Suppose $M$ is a countable transitive model of some fragment of $\mathbf{ZFC}$, $\mathbb{P}\in M$ is a forcing notion and $G, H$ are $\mathbb{P}$-generic such that $M[G]=M[H]$. Does it then follow that there is some automorphism $\pi:\mathbb{P}\longrightarrow\mathbb{P}$ such that $\pi\in M$ and $\pi[G]=H$?

If the answer is no, are there natural restrictions one could impose on $\mathbb{P}$ (maybe apart from $\mathbb{P}$ being finite) such that the above sentence holds?

  • $\begingroup$ I have recently found Lemma 25.5 in an old version of "Set Theory" by Thomas Jech, which states the exact result im looking for, but only in a version for complete boolean algebras (i.e. when we take $\mathbb{P}$ to be a complete boolean algebra). However, i am unsure how to adapt it to my case, even when $\mathbb{P}$ is separative and we can use the representation theorem. $\endgroup$ Jan 28, 2021 at 14:29
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    $\begingroup$ The adaptation to posets other than complete Boolean algebras is, as I said in my answer, messy. One problem is that two posets can produce isomorphic complete Boolean algebras even when the posets are nowhere near isomorphic. Another is that the $b$ and $b'$ in my answer cannot generally be found in the posets (you can get one or the other in the poset you want, but not generally both). $\endgroup$ Jan 30, 2021 at 16:48

1 Answer 1


This works much better in terms of complete Boolean algebras. If $\mathbb B$ and $\mathbb B'$ are complete Boolean algebras and a $\mathbb B$-generic filter $G$ and a $\mathbb B'$-generic filter $G'$ generate the same forcing extension, then there are $b\in G$ and $b'\in G'$ such that the part of $\mathbb B$ below $b$ is isomorphic to the part of $\mathbb B'$ below $b'$ by an isomorphism in the ground model, that sends the restriction of $G$ to the restriction of $G'$.

I believe this fact is in Serge Grigorieff's paper "Intermediate submodels and generic extensions in set theory" [Ann. Math. Second Series, Vol. 101, No. 3 (May, 1975), pp. 447-490].

It's certainly possible to rewrite this in terms of posets instead of complete Boolean algebras, but the result looks messy to me.

  • $\begingroup$ I hope you don't mind, i have "unaccepted" the answer because i managed to find a more suitable result in an old Version of Jechs "Set Theory" that is more closely related to my question. $\endgroup$ Jan 28, 2021 at 14:29

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