Classifying Vector Spaces without AC

Question

Using the axiom of choice we can give a simple classification of all vector spaces over a given field $K$ up to isomorphism: Any $K$-vector-space $V$ is just isomorphic to $\bigoplus_{i\in B}K$ where $B$ is a basis for $V$. Given AC we even know that there is some aleph $\aleph_\alpha$ such that $\vert B\vert = \aleph_\alpha$ and thus any vector-space-structure over $K$ appears in the sequence $\lbrace \bigoplus_{i\in \aleph_\alpha}K:\alpha\in Ord\rbrace$.

Now I was wondering whether a similar (weaker) classification still holds without AC. In this case not every vector space has a basis and not every cardinality is equal to some aleph, so clearly the classification above fails. However one could try something like this:

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Edit* Thanks to Asaf's answer I realised that my previous attempt at "covering all vector spaces" was flawed, so here is a more general approach:

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As far as I can tell "new" vector spaces can be created using the following "Gödel operations for vector spaces":

• taking the sum or direct product over a family of spaces,
• taking the quotient of two spaces,
• constructing the space of homomorphisms between two spaces
• considering the subspace of any space resulting from the above operations

We can thus construct a hierarchy which should be closed under all these operations: Let $X_0:=\lbrace K\rbrace$ and for $\alpha>0$ let $X_\alpha:=$

$\lbrace W:$ there exists $\beta < \alpha$ and $W_0,W_1\in \overline{X}_\beta$ such that $W=W_1/W_0$ or $W=Hom_K(W_0,W_1)$ or there exists $I\in V_\alpha$ (in the Von-Neumann-hierarchy) and a family $\lbrace W_i:i\in I\rbrace \subseteq \overline{X}_\beta$ such that $W=\sum_{i\in I}W_i$ or $W=\prod_{i\in I}W_i\rbrace$

where $\overline{X}_\alpha:=\lbrace W'\leq W: W\in X_\alpha\rbrace$. I suspect that every vector space over $K$ should be isomorphic to some element of $\bigcup_{\alpha\in Ord}\overline{X}_\alpha$. Is that correct and is there maybe a more "practical" approach to characterizing all vector spaces over $K$ in ZF?

Answer 1

To the edit, the answer is yes. For a simple reason.

Suppose that $V$ is a $K$-vector space, then $V$ is a quotient of $W=\bigoplus_{v\in V}K$ by mapping $(\alpha_v\mid v\in V)$ to $\sum_{v\in V}\alpha_v v$ in $V$ itself, this is well-defined since all but finitely many $\alpha_v$ must be $0$.

So by allowing sums and quotients, every vector space is reached in sum+quotient operations from the field $K$.

Unfortunately, this does not help us to give any practical structure to vector spaces. If anything, it highlights how badly we don't understand them when the axiom of choice is taken away from us.

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No. You can have a vector space over any given field (and indeed, over every field) that is not finitely generated, but every proper subspace is finitely generated.

Not only this vector space doesn't have a basis, it is not of the form of $\mathrm{Hom}_K(V,W)$ for any two vector spaces. Now, suppose that $\alpha$ is the least rank of such vector space, then for every $V,W$ of lower rank, including those generating our space, $V$ has an infinite dimensional subspace. Fix such $V_0\subsetneq V$ and consider $\{T\mid V_0\subseteq\ker T\}$. Easily, this is an infinite dimensional subspace, so it has to be equal to everything.

So that means that for any infinite subspace $V_0$, all operators must be $0$ on $V_0$. But now, if $v\notin V_0$, then $V_0\oplus\langle v\rangle$ is another infinite dimensional subspace. Therefore one of two options must hold:

1. $V=V_0\oplus\langle v\rangle$, and every operator $T\colon V\to W$ is just a function mapping $v$ somewhere in $W$, but that means that $W$ is isomorphic to $\operatorname{Hom}_K(V,W)$, and $W$ has a lower rank, so it has infinite dimensional proper subspaces. That's impossible.

2. For every $v\in V$, $v$ is in some infinite dimensional proper subspace of $V$, and therefore $\operatorname{Hom}_K(V,W)=\{0\}$, which is very much not infinite dimensional like we assumed. That's also impossible.

This argument should in principle work for every type of "terrible enough indecomposable" vector space, and so $\sf ZF+DC_\kappa$ will not solve your problems either, for any $\kappa$, as per my masters thesis.

Of course, one could ask if your statement implies choice. But there's a lot of really good questions about vector spaces without choice, and we don't really know how to solve most of them... throw this one on the pile.