# When is commutative reduced ring a finite direct product of domains?

All rings considered are unital and commutative.

Intro Direct products of domains are reduced rings. The opposite is not true but I must admit I have troubles finding counter-examples. It holds that reduced rings are subdirect products of domains. But this is not that useful as many interesting ring-theoretic properties are not closed under subalgebras (e.g. being Artinian).

Question What are some general, ring-theoretic properties that characterize when a reduced ring is a finite direct product of domains? Are there at least some special cases (rings of small dimension, with zero socle, zero radical...) for which this has some nice characterization?

I do not need this for a specific purpose, and I am not sure whether this has some good, clear answer. It is just a question that is intrinsically interesting to me.

I am not sure how to start. An obvious necessary condition is that socle of the ring is zero (because domains that are not fields have zero socle). But I can't prove that this is sufficient, neither can I find a counterexample.

EDIT: Thanks Eric for pointing out that direct sum of fields is counterexample. As a field is (the only) example of domain that has non-zero socle.

• It's not necessary for the socle to be $0$; consider a product of fields. – Eric Wofsey Feb 28 at 22:12
• Hi, thanks for pointing this out. I forget about this when typing the question. However, this is kind of pathological counterexample. As field is the only case when domain has non-zero socle. – dmk Feb 28 at 22:23
• It's also not sufficient for the socle to be $0$; for instance consider $k[x,y]/(xy)$. Really, I think the sort of conditions you're thinking about (which seem to come from a noncommutative algebra perspective) are not relevant here; this is something you want to think about from the perspective of algebraic geometry (as in my answer). – Eric Wofsey Feb 28 at 22:48

A ring $$R$$ is a finite product of domains iff $$R$$ has finitely many minimal primes and for each maximal ideal $$m\subset R$$, the localization $$R_m$$ is a domain. It is easy to see these properties are necessary (if $$R\cong\prod R_i$$ where each $$R_i$$ is a domain, the minimal primes are the kernels of the projections and every localization at a maximal ideal coincides with a localization of some $$R_i$$).
Conversely, suppose $$R$$ has finitely many minimal primes and its localization at each maximal ideal is a domain. Suppose $$p,q\subset R$$ are two distinct minimal primes and suppose $$p+q$$ is a proper ideal. Then we can extend $$p+q$$ to a maximal ideal $$m$$. Since $$p$$ and $$q$$ are both contained in $$m$$, they give two distinct minimal prime ideals in the localization $$R_m$$. But this is a contradiction, since $$R_m$$ is a domain.
Thus the minimal primes of $$R$$ are pairwise comaximal. By the Chinese remainder theorem, this gives an isomorphism $$R/\sqrt{0}\to \prod R/p_i$$ where $$p_i$$ ranges over the minimal primes of $$R$$. But the nilradical $$\sqrt{0}$$ is trivial, since $$R$$ is locally a domain and thus reduced.
(This is a basic and well-known result in scheme theory, though it is typically stated with "Noetherian" as a hypothesis rather than "finitely many minimal primes". In geometric terms, if $$\operatorname{Spec} R$$ has finitely many irreducible components, then as long as those irreducible components are disjoint, $$\operatorname{Spec} R$$ will be the coproduct of the irreducible components, and so this gives a finite direct product decomposition of $$R$$. But if two irreducible components intersect, then $$R$$ will not be locally integral at any point in the intersection.)