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Given posets $P,Q\in M$, I would like to know under what circumstances there are mutually generic filters $G\subseteq P$ and $H\subseteq Q$ (generic over $M$). Also, what are the characterizations of mutual genericity? And finally, what can we say about the relation between $M[G]$ and $M[H]$ in that case?

I have a slight difficulty finding references to the notion and properties of mutual genericity (whatever they are).

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up vote 8 down vote accepted

One of the basic facts of product forcing is the following, which appears in any of the standard accounts of forcing:

Theorem. If $M$ is a model of ZFC and $\mathbb{P},\mathbb{Q}$ are forcing notions in $M$, with $M$-generic filters $G\subset\mathbb{P}$ and $H\subset\mathbb{Q}$, then the following are equivalent:

  • $G$ is $M[H]$-generic for $\mathbb{P}$.

  • $H$ is $M[G]$-generic for $\mathbb{Q}$.

  • $G\times H$ is $M$-generic for the product forcing $\mathbb{P}\times\mathbb{Q}$.

And this is what it means for $G$ and $H$ to be mutually $M$-generic.

Thus, the first answer to your question is that there are mutually $M$-generic filters for $\mathbb{P}$ and $\mathbb{Q}$ precisely when there are any $M$-generic filters for the product forcing. For example, if $M$ is a countable model, then these always exist.

The following observation of Solovay gives one of the most important relations between $M[G]$ and $M[H]$.

Theorem. If $G$ and $H$ are mutually $M$-generic, then $M[G]\cap M[H]=M$.

Proof. Suppose that $x\in M[G]\cap M[H]$. Then $x=\tau_G=\sigma_H$ for $\mathbb{P}$-name $\tau$ and $\mathbb{Q}$-name $\sigma$. We may assume by $\in$-induction that $x\subset M$. Because of the equality, there must be a condition $(p,q)\in G\times H$ forcing that $\tau=\sigma$, and we may also assume $p$ and $q$ force $\tau\subset\check M$ and $\sigma\subset\check M$. (Note, we may canonically regard $\mathbb{P}$ and $\mathbb{Q}$-names as $\mathbb{P}\times\mathbb{Q}$-names.) It now follows, however, that $p$ decides $\check y\in \tau$ for every $y\in M$, for if not, then then there would be stronger conditions, some forcing some $\check y\in\tau$ and others forcing $\check y\notin \tau$. Let $p'$ be stronger than $p$, forcing an opposite answer to whether $y\in\sigma_H$, forced by some $q'\in H$. Thus, $(p',q')$ is stronger than $(p,q)$, and forces $\tau\neq\sigma$, a contradiction. QED

Finally, let me also mention an interesting fact about mutual genericity, which I heard years ago from Woodin.

Theorem. If $M$ is a countable model of ZFC, then there are $M$-generic Cohen reals $c$ and $d$ such that $M[c]$ and $M[d]$ have no common forcing extension. Thus, the models $M[c]$ and $M[d]$ are non-amalgamable, which makes them very far from mutually generic.

Proof. Fix a real $z$ that cannot possibly be added by forcing over $M$, such as a real coding all of $M$. Now, enumerate the dense subsets $D_n$ of Cohen forcing in $M$, and build $c$ and $d$ in stages, $c=\cup_n c_n$ and $d=\cup_n d_n$. First, let $c_0$ be a finite binary sequence in the first dense set, and let $d_0$ be all zeros to the same length, then $1$, then the first bit of $z$, and then extend to an element of $D_0$. Now extend $c_0$ to $c_1$ with all $0$s to the length of $d_0$, then $1$, then get into $D_1$, and so on. Each of $c$ or $d$ individually will be $M$-generic, but any model containing both of them will be able to define $z$. So $M[c]$ and $M[d]$ cannot be contained in any model $N\models$ZFC with the same ordinals. QED

One can easily arrange more complicated patterns, with three reals, for example, any two of which are mutually generic, but such that the three of them together are non-amalgamable.

Meanwhile, it is a theorem of mine (joint with G. Fuchs and J. Reitz) that if you have a countable family of forcing extensions $M[G_n]$ of a countable model $M$, whose forcing notions have uniformly bounded size in $M$, such that any finite subfamily is amalgamable, then there is a common forcing extension $M[H]$ with $M[G_n]\subset M[H]$ for every $n$. You can find the result in our recent paper on set-theoretic geology.

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Great! Thanks a lot for the clarification. I indeed liked the interesting fact of the non-amalgamable extensions. By the way, where could I find the theorem of yours regarding the positive side of amalgamating extensions? Thanks again! –  kvagk Jul 13 '11 at 12:52
    
I'm glad you like the answer. About the positive amalgamation result, an account is given in the soon-to-be-released paper "Set-theoretic geology," by Gunter Fuchs, myself and Jonas Reitz. We should post it on the arXiv within a few weeks. –  Joel David Hamkins Jul 13 '11 at 13:56
    
I added a link to our paper containing the amalgamation result. –  Joel David Hamkins Jul 30 '11 at 12:38
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Another fact to add to Joel's list, one which I found surprising when I first came across it, is that adding a single Cohen real adds a perfect set of reals such that any finitely many are mutually generic over each other. To see this simply consider the trace of a Cohen real on the members of an almost disjoint family and use the theorem Joel quotes about products.

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