Unbounded Ranks of Elliptic Curves over Q

Unbounded Ranks of Elliptic Curves over Q

Definitions and mathematical statement

For an elliptic curve over ${\mathbb{Q}}$, write $E({\mathbb{Q}})\simeq{\mathbb{Z}}^{r(E)}\oplus T$ with $T$ finite. The target is \[ \forall r\in{\mathbb{Z}}_{\ge0}\ \exists E/{\mathbb{Q}}:\qquad r(E)\ge r, \] or $\sup_{E/{\mathbb{Q}}}r(E)=\infty$. The rank is that of the actual rational-point group. Analytic ranks, Selmer upper bounds, and ranks after enlarging the base field are not interchangeable with this integer. No algorithm producing the curves, or prescribed rate of growth in their heights, is required.

Formulation and concept references: [S1].

Short English statement

Can elliptic curves over the rationals have arbitrarily many independent rational points of infinite order?

Research attempt

Current frontier.

The source paper develops searches for high-rank curves with specified isogenies; it does not establish unboundedness. [S1] The record table maintained by Dujella currently lists a 2026 example of Alpöge--Howell with rank at least $31$, together with rational point data. That is a reported finite rank certificate, not an assertion that the rank is exactly $31$; its full independence computation has not been audited in this notebook. [E2]

There is also substantive evidence pointing the other way. The Park--Poonen--Voight--Wood model predicts only finitely many curves of rank greater than $21$, while explicitly warning that special arithmetic families could depart from the model. This is a heuristic about actual curves, not a theorem bounding their ranks. A finite example of rank $29$ or $31$ does not contradict a prediction of finitely many exceptional curves. [E1] The attempt below therefore tests both a construction mechanism and the obstruction to descending it, without treating either expectation as an established answer.

Attack route.

One could attempt arbitrarily high generic rank over ${\mathbb{Q}}(t)$ and then specialize, but the arbitrarily-high-generic-rank input would itself be a major unresolved step. A second route creates points in independent quadratic extensions and tries to descend them; this has an accessible construction but an exact Galois obstruction. A third route searches for many points in one square class and certifies their actual independence by descent maps. I carry out the second route completely in its enlarged field, identify why it does not answer the target, and then implement the third route over ${\mathbb{Q}}$ with exact certificates.

Attempt.

1. Arbitrarily many independent points, but in the wrong field. Fix \[ E_0:\quad y^2=x^3-x, \qquad f(x)=x^3-x. \] For any prescribed $r\geq1$, choose distinct primes $q_1,\ldots,q_r\geq5$. For each $i$, the Chinese remainder theorem provides an integer $x_i\geq2$ such that \begin{equation}\label{wsref-0-33-30-30-30-36-37-33-31-3a-63-72-74} x_i\equiv q_i\pmod{q_i^2},\qquad x_i\equiv2\pmod{q_j}\quad(j\neq i). \end{equation} Put $d_i=f(x_i)>0$. Because $f'(0)=-1$ and, more directly, $f(q_i)\equiv-q_i\pmod{q_i^2}$, \[ v_{q_i}(d_i)=1,\qquad v_{q_j}(d_i)=0\quad(j\neq i). \] The second assertion uses $f(2)=6$ and $q_j\geq5$. Thus the classes of $d_1,\ldots,d_r$ are independent in ${\mathbb{Q}}^\times/{\mathbb{Q}}^{\times2}$: applying $v_{q_j}$ modulo two to a square relation detects its $j$th exponent. Consequently \[ K_r={\mathbb{Q}}(\sqrt{d_1},\ldots,\sqrt{d_r}) \] is multiquadratic of degree $2^r$, with an automorphism independently changing each square-root sign. Let $P_i=(x_i,\sqrt{d_i})\in E_0(K_r)$.

2. Nontorsion without an unverified numerical height calculation. The following elementary height argument applies to each $P_i$. For a point $P$ on $E_0$ whose $x$-coordinate is rational, the duplication formula is \begin{equation}\label{wsref-0-33-30-30-30-36-37-33-31-3a-64-6f-75-62-6c-65} x(2P)=\frac{(x(P)^2+1)^2}{4x(P)(x(P)^2-1)}. \end{equation} Write $x(P)=a/b$ in lowest terms and put $H(x)=\max(|a|,|b|)$. The numerator and denominator before cancellation in (30006731:double) are \[ (a^2+b^2)^2,\qquad 4ab(a^2-b^2). \] Their greatest common divisor divides $4$. An odd common prime could divide neither $a$ nor $b$, and would divide both $a^2+b^2$ and $a^2-b^2$, which is impossible. If $a,b$ are both odd, the numerator has $2$-adic valuation exactly $2$; if they have opposite parity, it is odd. Hence, for $H(x)\geq2$, \begin{equation}\label{wsref-0-33-30-30-30-36-37-33-31-3a-68-65-69-67-68-74} H(x(2P))\geq\frac{H(x(P))^4}{4}. \end{equation} The exceptional denominator values $x=0,1,-1$ have height one, so they cannot interrupt this iteration. With $h=\log H$ it follows that \[ h(x(2^nP))\geq 4^n\left(h(x(P))-\frac{\log4}{3}\right)+\frac{\log4}{3}. \] For $x(P)=x_i\geq2$, the coefficient of $4^n$ is positive. The iterated $x$-coordinates therefore have unbounded height, proving that $P_i$ is nontorsion. This argument is valid in the quadratic field of $P_i$: only its $x$-coordinate, and those of its doubles, are required to be rational.

Now suppose $\sum_i n_iP_i=O$ with integers $n_i$. Apply the automorphism that changes only the sign of $\sqrt{d_j}$ and subtract the original relation. This gives $2n_jP_j=O$. Since $P_j$ is nontorsion, $n_j=0$. Thus all $P_i$ are independent, and \begin{equation}\label{wsref-0-33-30-30-30-36-37-33-31-3a-77-72-6f-6e-67-66-69-65-6c-64} \operatorname{rank}E_0(K_r)\geq r. \end{equation} Every assertion up to (30006731:wrongfield) has been proved; the field is explicitly $K_r$, not ${\mathbb{Q}}$.

3. Why the apparent descent gives zero. For the sign character $\chi_i$ belonging to $d_i$, one has $\sigma P_i=\chi_i(\sigma)P_i$. In particular \[ \operatorname{Tr}_{K_r/{\mathbb{Q}}}(P_i) =\sum_{\sigma\in\operatorname{Gal}(K_r/{\mathbb{Q}})}\sigma P_i=O. \] The usual rational decomposition by characters also explains the failure: after tensoring the point group with ${\mathbb{Q}}$, different characters give direct summands, each identified with the rational point space on the corresponding quadratic twist. The constructed points occupy different summands. One point on each of many twists is not many independent points on one curve. Averaging the ranks over the $2^r$ characters gives no unbounded lower bound for one summand.

To make this concrete, write $d_i=e_i z_i^2$ with $e_i$ squarefree. The point $P_i$ corresponds to the rational point \[ (e_ix_i,e_i^2z_i)\quad\text{on}\quad Y^2=X^3-e_i^2X. \] The independent square classes force the $e_i$ to be different. Making all $e_i$ equal would remove the very sign-changing automorphisms used to prove independence. That is the exact missing bridge in this construction.

4. A same-twist search and a rank certificate using actual points. For a fixed squarefree $d>0$, put \[ E_d:\quad Y^2=X(X-d)(X+d). \] If $x\in{\mathbb{Q}}$ and $x^3-x=dz^2$ for $z\in{\mathbb{Q}}\setminus\{0\}$, then \begin{equation}\label{wsref-0-33-30-30-30-36-37-33-31-3a-70-6f-69-6e-74} (X,Y)=(dx,d^2z)\in E_d({\mathbb{Q}}). \end{equation} For $x=a/b>1$ in lowest terms, \[ x^3-x=\frac{a(a-b)(a+b)}{b^3} =\frac{ab(a-b)(a+b)}{b^4}. \] Thus the squarefree part of $ab(a-b)(a+b)$ identifies the twist. This makes it possible to group rational inputs by $d$ without mistaking an integer coordinate bound for an exhaustive search over rational points.

Here is the exact independence test. For $P=(X,Y)$ not a point of order two, set \[ \kappa(P)=([X],[X-d],[X+d]) \in({\mathbb{Q}}^\times/{\mathbb{Q}}^{\times2})^3, \] where brackets denote square classes, written multiplicatively. Put $\kappa(O)=(1,1,1)$, and for a root $e\in\{0,d,-d\}$ replace the vanishing coordinate of $\kappa((e,0))$ by $\prod_{e'\neq e}(e-e')$; retain the nonvanishing coordinates. In particular, \begin{equation}\label{wsref-0-33-30-30-30-36-37-33-31-3a-74-6f-72-73-69-6f-6e} \kappa(T_0)=([-1],[-d],[d]),\qquad \kappa(T_d)=([d],[2],[2d]), \end{equation} where $T_0=(0,0)$ and $T_d=(d,0)$.

Why this is a homomorphism. A nonvertical rational line meeting the cubic at $P,Q,R$ has $P+Q+R=O$. Substitute its equation into the cubic and evaluate the resulting monic cubic polynomial at a root $e$. If none of the three points is $(e,0)$, this gives \[ (X(P)-e)(X(Q)-e)(X(R)-e)=(\text{rational number})^2. \] If one point is $(e,0)$, differentiate that polynomial at $e$ instead. The product of the other two $X-e$ factors is $\prod_{e'\neq e}(e-e')$, which gives the same square-class identity using the prescribed replacement. Tangencies at non-two-torsion points are handled with intersection multiplicity. A vertical line pairs a point with its negative and gives the identity because their $X$-coordinates agree; doubling a point of order two is the same case. These cases cover the group law. Thus $\kappa(P+Q)=\kappa(P)\kappa(Q)$, and $2E_d({\mathbb{Q}})$ is in its kernel.

Since $E_d({\mathbb{Q}})\simeq{\mathbb{Z}}^r\oplus T$ and $E_d[2]({\mathbb{Q}})$ has dimension two over $\mathbb F_2$, \[ \dim_{\mathbb F_2}\left(E_d({\mathbb{Q}})/2E_d({\mathbb{Q}})\right)=r+2. \] Consequently a matrix of square-class columns obtained from actual points, of binary rank $R$, proves \begin{equation}\label{wsref-0-33-30-30-30-36-37-33-31-3a-72-61-6e-6b-63-65-72-74} \operatorname{rank}E_d({\mathbb{Q}})\geq R-2. \end{equation} No injectivity theorem for $\kappa$ is needed for this lower bound. It is not a Selmer upper bound, an analytic-rank guess, or a conditional BSD argument.

5. A fully checkable rank-at-least-four example. The rational search $1\leq b<a\leq600$, $\gcd(a,b)=1$, finds the squarefree integer \[ d=29274=2\cdot3\cdot7\cdot17\cdot41 \] and the following four points on $E_d$:

Call them $P_1,\ldots,P_4$ in this order. Direct substitution verifies their curve equations. Representatives for their square-class triples are \[ (287,246,42),\quad (123,1394,102),\quad (21,123,287),\quad (14637,123,119). \] Use columns $T_0,T_d,P_1,P_2,P_3,P_4$. Select the following six binary coordinates: the sign of the first square class, its valuations at $2,3,7,17$, and the valuation at $17$ of the second square class. The resulting matrix is \begin{equation}\label{wsref-0-33-30-30-30-36-37-33-31-3a-6d-61-74-72-69-78} \begin{pmatrix} 1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&1&0&1&1&1\\ 0&1&1&0&1&1\\ 0&1&0&0&0&1\\ 1&0&0&1&0&0 \end{pmatrix}. \end{equation} Its integer determinant is $-1$, so its rank over $\mathbb F_2$ is six. Equation (30006731:rankcert) proves that this curve has actual rational rank at least four. The two torsion columns are independent. Since $T/2T$ has dimension two, they span the entire image of the rational torsion subgroup under $\kappa$. The four point columns are independent modulo that span. In the free quotient $E_d({\mathbb{Q}})/T$, divide a hypothetical nonzero integer relation by the largest common power of two in its coefficients. Lifting back gives a relation modulo torsion with an odd coefficient, contradicting those four independent square-class columns. Thus the four points are independent modulo torsion as well. No claim is made that the curve or these points are new, or that its exact rank has been determined here.

The earlier integer-only search $2\leq x\leq30000$ produced certificates of rank at least two, for example $d=210$ and $x=6,15$. Allowing rational inputs was therefore a genuine enlargement of the experiment rather than a change in the target. The companion script reproduces both searches and verifies (30006731:matrix) using exact arithmetic.

Stress test.

The enlarged-field construction proves its own nontorsion and independence claims, but its trace calculation explicitly blocks the attempted descent. The same-twist search fixes this field issue, at the cost of losing the Galois-sign independence argument. Distinct points on a curve can be multiples of one point; bucket size alone is therefore not a rank bound. The matrix certificate, rather than the number of hits, is what proves the finite rank statement.

The asymptotic missing statement in the restricted search is:

For every $r$ there is a squarefree $d>0$ and a finite collection of rational solutions to $x^3-x=dz^2$ whose square-class columns, together with the rational two-torsion columns, have binary rank at least $r+2$.

This would imply the catalogue target, but it is stronger because it restricts to the fixed-$j$ family $j=1728$ and imposes a particular certificate mechanism. It is not known from the calculations. Even a proof that some ranks elsewhere are unbounded would not automatically show that this particular restricted family suffices. Conversely, failure of this route would not refute unboundedness among all elliptic curves.

The opposing boundedness model is not used as a theorem at any point. Neither small average rank nor finitely many computational records decides whether arbitrarily sparse high-rank exceptional curves exist. The model's caution about special families is relevant precisely to that distinction. [E1]

Outcome.

Outcome: Exploratory approach with unresolved gap.

This attempt establishes an elementary independent-point construction over explicit multiquadratic fields, proves exactly why its points do not descend as a high-rank rational family, and then obtains an unconditional, exact rank-at-least-four certificate for one rational curve by a reproducible same-twist search. These are checks of the proposed mechanism, not a new rank record or a proof of unbounded rational rank. The missing step is unbounded certificate rank on one rational curve at a time, or a genuinely different construction not subject to the same-square-class obstruction.

Sources

• S1 Harris B. Daniels and Hannah Goodwillie, On the ranks of elliptic curves with isogenies, arXiv1611.01329v2,11December2024,10pages. https://arxiv.org/pdf/1611.01329v2. Catalogue formulation source.
• E1 J. Park, B. Poonen, J. Voight, and M. Matchett Wood, A heuristic for boundedness of ranks of elliptic curves, arXiv:1602.01431v3 (10 July 2018), Section 1.1, Remark 1.1.2, and Section 8.2. https://arxiv.org/html/1602.01431v3. Consulted 27 September 2026; model theorems are distinguished from predictions for actual curves.
• E2 A. Dujella, History of elliptic curves rank records, and its Rank $\geq31$ data page, crediting Alpöge--Howell $2026$. https://web.math.pmf.unizg.hr/~duje/tors/rankhist.html; https://web.math.pmf.unizg.hr/~duje/tors/rk31.html. Consulted 27 September 2026. The record page and point list were inspected; their complete independence proof was not independently audited.