Why does an M-Ultrafilter on $\kappa$ imply that $\kappa$ is weakly compact?

Question

so far I´ve only been reading on this forum, this is my first question. I am trying to understand Kunens article on iterated ultrapowers from 1970 (Some applications of iterated ultrapowers) and somehow I dont get why $\kappa$ must be weakly compact in M, when there exists a M-Ultrafilter on $\kappa$. I do understand why it must be regular, I would be grateful for any tip how to continue to show that its weakly compact. Thank you in advance, John

Answer 1

I think Kunen's definition of an $M$-ultrafilter includes weak amenability meaning any $x\in M$ with $M\models |x|\leq \kappa$ satisfies that $x\cap U\in M$. From this we can deduce that the ultrapower $Ult(M, U)$ (or more precisely, the transitive collapse of it) has the same subsets of $\kappa$ as $M$.

$Ult(M, U)$ is well-founded simply because ${}^{<\kappa} M\subset M$ so $U$ is really countably complete.

Once we have the above facts, given any $\kappa$-tree $T=(\kappa, <_T)$ in $M$, let $j: M\to N\simeq Ult(M, U)$ be the ultrapower map (with critical point $\kappa$). Then $j(T)(\kappa)$ is a cofinal branch, but this branch is also in $M$. Hence $M\models \kappa$ has the tree property.