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Suppose that in a forcing extension $V[G]$ by some ccc forcing $P$ there is a Cohen real over $V$. By a general argument we can factor $P$ into an iteration $P=\mathrm{Add}(\omega,1)*\dot{Q}$ for some ccc quotient forcing $\dot{Q}$. Can we do better and factor $P$ as a product $\mathrm{Add}(\omega,1)\times R$ for some forcing $R$? Is every extension containing a Cohen real a Cohen extension of some intermediate model?

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  • $\begingroup$ Have you tried to consider a case where $\dot Q$ is something like $\rm Add(\omega_1,1)$? It seems to me that you could have encoded the generic Cohen into such $R$ if it existed. $\endgroup$
    – Asaf Karagila
    Commented Apr 27, 2013 at 22:05
  • $\begingroup$ @AsafKaragila Good point. I think that might be a problem, but in the case I have in mind $P$ (and therefore $\dot{Q}$) is ccc. I've edited the question. $\endgroup$ Commented Apr 27, 2013 at 22:30

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Suppose you first add a Cohen real and then, over the resulting model, force Martin's axiom and not CH in the usual way (by a ccc, finite-support iteration). The resulting model has no Souslin trees, because MA$(\aleph_1)$ implies Souslin's hypothesis. So this model is not of the form "add a Cohen real to some model", because adding a Cohen real always produces a Souslin tree.

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  • $\begingroup$ Thank you. Now why didn't I think of that? :) $\endgroup$ Commented Apr 28, 2013 at 0:50

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