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An ideal $I$ on $P(\kappa)$ is said to be $\kappa^+$-saturated if there is no sequence $\langle S_\alpha \mid \alpha<\kappa^+\rangle$ of $I^+$ sets such that $\alpha<\beta<\kappa^+\implies S_\alpha\cap S_\beta\in I$.

Theorem 3.5 in Baumgartner and Taylor, Saturation Properties of Ideals in Generic Extensions II, gives an example of a poset that forces that there are no $\omega_2$-saturated ideals on $\omega_1$ by adding a club subset of $\omega_1$ with finite conditions. Definition 4.2 in Cox and Eskew, Strongly Proper Forcing and Some Problems of Foreman, extends this to regular $\kappa=\mu^+$ with $\mu$ regular, $\mu^{<\mu}=\mu$ with a forcing $\mathbb{P}(\mu,\kappa)$.

Is there a known analogue poset for $\kappa$ inaccessible that forces there to be no $\kappa^+$-saturated ideals in the generic extension by adding a similar sort of club?

I don't believe that $\mathbb{P}(\mu,\kappa)$ as in Cox and Eskew (with choice of $\mu<\kappa$) will work, since their argument that $\mathbb{P}(\mu,\kappa)$ preserves $\kappa$ doesn't appear to go through for $\kappa>\mu^+$.

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  • $\begingroup$ Your question appears as Question 8.5 in our paper. It is an interesting problem, but I have no idea how to approach it. Note that any such forcing must destroy measurability of $\kappa$, since the dual of a normal measure on $\kappa$ is a trivial example of a saturated ideal. $\endgroup$
    – Sean Cox
    Commented Nov 21, 2018 at 2:15
  • $\begingroup$ Not only would you have to destroy measurability of $\kappa$, but you'd have to ensure that $\kappa$'s measurability cannot be resurrected in any further $\kappa^+$-cc extension (and, more generally, that $\kappa$ does not gain a $\kappa^+$-saturated ideal in any further $\kappa^+$-cc extension). $\endgroup$
    – Sean Cox
    Commented Nov 21, 2018 at 2:24
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    $\begingroup$ Regarding your question about recovering saturated ideals: this idea appeared (I think) in Kunen's paper on saturated ideals. Suppose $\mathbb{P}$ is $\kappa^+$-cc, and $\dot{J}$ is a $\mathbb{P}$-name for a $\kappa^+$-saturated ideal on $\kappa$. Let $I$ be the set of $A \subset \kappa$ such that $1_{\mathbb{P}}$ forces $\check{A} \in \dot{J}$. Then $I$ is an ideal on $\kappa$ (and is normal, if $\dot{J}$ is forced to be normal). Define a map from $P(\kappa)/I$ into $\text{ro}(\mathbb{P}) * \dot{P}(\kappa)/\dot{J}$ by sending $[S]_I$ (the $I$-equivalence class $\endgroup$
    – Sean Cox
    Commented Nov 22, 2018 at 2:17
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    $\begingroup$ of an $I$-positive set $S$) to the condition $([ \check{S} \notin \dot{J} ], [\check{S}]_{\dot{J}})$; here $[\check{S} \notin \dot{J}]$ denotes the $\mathbb{P}$-boolean value of that statement, and $[\check{S}]_{\dot{J}}$ is the $\mathbb{P}$-name for the $\dot{J}$-equivalence class of $\check{S}$. This map is $\perp$-preserving, and so the $\kappa^+$-chain condition of the 2-step target poset carries down to $P(\kappa)/I$. $\endgroup$
    – Sean Cox
    Commented Nov 22, 2018 at 2:19
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    $\begingroup$ In short, if you can force a saturated ideal on $\kappa$ with a $\kappa^+$-cc forcing, then you already have a saturated ideal in the ground model. $\endgroup$
    – Sean Cox
    Commented Nov 22, 2018 at 2:25

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