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Is it consistent to have a $(\kappa,\kappa,2)$-saturated ideal $I$ on $\kappa$ that is $\kappa$-complete and $\kappa$ is not weakly compact? Here $\kappa$ is inaccessible. An ideal is $(\kappa,\kappa, 2)$-saturated, if for any collection $\{A_i: i<\kappa\}\subset I^+$, there exists a sub collection of size $\kappa$ such that any two elements have $I$-positive intersection. In other words, $P(\kappa)/I$ is $\kappa$-Knaster.

The motivation comes from Kunen's result on the consistency of $\kappa$-saturated $\kappa$-complete ideal on inaccessible $\kappa$ but $\kappa$ is not weakly compact. In the model, $P(\kappa)/I$ is equivalent to forcing with a Suslin tree, so it's not $\kappa$-Knaster. Further restriction, if $\mathbb{P}$ is $\kappa$-Knaster and $\Vdash_{\mathbb{P}} \kappa$ is weakly compact, then $\kappa$ is weakly compact in $V$. So if the answer to the question is yes, $\kappa$ must not be weakly compact after forcing with $P(\kappa)/I$.

A more general question: forget about weak compactness, is it consistent to have a $(\kappa,\kappa, 2)$-saturated ideal at an inaccessible at all?

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Suppose $\kappa$ is regular uncountable and $I$ is a $\kappa$-aditive ideal on $\kappa$ such that forcing with $I$ is $\kappa$-Knaster. Force with $I$: Let $G$ be a generic filter and $j:V \to M \subseteq V[G]$ be the generic embedding with critical point $\kappa$ ($M$ is the well founded generic ultrapower). Let $T$ be a normal $\kappa$-tree in $V$. Let $B \subseteq T$ be a cofinal branch in $M$ (look at $j(T)$). Choose $\{p_i : i < \kappa\}$ such that $p_i$ decides the $i$th level node of $B$ and use $\kappa$-Knaster to get a branch in $V$. So $\kappa$ has the tree property. Add inaccessibility and you get weak compactness.

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  • $\begingroup$ By argument of Spencer Unger, you can weaken the assumption about the ideal $I$ to "$\mathbb{Q} \times \mathbb{Q}$ is $\kappa$-c.c. for $\mathbb{Q} = P(\kappa) / I$ and still get that the tree property holds at $\kappa$. $\endgroup$ – Yair Hayut May 26 '17 at 12:02
  • $\begingroup$ @YairHayut: I believe this result (at least the proof) appeared in Cummings' chapter in the handbook theorem 21.1 that the author said "we are not sure to whom the following result should be attributed" $\endgroup$ – Jing Zhang May 26 '17 at 15:08
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This should be a comment:

"Is it consistent to have a $(\kappa,\kappa, 2)$-saturated ideal at an inaccessible at all?"

This is well-known to be equiconsistent with "$\exists $ measurable" and could happen at small large cardinals - E.g., there could be a cardinal $\kappa < \mathfrak{c}$ and a $\kappa$-additive ideal $I$ on $\kappa$ such that forcing with $I$ is sigma-centered. The tree property continues to hold as explained.

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  • $\begingroup$ $\kappa<\mathfrak{c}$? But $\kappa$ here is inaccessible. Could you point me to some references? $\endgroup$ – Jing Zhang May 25 '17 at 18:37
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Sorry, I missed the inaccessible there. Assuming you want an atomless forcing here, I think this is impossible.

For let $I$ be an atomless $\kappa$-additive $\kappa$-Knaster ideal on $\kappa$. Contruct a tree $T = \langle t_{\sigma} : \sigma \in 2^{< \kappa}\rangle$ of $I$-positive sets as follows. At successor step, if $t_{\sigma}$ is $I$-positive, $t_{\sigma0}, t_{\sigma1}$ split it into $I$-positive sets. At limits take intersection. Some branches could die because of $I$-null intersection but inaccessibility of $\kappa$ lets the construction run for $\kappa$ levels. This gives us a $\kappa$-tree with no branch (otherwise $\kappa$-cc is violated). But this is impossible as we argued that $\kappa$ had the tree property.

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    $\begingroup$ Thanks for the answers. Are you the same person(just out of curiosity)? $\endgroup$ – Jing Zhang May 25 '17 at 19:16

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