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.



Whether the result that an ultraproduct which satisfies ACCP is automatically a field generalizes to ultrafilters on larger indexing 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.


