# Is there a minimal inner model for determinacy?

Assume $$\sf ZF+AD$$. Is there some inner model $$M$$ containing all the ordinals such that $$M\models\sf ZF+AD$$ as well?

What if we require $$\omega_1$$ and/or $$\omega_2$$ to be computed correctly?

Can we say anything about these models (e.g. $$M\models V=L(\Bbb R)$$)?

Is it at all consistent?

There's no real reason to expect a minimal model, of course, since Woodin cardinals are involved. But $$\sf AD$$ is also a very strong axiom and might have unexpected consquences.

Assuming AD, Woodin showed that no inner model that is missing a real correctly computes $$\omega_1$$. (This almost follows from Theorem 9 of Velickovic-Woodin's "Complexity of the set of reals of inner models of set theory.") Therefore if you take an inner model $$M$$ of AD + $$V = L(\mathbb R)$$ whose $$\omega_1$$ is as small as possible, this model has no proper inner model of AD. (This argument shows that inclusion is a wellfounded partial order of the inner models of $$\text{AD}+ V = L(\mathbb R)$$...)

If one generically wellorders the reals of $$M$$ without adding reals, one obtains a model of ZFC in which there is a minimum inner model of $$\text{AD}$$. Update: this might be true, but as Dmytro Taranovsky points out in the comments, it is far from clear.

In general, however, there need not be a minimum inner model of $$\text{AD}$$. Assume there is a model of $$\text{AD}$$ containing only countably many reals. (This hypothesis is justified in the following paragraph.) Let $$M$$ be a minimal model of $$\text{AD}$$ with this property. Then there is a real $$g$$ that is $$M$$-generic for Cohen forcing. Let $$N = L(\mathbb R)^{V[g]}$$. By another theorem of Woodin, there is an elementary embedding $$j : M\to N$$. (There are more details in Kechris-Woodin's "Generic codes for uncountable ordinals.") Therefore $$N$$ is a minimal model of $$\text{AD}$$, since this is first order expressible, and $$N$$ is obviously distinct from $$M$$.

Of course, after forcing with $$\text{Col}(\omega,\mathbb R)$$ over a model of $$\text{AD}^{L(\mathbb R)}$$, there is an inner model of $$\text{AD}$$ that contains countably many reals, so the hypothesis of the above paragraph is consistent. In fact, the hypothesis is true: using $$\mathbb R^\#$$, one can build an inner model containing only countably many reals that is elementarily embeddable into $$L(\mathbb R)$$. (Take a countable elementary substructure of the mouse $$\mathbb R^\#$$ and then iterate away the top measure.) Therefore large cardinals imply that there is no minimum inner model of $$\text{AD}$$.

• I, for one, would love to see this proof. I hope it is published somewhere... Jul 26, 2020 at 15:53
• Yes there will be. Start with a minimal model $M$ with countably many reals. Then there is a Cohen real $g$ over $M$. Now $N= L(\mathbb R)^{M[g]}$ is elementarily equivalent to $M$ (another theorem of Woodin, but this is one is published in Woodin-Kechris “Generic codes for uncountable ordinals”), so $N$ is a minimal model distinct from $M$, but $g\in M[g]\setminus M$. (It may be counterintuitive that $M$ is not contained in $N$.) Jul 26, 2020 at 16:11
• @FrançoisG.Dorais I found a reference for Woodin's theorem and updated the answer. Jul 26, 2020 at 22:55
• For a variation on the question, we have a strong antiminimality statement: If an inner model $M⊨\text{AD}$ and $|ℝ^M|=ω$, then there is an inner model $L(R')⊨\text{AD}$ with $ℝ^M=ℝ^{L(R')[g]}$ and $g∈ℝ$ Cohen generic over $L(R')$. Assuming the axiom of choice, does this also hold for uncountable $ℝ^M$ (which would contradict the second paragraph of the answer)? On the other hand, the minimal inner model with the reals closed under $M_ω^\#$ is in some ways the minimal canonical inner model of AD (even though it is not a minimal inner model of AD). Aug 1, 2020 at 20:27
• Using the converse of the Derived Model Theorem and $M⊨\text{AD}^+$, we get $M$ as a submodel of a generic extension. We can then 'factor out' one of the generic reals to get a generic $M'⊨\text{AD}$ (with $ℝ^{M'}⊂ℝ^M$) missing that real; and using $|ℝ^M|=ω$, $M'$ already exists in $V$. (P.S. Note that for a comment without names, only the author of the answer gets notified.) Aug 1, 2020 at 23:55