# If $M[G]\cap M[H]=M$, must $G\times H$ be generic?

Assume that $M$ is a countable transitive model of ZFC, $\mathbb{P}\in M$ is a forcing notion, and $G,H$ are two $\mathbb{P}$-generic filters over $M$. As it was shown in the answer to this question, if $G\times H$ is $\mathbb{P}\times\mathbb{P}$-generic over $M$ then $M[G]\cap M[H]=M$. Now the question is if the converse is true. Does $M[G]\cap M[H]=M$ imply that $G\times H$ is $\mathbb{P}\times\mathbb{P}$-generic over $M$?

No, and there is an example with Cohen forcing.

Consider the pairs $(s,t) \in 2^{<\omega} \times 2^{<\omega}$ with $\vert s \vert = \vert t \vert$ and for any $n < \vert s \vert$, $s(n) = 1 \rightarrow t(n) = 1$. Denote the tree of such $(s,t)$ with $T$.

Now forcing with $T$ gives a pair of Cohen reals $(c_0,c_1)$ which are not generic for $2^{<\omega} \times 2^{<\omega}$.

Now suppose $\tau_0$, $\tau_1$ are (Cohen forcing) names for sets of ordinals so that $\tau_0[c_0] = \tau_1[c_1]$. Then this is forced by $(s,t) \in T$, $s \subseteq c_0$, $t \subseteq c_1$.

Claim. Suppose $s' \leq s$ and $s' \Vdash \alpha \in \tau_0$. Then any $s'' \leq s$ with $\vert s'' \vert = \vert s' \vert$ forces $\alpha \in \tau_0$. Similarly for $\alpha \notin \tau_0$.

Proof. Let $s''$ be arbitrary with $\vert s'' \vert = \vert s' \vert$. Note that for any $t' \leq t$ so that $(s',t') \in T$, $t' \Vdash \alpha \in \tau_1$ (because $(s',t') \leq (s,t)$). Let $t' \leq t$ so that for $n \in [\vert t \vert,\vert s' \vert)$, $t'(n) = 1$. Then obviously $(s',t') \in T$ and $(s',t') \leq (s,t)$, so $t' \Vdash \alpha \in \tau_1$. Now notice again that for any $s_0 \leq s$ so that $(s_0,t') \in T$, $s_0 \Vdash \alpha \in \tau_0$. But now we can simply take $s_0 =s''$ and we have that $s'' \Vdash \alpha \in \tau_0$.

By the claim $\tau_0[c_0] \in M$.

Now to consider sets of ordinals is of course enough, because for an example $A \in (M[c_0] \cap M[c_1]) \setminus M$ with minimal rank we can consider a bijection in $M$ from $V_\delta$ to some ordinal for suitable $\delta$ and consider $A$ as a set of ordinals. Also the use of choice in $M$ can be avoided by considering names for subsets of $M$ instead of names for sets of ordinals.

• Finally an argument that I (more or less) understand, and which shows me what is important. Although I am slightly confused by the claim, I can imagine a Solovay-style argument ("A model of set-theory in which every set of real is Lebesgue measurable", Lemma 2.5). Apr 17, 2018 at 17:32
• @PeterElias I don't see what this has to do with Solovay's model. The argument here relies mostly on the way in which I defined $T$. Apr 17, 2018 at 19:25
• When trying to mimic the proof of Solovay's lemma I have finally understood what your claim is claiming and how it is proved. Nice trick, I like it a lot. Thank you! Apr 17, 2018 at 19:31
• Solovay's lemma just shows that a set of ordinals belongs to $M$ whenever it is in $M[G_1]\cap M[G_2]$ and $G_1$, $G_2$ are mutually generic. As you have proved, mutual genericity is not necessary, some weaker condition is enough. Apr 17, 2018 at 19:41
• In the claim we could equally well define $t_0(n)=1$ for each $n\ge |t|$, am I right? Apr 18, 2018 at 5:43

The answer is no. Consider any two forcings $\mathbb Q_1, \mathbb Q_2$ with generic filters $H_1,H_2$ such that $M[H_1] \cap M[H_2] = M$, yet $H_1 \times H_2$ is not $\mathbb Q_1 \times \mathbb Q_2$-generic. Now let $\mathbb P = \mathbb Q_1 \oplus \mathbb Q_2$ be the lottery sum.

• Do you also have an example of such $\Bbb Q_i$? Apr 16, 2018 at 21:48
• @AsafKaragila I'm certain there are simpler examples but taking $\mathbb Q_1$ to be Cohen forcing with mutually generic, disjoint Cohen reals $c,d$ and $\mathbb Q_2$ being a countable product of lottery sums of products that alter the GCH pattern high up in different ways works. Let $d$ determine in each factor which forcing is to be considered for the $\mathbb Q_2$-generic to ensure that $H_1 \times H_2$ is not $\mathbb Q_1 \times \mathbb Q_2$-generic. (Consider the dense set that says that for some index the lottery sum in $\mathbb Q_2$ was chosen according to the left forcing condition.) Apr 16, 2018 at 22:07
• Argh, I meant to say finite support product (of countably many factors) in the above comment. Apr 16, 2018 at 22:29
• Right. Probably a Cohen and a Random real work better. Or a Cohen real and a Suslin tree. Apr 16, 2018 at 23:10
• Or Cohen and Sacks. Or .. Apr 17, 2018 at 1:49