Do vector spaces without choice satisfy Cantor-Schroeder-Bernstein? If $V \hookrightarrow W$ and $W \hookrightarrow V$ are injective linear maps, then is there an isomorphism $V \cong W$?
If we assume the axiom of choice, the answer is yes: use the fact that every linearly independent set can be extended to a basis and apply the usual Schroeder-Bernstein theorem.
If we don't assume the axiom of choice, and we work in ZF, say (or some other formalism with excluded middle), then vector spaces don't necessarily have bases (in fact, Blass showed that there must be a vector space without a basis over some field), so we can't use the same proof strategy. Nevertheless, there's room for optimism, since Schroeder-Bernstein still holds for sets in ZF. So one might hope that it also holds for vector spaces in ZF.
Question: Work in ZF (or some other formalism with excluded middle but without choice). If $V \hookrightarrow W$ and $W \hookrightarrow V$ are injective linear maps of vector spaces over a field $k$, then is there an isomorphism $V \cong W$?
Variation 1: What if we assume that $k$ is finite, or even that $k = \mathbb F_p$ for a prime $p$?
Variation 2: What if we assume that $V$ is a direct summand of $W$ and vice versa?
The following consequence of Bumby's theorem appears to be constructive: If $k$ is a ring and every $k$-module is injective, then $k$-modules satisfy Schroeder-Bernstein. But the condition "every module over a field is injective" sounds pretty choice-ey to me. I suppose it's worth noting, though:
Variation 3: Does "Every vector space over any field is injective" imply choice? How about "Every vector space over $\mathbb F_p$ is injective"?
 A: Without the axiom of choice, it is possible that there is a vector space $U\neq 0$ over a field $k$ with no nonzero linear functionals.
Let $V$ be the direct sum of countably many copies of $U$, and $W=V\oplus k$.
Then each of $V$ and $W$ embeds in the other, but they are not isomorphic, since $V$ doesn’t have any nonzero linear functionals, but $W$ does.
I don't think there's any restriction on the field $k$, so this answers Variation 1 as well.
A: There are models of ZF+DC in which every subset of every Polish space has the property of Baire (I can try to add references later, I think to Solovay and Shelah, but these are pretty well known).  This implies that every linear map between Banach spaces is continuous.
So we can then take $\ell^\infty$ and $\ell^1$.  It is very easy to construct (continuous) linear injections either way: the identity map from $\ell^1$ into $\ell^\infty$, and to go the other way, map $x_n$ to $2^{-n} x_n$.
But if there were a linear isomorphism between them, it would be a homeomorphism, and this is impossible because $\ell^1$ is separable and $\ell^\infty$ isn't.
(As a tie-in to Jeremy's answer, in this model $\ell^1$ is reflexive, and $\ell^\infty / c_0$ is a Banach space with no nonzero linear functionals.)
