Proved
Rank equality at analytic rank 0 or 1 ↗
Source dated Feb 13, 2020 · Observed Oct 10, 2026
Theorem · version 2
Earlier version 2. View current version 4 →
For E/Q with ord_(s=1)L(E,s) equal to 0 or 1, rank E(Q) equals that analytic rank and Sha(E/Q) is finite.
Gross–Zagier and Kolyvagin, with modularity for all E/Q: rank equality and finite Sha in the low-analytic-rank regime.
Proved
Source dated Feb 13, 2020 · Observed Oct 10, 2026
In this problem · publication v4 · Selected context
Special case · Proved
The hypothesis is analytic rank ≤ 1, not merely algebraic rank ≤ 1. This does not alone prove the full leading-coefficient formula.
Scope: Every E/Q of analytic rank 0 or 1.
Limit: No unrestricted analytic-rank ≥ 2 theorem follows.
Relationship not independently reviewed. This concerns its use in this problem.
In this problem · publication v9
Special case · Proved
The hypothesis is analytic rank ≤ 1, not merely algebraic rank ≤ 1. This does not alone prove the full leading-coefficient formula.
Scope: Every E/Q of analytic rank 0 or 1.
Limit: No unrestricted analytic-rank ≥ 2 theorem follows.
The first starts from analytic rank ≤1; Skinner’s converse starts from algebraic rank one, finite Sha and specified semistable reduction.
Do not reverse the first implication without the additional converse hypotheses.
Proved · version 2
For E/Q with ord_(s=1)L(E,s) equal to 0 or 1, rank E(Q) equals that analytic rank and Sha(E/Q) is finite.
Scope: Gross–Zagier and Kolyvagin, with modularity for all E/Q: rank equality and finite Sha in the low-analytic-rank regime.
Proved · version 2
For semistable E/Q with nonsplit multiplicative reduction at an odd prime or split multiplicative reduction at two odd primes, algebraic rank 1 and finite Sha imply analytic rank 1.
Scope: Skinner’s converse in the stated semistable family; a proved implication with a finiteness assumption.
Relationship not independently reviewed.
Connection in Birch and Swinnerton-Dyer conjecture →Relationship not independently reviewed. This concerns its use in this problem.
For E/Q with ord_(s=1)L(E,s) equal to 0 or 1, rank E(Q) equals that analytic rank and Sha(E/Q) is finite.
paraphrase · No upstream full text or code is republished by this record.
For semistable E/Q with nonsplit multiplicative reduction at an odd prime or split multiplicative reduction at two odd primes, algebraic rank 1 and finite Sha imply analytic rank 1.
paraphrase · No upstream full text or code is republished by this record.
Christopher Skinner · Published
Annals of Mathematics 191 (2020), 329–354; published online 13 February 2020. Introduction also states the classical low-analytic-rank theorem.
Source revision: published-online-2020-02-13. Observed Oct 10, 2026.
The cited work does not report AI involvement; this is not an assertion that no AI was used.
Source updated Feb 13, 2020.
Wiles distinguishes the rank statement from the refined leading-coefficient formula and records the analytic-rank-zero-or-one theorem.
paraphrase · No upstream full text or code is republished by this record.
Andrew Wiles · Published
Official Clay problem description; separates rank equality from its refined formula.
Source revision: official-description. Observed Oct 10, 2026.
The cited work does not report AI involvement; this is not an assertion that no AI was used.
Official description PDF: official-description. External evidence linked; no OpenMath execution or independent reproduction is asserted.
Statement versions and problem publication versions are separate. A new publication can update context or sources while retaining the same statement.
Assessments retained in these publication records. Earlier statement versions retain their own source histories.
Proved
Source dated Feb 13, 2020 · Observed Oct 10, 2026 · Source passage →