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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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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. – Anjan Gupta Jan 30 '14 at 10:42

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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One might note that this does not need the ring $R$ to be reduced (as in the question). – Fred Rohrer Feb 1 '14 at 18:51

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