On SGn of orders
The main result of the paper is: If R is a Dedekind domain with field of fractions K, and M is an R-order in a separable K-algebra A, then, under certain conditions on R and M, SG n ( M) = ker( G n ( M) → G n ( A)) = 0 for all n⩾ 1. In particular, this theorem applies to the integral group ring of a...
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| Published in | Journal of algebra Vol. 133; no. 1; pp. 125 - 131 |
|---|---|
| Main Authors | , |
| Format | Journal Article |
| Language | English |
| Published |
Elsevier Inc
15.08.1990
|
| Online Access | Get full text |
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| Summary: | The main result of the paper is: If
R is a Dedekind domain with field of fractions
K, and
M is an
R-order in a separable
K-algebra
A, then, under certain conditions on
R and
M,
SG
n
(
M) = ker(
G
n
(
M) →
G
n
(
A)) = 0 for all
n⩾ 1. In particular, this theorem applies to the integral group ring of a finite group
G. If
M is regular, e.g., if
M is maximal, then
SK
n
(
M) = 0. |
|---|---|
| ISSN: | 0021-8693 1090-266X |
| DOI: | 10.1016/0021-8693(90)90073-W |