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I recently came across a property of commutative rings which I could prove only for rings that are (isomorphic to) a direct product of (possibly infinitely many) local rings.

It might be that my proof can be generalized to other kinds of rings, but nevertheless I am curious as to which commutative rings satisfy this property, i.e. are a direct product of local (commutative) rings.

For finite products of rings, I reckon it is at least necessary that no prime ideal is contained in more than one maximal ideal, as Spec$(A \times B)$ consists of ideals of the form $\mathbb{p} \times B$ or $A \times \mathbb{q}$ for $\mathbb{p} \in $ Spec$(A)$, $\mathbb{q} \in $ Spec$(B)$. I don't know whether this is sufficient or whether this would generalize to arbitary products of rings.

Any thoughts you have on this are welcome!

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    $\begingroup$ It sounds like you are asking about the spectrum of an infinite product of rings (here, local ones), since the condition you give in the finite product case is basically "the irreducible components of $Spec(R)$ have only one closed point". It is known to be a difficult question ; you may check math.stackexchange.com/questions/1529988/… $\endgroup$ – Captain Lama Mar 28 '16 at 13:26
  • $\begingroup$ Do you have any thoughts on the finite case? Is it sufficient that every prime ideal is contained in exactly one maximal ideal, to conclude that the ring is a finite product of local rings? Or what other conditions are needed? $\endgroup$ – Bib-lost Mar 28 '16 at 13:59
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    $\begingroup$ Finite case: math.stackexchange.com/questions/834213/… $\endgroup$ – user26857 Mar 28 '16 at 14:08
  • $\begingroup$ This gives sufficient conditions, but not necessary. Surely, not even all local rings have a nilpotent nilradical? $\endgroup$ – Bib-lost Apr 1 '16 at 22:18

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