## How much are reduced powers different ?

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Given two infinite sets $X$ and $I$, and a filter ${\cal F}$ on $I$, one defines as usual the equivalence relation $\approx_{\cal F}$ on $X^I$ and obtains the reduced power $Y = X^I / \approx_{\cal F}$.

Question 1 : to what extent do such reduced powers differ when one filter on $I$ is changed to another filter on $I$ ?

Question 2 : consider question 1 in the case of different ultrafilters on $I$, thus in the case of ultrapowers.

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Are you interested in reduced products of bare naked sets or of structures such as groups etc? – Simon Thomas Jul 3 2010 at 14:58
Actually, I am interested in reduced powers and ultrapowers of the field \mathbb{R} of usual real numbers, but also more generally, of rings. – Elemer E Rosinger Jul 3 2010 at 16:26

## 6 Answers

Easy differences arise if one allows principal ultrafilters, since the ultrapower of $X$ by a principal filter is canonically isomorphic to $X$, but other ultrapowers are not. Another easy difference arises when $I$ is uncountable, since one filter might concentrate on a countable subset of $I$ and others might not, and this can dramatically affect the size of the reduced power, making them different.

So the question is more interesting when one considers only non-principal filters and also only uniform filters, meaning that every small subset of $I$ is measure $0$.

In this case, under the Generalized Continuum Hypothesis, the ultrapower of any first order structure is saturated, and thus any two of them will be canonically isomorphic by a back-and-forth argument. Without the GCH, it is consistent with ZFC to have ultrafilters on the same set leading to nonisomorphic ultrapowers.

Also relevant is the Keisler-Shelah theorem, which asserts that two first order structures---such as two graphs, groups or rings---are elementarily equivalent (have all the same first order truths) if and only if they have an isomorphic ultrapowers.

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What´s "small"? – Mariano Suárez-Alvarez Jul 3 2010 at 14:53
In this case, it means size less than the cardinality of $I$. For a filter to give measure $1$ to a strictly smaller set, means in a sense that you have the wrong index set. – Joel David Hamkins Jul 3 2010 at 14:56
Thank you for the answer. As I commented above, I am interested only in non-principal ultrafilters and in filters which contain the Frechet filter. – Elemer E Rosinger Jul 3 2010 at 16:29
Elemer, if by the Frechet filter, you mean the filter of finite sets, this is not good enough, when $I$ is uncountable, since one filter might still concentrate on a countable set while another does not. What you want is the filter of all co-small sets, making your filter uniform in the sense I mentioned. – Joel David Hamkins Jul 3 2010 at 16:32

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In fact, the existence of non-isomorphic ultrapowers of the linearly ordered sets $\mathbb{N}$ or $\mathbb{R}$ over a countable index set is not just consistent with $\neg CH$. It is actually equivalent to $\neg CH$. – Simon Thomas Jul 3 2010 at 16:50
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Given that you are interested in ultrapowers of $\mathbb{R}$, you might like the following which appears in a joint paper with Kramer, Shelah and Tent.

Theorem: Up to isomorphism, the number of ultrapowers $\prod_{\mathcal{U}} \mathbb{R}$, where $\mathcal{U}$ is a nonprincipal ultrafilter over $\mathbb{N}$, is 1 if $CH$ holds and $2^{2^{\aleph_{0}}}$ if $CH$ fails.

Here $CH$ is the Continuuum Hypothesis. (In the case when $CH$ fails, the relevant ultrapowers are already non-isomorphic merely as linearly ordered sets.) The relevant reference is:

L. Kramer, S. Shelah, K. Tent and S. Thomas Asymptotic cones of finitely presented groups, Advances in Mathematics 193 (2005), 142-173.

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Since the question was about reduced powers, not just ultrapowers, it seems worthwhile to point out that, when $F$ is a filter on $I$ but not an ultrafilter, then the reduced power of a field $k$ with respect to $F$ will not be a field, not even an integral domain. Proof: Since $F$ isn't an ultrafilter, $I$ can be partitioned into two pieces $A$ and $B$, neither of which is in $F$. Then the characteristic functions of these pieces represent nonzero elements of the reduced power $k^I/F$ whose product is zero.

Thus, reduced powers of fields modulo non-ultra-filters differ very strongly from ultrapowers, since the latter are fields.

A similar argument shows, for example, that if $X$ is a linearly ordered set with at least two elements, then the reduced power $X^I/F$ is linearly ordered if and only if $F$ is an ultrafilter.

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The (latest version of the) question is probably more general than it should be. Since $X$ is allowed to change and since the filters could be principal ultrafilters, the answer is that absolutely anything can happen. Use a principal ultrafilter $F$ on any set $I$, and vary $X$ at will. If the intention was to prohibit principal ultrafilters, then I can't say that absolutely anything can happen, but quite a lot can. For example, given any ultrapower of any $X$, one could change $X$ to a set bigger than that ultrapower; any ultrapower (or any reduced power) of the new $X$ would be bigger than the ultrapower you started with. So it's not clear that anything useful can be said at (or even near) the level of generality of the question.

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Let me specify my question. Given two ultrapowers $X^I/F$ and $Y^J/G$ where $X,I,Y,J$ are arbitrary infinite sets, while $F,G$ are ultrafilters on $I$ and $J$< respectively, the questions is to what extent :

1) the cardinals of those two ultrapowers can differ ?

2) those two ultrapowers are not isomorphic when $X$ and $Y$ are fields ?

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If you want to add information to the question, you should edit the question to include the information. – JBL Nov 2 2010 at 14:52

A more precise, and at the same time, more general form of the question is as follows : Given a reduced power X^I / F, where X and I are arbitrary infinite sets, while F is an arbitrary filter on I, to what extent does the reduced power X^I / F change, when X, I or F are changed ?

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