[FOM] inverse forcing

Monroe Eskew meskew at math.uci.edu
Tue Aug 24 20:52:49 EDT 2010

On July 22, I posed the question of whether after forcing with P to
get M[G], is the ground model M uniquely determined by the parameters
P, G.  Ali Enayat suggested on July 23 that a counterexample could be
found by 2-step iteration of Cohen forcing.  I replied that I wasn't
sure it works with Cohen forcing.  However, there is a way to do it:

There exist x,y,z mutually generic Cohen reals such that M[x][y] =
M[x][z] = M[y][z].

The trick is to start with x generic over M, then z generic over M[x].
We can use z to code an automorphism \phi of Cohen forcing (the
binary tree).  Namely the nth digit in all nodes is switched iff z(n)
= 1.  Let y be the image of x under \phi.  One can show that y is
generic over M[x].  The equalities above follow from the fact that any
real in {x,y,z} is definable from the other two.

Monroe

On Fri, Jul 23, 2010 at 4:52 PM, Ali Enayat <ali.enayat at gmail.com> wrote:
>
> In a more recent posting (July 23), Eskew noted that the answer to (1)
> in negative by choosing P as a collapsing poset.
>
> Alternatively, one can choose mutually generic Cohen reals r and s
> over some M, and observe that if N = M(r,s), then N is a generic
> extension of each of M(r) and M(s), via the same notion of forcing.