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For any $n \geq 0$, does there exist a ring $A$, commutative with unity, whose primes form a single chain of length $n$ under strict inclusion?

Asking this question out of pure curiosity. If it's true, I can't seem to construct an example: we would have $\dim{A} = n$, and the open sets of $\operatorname{Spec}{A}$ would form a single chain of inclusions. We can also assume $A$ to be an integral domain, since it has a single minimal prime equal to the nilradical, which we can quotient out.

Emory Sun
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  • @DietrichBurde The inclusions must be strict—sorry if that wasn't clear from the question. – Emory Sun Jan 29 '23 at 14:36
  • $A$ is local. If it is Noetherian, then any non-invertible element is contained in the prime of height $1$ (or $0$, but this one is contained in the prime of height $1$). So we must have $n=1$. So for $n \geq 2$, examples must be non-Noetherian. Note also that if it’s impossible with $n=2$, then it’s impossible for any $n \geq 2$. – Aphelli Jan 29 '23 at 14:39
  • I’m not familiar at all with general valuation rings, but maybe a construction like in the accepted answer to https://math.stackexchange.com/questions/6994/how-badly-can-krulls-hauptidealsatz-fail-for-non-noetherian-rings can work. – Aphelli Jan 29 '23 at 14:48
  • @Aphelli Could you elaborate on why $A$ local and Noetherian in this case implies every nonunit is in the prime of height $1$? Of course every nonunit is contained in some prime on the chain, but why can you guarantee it to be the "smallest" non-minimal prime? – Emory Sun Jan 29 '23 at 14:58
  • @Aphelli I just realized this follows from Krull—sorry. Thank you very much for your answer and the link you shared. – Emory Sun Jan 29 '23 at 14:59

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