# Prime ideals in an arbitrary direct product of rings

By ring I mean commutative unital ring.

The prime ideal structure of a finite direct product of rings is well known: For $\prod_{i=1}^n R_i$, it is of the form $\prod_{i=1}^n P_i$ where only one $P_j$ is a proper prime ideal of $R_j$ and for $i\neq j$, $P_i=R_i$.

I could not find any information for arbitrary direct products. Certainly, an ideal of the form above is prime. I have a feeling that not all prime ideals are of this form. Has the prime ideal structure of an arbitrary direct product been completely determined in general? Or does one need to construct some pathological example to prove that not all prime ideals are of this form?

• Can you refer me to a proof of the theorem for finite products, please? – Sam Dec 15 '14 at 21:30
• @Itried I just posted the answer on this site. – Andrew Chiriac Apr 1 '15 at 19:09

## 2 Answers

The prime ideal structure of $\prod_{i \in I} R_i$ is very complicated. If each $R_i$ is a field, then it is a well-known fact (or exercise, if you want) that $F \mapsto \{a \in \prod_i R_i : \{i \in I : a(i)=0\} \in F\}\}$ sets up an order-preserving bijection between filters on $I$ and ideals of $\prod_{i \in I} R_i$. Since $\prod_{i \in I} R_i$ is von Neumann regular, the prime ideals are exactly the maximal ideals. Hence, the bijection restricts to a bijection (actually, homeomorphism) between ultrafilters on $I$ and prime ideals of $\prod_{i \in I} R_i$. The principal ultrafilters correspond to the "obvious" maximal ideals $\ker(\prod_{i \in I} R_i \to R_i)$, but using the axiom of choice one constructs lots of non-principal ultrafilters on $I$ and therefore "complicated" maximal ideals of $\prod_{i \in I} R_i$. The residue fields are also hard to control. If $\mathfrak{m}$ is a non-principal maximal ideal of $\prod_{n \in \mathbb{N}} \mathbb{R}$, then the residue field $(\prod_{n \in \mathbb{N}} \mathbb{R})/\mathfrak{m}$ is known as "the" field of hyperreal numbers. It does not depend of the choice of $\mathfrak{m}$ if and only if the Continuum Hypothesis holds! I don't dare to think about the prime ideals of $\prod_{n \in \mathbb{N}} \mathbb{Z}$.

• Can you please reference a proof of the theorem for a finite direct product? – Sam Dec 15 '14 at 22:21

The direct sum of the rings $R_i$ - i.e. elements of the direct product with only finitely many non-zero entries - is a proper ideal of the direct product not of that form.

• Yes but it is also not a prime ideal. To extend this ideal to a maximal ideal one needs the axiom of choice (or something similar, at least). – Martin Brandenburg Dec 3 '14 at 14:14