Is it possible to take a forcing extension which is elementarily equivalent to the ground model? Here I'm assuming the extension is proper, that is, it adds a new set.

It's clear it can't be an elementary extension (the forcing notion has a generic filter in the extension but not the bottom), which is why I ask about equivalence. Some hypotheses would prevent this (no forcing extension of a model of V=L is still a model of V=L), so this is really a question about consistency.

It's also clear by a pigeonhole principle argument that since there are many forcing extensions but few complete extensions of ZFC, some pair of them must have the same theory. But this doesn't mean they have the same theory as the ground model.

Any thoughts?

  • $\begingroup$ I am not sure I understand your pigeonhole principle argument. Are you talking about forcing extensions of $V$? Because then mentioning extensions of $\mathsf{ZFC}$ is out of place. Are you talking instead about forcing extensions of set models of $\mathsf{ZFC}$? Because then I do not understand how you conclude that there are "many" extensions (many forcing posets?). There could just be countable models, and countably many posets. $\endgroup$ Nov 4, 2015 at 0:57
  • $\begingroup$ Fair enough, I may have been too quick. My thought was, there are continuum many possible theories of forcing extensions, but class many posets, so class many forcing extensions. Consequently some two of them must have the same theory. But this argument lives in some model of ZFC, I guess, so it's no longer true to say there are class many posets trivially. Also, different posets may yield equal (!) forcing extensions, which had not occurred to me. $\endgroup$ Nov 4, 2015 at 2:15
  • $\begingroup$ Note that your argument that $V\not\prec V[G]$ is incomplete: we need that $V$ is definable in $V[G]$ (from parameters). This was proved by Laver and Woodin independently, but is far from trivial. $\endgroup$ Nov 4, 2015 at 2:25
  • $\begingroup$ Why do we need that? It's simply true that there is a sentence (with parameters in V) which is true in V but false in V[G] $\endgroup$ Nov 4, 2015 at 2:27
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    $\begingroup$ @NoahSchweber For some $\alpha$, $V_\alpha^V\ne V_\alpha^{V[G]}$. That's all. $\endgroup$ Nov 4, 2015 at 2:29

1 Answer 1


Sure. Consider when $V=L[c]$ where $c$ is a Cohen real over $L$. If we force to add a second Cohen real $r$ over $V$, the result is again a Cohen extension of $L$.

And because the Cohen forcing is homogeneous, every two extensions are elementarily equivalent. So both $V$ and $V[r]$ are elementarily equivalent.

  • $\begingroup$ Note that $L$ isn't doing anything here: if $M$ is any model of $ZFC$, and $G, H$ are mutually Cohen generic over $M$, then $M[G]\equiv M[H]\equiv M[G][H]$. $\endgroup$ Nov 4, 2015 at 2:26
  • $\begingroup$ Right. In fact it looks like these extensions are actually isomorphic, as opposed to just equivalent, which is interesting. $\endgroup$ Nov 4, 2015 at 2:32
  • $\begingroup$ @RichardRast They're not isomorphic as long as your ground model is actually well-founded, since well-founded models of ZFC are rigid. $\endgroup$ Nov 4, 2015 at 2:35
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    $\begingroup$ Hmm, that seems to be an interesting question, actually. Assuming $\mathsf{ZFC}$ is consistent, are there models $M\subset M[c]$ where $c$ is Cohen over $M$ and such that $M$ and $M[c]$ are isomorphic? This sort of thing has probably been looked at before... $\endgroup$ Nov 4, 2015 at 4:02
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    $\begingroup$ (Joel Hamkins provided me with some references showing that the result I asked about in the comment above indeed holds. The argument goes by appropriately generalizing some facts previously known in the context of nonstandard models of arithmetic. To get started, see Lemmas 5 and 6 here.) $\endgroup$ Nov 4, 2015 at 14:42

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