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...)



Ultrapowers by normalized ultrafilters
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 $j\_U(\kappa)$.


