as a first approximations, than trying to get a global lift when all local lifts do not prohibit it. This step is a priori a problem captured by the Sha-group obstruction, the one which is making the whole procedure analgorithmic so far. In our case we are lucky, each locally liftable Selmer group element is globally liftable. (The final step of getting smaller heights is of cosmetical nature, but it gives the generators from the post.)