2018
DOI: 10.5802/aif.3179
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On the local-global divisibility over abelian varieties

Abstract: Let p ≥ 2 be a prime number and let k be a number field. Let A be an abelian variety defined over k. We prove that if Gal(k(A[p])/k) contains an element g of order dividing p − 1 not fixing any non-is trivial, then the local-global divisibility by p n holds for A(k) for every n ∈ N. Moreover, we prove a similar result without the hypothesis on the triviality of H 1 (Gal(k(A[p])/k), A[p]), in the particular case where A is a principally polarized abelian variety. Then, we get a more precise result in the case w… Show more

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Cited by 12 publications
(19 citation statements)
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“…In the following proposition we put together the main results of [PRV1] and [PRV2] and we use some results of [GR2].…”
Section: Known Resultsmentioning
confidence: 99%
See 1 more Smart Citation
“…In the following proposition we put together the main results of [PRV1] and [PRV2] and we use some results of [GR2].…”
Section: Known Resultsmentioning
confidence: 99%
“…Many mathematicians got criterions for the validity of the local-global divisibility principle for many families of commutative algebraic groups, as algebraic tori ( [DZ1] and [Ill]), elliptic curves ( [Cre1], [Cre2], [DZ1], [DZ2], [DZ3], [GR1], [Pal1], [Pal2], [PRV1], [PRV2]), and very recently polarized abelian surfaces ( [GR2]) and GL 2 -type varieties ( [GR3]).…”
Section: Introductionmentioning
confidence: 99%
“…Proof In [, Section 5], it was proved that H cyc 1false(G2,false(double-struckZ/p2double-struckZfalse)2false)0. Let N:=ker(π) be the subgroup of matrices congruent to identity modulo p2. By construction, we have an exact sequence 12emN2emG32emG22em1. Let M3:=(Z/p3Z)2 and M2=M3false[p2false].…”
Section: Non‐direct Summandsmentioning
confidence: 99%
“…This implies that H1false(G2,M2false)=H cyc 1false(G2,M2false), hence one can deduce from the previous discussion, combined with the fact that the inflation sends H cyc 1false(G2,M2false) to H cyc 1false(G3,M3false), that H cyc 1false(G2,false(double-struckZ/p2double-struckZfalse)2false)=H cyc 1false(G3,false(double-struckZ/p3double-struckZfalse)2false). Let g:=1312,which has order 3. Because gH2=H2g (see [, Section 5]), H2 is a normal subgroup of G2, and we have an exact sequence 12emH22emG22emfalse⟨gfalse⟩2em1. As p3, given a cyclic subgroup ΓG2, its p‐Sylow subgroup is a subgroup of H2, hence of the form h…”
Section: Non‐direct Summandsmentioning
confidence: 99%
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