Suppose $j\colon V\to M$ is an elementary embedding and $\kappa$ is the critical point of $j$, then $\kappa$ is measurable, and we can define the ultrafilter $U$ over $\kappa$ as: $$A\in U\iff \kappa\in j(A)$$

This is a normal ultrafilter. Despite not seeing an actual example, I am aware that $\operatorname{Ult}(V,U)$ may not be $M$ itself (where $\operatorname{Ult}(V,U)$ is the transitive collapse of the ultrapower of $V$ by $U$).

I have two questions in this regard:

  1. When does $\operatorname{Ult}(V,U)=M$? (Except, of course, the trivial case where $M$ was already defined as that ultrapower)
  2. Suppose $\kappa$ is supercompact, and $M$ witnesses some $\lambda$-supercompactness, we can of course repeat the above construction, is there any extra properties we can say about $\operatorname{Ult}(V,U)$ in relation to $M$? (some further closure properties, or so...)

2 Answers 2


I would answer question 1 by saying that $M=\text{Ult}(V,U)$ if and only if $j$ is isomorphic to an ultrapower by some normal measure. This is another way of saying that normal measures are minimal with respect to the Rudin-Kiesler order. The point is that an embedding is the ultrapower by a normal measure if and only if it is the ultrapower by its induced normal measure, if and only if $\kappa$ as a seed generates the whole embedding, in the sense that every element of $M$ has the form $j(f)(\kappa)$.

For question 2, suppose that $j:V\to M$ witnesses the $\lambda$-supercompactness of $\kappa$, so that $\kappa$ is the critical point of $j$ and $M^\lambda\subset M$. Your ultrafilter $U$ is what is sometimes called the induced normal measure of $j$, and since the seed hull $X_\kappa=\{j(f)(\kappa)\mid f:\kappa\to V\}$ is an elementary substructure of $M$, we may collapse it and form a commutative diagram, with $V\to M_U\to M$, where $j_U:V\to M_U=\text{Ult}(V,U)$ is the ultrapower by $U$ and $k:M_U\to M$ is the inverse collapse of $X_\kappa$ and $j:V\to M$ is the composition $j=k\circ j_U$.

  • insert triangular commutative diagram here

Since $M^\lambda\subset M$ in $V$ and $M_U\subset V$, it follows immediately that $M^\lambda\cap M_U\subset M$. Thus, $M$ remains $\lambda$-closed for sequences in $M_U$. However, when $\lambda\gt\kappa$, then it will never be the case that $M\subset M_U$. This is because $M_U$ is an ultrapower by a filter on $\kappa$ and therefore $j_U$ is continuous at ordinals of cofinality $\kappa^+$. Therefore, $j_U(\kappa^+)$ has true cofinality $\kappa^+$ in $V$ and hence also in $M$, but it is regular in $M_U$, being a successor cardinal there. Thus, although the quotient map $k:M_U\to M$ is a $\lambda$-closed embedding, it can never be an internal embedding from the perspective of $M_U$, since $M$ is not a subclass of $M_U$.

However, if you are willing to consider the case $\lambda=\kappa$, then it is possible for $k:M_U\to M$ to be internal to $M_U$, for we may simply let $j$ be the ultrapower by $U\times U_1$ for any other measure on $\kappa$, so that $k$ is simply the ultrapower in $M_U$ by $j_U(U_1)$. In this case, $M$ is $j_U(\kappa)$-closed in $M_U$, since it is an internal ultrapower there by a measure on $j_U(\kappa)$.

  • $\begingroup$ @Joel: Thanks a lot. If I get it correct then in the supercompact case we have that it is never an internal embedding. But if we reduce ourselves to $\lambda=\kappa$ (which is essentially the measurable case again) then it is possible for the embedding to be internal, can we deduce anything on the relationship in the Mitchell order between the original measure and the normal one? (Assuming, of course, the original $M$ is an ultrapower itself) $\endgroup$
    – Asaf Karagila
    Commented Aug 9, 2011 at 12:10

This "answer" to question 1 is more or less tautologous, but I hope it might be of some use anyway. $\text{Ult}(V,U)=M$ just in case $M$ is generated by the range of $j$ plus the single element $\kappa$ --- roughly speaking, $M$ is obtained from (a copy of) $V$ by adjoining a single element that realizes the 1-type $U$. Here "generated" and "adjoining" can be understood simply as applying functions (in $M$) of the form $j(f)$ (with $f\in V$) to $\kappa$.

Notice also that, once you have an elementary embedding $j:V\to M$, you can compose it with any elemetary $k:M\to N$ whose critical point is strictly above $\kappa$, and the composition will be another elementary embedding $j':V\to N$ inducing the same ultrafilter $U$. So there can be considerable flexibility about $M$ even when $U$ is fixed.

  • $\begingroup$ Now that you mention that, I think that I may have seen this sort of example of why $M$ need not be equal to $\operatorname{Ult}(V,U)$. Many thanks for the answer! $\endgroup$
    – Asaf Karagila
    Commented Aug 4, 2011 at 19:38
  • 1
    $\begingroup$ @Andreas: your characterization can be fine-tuned by adding $Ult(V,U)=M$ precisely when every element of $M$ is definable in $M$ by a first order formula whose parameters are allowed to come from {$\kappa$} $\cup$ the range of $j$. $\endgroup$
    – Ali Enayat
    Commented Aug 4, 2011 at 20:15

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