Compatibility of $\mathsf{SVC}$ and Reinhardtness

Question

Can we prove the consistency of $\mathsf{ZF+SVC}$ + "There is a Reinhardt cardinal?" (Preferably from the consistency of $\mathsf{ZF}$ with a Reinhardt cardinal, but using a stronger assumption is also okay.)

Here $\mathsf{SVC}$ means the Small Violation of Choice, claiming the axiom of choice is forcible by a set forcing.

Here are some easy observations.

• Let $\kappa$ be a critical point of $j\colon V\to V$. Then no forcing $\mathbb{P}\in V_\kappa$ can force $\mathsf{AC}$. Similarly, no forcing $\mathbb{P}\in V_{j^\omega(\kappa)}$ can force $\mathsf{AC}$ (as $j^n(\kappa)$ is also a critical point of some elementary embedding $V\to V$.)

• If there is a super Reinhardt cardinal $\kappa$, then $\mathsf{SVC}$ fails: if there is a $\mathbb{P}$ forcing $\mathsf{AC}$, then we can pull it back to $V_\kappa$ by using $j\colon V\to V$ satisfying $j(\kappa)>\operatorname{rank} \mathbb{P}$.

Answer 1

No, a Reinhardt cardinal implies SVC is false.

First, if there is a Reinhardt cardinal, then by Woodin's proof of the Kunen inconsistency theorem, for sufficiently large regular cardinals $\delta$, the set $S^\delta_\omega$ of ordinals of cofinality $\omega$ cannot be partitioned into $\delta$-many disjoint stationary sets.

On the other hand, SVC implies that for all sufficiently large regular cardinals $\delta$, every stationary subset of $\delta$ can be partitioned into $\delta$-many stationary sets: to see this, let $\mathbb P$ be a partial order that forces choice, and let $\delta$ be a regular cardinal such that $|\mathbb P| < \delta$ in $V^{\mathbb P}$. Fix any stationary subset $S$ of $\delta$. In $V^\mathbb P$, $S$ remains stationary (by a standard argument, a club in the extension contains a club in the ground model) and there is a partition of $S$ into stationary sets $\langle S_\alpha\rangle_{\alpha < \delta}$ by Solovay's theorem. For each $p\in \mathbb P$, let $S^p_\alpha = \{\xi < \delta : p\Vdash \xi\in S_\alpha\}$ (fixing a name, etc, etc). We claim there is some $p\in \mathbb P$ such that $A_p = \{\alpha < \delta : S^p_\alpha\text{ is stationary}\}$ has cardinality $\delta$. If not, $|\bigcup_{p\in \mathbb P} A_p| < \delta$, so there is some $\alpha\in \delta\setminus \bigcup_{p\in \mathbb P} A_p$. But then in $V^{\mathbb P}$, $S_\alpha \subseteq \bigcup_{p\in \mathbb P} S^p_\alpha$ is a stationary set which is contained in the union of fewer than $\delta$ nonstationary sets, which is a contradiction. So fix $p$ such that $|A_p| = \delta$. Then in $V$, we have partitioned $S$ into $\delta$-many stationary sets of the form $S_\alpha^p$.