Timeline for Is a field uniquely determined by its multiplicative group/how much knows K_1 about fields?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Apr 23, 2010 at 13:26 | comment | added | Torsten Ekedahl | @Robin: Right I saw potential problems only because for some strange reason I had confused what was subgroup of what. | |
| Apr 23, 2010 at 12:29 | comment | added | Robin Chapman | The class number is irrelevant. If $K$ is a number field, then $K^*$ is isomorphic to the direct product of a finite cyclic group (whose order is the number of roots of unity in $K$) with a free abelian group of infinite countale rank. | |
| Apr 23, 2010 at 11:04 | comment | added | Torsten Ekedahl | You are right, what I wrote is true but invalidates the following arguments. Corrected. | |
| Apr 23, 2010 at 11:02 | history | edited | Torsten Ekedahl | CC BY-SA 2.5 |
Put "countable" in the right position.
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| Apr 23, 2010 at 10:42 | comment | added | naf | I suppose you meant a free abelian group with a countably infinite basis... | |
| Apr 23, 2010 at 10:00 | history | answered | Torsten Ekedahl | CC BY-SA 2.5 |