10
$\begingroup$

It is well known that the Axiom of Choice is needed to prove that every vector space has a basis (in fact this statement is equivalent to the AC). But what about the apparently weaker statement that every subspace of a vector space has a complementary subspace, or, equivalently, every short exact sequence of vector spaces splits? (This may actually not be a weaker statement after all, since it would apply also to a vector space whose cardinality is not an aleph.)

$\endgroup$
14
  • 2
    $\begingroup$ Formally speaking, "The axiom of choice is needed to prove X" means "X implies the axiom of choice (in ZF)", and does not mean "X is not a Theorem of ZF". $\endgroup$
    – YCor
    Nov 25, 2021 at 17:20
  • 5
  • 2
    $\begingroup$ @BugsBunny I don’t think it’s that simple. Assume there is a set $Y$ of disjoint pairs without a choice function. Let $X=\bigcup Y$, and let $f\colon X\to Y$ map each element to the pair it is in. Then $f$ is a surjection without right inverse. However, if $K$ is a field of characteristic $\ne2$, then $Kf$ has a splitting that maps each pair $\{u,v\}\in Y$ to $\frac12u+\frac12v$. (If $K$ has characteristic $2$, do it with triples instead of pairs.) $\endgroup$ Nov 25, 2021 at 17:57
  • 4
    $\begingroup$ @AsafKaragila As your comment fully answers the question, you should post it as an answer. $\endgroup$ Nov 25, 2021 at 18:05
  • 3
    $\begingroup$ @MaximeRamzi Do you have a reference for that? The last time I heard, the proof of existence of bases implying AC (over ZF) requires considering some large auxiliary fields. How can you do it over $\mathbb F_2$ only? $\endgroup$
    – Wojowu
    Nov 25, 2021 at 18:25

1 Answer 1

6
$\begingroup$

Your question is essentially answered in a math.SE post by Asaf Karagila, but I think it is worth spelling out a subtle point. The axiom of multiple choice (or MC for short) says:

For every set $X$ of nonempty sets, there exists a function $f$ on $X$ such that for every $x\in X$, $f(x)$ is a finite nonempty subset of $x$.

In Lemma 2 of Some theorems on vector spaces and the axiom of choice (Fund. Math. 54 (1964), 95–107), M. N. Bleicher showed that MC is implied by the following statement:

For some field $F$, every subspace of a vector space over $F$ has a complementary subspace.

In fact, Bleicher's results combined with a result of M. C. Armbrust (An algebraic equivalent of a multiple choice axiom, Fund. Math. 74 (1972), 145–146) imply that the above statement is equivalent to MC.

The subtle point is that the above equivalence can be proved not just in ZF, but in some other set theories such as ZFU (a variant of ZF with "urelements" or "atoms"). So this equivalence is "robust" in some sense. On the other hand, ZF proves the equivalence of MC with the axiom of choice (for a non-paywalled proof, see Theorem 2.18 in Kevin Barnum's note, The axiom of choice and its implications), but this equivalence is "delicate" because if you switch to a slightly different set theory, MC might be strictly weaker than AC.

So the answer to your question is that your "apparently weaker" statement is equivalent to AC if you work over ZF. But over some other closely related set theories, I'm not sure the answer is known. Some further information may be found in Paul Howard's paper, Bases, spanning sets, and the axiom of choice (Math. Logic Q. 53 (2007), 247–254), and Marianne Morillon's paper, Linear extenders and the axiom of choice (Comm. Math. Univ. Carol. 58 (2017), 419–434).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.