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Let's gather up all the countable integral domains into a set $\vec{R}$ such that each countable integral domain appears exactly once up to isomorphism in $\vec{R}$. Let $\vec{R}$ be indexed by $I$.

Let's then define $A$ as $\prod_{i \in I} \vec{R}_i$.

We know from this question and its answers that classifying the prime ideals of this object is probably hopeless, since even the simpler object $\prod_{i \in \mathbb{N}} \mathbb{Z}$ has really complicated prime ideals. I am curious, however, what prime ideals are lurking inside this object.

With that in mind, what are some prime ideals of $A$?

I have found one kind of prime ideals so far.

Ideals that are prime in exactly one component

Let $I$ be the product of $\vec{R}_i$ for all components of $\vec{R}$ but one and a prime ideal $J$ for the remaining component.

The complement of $I$ is isomorphic to the complement of $J$ and hence multiplicatively closed.

Greg Nisbet
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  • Is there any reason that the collection you're proposing to form the product with is not a proper class? I'm curious to know if something clever is going on there, because if it's a proper class then the product might not really even have an underlying set... But OTOH I would have expected Martin to object immediately if this was an issue. – rschwieb May 06 '25 at 15:48
  • @rschwieb I think it works. The once up to isomorphism-thing is what's doing the heavy lifting. Each integral domain only appears once up to isomorphism in the set $\vec{R}$. Very concretely, if we took $\mathbb{N}$ as our carrier and took all possible structures over it in the language of rings and got $\vec{S}$, we could build $\vec{R}$ by well-ordering $\vec{S}$ and then throwing out everything that isn't an integral domain or is isomorphic to something we've already seen. – Greg Nisbet May 06 '25 at 16:05
  • You do need both the cardinality restriction and the once-up-to-isomorphism thing to avoid blowing up and including proper-class-many things though. I think. – Greg Nisbet May 06 '25 at 16:07

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There won't be much you can say. There is a surjective homomorphism $\vec{R} \to \prod_{d > 1} \mathbb{Z}[\sqrt{d}]$ and I bet that even that simpler product ring has no known prime ideal structure. See SE/1533237 for the already quite hard case $\prod_{n \in \mathbb{N}} \mathbb{Z}$.

But since you have asked for examples. Let $(R_i)_{i \in I}$ be any family of integral domains. If $\mathcal{U}$ is an ultrafilter on the set $I$, then $\{a \in \prod_{i \in I} R_i : \{i : a(i)=0\} \in \mathcal{U}\}$ is a prime ideal in $\prod_{i \in I} R_i$. This applies in particular to your example where $I = $ all countable integral domains up to $\cong$ and $R_{i}=i$.

Martin Brandenburg
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