If $U,D$ are $\kappa$-complete nonprincipal ultrafilters on $\kappa$ and $j_U(U) = j_D(D)$, is $U=D$?

Question

Here, $\, j_U, \, j_D$ are the canonical elementary embeddings induced by $U,D$ respectively.

I note that it is consistent with the existence of a measurable that the answer be yes: it is true in the model $L[D]$ for $D$ a measure on $\kappa$.

Answer 1

It is consistent with ZFC that the answer is no, but under the Ultrapower Axiom, the answer is yes, not only for $\kappa$-complete ultrafilters on $\kappa$, but also for arbitrary countably complete ultrafilters.

First I'll show that in the Kunen-Paris model, there exist distinct normal ultrafilters $U_0$ and $U_1$ such that $j_{U_0}(U_0) = j_{U_1}(U_1)$. Moreover, $M_{U_0} = M_{U_1}$.

Let $U$ be a normal ultrafilter on a measurable cardinal $\kappa$ and let $j : V\to M$ denote its ultrapower. Assume $2^\kappa = \kappa^+$. Let $\mathbb P$ be the Easton product $\prod_{\delta\in I}\text{Add}(\delta,1)$ where $I\subseteq\kappa$ is a $U$-null set of cardinals. Let $\mathbb Q = j(\mathbb P)$ and let $\mathbb Q/\mathbb P$ denote the product $\prod_{\delta\in j(I)\setminus \kappa}\text{Add}(\delta,1)$ as computed in $M$. Thus $\mathbb Q \cong \mathbb P\times (\mathbb Q/\mathbb P)$. Since $\kappa\notin j(I)$, $\mathbb Q/\mathbb P$ is ${\leq}\kappa$-closed and $j(\kappa)$-cc in $M$, so by standard arguments, one can construct an $M$-generic filter $G\subseteq \mathbb Q/\mathbb P$ in $V$.

Let $H\subseteq \mathbb P$ be a $V$-generic filter. Let $j_0:V[H]\to M[H\times G]$ be the unique lift of $j$ such that $j_0(H) = H\times G$. Let $\sigma_{\alpha,\kappa}$ denote the automorphism of $\mathbb P$ given by $$\sigma_{\alpha,\kappa}((p_\delta)_{\delta\in I}) = (p_\delta)_{\delta\in I\cap \alpha}{}^\frown(p^*_\delta)_{\delta\in I\setminus \alpha}$$ where for $q\in \text{Add}(\delta,1)$, $q^*$ denotes the result of flipping the bits in $q$. Denote the similar automorphism of $\mathbb Q$ by $\sigma_{\alpha,j(\kappa)}$. Let $j_1 : V[H]\to M[H\times G]$ be the lift of $j$ such that $j_1(H) = \sigma_{\kappa,j(\kappa)}(H\times G)$.

Now it's time to show $j_0(j_0) = j_1(j_1)$. Since $j_0(j_0)\restriction M = j_1(j_1)\restriction M$, it suffices to show that $j_0(j_0)(H\times G) = j_1(j_1)(H\times G).$ This follows from a long fun computation:

\begin{align*} j_0(j_0)(H\times G) &= j_0(j_0)(j_0(H))\\ &= j_0(j_0(H))\\ &= j_0(H\times G)\\ &= j_0(H)\times j(G)\\ &= H\times G\times j(G)\\ &= \sigma_{\kappa,j(j(\kappa))} \circ \sigma_{\kappa,j(j(\kappa))}(H\times G\times j(G))\\ &= \sigma_{\kappa,j(j(\kappa))}\circ \sigma_{j(\kappa),j(j(\kappa))} (\sigma_{\kappa,j(\kappa)}(H\times G)\times j(G))\\ &= \sigma_{\kappa,j(j(\kappa))}\circ \sigma_{j(\kappa),j(j(\kappa))} (j_1(H\times G))\\ &= \sigma_{\kappa,j(j(\kappa))} (j_1(\sigma_{\kappa,j(\kappa)}(H\times G)))\\ &= \sigma_{\kappa,j(j(\kappa))}(j_1(j_1(H)))\\ &= \sigma_{\kappa,j(j(\kappa))}(j_1(j_1)(j_1(H)))\\ &= j_1(j_1)(\sigma_{\kappa,j(\kappa)}(j_1(H)))\\ &= j_1(j_1)(j_0(H))\\ &= j_1(j_1)(H\times G) \end{align*}

Finally, let $U_0$ and $U_1$ be the normal ultrafilters derived from $j_0$ and $j_1$. Since $j_0(j_0) = j_1(j_1)$, $j_0(U_0) = j_1(U_1)$, as desired.

Second I'll sketch a proof that under the Ultrapower Axiom, the answer to your question is yes for arbitrary countably complete ultrafilters.

I'll do this by answering Trevor's question from the comments:

Fact (UA). Suppose $U_0$ and $U_1$ are countably complete ultrafilters on ordinals $\delta_0$ and $\delta_1$. Let $j_0 :V\to M_0$ and $j_1:V\to M_1$ denote their ultrapower embeddings, and assume $j_{0}(P(\delta_0)) = j_{1}(P(\delta_1))$. Then $j_0 = j_1$.

The fact suffices, since if $j_0(U_0) = j_1(U_1)$, the hypotheses of the fact hold, and hence $j_0 = j_1$, which means $j_0(U_0) = j_0(U_1)$, so $U_0 = U_1$. I'll sketch a direct proof of the fact assuming $2^{{<}\delta_0} = \delta_0$, although with significantly more work, one can do without.

Proof of fact. Apply UA to obtain internal ultrapower embeddings $i_0 : M_0\to N$ and $i_1:M_1\to N$ such that $i_0\circ j_0 = i_1\circ j_1$. Let $\alpha_0 = [\text{id}]_{U_0}$, $\alpha_1 = [\text{id}]_{U_1}$, and $\delta_* = j_0(\delta_0) = j_1(\delta_1)$. Note that $U_0$ is the ultrafilter derived from $i_0\circ j_0$ using $i_0(\alpha_0)$ and likewise for $U_1$, so if $i_0(\alpha_0) = i_1(\alpha_1)$, then $U_0 = U_1$, and we're done. So assume without loss of generality that $i_0(\alpha_0) < i_1(\alpha_1)$. Let $D$ be the $M_1$-ultrafilter on $\alpha_1$ derived from $i_1$ using $i_0(\alpha_0)$, let $k_1 : M_1\to P$ be its ultrapower, and let $\ell:P \to N$ be the factor map. One can define $k_0 : M_0\to P$ by $k_0([f]_{U_0}) = [j_1(f)]_D$. Since $\ell\circ k_0 = i_0$ is an internal ultrapower embedding, one can conclude that $k_0$ is too. Since $D$ is an ultrafilter on an ordinal less than $\delta_*$, $D$ is coded in $M_1$ by a subset of $\delta_*$ (using that $2^{<\delta_*} = \delta^*$ in $M_1$), so $D\in M_0$. One can show $j_D^{M_0}\restriction \text{Ord} = k_1\restriction \text{Ord}$, and as a consequence $M_1\subseteq M_0$: if $A$ is a set of ordinals in $M_1$, then $k_1(A)\in P\subseteq M_0$, so $k_1(A)\in M_0$, so $A = (k_1\restriction \text{Ord})^{-1}[k_1(A)]\in M_0$. Under UA, $M_1\subseteq M_0$ implies that there is an internal ultrapower embedding $h : M_0\to M_1$ such that $h\circ j_0 = j_1$. (See Corollary 5.4.21 here.) From the perspective of $M_0$, the embedding $h$ preserves the powerset of $\delta_*$. But $h(\delta_*) = h(j_0(\delta_0)) = j_1(\delta_0)\leq j_1(\delta_1) = \delta_*$, so by the Kunen inconsistency theorem, $j\restriction\delta_*$ is the identity. Therefore $h$ is surjective: if $a\in M_1$, $a = j_1(f)(\alpha_0) = h(j_0(f)(\alpha_1))$. So $h$ is the identity, and since $h\circ j_0 = j_1$, we finally conclude that $j_0 = j_1$.