Proving that $\mathbb{Q}$ is an injective abelian group without the Axiom of Choice

abelian-categoriesabelian-groupsaxiom-of-choicehomological-algebra

The question is simple: Is there a proof that $\mathbb{Q}$ is an injective abelian group that does not invoke the axiom of choice?

I expect the anser to be that there is no such proof, so I also leave here the following two questions:

Is there a proof using only a weak form of choice, like dependent choice? Is there a subcategory of abelian groups (maybe countably generated abelian groups) where $\mathbb{Q}$ is injective?

I know there is no point in doing homological algebra without choice, but this is just curiosity, and as far as I know there is some interest for constructive algebraic geometry/ homological algebra.

A few comments:

For me, an injective abelian group is a group $G$ satisfying one of the two equivalent (at least they are equivalent in an abelian category) conditions:

  • For any subgroup $A \subset B$, any homomorphism $f : A \to G$ can be extended to $B$

  • Every short exact sequence $0 \to G \to A \to B \to 0$ splits

I am not asking for a proof without Baer'scriterion, I am asking for a proof without Axiom of Choice. In particular, one cannot use that every short exact sequence of vector spaces splits, or that free groups are projective, or that subgroups of free (abelian) groups are free, or even that there are enough projectives in the category of abelian groups.

$\bf EDIT$: (After Mark Saving's answer) $\mathbb{Q}$ may fail to be injective even if dependent choice is assumed to be true.

$\bf EDIT 2$: (After Asaf Karagila's comment) Andreas Blass proved, in a paper from 1979, that the axiom of choice is equivalent to:

  • Every divisible abelian group is injective.
  • Every free abelian group is projective.

He also shows that one cannot prove the existence of an injective abelian group without choice, but that under a weaker axiom (SVC) then there are enough injective abelian groups

Best Answer

Recall that $\mathbb{Q}$ being an injective Abelian group means that for all Abelian groups $B$ and all subgroups $A \subseteq B$, every group homomorphism $A \to \mathbb{Q}$ extends to a homomorphism $B \to \mathbb{Q}$.

Suppose that $\mathbb{Q}$ is an injective Abelian group. Then consider $1_\mathbb{Q} : \mathbb{Q} \to \mathbb{Q}$. Extend this to a group homomorphism $f : \mathbb{R} \to \mathbb{Q}$.

We thus have a group homomorphism $f : \mathbb{R} \to \mathbb{R}$, which is not $\mathbb{R}$-linear since $f(\sqrt{2} \cdot 1) \neq \sqrt{2} = \sqrt{2} \cdot f(1)$.

But according to this answer, it is equiconsistent with ZF + dependent choice that all sets of reals have the property of Baire. And if all sets have the property of Baire, then that all group homomorphisms $\mathbb{R} \to \mathbb{R}$ are $\mathbb{R}$-linear. So it is equiconsistent with ZF + dependent choice that $\mathbb{Q}$ is not an injective object.

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