PreprintCC BY 4.0arXiv
The silp--spli equality for group rings
Abstract
Let R be an arbitrary associative ring with an identity. In this paper, we show that the equality silp(R)=spli(R) implies silp(RG)=spli(RG) for any group G.
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BibTeX
@misc{zhang2026silpspli,
title = {{The silp--spli equality for group rings}},
author = {Xiaolei Zhang and Guocheng Dai},
howpublished = {arXiv},
year = {2026},
month = oct,
doi = {10.48550/arxiv.2610.34406},
eprint = {2610.34406},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2610.34406},
note = {CC BY 4.0},
}
RIS
TY - UNPB
TI - The silp--spli equality for group rings
AU - Xiaolei Zhang
AU - Guocheng Dai
PY - 2026
DA - 2026/10/25
PB - arXiv
DO - 10.48550/arxiv.2610.34406
UR - https://arxiv.org/abs/2610.34406
AB - Let R be an arbitrary associative ring with an identity. In this paper, we show that the equality silp(R)=spli(R) implies silp(RG)=spli(RG) for any group G.
ER -