TITLE: $(\kappa, \kappa, 2)$-saturated ideals?
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: 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.