corner of a ring
Does there exist a subset of a ring which is a ring with a multiplicative identity, but not a subring of ?
Let be a ring without the assumption that has a multiplicative identity. Further, assume that is an idempotent
of . Then the subset of the form is called a corner of the ring .
It’s not hard to see that is a ring with as its multiplicative identity:
-
1.
,
-
2.
,
-
3.
is the additive inverse of in ,
-
4.
, and
-
5.
, with , for any .
If has no multiplicative identity, then any corner of is a proper subset![]()
of which is a ring and not a subring of . If has 1 as its multiplicative identity and if is an idempotent, then the is not a subring of as they don’t share the same multiplicative identity. In this case, the corner is said to be proper. If we set , then is also a proper corner of .
Remark. If has 1 with an idempotent. Then corners and , where , are direct summands![]()
(as modules over ) of via a Peirce decomposition
![]()
.
References
- 1 I. Kaplansky, Rings of Operators, W. A. Benjamin, Inc., New York, 1968.
| Title | corner of a ring |
|---|---|
| Canonical name | CornerOfARing |
| Date of creation | 2013-03-22 15:43:56 |
| Last modified on | 2013-03-22 15:43:56 |
| Owner | CWoo (3771) |
| Last modified by | CWoo (3771) |
| Numerical id | 9 |
| Author | CWoo (3771) |
| Entry type | Definition |
| Classification | msc 16S99 |
| Related topic | UnityOfSubring |