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Let $R$ be a reduced commutative non noetherian ring of dimension $d$ and $a$ a non zero divisor. Can I say that Krull dimension of $R/(a)$ is at most $d - 1$?

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    $\begingroup$ Just now I realized that in any arbitrary reduced commutative ring set of zero divisors is the union of minimal primes. So I got the answer of the first part of my question. $\endgroup$
    – Anjan Gupta
    Commented Jan 30, 2014 at 10:42

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Since $a$ is a non-zerodivisor of $R$, it does not lie in any of the minimal primes of $R$. Therefore any chain of primes in $R$ that all contain $(a)$ has no minimal prime in the chain and can therefore be extended to a larger chain of primes in $R$, namely, by including a prime properly contained in the smallest prime in the chain. So $\dim R/(a) < \dim R$.

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    $\begingroup$ One might note that this does not need the ring $R$ to be reduced (as in the question). $\endgroup$
    – Fred Rohrer
    Commented Feb 1, 2014 at 18:51
  • $\begingroup$ And it also holds for Noetherian rings. Place haha emoticon here. $\endgroup$
    – Jesse Elliott
    Commented Nov 2, 2023 at 2:38

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