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Solvability of isotropic $ \mathrm{K}_1 $-functor over semilocal rings
Abstract
We show that the $ \mathrm{K}_1 $-functor modeled on simple reductive groups over semilocal rings is solvable if the isotropic rank is at least $ 2 $ and that the Tits index is neither $ {}^{2} \mathsf{E}_{6, 2}^{16'} $ nor $ \mathsf{E}_{8, 2}^{78} $. For these two Tits indices the result is already known, but assuming that the base ring contains a field. Our result implies that the elementary subgroup (or its derived subgroup) is the maximal perfect subgroup of the reductive group.
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BibTeX
@misc{voronetsky2026solvability,
title = {{Solvability of isotropic $ \mathrmK_1 $-functor over semilocal rings}},
author = {Egor Voronetsky},
howpublished = {arXiv},
year = {2026},
month = sep,
doi = {10.48550/arxiv.2609.25124},
eprint = {2609.25124},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2609.25124},
}
RIS
TY - UNPB
TI - Solvability of isotropic $ \mathrm{K}_1 $-functor over semilocal rings
AU - Egor Voronetsky
PY - 2026
DA - 2026/09/20
PB - arXiv
DO - 10.48550/arxiv.2609.25124
UR - https://arxiv.org/abs/2609.25124
AB - We show that the $ \mathrm{K}_1 $-functor modeled on simple reductive groups over semilocal rings is solvable if the isotropic rank is at least $ 2 $ and that the Tits index is neither $ {}^{2} \mathsf{E}_{6, 2}^{16'} $ nor $ \mathsf{E}_{8, 2}^{78} $. For these two Tits indices the result is already known, but assuming that the base ring contains a field. Our result implies that the elementary subgroup (or its derived subgroup) is the maximal perfect subgroup of the reductive group.
ER -