Let $F$ be an ultrafilter on some set $X$, $R$ an integral domain and $R_F$ the resulting ultraproduct ring. For an element $(a)$ in the product ring of $R$ indexed by $X$, denote its equivalence class in the ultraproduct by $(a)_F$.

It is a known result that if $X = \mathbb{N}$, and $R_F$ satisfies the ascending chain condition on principal ideals, then $R_F$ is automatically a field.

To see this, note that if $R_F$ is not a field, then neither is $R$, so it is possible to pick some nonzero nonunit $d \in R$. We then can construct an infinite sequence of elements in $R_F$, as $(a_1)_F = (1, d, d^2, d^3, ...)_F$; $(a_2)_F = (1, 1, d, d^2, ...)_F$; $(a_3)_F = (1, 1, 1, d, ...)_F$.

Clearly $(a_1)_F$ is properly divided by $(a_2)_F$ is properly divided by $(a_3)_F$ is ... whence we have the infinite chain of ascending principal ideals $( (a_1)_F ) \subset ( (a_2)_F ) \subset ( (a_3)_F )...$, so $R_F$ will not satisfy ACCP.

I have tried to prove the general result that when $X$ is any infinite set: "$R_F$ cannot satisfy ACCP unless it is a field," but I have been unable to do so. Does anyone think this result is true when $X$ is any infinite set? Is there no way to generalize the proof given to an arbitrary well-ordered set? Does anyone have any ideas for constructing a counterexample?

This result is of interest to me because when $X = \mathbb{N}$, an integral domain is either "very nice" (a field) or "not so nice" (does not satisfy ACCP), with nothing in between. If some counterexample could be given to the general conjecture, it will be of interest to look at ultraproducts indexed by other infinite sets.


The result you want is true for ultrapowers on an arbitrary set, provided that the ultrafilter is not $\sigma$-complete. This includes all nonprincipal ultrafilters on any set, unless there is a measurable cardinal, and in any case includes all nonprincipal ultrafilters on any set of size less than the least measurable cardinal. Finally, if there is a measurable cardinal, then the property can fail for those exotic ultrapowers.

Theorem. If $R$ is an integral domain and $F$ is an ultrafilter on a set $X$, but is not $\sigma$-complete, then $\Pi R/F$ is a field if and only if it satisfies ACCP.

Proof. Suppose that $F$ is an ultrafilter on $X$, but it is not $\sigma$-complete. This means that we may partition $X=\bigsqcup_n X_n$ with $X_n\notin F$. Now, if the ultrapower $\Pi R/F$ is not a field, let $d$ be any nonzero non-unit, and let $a_1=(1\upharpoonright X_0,d\upharpoonright X_1,d^2\upharpoonright X_2,\ldots)$ and similarly $a_2=(1\upharpoonright X_0,1\upharpoonright X_1,d\upharpoonright X_2,\ldots)$ and with $a_n$ just as in your argument, but using constant values on the pieces of the partition, instead of just on one point. Since $d$ is not a unit, the principal ideals $(a_1)\subsetneq (a_2)\subsetneq\cdots$ are strictly ascending, just as in your answer, and so the ascending chain condition is violated. QED

So the result is true for all $\sigma$-incomplete ultrafilters. But observe that this actually includes all nonprincipal ultrafilters, if there is no measurable cardinal, since the existence of a measurable cardinal is equivalent to the existence of an $\omega$-complete nonprincipal ultrafilter on a set. Furthermore, even if there is a measurable cardinal, then the result is still true for all ultrafilters on any set of size less than the least measurable cardinal (which is quite large), since no nonprincipal ultrafilter on any set of size less than the least measurable cardinal can be $\sigma$-complete.

Thus, in any case, te result is true for all ultrafilters on $\aleph_1$, $\aleph_2$, on the continuum $2^{\aleph_0}$ and so on for a long way past the continuum and more, just because all these cardinals are definitely less than the least measurable cardinal, if any.

Meanwhile, if there does happen to be a measurable cardinal, then the result can fail. The reason is that if $\kappa$ is a measurable cardinal, with a $\kappa$-complete nonprincipal ultrafilter $\mu$ on $\kappa$, then the ultrapower $\Pi R/\mu$ of any structure $R$ is actually isomorphic to $R$, if $R$ is smaller in size than $\kappa$. The point is that these ultrapowers only affect extremely large objects. Since there are non-field integral domains with the ACPP, it follows that there can be ultrapowers by nonprincipal ultrafilters that are not fields, but still have ACCP.

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    $\begingroup$ The complementary characterization is that if $F$ is $\sigma$-complete, then $\prod R/F$ satisfies ACCP iff $R$ does. $\endgroup$ – Emil Jeřábek Jul 1 '14 at 10:30

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