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Assume That $U,V$ are two filters on the natural number $\mathbb{N}$. We say that $U$ is equivalent to $V$ if there is a bijection $\phi: \mathbb{N} \to \mathbb{N}$ such that $\tilde{\phi}(U)=V$ where $\tilde{\phi}:P(\mathbb{N}) \to P(\mathbb{N})$ is the natural extension of $\phi$ to the power set $P(\mathbb{N})$. Let $U,V$ be two non principal ultra filter on $\mathbb{N}$.

Let $\mathbb{R}^*_{U}$ and $\mathbb{R}^*_{V}$ be the corresponding nonstandard extension of real numbers associated with $U$ and $V$, respectively.

Assume that $\mathbb{R}^*_{U}$ and $\mathbb{R}^*_{V}$ are isomorphic as fields. Does this imply that $U$ and $V$ are equivalent filters?

My apology in advance, if the question is elementary. The question arose me about 17 years ago when I was trying to understand the application of non standard analysis to ordinary differential equations.

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    $\begingroup$ I think it's known that under CH, all ultrapowers $\mathbb{R}_U$ are isomorphic. Maybe somebody more familiar known more about this. $\endgroup$
    – YCor
    Commented Apr 11, 2017 at 9:34
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    $\begingroup$ A question that addresses the result mentioned by YCor can be found here. $\endgroup$
    – Michael Greinecker
    Commented Apr 11, 2017 at 11:03
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    $\begingroup$ Just to finish the proof, there are always nonequivalent nonprincipal ultrafilters (since there are $2^{2^{\aleph_0}}$-many nonprincipal ultrafilters on $\mathbb{N}$ and only $2^{\aleph_0}$-many bijections from $\mathbb{N}$ to itself), so under CH the answer is no. But I think the answer is probably no just in ZFC. $\endgroup$
    – Paul McKenney
    Commented Apr 12, 2017 at 18:11
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    $\begingroup$ Some articles related to this can be found in this MSE post. There are $2^c$ ultrafilters but also $2^c$ non-isomorphic fields when $\neg CH$, so straightforward "counting" does not seem to allow one to decide. $\endgroup$
    – Mikhail Katz
    Commented Apr 19, 2017 at 12:02
  • $\begingroup$ math.stackexchange.com/a/280935/72694 is a related answer by @AndreasBlass. $\endgroup$
    – Mikhail Katz
    Commented Apr 19, 2017 at 13:38

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As far as I understand the question is still open under $\neg CH$, namely whether isomorphism of hyperreal fields implies equivalence of filters (up to permutation of index set). Perhaps one can try the following approach.

In a universe $V$ satisfying $CH$, we can take two inequivalent filters and obtain fields that are automatically isomorphic by the result of Erdos &Co mentioned in the MSE post linked in the comments above. Now the idea is to take a forcing extension $V^F$ satisfying $\neg CH$.

One can't transfer naively the construction of the hyperreal field to $V^F$ because the ultrafilter is not definable, but perhaps one can work with the definable hyperreal field of Kanovei and Shelah (which exploits a huge index set using all ultrafilters simultaneously, thereby defeating non-definability). Perhaps one can specify two variants of the Kanovei-Shelah construction whether the ideals are not equivalent but the quotient fields will be by the Erdos-type argument, and then take a forcing extension to exhibit a similar phenomenon under $\neg CH$.

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