When it is said that Kunen inconsistency theorem proves that given $\sf ZFC$ no elementary embedding can exist from the universe to itself. Most references quote full choice in stating that result, and contemplate a salvage by forsaking choice altogether like in Reinhardt's cardinals setting.
Is there a known weaker form of choice, like dependent or countable choice, that can evade this theorem?
Work of Usuba combined with work of Woodin shows that if there is a Reinhardt cardinal $\kappa$ that is a limit of Lowenheim-Skolem cardinals, then there is a forcing extension in which $\kappa$ remains a Reinhardt cardinal but $\text{DC}_\kappa$ holds. This means one has $\text{DC}_\lambda$ where $\lambda = \sup_{n < \omega} j^n(\kappa)$ where $j : V\to V$ is elementary with critical point $\kappa$. In this context, $\lambda^+$ must be a measurable cardinal, so one really cannot have much more choice than $\text{DC}_\lambda$; e.g., $\text{DC}_{\lambda^+}$ and even $\text{AC}_{\lambda^+}$ fail.
The consistency proof uses Woodin's Easton iteration of collapse forcings as in SEM 1 Theorem 226 but substituting Usuba's Proposition 4.7. One only iterates up to $\lambda$ (so one is not forcing full choice, which of course would kill the Reinhardt). Then one has to lift $j$ to the forcing extension, which uses the master condition argument for rank-to-rank embeddings which can be found in Section 5 of Hamkins's "Fragile measurability" paper, in the context of $I_1$.
There are also "global" forms of choice that are consistent with Reinhardts relative to stronger principles. For example, the axiom WISC follows from a Reinhardt cardinal plus a proper class of Lowenheim-Skolem cardinals, and so in particular, Reinhardt plus WISC follows from a super-Reinhardt cardinal (or just a Reinhardt and a proper class of supercompacts).
In particular, from choiceless large cardinal axioms beyond a Reinhardt (e.g., a Berkeley cardinal) one can prove the consistency of Reinhardts with weak forms of choice. It is an open question whether, for example, Reinhardt plus DC can be proved consistent starting from a single Reinhardt.
