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.

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    $\begingroup$ It's not necessary for the socle to be $0$; consider a product of fields. $\endgroup$ – Eric Wofsey Feb 28 at 22:12
  • $\begingroup$ 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. $\endgroup$ – dmk Feb 28 at 22:23
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    $\begingroup$ 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). $\endgroup$ – 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.)

  • $\begingroup$ Why the downvote? Did I say something wrong? $\endgroup$ – Eric Wofsey Feb 28 at 22:56

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