2018
DOI: 10.1093/imrn/rny274
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No Singular Modulus Is a Unit

Abstract: A result of the second-named author states that there are only finitely many CM-elliptic curves over C whose j-invariant is an algebraic unit. His proof depends on Duke's Equidistribution Theorem and is hence noneffective. In this article, we give a completely effective proof of this result. To be precise, we show that every singular modulus that is an algebraic unit is associated with a CM-elliptic curve whose endomorphism ring has discriminant less than 10 15 . Through further refinements and computerassiste… Show more

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Cited by 14 publications
(42 citation statements)
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“…This was an improvement over an earlier result for all (ineffectively) large discriminants in [Hab15]. By specializing Theorem 1.1 to m = 1 and d 1 = −3, we recover the main result of [BHK20]. By allowing both CM points to vary, we actually have the following more general result.…”
Section: Singular Units and Isogenies Between CM Elliptic Curvesmentioning
confidence: 50%
See 3 more Smart Citations
“…This was an improvement over an earlier result for all (ineffectively) large discriminants in [Hab15]. By specializing Theorem 1.1 to m = 1 and d 1 = −3, we recover the main result of [BHK20]. By allowing both CM points to vary, we actually have the following more general result.…”
Section: Singular Units and Isogenies Between CM Elliptic Curvesmentioning
confidence: 50%
“…Remark 1.4. When one of the discriminants is fixed, the proof in [BHK20] can be adapted to prove the result above. However, this involves eliminating finitely many cases by computer calculation, and it is not clear if the same strategy works with both discriminants varying.…”
Section: Singular Units and Isogenies Between CM Elliptic Curvesmentioning
confidence: 99%
See 2 more Smart Citations
“…Indeed, assume the contrary: k 1 ≥ 3 and k 1 + k 2 ≥ 5. Writing ∆ = −p 2 1 m, Proposition 3.3 implies that min{p 2 1 , (m + (p 1 − 2) 2 )/4} is suitable for ∆. However, since k 1 + k 2 ≥ 5, we have m > 4p 2 1 , which implies that the minimum is, actually, p 2 1 .…”
Section: We Have K 1 ≤mentioning
confidence: 99%