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](https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_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](http://www.math.lsa.umich.edu/%7Eablass/bases-AC.pdf) 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](https://doi.org/10.1007/BF01220018) 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"?



Do vector spaces without choice satisfy Cantor-Schroeder-Bernstein?
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.


