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Let $\mathsf{ZFC}^-$ be the Zermelo-Fraenkel set theory without power set axiom. For a transitive model $M$ of $\mathsf{ZFC}^-$ and an cardinal $\kappa\in M$ in the sense of $M$, an unary predicate $U$ over $M$ is an $\kappa$-complete ultrafilter if $M$ thinks it is $\kappa$-complete and $U$ is weakly amenable, that is, for any $F\in {^\kappa}M\cap M$, $\{\xi<\kappa : F(\xi)\in U\}\in M$.

$M$ thinks $\kappa$ is weakly compect. I think in fact we can show $M$ thinks $\kappa$ is Ramsey. However it does not tell about their consistency strength as $\mathsf{ZFC}^-$ is far more weaker than $\mathsf{ZFC}$.

My question is, the exact consistency strength of the existence of the model $M\models \mathsf{ZFC}^-$ with a $\kappa$-complete ultrafilter over $M$ is known? Theorem 1.1 of an article by Gitman and close relation between $\kappa$-complete ultrafilter and elementray embedding suggests its consistency strength may be related to the existence of a weakly compact cardinal.

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    $\begingroup$ Do you require that the ultrapower is well-founded? Also, do you want $M$ to satisfy that the ultrafilter is normal? $\endgroup$ – Victoria Gitman Mar 6 '18 at 16:33
  • $\begingroup$ @VictoriaGitman I did not consider them when I wrote the question, but do not matter about it. $\endgroup$ – Hanul Jeon Mar 7 '18 at 11:06
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Let's make some additional assumptions on the ultrafilter $U$. Suppose $M\models{\rm ZFC}^-$ and $\kappa$ is a cardinal in $M$. We say that $U$ is an $M$-ultrafilter if $\langle M,\in,U\rangle$ satisfies that $U$ is a $\kappa$-complete normal ultrafilter on $\kappa$. Because $U$ is only $\kappa$-complete for sequences in $M$ and $M$ might be missing even countable sequences, the ultrapower of $M$ by $U$ may not be well-founded. So let's say that $U$ is good if the ultrapower by $U$ is well-founded.

The consistency of the existence of $M\models{\rm ZFC}^-$ for which there is a good weakly amenable $M$-ultrafilter $U$ on a cardinal $\kappa$ in $M$ is between an ineffable cardinal and an $\omega$-Erdős cardinal, which is much weaker than a Ramsey cardinal. In particular, this assumption is consistent with $V=L$.

Here is a sketch of the argument. Suppose $j:M\to N$ is the ultrapower by $U$ (with $N$ transitive by assumption) with critical point $\kappa$ (by normality). By weak amenability, in $N$, $\kappa$ has the property that every collection of $\kappa$-many subsets of $\kappa$ has a $\kappa$-complete filter. Thus, by elementarity $M$ satisfies that this property holds for cofinally many $\alpha<\kappa$. Now, we argue that each such $\alpha$ is weakly compact in $L_\kappa\models{\rm ZFC}$. Next, we can argue that $\kappa$ has the ineffability property in $N$, and therefore there are cofinally many $\alpha$ that are ineffable in $L_\kappa$. The reason I am going down to $L$ is that we did not assume that $M$ has any powersets and therefore even the notion of inaccessible cardinal does not make sense for $M$. For the upper bound, an $\omega$-Erdős cardinal implies the consistency of what I call a $1$-iterable cardinal $\kappa$ having the property that every $A\subseteq\kappa$ is contained in a model $M\models{\rm ZFC}^-$ of size $\kappa$ with $\kappa\in M$ for which there is a good weakly amenable $M$-ultrafilter on $\kappa$.

If we don't assume that $U$ is normal or that the ultrapower by $U$ has to be well-founded, the consistency strength goes down significantly. At most a weakly compact suffices. If $\kappa$ is weakly compact and $M_0\models{\rm ZFC}^-$ has size $\kappa$, then we can find a $\kappa$-complete ultrafilter $U_0$ for $M_0$ with a well-founded ultrapower. Moreover, if $M_1\models{\rm ZFC}^-$ has size $\kappa$ and extends $M_0$, then we can extend $U$ to a $\kappa$-complete ultrafilter $U_1$ for $M_1$. Thus, in $\omega$-many steps, we can obtain $M=\bigcup_{n\in\omega}M_n$ and a weakly amenable $U$ that is $\kappa$-complete for sequences from $M$. The argument for the "moreover" part can be found here in an article of Peter Holy and Philipp Schlicht. I suspect that you don't even need a weakly compact. (I am assuming $\kappa$ uncountable here).

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    $\begingroup$ Hi Vika. In connection with your penultimate sentence: when $M$ is a model of full ZFC, weak compactness of $\kappa$ can be "purchased" from the existence of $U$ since the ultrapower mod $U$ will not add any new subsets of $\kappa$, which in turn can be used to show the tree property of $\kappa$. The strong inaccessibility of $\kappa$ can also be verified using an ultrapower argument. I do not know off the top of my head what happens when $M$ is a model of $ZFC^-$. $\endgroup$ – Ali Enayat Mar 8 '18 at 19:07
  • $\begingroup$ Ali, I am just seeing the comment because I have been away from MO for a while. So you are saying that even an ill-founded ultrapower by a non-normal filter as long as it is weakly amenable can be used to show that kappa is weakly compact in M? $\endgroup$ – Victoria Gitman Mar 17 '18 at 17:03
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    $\begingroup$ No problem Vika, I have been away too. The answer to your question is yes, at least when M is a model of ZFC, I have not had the chance to revisit the proof (which I last did back in the mid 1980s) to see if it uses the power set axiom in a significant way or not, I will try to do so. $\endgroup$ – Ali Enayat Mar 19 '18 at 0:37
  • $\begingroup$ This is really very nice! $\endgroup$ – Victoria Gitman Mar 19 '18 at 13:19

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