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I've read, that the accessibility of the category of all free abelian groups is independent on basic set theory (say ZFC). What is the reason for that? And how can I interpret this result? Does it mean that the property of been a free abelian group is not a first order property?

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    $\begingroup$ The property of being a free abelian group is not first-order. Indeed, consider a non-trivial ultrapower of $\mathbb{Z}$. It contains an element $(1, 2, 6, 24, 120, \ldots)$ that is divisible by every integer, hence contains a copy of $\mathbb{Q}$. $\endgroup$ – Zhen Lin Mar 21 '14 at 8:44
  • $\begingroup$ Thanx, so far my intuition goes that I didn't expect it to be first order property. Does it, however, follow from the category not being accessible? $\endgroup$ – bbxlmnistvii Mar 21 '14 at 9:40
  • $\begingroup$ In principle, yes. But as you say, we do not know whether it is accessible or not. $\endgroup$ – Zhen Lin Mar 21 '14 at 9:55
  • $\begingroup$ Yes, it holds true on assumption of existence of some large cardinals, I believe the compact one. On the other hand it does not if axiom of constructibility is assumed. However the second axiom is completely new to me. So, if the fact that to be a free abelian is not a first order property cannot be proved in ZFC, what other properties are of this sort? Is this a general pattern? $\endgroup$ – bbxlmnistvii Mar 21 '14 at 10:10
  • $\begingroup$ I just gave you a ZFC proof that being a free abelian group is not a first-order property! $\endgroup$ – Zhen Lin Mar 21 '14 at 11:44
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The Question:

"Is the category of free abelian groups accessible?"

There's an ambiguity: what should we take to be the morphisms of the category of free abelian groups? The following discussion will work if we consider any of the following categories:

  • $\mathsf{Free}$: free abelian groups and all homomorphisms
  • $\mathsf{Free}_\mathrm{inj}$: free abelian groups and all injective homomorphisms
  • $\mathsf{Free}_\mathrm{pure}$: free abelian groups and all pure homomorphisms
  • $\mathsf{Free}_\mathrm{elem}$: free abelian groups and all elementary embeddings

The undecidability of the question is stated in Makkai and Paré's Accessible Categories: The Foundations of Categorical Model Theory, Section 5.5. They actually say that "it is easy to conclude" from "known results" in Eklof and Mekler, On Constructing Indecomposable Groups in L. I can't quite reconstruct all of this, but I can reconstruct enough to get to the conclusion that the question is undecidable (assuming large cardinal hypotheses are consistent).

The Affirmative Answer:

Makkai and Paré claim that the affirmative answer follows easily from the existence of a compact cardinal using Eklof and Mekler's work. This is the part I can't confirm, but Makkai and Paré do go on to prove a more general result under slightly stronger hypotheses.

Theorem (5.5.1 in Makkai and Paré): Assume there are arbitrarily large compact cardinals. Let $\mathcal{C}$ and $\mathcal{D}$ be accessible categories, and let $F: \mathcal{C} \to \mathcal{D}$ be an accessible functor. Let $F_!(\mathcal{C})$ denote the full subcategory of $\mathcal{D}$ whose objects are all subobjects of objects $F(C)$ for $C \in \mathcal{C}$. Then $F_!(\mathcal{C})$ is an accessible category.

Makkai and Paré call the category $F_!(\mathcal{C})$ the powerful image of $F$, but I just made up the notation. Since every subgroup of a free abelian group is free, we have (accessible) "free abelian group" functors in each of the following contexts:

  • $\mathsf{Set} \to \mathsf{Ab}$, powerful image is $\mathsf{Free}$.
  • $\mathsf{Set}_\mathrm{inj} \to \mathsf{Ab}_\mathrm{inj}$, powerful image is $\mathsf{Free}_\mathrm{inj}$.
  • $\mathsf{Set}_\mathrm{inj} \to \mathsf{Ab}_\mathrm{pure}$, powerful image is $\mathsf{Free}_\mathrm{pure}$.
  • $\mathsf{Set}_\mathrm{inj} \to \mathsf{Ab}_\mathrm{elem}$, powerful image is $\mathsf{Free}_\mathsf{elem}$.

(For the last two, you need to know that the homomorphism of free abelian groups induced by an inclusion of basis elements is elementary -- this takes a bit of quantifier elimination. We might have to restrict to infinite sets, but this should be fine.) So if there are arbitrarily large compact cardinals, we get the affirmative answer in each case: the category of free abelian groups is accessible.

The Negative Answer:

The negative answer follows from the following result of Eklof and Mekler (a consequence of their "Main Theorem"):

Theorem Assume $V=L$. Then for any successor cardinal $\kappa$, there is an abelian group of cardinality $|A| = \kappa$ such that

  1. Every subgroup $S \subset A$ with $|S| < \kappa$ is free.
  2. $A$ is not free.

From this, we deduce (under $V=L$) that the category $\mathsf{Free}$ of free abelian groups is not accessible as follows. Suppose for contradiction that $\mathsf{Free}$ is accessible. Then every object of $\mathsf{Free}$ is presentable. Let $\kappa$ be a successor cardinal such that $\mathbb{Z}$ is $\kappa$-presentable and $\mathsf{Free}$ is $\kappa$-accessible. By the theorem, there is an non-free abelian group $A$ of cardinality $|A| = \kappa$ such that every subgroup $S \subset A$ with $|S| < \kappa$ is free. Let $P$ be the poset of subgroups of $A$ of size $< \kappa$, and let $F: P \to \mathsf{Free}$ be the obvious functor. Since $P$ is $\kappa$-directed and $\mathsf{Free}$ is $\kappa$-accessible, $\varinjlim F$ exists in $\mathsf{Free}$, and since $\mathbb{Z}$ is $\kappa$-presentable, the underlying set of $\varinjlim F$ can be identified with $A$ in the obvious way. Now the fact that the colimit inclusions are abelian group homomorphisms allows us to compute that the addition of $\varinjlim F$ is exactly that of $A$ under the identification we've made, i.e. $\varinjlim F \cong A$ as abelian groups. But $\varinjlim F$ is free, while $A$ is not, a contradiction.

This completes the proof for $\mathsf{Free}$ and $\mathsf{Free}_\mathrm{inj}$-- to get the others we just modify $P$ and $F$ to use just elementary subgroups -- the Downward Lowenheim-Skolem theorem tells us that these posets are still $\kappa$-directed.

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  • $\begingroup$ Let me just add that I find this result truly astounding. I had always assumed that for any "reasonable" category the question of its accessibility should not be independent of ZFC. $\endgroup$ – tcamps Jul 6 '16 at 22:33
  • $\begingroup$ Maybe because I'm well-trained at independence; maybe because a lot of the results on free abelian groups gew out of Jerusalem. I don't know. But it's not very surprising. It's very nice, and it's a huge "in your face" to mathematicians that feel that set theorists are like children playing with unimportant nonsensical "independence" of uninteresting statements. I love it. There are some results about free abelian groups whose only known proof is via heavy set theoretic tools, that's even better! $\endgroup$ – Asaf Karagila Jul 7 '16 at 6:42
  • $\begingroup$ Also, when you/they say "compact cardinal", do you mean weakly/strongly/super-? $\endgroup$ – Asaf Karagila Jul 7 '16 at 6:43
  • $\begingroup$ Makkai and Paré just say "compact", but from these slides I gather they mean "strongly compact". If you look a couple slides further, you'll also see that this has been improved a bit by Andrew Brooke-Taylor and Jiří Rosický to "strongly $\gamma$-compact". Makkai and Paré's proof of the powerful image theorem is very short, I think I'm going to sit down with it today. $\endgroup$ – tcamps Jul 7 '16 at 12:01
  • $\begingroup$ Thanks. I suspected as much, but wasn't sure. $\endgroup$ – Asaf Karagila Jul 7 '16 at 12:08

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