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"?


2 Answers 2


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.

  • $\begingroup$ Thanks! Where can I read about such vector spaces $U$? And when you say "it is possible", do you mean "If every vector space admits a linear functional, then choice holds", or just that "there are models of ZF where not every vector space admits a linear functional"? I think this touches on Variation 3 as well, since if $U$ doesn't admit any linear functionals, then it is certainly not injective! $\endgroup$
    – Tim Campion
    Oct 23, 2020 at 19:53
  • 1
    $\begingroup$ @Tim: to be explicit, I think something along the lines of $\prod k / \oplus k$ (countable product and sum) already may not admit nonzero linear functionals in, say, a model of ZF with no non-principal ultrafilters. $\endgroup$ Oct 23, 2020 at 20:01
  • 7
    $\begingroup$ @TimCampion I meant "there are models ..." Take a look at this MO thread, where people more expert than me explain why every nonzero vector space having a linear functional doesn't imply choice. $\endgroup$ Oct 23, 2020 at 20:03
  • 1
    $\begingroup$ Sounds like something someone gave a short talk about in Bristol just before lockdown... :-) $\endgroup$
    – Asaf Karagila
    Oct 24, 2020 at 1:07
  • $\begingroup$ @Jeremy: that thread's a lovely find and resolved a question I'd been wondering about on math.SE! To pull an example out of it explicitly: there's a model of ZF in which $\mathbb{R}$ considered as a $\mathbb{Q}$-vector space has no nonzero linear functionals, and in fact in which every homomorphism $\mathbb{R} \to \mathbb{R}$ is continuous (which implies this). $\endgroup$ Oct 24, 2020 at 7:00

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.)

  • $\begingroup$ Did you mean "every subset of a Polish space has the property of Baire"? $\endgroup$
    – tomasz
    Oct 26, 2020 at 12:48
  • $\begingroup$ @tomasz: I did, thanks. Fixed. $\endgroup$ Oct 26, 2020 at 14:20

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