Tuesday, November 24, 2015 — 11:30 AM EST

Christopher Schafhauser, Department of Pure Mathematics, University of Waterloo

“Goldie’s Theorem on semiprime rings”

Recall that R is a right Ore ring if its set S = C(R) of cancelable elements satisfies the right Ore condition. In this case we can form the right classical ring of fractions Q = RS1, and there is a natural inclusion Q such that S Q× and each element of Q has the form as1,forsomeaRandsS. WesawthatifRisadomainthenQisadivisionring,but this fails if R is not right Ore — in fact, in the non-Ore case we can’t form the right ring of fractions. But we can still ask whether or not R embeds in some ring Q such that S Q× and each element of Q has the form as1. In this case R is called a right order in Q.

Our aim now is to determine when R is a right order in a semisimple ring. For this, there are necessary and sufficient conditions given by Goldie’s Theorem.

MC 5403

S M T W T F S
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
1
2
3
  1. 2023 (167)
    1. June (1)
    2. May (34)
    3. April (21)
    4. March (51)
    5. February (33)
    6. January (27)
  2. 2022 (179)
    1. December (8)
    2. November (31)
    3. October (24)
    4. September (17)
    5. August (9)
    6. July (15)
    7. June (14)
    8. May (13)
    9. April (14)
    10. March (15)
    11. February (12)
    12. January (7)
  3. 2021 (135)
  4. 2020 (103)
  5. 2019 (199)
  6. 2018 (212)
  7. 2017 (281)
  8. 2016 (335)
  9. 2015 (211)
  10. 2014 (235)
  11. 2013 (251)
  12. 2012 (135)