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In model theory, two structures $\mathfrak{A}, \mathfrak{B}$ of identical signature $\Sigma$ are said to be elementarily equivalent ($\mathfrak{A} \equiv \mathfrak{B}$) if they satisfy exactly the same first-order sentences w.r.t. $\Sigma$. An astounding theorem giving an algebraic characterisation of this notion is the so-called Keisler-Shelah isomorphism theorem, proved originally by Keisler (assuming GCH) and then by Shelah (avoiding GCH), which we state in its modern strengthening (saying that only a single ultrafilter is needed):

$\mathfrak{A} \equiv \mathfrak{B} \ \iff \ \exists \mathcal{U} \text{ s.t. } (\Pi_{i\in\mathcal{I}} \ \mathfrak{A})/\mathcal{U} \cong (\Pi_{i\in\mathcal{I}} \ \mathfrak{B})/\mathcal{U},$

where $\mathcal{U}$ is a non-principal ultrafilter on, say, $\mathcal{I} = \mathbb{N}$. That is, two structures are elementarily equivalent iff they have isomorphic ultrapowers.

My question is the following (admittedly rather vague): Does anyone know of constructions in which an ultrafilter is chosen by an appeal to this characterisation and then used for other means? An example of what I have in mind would be something like this (using the fact that any two real closed fields are elementarily equivalent w.r.t. the language of ordered rings): In order to perform some construction $C$ I ``choose'' a non-principal ultrafilter $\mathcal{U}$ on $\mathbb{N}$ by specifying it as a witness to the following isomorphism induced by Keisler-Shelah:

$\mathbb{R}^\mathbb{N}/\mathcal{U} \cong \mathbb{R}_{alg}^\mathbb{N}/\mathcal{U},$

where $\mathbb{R}_{alg}$ is the field of real algebraic numbers. So the construction $C$ should be dependent upon the fact that $\mathcal{U}$ is a non-principal ultrafilter bearing witness to the Keisler-Shelah isomorphism between some ultrapower of the reals and the algebraic reals, resp.

Also, a follow-up question: Let's say I'd like to ``solve'' the above isomorphism for $\mathcal{U}$. Are there interesting things in general known about the solution space, e.g., the set of all non-principal ultrafilters bearing witness to the Keisler-Shelah isomorphism for two fixed elementarily equivalent structures such as $\mathbb{R}$ and $\mathbb{R}_{alg}$? What machinery is useful in investigating this?

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up vote 9 down vote accepted

Under the Continuum Hypothesis, your solution space is all nonprincipal ultrafilters. This is because under CH, the ultrapower $M^N/U$ of a mathematical structure $M$ of size at most continuum does not actually depend on the (nonprincipal) ultrafilter $U$. One can see this by using the fact that the ultrapower will be saturated, and so one can run a back-and-forth argument to achieve the isomorphism. In particular, it follows under CH that any $U$ will witness your desired isomorphism for $R^N/U\cong (R_{alg})^N/U$. (See Corollary 6.1.2 in Chang-Keisler's book Model Theory.)

A similar fact holds for larger cardinals and larger structures under GCH, but here, one needs an additional assumption on the ultrafilter. Namely, Theorem 6.1.9 in Chang-Keisler asserts that if $2^\alpha=\alpha^+$ and $A$ and $B$ are two structures of size at most $\alpha^+$, then they are elementarily equivalent if and only if $\Pi_DA\cong\Pi_D B$ for any $\alpha^+$-good incomplete ultrafilter $D$ on $\alpha$. The proof uses the same saturation idea, and this establishes the Keisler-Shelah theorem in the case that GCH holds.

Chang-Keisler states (page 393-394) that it is open whether the assertion of Theorem 6.1.9 stated above holds under $\neg CH$.

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This is so helpful, thanks so much! –  Grant Olney Passmore May 13 '10 at 20:55
    
Thanks for accepting. Meanwhile, I'd still like to hear how flexible the choice of U is when CH fails... –  Joel David Hamkins May 13 '10 at 21:03
    
I'm very interested in the not-CH case as well, though the question seems like it might be difficult. If we do eventually get a nice answer for the set of solutions in the context of not-CH, what is your feeling on the right machinery for examining structure in the solution space? Is the Rudin-Blass ordering on ultrafilters here the standard machinery? –  Grant Olney Passmore May 13 '10 at 21:56
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I've realized that the saturation+back-and-forth argument also gives you a quick proof of your instance of the Keisler-Shelah theorem in the case that CH holds, since the two ultrapowers will be elementary equivalent and saturated of size continuum. When CH fails, I'm not sure what happens, but the class of ultrafilters is intensely studied. The Rudin-Keisler order is equivalent to commutative diagrams factoring the ultrapower maps, so this may interest you. –  Joel David Hamkins May 13 '10 at 22:37
    
Ah beautiful! And thank you for suggesting the Rudin-Keisler order + pointing out that equivalence. Again, this is very helpful. –  Grant Olney Passmore May 13 '10 at 23:45
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As you might expect, things are consistently much more interesting if $CH$ fails. This has been explored by Shelah in a fascinating series of papers "Vive la difference I - III". For example, it is consistent that there is a nonprincipal ultrafilter $\mathcal{U}$ on $\omega$ such that if $(R_{n})$ and $(S_{n})$ are sequences of discrete rank 1 valuation rings having countable residue fields, then any isomorphism $\varphi: \prod_{\mathcal{U}}R_{n} \to \prod_{\mathcal{U}}S_{n}$ is an ultraproduct of isomorphisms $f_{n}: R_{n} \to S_{n}$. In particular, $\mathcal{U}$-almost all $R_{n}$ are isomorphic to the corresponding $S_{n}$ and so the Ax-Kochen isomorphism theorem doesn't hold with respect to $\mathcal{U}$.

If you are only interested in ultraproducts of fixed structures $A$, $B$, then I should mention that it is also consistent that there exists an ultrafilter $\mathcal{A}$ on $\omega$ such that if $A$ and $B$ are countable structures which satisfy the strong independence property, then the corresponding $\mathcal{A}$-ultraproducts are isomorphic iff $A \cong B$.

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This is fascinating and extremely helpful. Thanks so much! –  Grant Olney Passmore May 13 '10 at 23:44
    
I didn't know about the connection between these questions and the strong independence property! Is this "recent" (well, mod the fact you wrote this answer two years ago) or is this part of the Vive la différance I-III series? –  Andrés Villaveces Mar 16 '12 at 4:35
    
Yes, it is part of the Vive la différance I-III series. –  Simon Thomas Mar 16 '12 at 13:24
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