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Is it consistent relative to large cardinals that there is a precipitous ideal on $\omega_1$ forcing a generic elementary embedding $j : V \to M \subseteq V[G]$, such that $j(\omega_1) = \omega_n^V$ for some $n > 2$?

A little background: The first thought may be start with a measurable $\kappa$ for which $2^\kappa = \kappa^{+n}$ for some large $n$. Collapse $\kappa$ to $\omega_1$ and look at the ideal induced by the dual of the measure. But it is easy to show that $2^\kappa < j(\kappa) < (2^\kappa)^+$. In order to get an ideal forcing $j(\omega_1) = \omega_2$, usually stronger assumptions are used to produce a pre-saturated ideal. It is known how we can get ideals forcing $j(\omega_1) = \omega_n$ for whatever $n$, but these are normal ideals on something like $\mathcal P_{\omega_1}(\omega_{n-1})$.

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  • $\begingroup$ Not being fluent in these type of arguments, what is the "obvious difficulty" (e.g. why won't taking some $j(\kappa)=\kappa^{+4}$ and collapsing $\kappa$ to be $\omega_1$ in a way where the embedding can be extended generically work)? $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 21, 2016 at 19:06
  • $\begingroup$ This might work, but what large cardinal assumption gets such an embedding? $\endgroup$
    – Monroe Eskew
    Commented Mar 21, 2016 at 19:20
  • $\begingroup$ I don't know. What large cardinals gives you $j(\kappa)=\kappa^{+4}$? Something like $\kappa^{+4}$-supercompact? Or something like that? $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 21, 2016 at 19:24
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    $\begingroup$ Typically the target of $\kappa$ is not a cardinal, unless you're using a very strong assumption where it's actually inaccessible. $\endgroup$
    – Monroe Eskew
    Commented Mar 21, 2016 at 19:26
  • $\begingroup$ What about $j^2$, where $j$ is a 2-huge embedding? $\endgroup$
    – Sean Cox
    Commented Mar 21, 2016 at 20:06

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