Timeline for Linear Algebra without Choice
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Nov 8, 2014 at 4:26 | comment | added | LSpice | @FrancoisG.Dorais, isn't 'yes and no' the 'law of the included middle'? :-) | |
| Sep 20, 2013 at 7:19 | vote | accept | Qfwfq | ||
| Sep 20, 2013 at 7:14 | history | edited | Qfwfq | CC BY-SA 3.0 |
added 281 characters in body
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| Sep 20, 2013 at 3:47 | comment | added | François G. Dorais | @Qiaochu: Yes and no, the law of excluded middle is true in ZF, so fields are indeed fields. | |
| Sep 19, 2013 at 22:42 | comment | added | Ryan Reich | @Toink: getting a basis from a generating set is a matter of well-ordering that set (and removing dependent vectors from the top), so in finite dimensions there is no problem. Likewise, invariance of dimension can be proven using row reduction or the replacement lemma (I think in infinite dimensions you would need well-ordering for this as well). | |
| Sep 19, 2013 at 22:40 | comment | added | Qiaochu Yuan | Linear algebra without choice should be like doing linear algebra in a topos, which in turn should be like doing linear algebra over a ring that isn't necessarily a field. So you would talk about finitely generated projective modules, wouldn't you? | |
| Sep 19, 2013 at 22:31 | comment | added | Asaf Karagila♦ | @Goldstern: Not quite. As it turned out, if you remember, the baseless part had a direct complement which was isomorphic, of course, to the quotient. It had a basis, so there were many endomorphisms. On the other hand, it is possible to have a vector space where "every injective endomorphism is an isomorphism", simply by constructing a vector space which only has scalar-multiplication endomorphisms. | |
| Sep 19, 2013 at 22:28 | comment | added | Asaf Karagila♦ | @Andreas: Of course not, it's in Lauchli's paper that he constructs a vector space with two bases: one which is Dedekind finite and another which is Dedekind infinite (if I recall correctly). This also appears as one of the problems in Jech's Axiom of Choice book, somewhere in Ch. 10, if I recall correctly. | |
| Sep 19, 2013 at 22:24 | answer | added | Asaf Karagila♦ | timeline score: 38 | |
| Sep 19, 2013 at 17:47 | comment | added | Andreas Blass | Is having a basis really enough to make a vector space "dimensional"? Couldn't it have bases of different cardinalities? | |
| Sep 19, 2013 at 17:44 | comment | added | Toink | you can still define a vector space to be finite dimensional if it is finitely generated. I think it is a basic theorem of linear algebra that such a vector space automatically has a (finite) basis, even without the axiom of choice. | |
| Sep 19, 2013 at 17:05 | comment | added | Goldstern | An example of an "exotic" phenomenon appears here: mathoverflow.net/questions/80765 . If I show that property $(\ast)$ is not inherited by subspaces and/or quotient spaces, would that count as a "no" for question 3? I think that some variant of the "basis=sequence of pairs of socks" space can be used as a conterexample. | |
| Sep 19, 2013 at 15:23 | history | asked | Qfwfq | CC BY-SA 3.0 |