Density-zero localizations do not force rank-small integral points
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Definitions and mathematical statement
Fix an integral affine Weierstrass model $U=E\setminus\{O\}$ of an elliptic curve over ${\mathbb{Q}}$. For a set $S$ of rational primes, let $R_S={\mathbb{Z}}[S^{-1}]$ and \[ U(R_S)=\{P\in U({\mathbb{Q}}):x(P),y(P)\in R_S\},\qquad H_S=\langle U(R_S)\rangle\le E({\mathbb{Q}}). \] The source's notation $E(R_S)$ is interpreted as affine coordinate integrality: the Siegel--Mahler finiteness assertion is not about all points on a proper projective model. Write $D(P)$ for the finite set of primes occurring in the denominators of $x(P)$ or $y(P)$, so $P\in U(R_S)$ iff $D(P)\subseteq S$.Question 21, Shlapentokh. For a rank-two elliptic curve, can an infinite prime set $S$ be chosen with $\operatorname{rank}H_S=1$? If $S$ is infinite of density zero, must its affine integral points remain "small"? The source cites Siegel--Mahler finiteness for finite $S$, but finite cardinality of the point set is distinct from rank of the subgroup it generates. "Small" is not given a precise cardinality or height-growth meaning.
Supplied density-zero full-rank construction. For any fixed rank-$r$ curve/model, choose rational points $P_1,\ldots,P_r$ whose classes form a Mordell--Weil free basis, and let $S_0=\bigcup_iD(P_i)$. If $p_j$ is the $j$th prime, put \[ S=S_0\cup\{p_{2^n}:n\ge1\}. \] The aid asserts that $S$ is infinite and recursive and has both relative natural prime density and Dirichlet prime density zero, while $P_i\in U(R_S)$ and hence $\operatorname{rank}H_S=r$. Natural density here means $\#\{p\in S:p\le X\}/\pi(X)\to0$, not merely ordinary integer density of a prime subset. Recursiveness is relative to the fixed finite data; no uniform algorithm for finding a Mordell--Weil basis from an arbitrary curve equation is inferred. This construction concerns generated rank and does not prove that $U(R_S)$ itself is infinite or large by a height-counting criterion.
Supplied fixed-model obstruction. On the model labelled 389a1, \[ y^2+y=x^3+x^2-2x, \] the aid states that $(0,0)$ and $(1,0)$ are independent Mordell--Weil generators. Both are integral and therefore lie in $U(R_S)$ for every $S$. Thus $\operatorname{rank}H_S=2$ for every prime set: no infinite $S$ can give rank one on this model. This refutes a universal assertion for every fixed integral rank-two model; it is not a claim about every alternative model or every refined existence question.
Supplied denominator-support criterion. Let $A=E({\mathbb{Q}})/E({\mathbb{Q}})_{\mathrm{tors}}\cong{\mathbb{Z}}^2$ and choose a primitive rank-one line $L\subset A$. Then $H_S$ has nonzero free rank contained in $L$ exactly when:
- some non-torsion $P$ with class in $L$ has $D(P)\subseteq S$;
- every $Q$ whose class is outside $L$ satisfies $D(Q)\cap(\mathcal P\setminus S)\ne\varnothing$, where $\mathcal P$ is the set of all primes.
Short English statement
Density zero alone does not force small generated Mordell--Weil rank: the aid gives sparse prime sets containing a finite basis-denominator seed. Its explicit rank-two model already has two integral independent points, so no prime set gives generated rank one there. The exact remaining rank-one selection problem is expressed through denominator supports; other meanings of smallness are unspecified.Sources
- Local UnsolvedMath registry, permanent ID 20002358, source code catalogue entry; canonical
aim-logic-notes.json, notes[133]. - AIM 2005 list Extensions of Hilbert's tenth problem, Question 21 (Shlapentokh): https://aimath.org/WWN/hilberts10th/hilberts10th.pdf.
- Siegel--Mahler theorem named in the original question. The aid supplies the model label 389a1, its equation and generator claim, but no database link or separate proving citation. None is invented.
- Registry research attempt 1, dated 2026-08-14: partial result, rigor "proved", novelty confidence low. The explicitly machine-generated, non-peer-reviewed aid requires independent verification before citation.
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