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Let's recall that:

(1): The Ax-Kochen principle says that if $\mathcal{U}$ is a non-principal ultrafilter over prime numbers, then $\prod_{\mathbb{U}} \mathbb{F}_p((t)) \equiv \prod_{\mathbb{U}} \mathbb{Q}_p$.

(2): The Ax-Kochen isomorphism theorem says that if $CH$ (the Continuum hypothesis) holds, and if $\mathcal{U}$ is a non-principal ultrafilter over prime numbers, then $\prod_{\mathbb{U}} \mathbb{F}_p((t)) \cong \prod_{\mathbb{U}} \mathbb{Q}_p$.

It is easily seen that (1) implies (2), as under $CH$, both of the structures $\prod_{\mathbb{U}} \mathbb{F}_p((t))$ and $ \prod_{\mathbb{U}} \mathbb{Q}_p$ have size $\aleph_1$ and are saturated, hence are isomorphic.

On the other hand (2) implies (1) by a forcing argument and absoluteness. To see this, let $V$ be an arbitrary model of $ZFC$ and let $\mathcal{U}$ be a non-principal ultrafilter over prime numbers. Let $G$ be $Add(\aleph_1, 1)$-generic over $V$. Then $V$ and $V[G]$ have the same reals, in particular $\mathcal{U}$ remains an ultrafilter in $V[G]$. Also $CH$ holds in $V[G]$, so $V[G]\models\prod_{\mathbb{U}} \mathbb{F}_p((t)) \cong \prod_{\mathbb{U}} \mathbb{Q}_p$. Hence $V[G]\models\prod_{\mathbb{U}} \mathbb{F}_p((t)) \equiv \prod_{\mathbb{U}} \mathbb{Q}_p$, and by absoluteness $\prod_{\mathbb{U}} \mathbb{F}_p((t)) \equiv \prod_{\mathbb{U}} \mathbb{Q}_p$ holds in $V$.

In Vive la différence II. The Ax-Kochen isomorphism theorem, Shelah has shown $CH$ can not be removed from the Ax-Kochen isomorphism theorem.

Motivated by these results, I would like to ask the following:

Question Is the Ax-Kochen isomorphism theorem consistent with the negation of Continuum hypothesis?

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  • $\begingroup$ I'm confused: if Shelah showed that $\mathsf{CH}$ can't be removed from the theorem, then that means the theorem isn't consistent with $\neg\mathsf{CH}$? Am I missing something? $\endgroup$
    – David Fernandez-Breton
    Commented Jul 15, 2019 at 13:28
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    $\begingroup$ It is a consistency result, there is a model of $ZFC$ in which CH fails and the isomorphism theorem does not hold. It does not say that the failure of CH implies the negation of the theorem. $\endgroup$
    – Mohammad Golshani
    Commented Jul 15, 2019 at 14:04
  • $\begingroup$ It's been a long time since I've thought about model theory, but can this be addressed by techniques of Ellentuck and Rucker "Martin's Axiom and Saturated Models"? They show that under MA there is an ultrafilter $F$ on $\omega$ such that the $F$-ultrapower of any countable structure is saturated. $\endgroup$
    – Todd Eisworth
    Commented Jul 17, 2019 at 14:33
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    $\begingroup$ Also interesting : is the absoluteness of the result of Moloney mentioned in Shelah's paper still open? $\endgroup$
    – Paul Larson
    Commented Jul 17, 2019 at 23:19
  • $\begingroup$ @ToddEisworth Thanks for introducing this nice reference. However, it does not answer the question, as for the isomorphism theorem, we require it holds for all non-principal ultrafilter over prime numbers $\endgroup$
    – Mohammad Golshani
    Commented Jul 18, 2019 at 4:29

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