11
$\begingroup$

It is known that the existence of nontrivial injective abelian groups is independent of choice in ZF (or, rather, ZFA). In particular, $\mathbb{Q}$ is not provably injective, much less $\mathbb{R}$, so let's work in ZF and suppose injectivity of $\mathbb{Q}$. It is also known, though (and this may just reflect my ignorance of homological algebra) seems to be more folkloric, that (at least with choice) the injective abelian groups are precisely the direct sums of $\mathbb{Q}$'s and Prüfer $p$-groups. In particular, as every vector space has a basis (with choice!) and $\mathbb{R}$ is always a vector space over $\mathbb{Q}$, $\mathbb{R}$ is injective with choice. My question then is "how much choice" is needed - say, can we get injectivity as long as $\mathbb{Q}$ is injective? or do we need more machinery?

Improve this question
$\endgroup$
13
  • 3
    $\begingroup$ Just to have it said, if you want to use a $\mathbb Q$-basis for $\mathbb R$ to argue injectivity, then that is itself a use of choice. (Of course, that doesn't mean that there isn't some other way to argue from the injectivity of $\mathbb Q$.) $\endgroup$
    – LSpice
    Commented Oct 12 at 17:33
  • 3
    $\begingroup$ I haven't checked Blass' proof for the implication of AC from "every $\mathbb{Q}$-linear space is injective" (the internet in my home isn't working, so I can't check his paper without risking using much mobile data) but it is possible that "$\mathbb{R}$ as a $\mathbb{Q}$-linear space is injective" may imply a fragment of choice (like, choice for a limited size of family of the limited size of sets under some cardinality.) Also, I recommend adding a set-theory tag. $\endgroup$
    – Hanul Jeon
    Commented Oct 13 at 1:31
  • 3
    $\begingroup$ @HanulJeon I don’t know if this is the proof you’re referring to, but the proof outline I had in mind is as follows: by injectivity, any s.e.s. of $\mathbb{Q}$-linear spaces splits, so in particular any surjective $\mathbb{Q}$-linear map has a right inverse. Then, given a surjection $\pi:J\to I$, it induces a surjective $\mathbb{Q}$-linear $f:\mathbb{Q}[J]\to\mathbb{Q}[I]$, which then admits a right inverse $g$. For each $i\in I$, $g(e_i)$ has finite support and the support must contain some elements of $\pi^{-1}(i)$, so $i\mapsto\text{supp}(g(e_i))\cap\pi^{-1}(i)$ is… $\endgroup$
    – David Gao
    Commented Oct 13 at 2:38
  • 3
    $\begingroup$ … a multiple choice function. As axiom of multiple choice is equivalent to AC under ZF, this proves the implication. (As far as I can tell, I don’t see a way to adapt this proof to show any fragment of AC using just $\mathbb{Q}$ and $\mathbb{R}$ are injective.) $\endgroup$
    – David Gao
    Commented Oct 13 at 2:42
  • 3
    $\begingroup$ @DavidGao Since the Blass paper works within ZFA I think it's worth bearing that, in ZFA, multiple choice does not entail full choice, at least according to the nlab. Doesn't change anything morally as we're still talking about a fragment of AC, but feels worth noting for the record. $\endgroup$
    – Garrett Figueroa
    Commented Oct 14 at 22:02

0

Reset to default

You must log in to answer this question.