Minimal model of ZFC without power set axiom

Question

We know that $L$ is the minimal standard model of ZFC.

The question is, what is the minimal "standard" model of ZFC$^-$ (meaning ZFC without the Power Set axiom)?

This is really two questions:

1. Is there such a minimal model (this is non-trivial, but sound intuitively true to me)? If so, let's call it $L^-$.
2. Assuming $L^-$ exists, is it equal to, or is it a proper subset of $L$ (obviously $L$ itself is a model of ZFC$^-$ so $L^- \subseteq L$)?

Both possible answers for (2) seem puzzling to me. If $L^-=L$ then somehow the Power Set axiom is sufficiently embedded in ZFC$^-$, that even if we don't assume it, it kind of crops up by itself in the minimal model (kind of the way AC and GCH do in $L$).

If $L^-\subsetneq L$ then a whole host of new questions pop up - like whether we have AC if we only assume ZF$^-$ to begin with, and what happens to all the other nice properties $L$ has. Furthermore, if $L^-$ doesn't have the exact same cardinals as $L$, then it appears to me we have some ordinal in $L$ that the power set axiom was necessary to show it has the same cardinality (which sounds peculiar because the bijection has the same cardinality, so why did we "need" anything bigger to construct it? I would have thought the axiom of replacement would be enough).

Answer 1

I posted the question (with more precise phrasing) in MathOverflow and Prof. Hamkins answered it quite elegantly. Basically, you can construct $L$ in ZF$^-$. Therefore $L = L^-$ and it is the unique least inner model.

Bottom line: If $V$ satisfies ZFC then $L^- = L$.

https://mathoverflow.net/questions/235896/least-inner-model-of-zf-without-power-set-axiom

Answer 2

In ZFC you can not prove the existence of a minimal model, even assuming ZFC is consistent. Since you would have a minimal model of ZFC if you had your $L^-$, by taking closure under the operation of power set, you can not have $L^-$ either (as a consequence of ZFC).

So, basically, we can't know the answer to your question without assuming consistency of a theory not strictly contained in ZFC. I'd be weary of any such proof.

Answer 3

This isn't an answer, but a bit long for a comment; hope it helps.

First, there is nothing 'obvious' about the existence of a minimal model. In practice, you could very well be able to prove the existence of some set $X$ such that you can't prove what is exactly its structure, such that there are two models of $X$, one being 'large' and the other 'deep' (in Hamkins words, one fat and one tall, more on that below), so that there is no embedding of one in the other, and both are minimal. So there's some (nontrivial) work to do to show that.

You could generalize this question of non-embedability to "are there two models that can't be embedded into each others?". Turns out that this question was answered negatively for ZFC by Joel D. Hamkins [0]. Now, since ZFC minus powerset is a strictly weaker axiomatic system, the result could not apply. It turns out this specific axiomatic system was also studied by Hamkins and co-authors [1], and results in the paper include reasons to be worried. Now your best bets are to either work the answer as a corollary of these papers, or contact him directly.

[0] J. D. Hamkins, “Every countable model of set theory embeds into its own constructible universe,” J. Math. Log., vol. 13, iss. 2, p. 1350006, 27, 2013.

[1] Gitman, Victoria, Joel David Hamkins, and Thomas A. Johnstone. "What is the theory ZFC without power set?." arXiv preprint arXiv:1110.2430 (2011).