$\kappa$ is measurable if $\kappa$ is the critical point of a non trivial elementary embedding $j:V \rightarrow M$ of $V$ into a transitive class $M$ (with $j$ a class function on $V$).
Given cardinal $\kappa$ and ordinal $\lambda$, then $\kappa$ is $\lambda$-supercompact if there is an elementary embedding $j:V \rightarrow M$ with $\operatorname{cp}(j)=\kappa$, $j(\kappa)> \lambda$, and $M^\lambda \subseteq M$ (i.e. $M$ is closed under $\lambda$-sequences).
So if $\kappa$ is $\kappa$-supercompact, clearly $\kappa$ is measurable. I'm wondering if this statement is if and only if. I'm wondering does the following hold?
$ \kappa $ is $\kappa $-supercompact if and only if $\kappa$ is measurable.
The main issue is as follows: given $\kappa$ measurable let $j: V \rightarrow M$ witness this with critical point $\kappa$, does it then follow that $M^\kappa \subseteq M$, i.e. is $M$ closed under $\kappa$-sequences? There is Theorem 21.9 from Jech which says forcing extensions $M[G]$ with $|\mathbb{P}|<\kappa$ on are closed under $\kappa$-sequences, but I'm asking about $M$ not $M[G]$.
Best Answer
Not every elementary embedding of $V$ with critical point $\kappa$ is closed under $\kappa$-sequences, but any measurable ultrapower embedding (i.e. by a normal, or even merely $\kappa$-complete and nonprincipal ultrafilter on $\kappa$) is. So, since for any embedding $j$ with critical point $\kappa$, $D=\{Z\subseteq \kappa: \kappa\in j(Z)\}$ is a normal ultrafilter, we can take the ultrapower by that to get a $\kappa$-supercompact embedding.
The result that ultrapower embeddings are closed under $\kappa$-sequences is a one-liner (explained in a bit more detail in Jech chapter 17): If $[g_\xi]\in M$ for $\xi < \kappa,$ and we take $F=\{(\alpha, \langle g_\xi :\xi < h(\alpha)\rangle): \alpha < \kappa\}$ where $[h] = \kappa$, then $[F] = \langle[g_\xi]:\xi < \kappa\rangle.$