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A reliable source made the following claim:

Suppose CH there is an $\omega_2$-saturated ideal on $\omega_1$. Then this is preserved by $\mathrm{Add}(\omega_1,\omega_2)$.

Question 1: How do you show this?

Question 2: Is there an example where adding a Cohen subset of $\omega_1$ changes the truth-value of, "There is a saturated ideal on $\omega_1$"? Perhaps under MM?

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  • $\begingroup$ Monroe, this query is out of my ignorance but can you show that adding one Cohen subset of $\omega_1$ preserves this? since this forces diamond, it must possibly kill old (say e.g., the non stationary ideal) saturated ideals. $\endgroup$
    – Ashutosh
    Commented Jul 7, 2015 at 19:20
  • $\begingroup$ I was considering the possibility that the ideal generated by the old saturated ideal stays saturated. Is this ever possible after adding a Cohen subset of $\omega_1$? I will stop blabbering now. $\endgroup$
    – Ashutosh
    Commented Jul 7, 2015 at 19:26
  • $\begingroup$ If $Col(\omega,\omega_1)$ completely embeds into $\mathcal P(\omega_1)/I$, then we can use a variant of Foreman's Duality Theorem to compute the generated ideal and show that it is not saturated. However in this situation there is another saturated ideal in the extension. Actually I don't know of a model of CH + saturated ideal + the quotient algebra doesn't have this factor. $\endgroup$
    – Monroe Eskew
    Commented Jul 7, 2015 at 19:31
  • $\begingroup$ Can you tell me the simplest way to get a model of diamond plus there is a saturated ideal on $\omega_1$? $\endgroup$
    – Ashutosh
    Commented Jul 7, 2015 at 19:49
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    $\begingroup$ If you Levy collapse a Woodin cardinal to be $\omega_2$ then there is a saturated ideal in the extension. Also diamond holds because it is always forced by this countably closed collapse. $\endgroup$
    – Monroe Eskew
    Commented Jul 7, 2015 at 19:57

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