# Indecomposable vector spaces and the axiom of choice

It is a known result by A. Blass that the axiom of choice is equivalent to the assertion that every vector space has a basis. (Rubin's Equivalents of the Axiom of Choice: form B)

It is also known that the axiom of choice equivalent to the existence of a field $F$ such that for every vector space $V$ over $F$ has the property that if $S\subseteq V$ is a subspace of $V$ then there is $S'\subseteq V$ such that $S\oplus S'=V$.

In this case, assuming the axiom of choice does not hold, we have a vector space without a basis somewhere in the model, and over every field there is a vector space $V$ that has a subspace without a complement.

My question: Can we derive from the absence of choice that there will be a vector space (over some field? any field?) such that for any nontrivial subspace $S$ of $V$ there is no subspace $S'$ of $V$ such that $S\oplus S'=V$?

My initial instinct was to try and take a vector space without a basis and start decomposing it, somehow deriving if we stop at a successor stage we have found such subspace, and if we happened to hit a limit stage with an empty subspace then we can construct a basis. I think I overlooked something because I couldn't finish my argument.

Edit: Just to be clear, I am not looking whether or not this is consistent with ZF, but rather if the negation of choice implies there exists such vector space.

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I think the tag linear-algebra would also be appropriate. –  SNd May 7 '11 at 17:58

H. Läuchli [Auswahlaxiom in der Algebra, Comment. Math. Helv. 37 1962/1963, MR143705] constructed a permutation model wherein there is a vector space which is not finite dimensional and such that all of its proper subspaces are finite dimensional. This is an example of a vector space with the property you want.

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This example appears in Jech's The Axiom of Choice. I am well aware of this construction, my question is whether or not this sort of behaviour (where you can replace "finite subspaces" by "subspaces of cardinality $<\kappa$" for a fixed $\kappa$) can always be found in the absence of choice. –  Asaf Karagila May 7 '11 at 20:42
Ah! That's not the way I read your question. I will have to think about the situation you have in mind. –  François G. Dorais May 8 '11 at 1:34
I will clarify my question as well. Thank you! –  Asaf Karagila May 8 '11 at 5:35

This is not meant to be a complete answer to the question (which I find very interesting), but rather an explanation for why the particular approach you described is bound to run into difficulties. I'm new here, and I'm not sure this belongs as an "answer;" please let me know if this is inappropriate.

It is consistent with ZF that there is a vector space (over the field of two elements) which does not have a basis and yet every nontrivial subspace admits a decomposition into two complementary subspaces. In fact, one such example is quite familiar.

Set $V = \mathcal{P}(\omega)/\mathrm{FIN}$, viewed as a vector space over $\mathbb{Z}/2\mathbb{Z}$ (with, of course, symmetric difference serving as the addition operation). A fairly straightforward Baire category argument shows that $V$ does not have a basis in a model of ZFDC + all sets of reals have the Baire property. However, any subspace $W \subseteq V$ with more than two elements can be decomposed as $S_0 \oplus S_1$: simply fix some element $A \in W$ which does not almost contain every element of $W$ and set

$S_0 = \{B \in W : B \mbox{ is almost contained in A}\}$ and $S_1 = \{B \in W : B \mbox{ is almost disjoint from A}\}$.

I'm sure the same argument actually works for $\mathcal{P}(\omega)$ itself, but I hate thinking about finite sets.

Edit: OK, now I'm worried about offending finite sets. I don't hate thinking about you in general, but I'd rather not think about you in the particular Baire category argument I have in mind.

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This example is very interesting as itself, and it does explain why my approach ran into some difficulties. Do note that this works for the vector space Francois mentioned, as every subspace is of finite dimension, and spaces of finite dimension can easily be decomposed even without the axiom of choice. –  Asaf Karagila May 8 '11 at 11:27
Right. I suppose I should've emphasized that I was going for a simple example (working in ZFDC) with lots of subspaces. –  Clinton Conley May 8 '11 at 11:41