Timeline for Rank of a free module without the axiom of choice
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Dec 10, 2010 at 2:49 | comment | added | Andrés E. Caicedo | @Robin : I mean, the issue is that you want the sets in the unions to be finite. Otherwise, I know how to produce several examples; since you are asking that ${}|A|\ne|B|$ but there are surjections in both directions, see mathoverflow.net/questions/38771/… | |
| Dec 10, 2010 at 2:44 | comment | added | Andrés E. Caicedo | "there are models of ZF with non-equinumerous infinite sets such that $A$ is the union of ${}|B|$ finite sets and vice versa." I would think so... I'll think about it. | |
| Dec 9, 2010 at 18:06 | comment | added | Robin Chapman | For finitely generated free modules over commutative rings, the result is elementary. For finitely generated free modules over non-commutative rings, the assertion can fail. | |
| Dec 9, 2010 at 18:02 | comment | added | Laurent Moret-Bailly | This only works for modules with an infinite basis. On the other hand, the ring does not have to be commutative in this case. | |
| Dec 9, 2010 at 17:47 | history | answered | Robin Chapman | CC BY-SA 2.5 |