Does a direct sum decomposition of an infinite-dimensional vector space require Zorn's lemma?

Question

Let $V$ be an infinite-dimensional vector space and $V'\subset V$ a subspace. Does it require Zorn's lemma to write $V=V'\oplus V''$ for some other subspace $V''\subset V$?

Answer 1

Here is some context for thinking about this question. Suppose, more generally, that

$$0 \to A \to B \to C \to 0$$

is a short exact sequence of modules over some ring $R$, and you want to know whether this short exact sequence splits in the sense that it is isomorphic to the short exact sequence

$$0 \to A \to A \oplus C \to C \to 0$$

where the first map is the standard inclusion and the second map is the standard projection. For fixed $C$, this condition holds for all possible choices of $A$ and $B$ iff $C$ is projective, meaning that if $e : D \to E$ is surjective and $f : C \to E$ is a map, then there exists a lift $\tilde{f} : C \to D$ such that $f = e \circ \tilde{f}$. More formally this means that $\text{Hom}(C, -)$ preserves epimorphisms.

The axiom of choice implies that if $k$ is a field, then all $k$-modules (vector spaces over $k$) are projective, which in turn means precisely that all subspaces of a vector space have a complement. One way to organize the proof is as follows:

1. The axiom of choice implies, and is in fact equivalent to, the statement that all sets are projective.
2. The free module functor $\text{Set} \to \text{Mod}(k)$ preserves projectives (because it has a right adjoint which preserves epimorphisms). Hence the axiom of choice implies that all free $k$-modules are projective.
3. The axiom of choice also implies that all $k$-modules are free.

Without the axiom of choice, two steps in this proof fail. So if you want some intuition for how vector spaces behave in the absence of choice, spend some time studying modules over more general rings: vector spaces will behave more like arbitrary modules in the absence of choice.

It's also interesting to observe that if you state the axiom of choice as "all sets are projective," then "all vector spaces are projective" can be thought of as a linear analogue of choice. (Maybe it's also equivalent? I wouldn't know.)

Answer 2

Yes, and not only that, it is in fact equivalent to Zorn's lemma.

Just to give an example of something that might happen, it could be that there is a vector space over, say $\Bbb Q$, that every proper subspace is finitely dimensional, but the space itself is not finitely generated.

In particular no nontrivial subspace has a complement!