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Suppose that $M$ is a model of ZFC, $P\in M$ is a notion of forcing and $G$ is a $P$-generic filter over $M$.

It is a well-known theorem that if $M\subseteq N\subseteq M[G]$, and $N$ is a model of ZFC then $N=M[H]$ for some generic set $H$, and there is some $H'$ which is generic over $M[H]$ such that $M[G]=M[H][H']$.

But the proof I know uses Boolean valued models and is not particularly insightful about the following question:

Suppose that $M$ is a model of ZFC, $P\in M$ is a notion of forcing, and $\dot x$ is a $P$-name, such that if $G$ is a $P$-generic over $M$, and $x=\dot x^G$ then $G\notin M[x]$.

Can we give an explicit $Q\in M$ such that $x$ is $Q$-generic over $M$?

An additional question which is relevant to one particular case:

Suppose that $P$ is the Cohen forcing with $2^{<\omega}$. In such case every sub-forcing is isomorphic to $P$, what does that mean for us? Does it mean that the generic needs to be particularly chosen, or what?

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I feel that "sub-forcing" is the wrong term, but I can't recall any better term at the moment. Any suggestions on terminology (or approvals) would be welcomed. –  Asaf Karagila Jan 8 '13 at 23:33
    
Sub-forcing is fine, but I take this to mean simply the forcing extension obtained by a complete sub-algebra of the complete Boolean algebra giving rise to the extension, and this is the same thing as an intermediate ZFC model $M\subset N\subset M[G]$. I don't understand your second question---could you clarify what you want? –  Joel David Hamkins Jan 9 '13 at 0:59
    
Joel, it's simple. Suppose $P$ was the product of two random reals, which adds a Cohen. We can consider the sub-forcing which only adds this Cohen. Clearly this subforcing is very simple and the full forcing is... less simple. But if we start with a Cohen forcing then the subforcing is again a Cohen forcing, for obvious reasons. So now how do we discern between the two results? Those would have to be two different generics, in some sense, but surely there is some way to say that $G$ is "more generic" than $x$. –  Asaf Karagila Jan 9 '13 at 1:02
    
I see. In general, since adding a Cohen real is countably-dense forcing, it is not difficult to see that every sub-forcing of it is also countably-dense and hence the same as Cohen real forcing. Furthermore, the quotient forcing is also countably-dense, and hence also Cohen-real forcing. So every factorization of Cohen real forcing is just adding two Cohen reals. –  Joel David Hamkins Jan 9 '13 at 1:10
    
Exactly. This is why I wanted to try and understand how we can better discern between those two Cohens, the first is just a regular Cohen at your local temple, but the other is the great Cohen at the temple in Jerusalem. (And this is as far I will go with this analogy for tonight! :-)) –  Asaf Karagila Jan 9 '13 at 1:18

1 Answer 1

The intermediate model proof for $M\subset N\subset M[G]$ is not so inexplicit. The situation is that, if $A\subset\text{Ord}$ is a set of ordinals coding $P(\mathbb{B})^N$, the $N=M[G\cap \mathbb{B}_0]$, where $\mathbb{B}_0$ is the Boolean algebra generated by the $\mathbb{B}$-Boolean values of membership in $\dot A$, where $\dot A$ is a $\mathbb{B}$-name for $A$. (This is proved in Jech.) After all, if you know $G$ on this $\mathbb{B}_0$, then you know $P(\mathbb{B})^N$ and hence all of $N$ (as in Jech), and conversely, $N$ certainly knows the right answers for $\dot A$.

In your case, you have $N=M[x]$, and so $A$ is any set of ordinals coding $P(\mathbb{B})^{M[x]}$ in $M[x]$, with $\mathbb{B}_0$ the Boolean values generated by the values $\[\check\alpha\in\dot A\]^{\mathbb{B}}$.

Meanwhile, one cannot get too explicit about it, since the proof in Jech uses AC and the fact is not generally true for ZF as opposed to ZFC models. First of all, note that there can be intermediate ZF models. For example, one might start with $M=L$, say, and add $\omega_1$ many Cohen reals, to form $M[G]$. Let $N=L(\mathbb{R})^{M[G]}$, which is intermediate between $M$ and $M[G]$, but it doesn't have the form $M[G\cap\mathbb{B}_0]$ for any complete subalgebra, since it violates AC. In general, no intermediate $\text{ZF}+\neg\text{AC}$ model can be $M[G\cap\mathbb{B}_0]$, since these satisfy ZFC.

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Oh, about that counterexample, I am actually working towards reconstructing, generalizing, and further development of such counterexample. :-) –  Asaf Karagila Jan 9 '13 at 0:37
    
I should also say that I don't mind having this poset definable from the name itself (either by a general choicey algorithm, or explicitly if the name is nice enough. At the moment, however, I don't know what is the name, so I can't really say more...) –  Asaf Karagila Jan 9 '13 at 0:39
    
I was just looking at such a counterexample recently, but I don't recall now... –  Joel David Hamkins Jan 9 '13 at 0:39
    
Joel, when you write "Boolean values of membership in $\dot A$", Do you mean the subalgebra whose $1$ element is $\sum[\dot a\in\dot A]$ over all the elements which $1\not\Vdash\dot a\notin\dot A$ or something like that? –  Asaf Karagila Jan 9 '13 at 0:47
    
I mean the complete Boolean subalgebra of $\mathbb{B}$ generated by the elements of the form $\[\check\alpha\in\dot A\]$. The one element of the subalgebra will be the same as the one element of $\mathbb{B}$. –  Joel David Hamkins Jan 9 '13 at 0:51

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