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We consider the field of "usual" linear algebra.

Q. Which aspects of it can be carried out without the Axiom of Choice?

Q. Do interesting "exotic" phenomena appear in presence of (some instance of) the negation of the Axiom of Choice?

Without Choice, vector spaces may have a basis (hence, in particular, be dimensional) or not and hence be adimensional. [As Andreas Blass observes in the comments, the terminology "dimensional/adimensional" should rather be used to denote the property of having all bases of the same cardinality, rather than just having a basis, as there are vector spaces with two bases of different cardinality]

Q. Could the following property of a vector space $V$

Property ($\star$) Every injective endomorphism of $V$ is an automorphism.

be a valid substitute for finite-dimensionality for the class of not-necessarily-dimensional vector spaces over a field? Would the linear algebra of vector spaces verifying ($\star$) be reasonably similar to the usual one for finite dimensional spaces?

Qfwfq
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    An example of an "exotic" phenomenon appears here: http://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. – Goldstern Sep 19 '13 at 17:05
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    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. – Toink Sep 19 '13 at 17:44
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    Is having a basis really enough to make a vector space "dimensional"? Couldn't it have bases of different cardinalities? – Andreas Blass Sep 19 '13 at 17:47
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    @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. – Asaf Karagila Sep 19 '13 at 22:28
  • @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. – Asaf Karagila Sep 19 '13 at 22:31
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    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? – Qiaochu Yuan Sep 19 '13 at 22:40
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    @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). – Ryan Reich Sep 19 '13 at 22:42
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    @Qiaochu: Yes and no, the law of excluded middle is true in ZF, so fields are indeed fields. – François G. Dorais Sep 20 '13 at 03:47
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    @FrancoisG.Dorais, isn't 'yes and no' the 'law of the included middle'? :-) – LSpice Nov 08 '14 at 04:26

1 Answers1

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Some things about vector spaces which are consistent with the failure of choice:

  1. Vector spaces may have bases of different cardinality. In particular, this means that the notion of "dimension" is not well-defined. It follows from the Boolean Prime Ideal theorem (which is strictly weaker than $\sf AC$ itself) that if there is a basis, then its cardinality is unique. See Sizes of bases of vector spaces without the axiom of choice for more details.

  2. The existence of a basis is no longer hereditary. That is, it is consistent that there is a vector space which has a basis, but it has a subspace which doesn't have a basis. You can find the example in Goldstern's answer If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?, and what is even more interesting is the fact that the vector space without a basis has a direct complement which has a well-ordered basis.

  3. It is consistent that there is a vector space, that all its endomorphisms are scalar multiplications (which is not $(0)$ or the field itself). In particular every non-zero endomorphism is an automorphism, and this answers yours final question. Indeed every non-zero endomorphism is an injective endomorphism and an automorphism. These spaces were the main topic of my masters thesis, where I somewhat extended Lauchli's original result (and construction) of such spaces. You can find somewhat of an outline of the general result here: Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?

  4. It is consistent that there is a vector space, which is not finitely generated, which is (naturally) isomorphic to its algebraic double dual. In particular this can be $\ell_2$. See my answer at Does the fact that this vector space is not isomorphic to its double-dual require choice? for details.

There are other properties which fail for non-finitely generated vector spaces in $\sf ZFC$ which are consistent with the failure of choice. The list is long, and these just a few I could write about from the top of my head.

Whether or not any of them is equivalent to the axiom of choice is usually an open (and a difficult) question. But it is usually the case that if a property requires some form of choice (like a basis, or extension of functinoals, etc.) then it can fail in suitable models of $\sf ZF+\lnot AC$.

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Asaf Karagila
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    Although I know that there is a tradition in set theory to use this "consistently" phraseology as you do, it has always sounded off to my ear, for to say that something is "consistently happening" is to suggest that it happens quite reliably or even every single time, as in, "the maid consistently left mints on the bed pillows." So I have never liked this usage of "$x$ is consistently true" to mean "it is consistent that $x$ is true." I suggest saying, "Some things that are consistent with the failure of choice are..." and "there are other properties that are consistent with not AC...". – Joel David Hamkins Sep 19 '13 at 22:33
  • Just to echo that thought, it hadn't occurred to me until reading Joel's comment that Asaf was using the word in the technical sense (as opposed to the mints sense). – Cam McLeman Sep 19 '13 at 22:39
  • @Joel: Well, as good (or bad) as my English is, it will never be the same as a native speaker. But if I recall correctly, the standard language of mathematics (and science) is "broken English" anyway. :-) – Asaf Karagila Sep 19 '13 at 22:58
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    Asaf, your English is outstanding, so please don't worry about that, and as I've said, you're not the only one to use "consistently" this way, although I'm not entirely sure whether I've ever heard it from native speakers. (Meanwhile, you have another instance of "consistently true" in the last paragraph.) – Joel David Hamkins Sep 19 '13 at 23:18
  • @Joel: Thank you for the compliment. I wasn't worried about that at all, I was merely pointing it out. (There was even a smiley face at the end to indicate this is a lighthearted comment! :-)) – Asaf Karagila Sep 20 '13 at 08:17
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    Do you mean "vector space not isomorphic to $0$ or $k$" in 3. and 5.? What does "naturally" mean in 4.? – Martin Brandenburg Sep 20 '13 at 08:34
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    @Martin: Of course that's what I mean in (3) and (5); in (4) perhaps I should have written naturally isomorphic with its double dual (by evaluation functionals). But the isomorphism with the dual doesn't require a choice of basis either (I mean, hey... there is no basis!) – Asaf Karagila Sep 20 '13 at 08:49
  • In 5, what inverse is false? It seems to me that the converse, which states that every vector space such that every endomorphism is scalar multiplication must be indecomposable, is true - in a decomposable vector space, a projection onto one of the factors is a non-scalar endomorphism. Do you mean something else? – Will Sawin Sep 26 '13 at 03:47
  • @Will: Yes, of course, you're right. I was thinking on something else (more specifically, the fact that the dual space is trivial). However restricting to injective endomorphims the statement is true. We can show the consistency of a decomposable space where every injective endomorphism is a scalar multiplication (take two non-isomorphic indecomposable spaces). – Asaf Karagila Sep 26 '13 at 05:58
  • @Will: Suddenly, I'm not sure if that's true (I have a good feeling that it is, but I can't prove that outright). I'll remove that part. Thank you! – Asaf Karagila Oct 01 '13 at 05:43