Lifting all elements in $\mathrm{SL}_{n}(\mathbb{Z}/q\mathbb{Z})$

We show that every element of \mathrm{SL}_{n}(\mathbb{Z}/q\mathbb{Z}) can be lifted to an element of \mathrm{SL}_{n}(\mathbb{Z}) of norm at most Cq^{2}\log q , while there exists an element such that every lift of it is of norm at least q^{2+o(1)} . This should be compared to the recent result that almost every element has a lift of norm bounded by q^{1+1/n+o(1)} . The main step in the proof is showing that for every q , there is a small element in (\mathbb{Z}/q\mathbb{Z})^{\times} with a large n -th root, which is a result of independent interest. In the proof we use tools from additive combinatorics including Bohr sets.

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Publication Details

Journal
Journal of the European Mathematical Society
Published
2026-05-11
DOI
https://doi.org/10.4171/jems/1790
Primary Topic
Analytic Number Theory Research
Type
preprint
Field-Weighted Citation Impact
0.00

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preprint

Lifting all elements in $\mathrm{SL}_{n}(\mathbb{Z}/q\mathbb{Z})$

Amitay Kamber, Péter P. Varjú
Journal of the European Mathematical Society
Analytic Number Theory Research
preprint

Lifting all elements in $\mathrm{SL}_{n}(\mathbb{Z}/q\mathbb{Z})$

Amitay Kamber, Péter P. Varjú
preprint en

Abstract

We show that every element of \mathrm{SL}_{n}(\mathbb{Z}/q\mathbb{Z}) can be lifted to an element of \mathrm{SL}_{n}(\mathbb{Z}) of norm at most Cq^{2}\log q , while there exists an element such that every lift of it is of norm at least q^{2+o(1)} . This should be compared to the recent result that almost every element has a lift of norm bounded by q^{1+1/n+o(1)} . The main step in the proof is showing that for every q , there is a small element in (\mathbb{Z}/q\mathbb{Z})^{\times} with a large n -th root, which is a result of independent interest. In the proof we use tools from additive combinatorics including Bohr sets.

Journal of the European Mathematical Society
University of Cambridge (GB)
Royal Society, European Commission
Openalex Percentile: Top 99%
Analytic Number Theory Research
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