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.