$(\kappa, \kappa, 2)$-saturated ideals?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.