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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.

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    $\begingroup$ 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: mathoverflow.net/questions/12943/… $\endgroup$
    – Alex Kruckman
    Sep 11, 2021 at 16:24
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    $\begingroup$ 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) $\endgroup$
    – Atticus Stonestrom
    Sep 11, 2021 at 17:24
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    $\begingroup$ 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. $\endgroup$
    – Alex Kruckman
    Sep 14, 2021 at 1:59
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    $\begingroup$ 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. $\endgroup$
    – Alex Kruckman
    Sep 14, 2021 at 2:06
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    $\begingroup$ (@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) $\endgroup$
    – Atticus Stonestrom
    Sep 15, 2021 at 2:48

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