Timeline for Cardinality vs. isomorphism type of vector spaces without choice
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 21, 2018 at 19:20 | comment | added | Asaf Karagila♦ | @Gerald: Ultimately uneventful search, I'm afraid. | |
| Sep 21, 2018 at 19:20 | history | edited | Asaf Karagila♦ | CC BY-SA 4.0 |
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| Sep 21, 2018 at 18:52 | history | edited | Asaf Karagila♦ | CC BY-SA 4.0 |
added 28 characters in body
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| Sep 21, 2018 at 17:43 | comment | added | Asaf Karagila♦ | @AlexKruckman: Boy was I tired when I posted this... :P | |
| Sep 21, 2018 at 17:41 | comment | added | Alex Kruckman | Any two uncountable fields of the same characteristic are isomorphic as additive groups iff they have the same cardinality. In fact, you get something better in finite characteristic: any two infinite fields of characteristic $p\neq 0$ are isomorphic as additive groups iff they have the same cardinality. | |
| Sep 21, 2018 at 0:30 | comment | added | Asaf Karagila♦ | Yes, that's usually a good choice. But if my memory serves me right, this isn't there. I'll take a look tomorrow morning, see what I find there. Thanks. | |
| Sep 21, 2018 at 0:29 | comment | added | Gerald Edgar | I would say, first step: consult Rubin & Rubin, Equivalents of the Axiom of Choice | |
| Sep 21, 2018 at 0:23 | history | edited | Asaf Karagila♦ | CC BY-SA 4.0 |
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| Sep 21, 2018 at 0:22 | history | edited | Theo Johnson-Freyd | CC BY-SA 4.0 |
replaced "existence of vector spaces" with "axiom of choice"
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| Sep 20, 2018 at 23:50 | comment | added | Asaf Karagila♦ | Usually, I'd ask myself that question. But my mind is elsewhere these days, and I feel that I'm missing something obvious. | |
| Sep 20, 2018 at 23:49 | history | asked | Asaf Karagila♦ | CC BY-SA 4.0 |