A friend of mine and I ran into the following question while reading about proper forcing, and have been unable to resolve it:

*Definition.* A cardinal $\kappa$ is *supercompact* if for all ordinals $\lambda$, there exists a transitive inner model $M_\lambda$ of $V$ and an elementary embedding $j_\lambda: V\rightarrow M_\lambda$ such that $crit(j_\lambda):=\min\lbrace\alpha\in ON: j(\alpha)\not=\alpha\rbrace=\kappa$ and $M_\lambda$ is closed under $\lambda$-sequences (that is, $^\lambda M_\lambda\subseteq M_\lambda$).

Now in the definition of supercompactness, the inner models $M_\lambda$ are allowed to vary with $\lambda$. We could define a large cardinal notion, *uniform supercompactness*, by removing this possibility: $\kappa$ is uniform supercompact if there is a single $M$ and $j$ such that $M$ is closed under all sequences of ordinal length and $j: V\rightarrow M$ is an elementary embedding with critical point $\kappa$. But in ZFC, this is not an interesting notion. First, assuming choice, $M$ is closed under sequences of ordinal length if and only if $M=V$ (since two models with the same ordinals and sets of ordinals are equal), so $ZFC\models$ "$\kappa$ is uniformly supercompact $\iff$ $\kappa$ is Reinhardt;" second, Kunen showed that Reinhardt cardinals are inconsistent with $ZFC$. (EDIT: as Joel points out, this statement doesn't actually make sense, as Reinhardt-ness is not a first-order property. One needs to pass to some extension of ZFC which can talk about proper classes directly, like Morse-Kelly set theory, which is the setting Kunen used for his proof.)

However, assuming $\neg$Choice, both of these points seem suspect. It is still open whether Reinhardt cardinals are consistent with $ZF$, and it is unclear to me that $ZF$ alone proves that uniformly supercompact cardinals are Reinhardt. So my questions are the following:

Question 1: Does $ZF\models$ every uniformly supercompact cardinal is Reinhardt?

Question 2: What is the consistency strength of $ZF+$ there exists a uniformly supercompact cardinal?

Obviously $ZF$ proves that every Reinhardt cardinal is uniformly supercompact; also obviously, Question 2 is really only interesting in lieu of a positive answer to Question 1. The most basic obstruction to a positive answer to Question 1 is the fact that given an elementary embedding $j: V\rightarrow M$ with $M$ closed under ordinal-length sequences and $crit(j)=\kappa$, the restriction of $j$ to $M$, $j\upharpoonright M: M\rightarrow M$ is not obviously (to me at least) an elementary embedding.

The Axiom of Choice, Ch. 5, Problem 17). This, however is "going up" while inner models are "going down" so there might be some difference there. $\endgroup$ – Asaf Karagila Apr 23 '12 at 6:21sets$A$, then it follows easily that $M=V$ even in ZF, since if $A$ is transitive, then $A$ is the Mostowski collapse of $j"A$. $\endgroup$ – Joel David Hamkins Apr 23 '12 at 14:34