Is it possible to have a saturated ideal on a successor cardinal which does not extend the nonstationary ideal? (i.e. some nonstationary set is positive for this ideal)
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asked Jan 19, 2012 at 1:24
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Yes, it is. The reason is that an ideal $I$ on $P(\kappa)$ is saturated just in case the quotient Boolean algebra $P(\kappa)/I$ satisfies the $\kappa^+$-chain condition, and this is a property that is preserved by permutations of the underlying set $\kappa$. But the property of extending the non-stationary ideal is not preserved by such permutations, since we can perform a permutation of $\kappa$ that takes a nonstationary set to a club set. Thus, there are isomorphic versions of any saturated ideal that remain saturated, but which do not extend the non-stationary ideal.
answered Jan 19, 2012 at 1:35
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$\begingroup$ But if you add a normality condition, the story becomes more complicated. $\endgroup$ Commented Jan 19, 2012 at 1:42
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$\begingroup$ I also just realized, if you take the induced ideal: $\{ X : 1 \Vdash \delta \in j(X) \}$, then this is saturated for any choice of delta, so we may pick a successor ordinal less than $\kappa^+$ and the get set of limit ordinals to be in the induced ideal. $\endgroup$ Commented Jan 19, 2012 at 1:44
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$\begingroup$ Yes, I think that is right. $\endgroup$ Commented Jan 19, 2012 at 1:48