5

Which first-order theories have models uniquely determined (up to isomorphism) by their automorphism groups?

For the purposes of this question, I'm assuming that a model cannot be empty.

I will nonstandardly call a ring that is not necessarily commutative an ordinary ring.

I was wondering the other day whether models of Peano Arithmetic are uniquely determined by their automorphisms. The standard model of arithmetic has no nontrivial automorphisms (because zero and all of its successors are fixed), but the other models have nonstandard elements that seem less nailed down at first glance. It turns out that this is not true by the argument in this answer and there are many nonstandard models of arithmetic with no nontrivial automorphisms.

Then I thought about theories where models are uniquely determined (up to isomorphism) by their automorphism group and came up with the empty theory $T = \varnothing$. A model of the empty theory is essentially just a set, and two sets are isomorphic iff they have the same cardinality. Their automorphism groups are the symmetric groups, which are also isomorphic iff they have the same cardinality.

Groups are not uniquely determined by their automorphism groups, by the argument in this answer.

Ordinary rings are not uniquely determined by their automorphism groups. By the argument in this answer there is a non-commutative ring with a trivial automorphism group. The ordinary ring $\{0, 1\}$ also has a trivial automorphism group.

Greg Nisbet
  • 11,657
  • 5
    It's true that sets are determined up to isomorphism by their automorphism groups, but not quite for the reason you said: for example, it's consistent with ZFC that $|S_{\aleph_0}|=2^{\aleph_0}=2^{\aleph_1}=|S_{\aleph_1}|$, so it's not provable that symmetric groups are isomorphic if and only if they have the same cardinality. But ZFC does prove that if $S_X \cong S_Y$, then $X$ and $Y$ have the same cardinality. See the discussion here: https://mathoverflow.net/questions/12943/can-the-symmetric-groups-on-sets-of-different-cardinalities-be-isomorphic – Alex Kruckman Sep 11 '21 at 16:24
  • Thanks for the link. It's also not true unless I rule out the empty model, as pointed out by this comment. – Greg Nisbet Sep 11 '21 at 17:12
  • 5
    nice question! this is a highly pedantic point, but I'm not sure if I would call the ring $\mathbb{F}_2$ the trivial ring – normally a trivial object is at least a terminal or initial object in the category of interest, and $\mathbb{F}_2$ is neither terminal nor initial in the category of rings. I think the best choice for a "trivial ring" would be the zero ring ${0}$, which is the terminal object in the category of rings. (this ring does of course have trivial automorphism group, so as an example it still works) – Atticus Stonestrom Sep 11 '21 at 17:24
  • 2
    How are you construing ${0,1}$ as a non-commutative ring, exactly? – Noah Schweber Sep 11 '21 at 17:30
  • 1
    I meant non-commutative as not necessarily commutative. – Greg Nisbet Sep 11 '21 at 17:59
  • 2
    I fixed up the post. I'm not calling $\mathbb{F}_2$ trivial anymore and I'm calling a not-necessarily-commutative ring an "ordinary ring". I'm really not sure how to describe a not-necessarily-commutative ring in a way that's less verbose than "not necessarily commutative ring". – Greg Nisbet Sep 11 '21 at 19:31
  • 3
    This is a great question! Already the following very special case seems non-trivial and interesting: Let $T$ be a strongly minimal theory. For models $M,N\models T$, does $\text{Aut}(M)\cong \text{Aut}(N)$ imply $M\cong N$? The answer is yes for sets and vector spaces over a fixed field, and I believe the answer should be yes in general, but I don't see an easy proof. – Alex Kruckman Sep 14 '21 at 01:59
  • 3
    Does anyone know the answer for algebraically closed fields? Note that there are two versions of the question, depending on whether the automorphism groups are required to be isomorphic as groups or as topological groups. – Alex Kruckman Sep 14 '21 at 02:06
  • @AlexKruckman the ACF case seems very natural; is it even clear that it holds in the countable case? (eg is there an easy way to show for example that $\overline{\mathbb{Q}}$ and $\overline{\mathbb{Q}(t)}$ have non-isomorphic automorphism groups?) – Atticus Stonestrom Sep 14 '21 at 09:37
  • I guess for the more general strongly minimal case, do we want to make some assumption that $L$ is finite? otherwise I think we can get silly counterexamples eg by taking $L=(c_i)_{i\in\omega}$ to just consist of consist symbols and taking $T$ to just assert that $c_i\neq c_j$ for each $i\neq j$. then the models $M=\omega$ and $N=\omega\cup{\star}$, with each $c_i$ just interpreted as $i\in\omega$, are both rigid, but not isomorphic as $L$-structures – Atticus Stonestrom Sep 14 '21 at 09:40
  • 1
    @AtticusStonestrom Ha! Nice counterexample. Sure, we can specify finite language, but your example makes me doubt that the conjecture is true, even with this adjustment. – Alex Kruckman Sep 14 '21 at 11:47
  • 2
    I think the theory of a single equivalence relation $\simeq$ works. If all the elements of the domain are equivalent to each other, then this works by the same argument as for the empty theory. If there's more than one equivalence class, then any automorphism factors into an automorphism that preserves each equivalence class, and one that "swaps" elements between equivalence classes of the same cardinality. (This might already be included in the more general classes of theories that others have mentioned.) – Greg Nisbet Sep 14 '21 at 20:18
  • Also also, I think there's a broad class of not-that-interesting examples that we can get by imposing the requirement that no model has cardinality $\ge n$ for some fixed $n \in \mathbb{N}$, which is a first-order condition. We may want to impose the requirement that $T$ has arbitrarily large finite models or at least one infinite model or something similar. – Greg Nisbet Sep 14 '21 at 21:57
  • 3
    (@AlexKruckman alternatively, maybe the "right" specialization would be to only consider models of $T$ of size $>\aleph_0$, where $T$ is countable and strongly minimal? then the question would just reduce to whether automorphism groups can detect the cardinality of sufficiently large models, which seems closest in spirit to Gregory's original example of the theory of infinite sets) – Atticus Stonestrom Sep 15 '21 at 02:48

0 Answers0