Let $(A,m)$ be a Noetherian local ring and let $x \in m$. If $x$ is a non-zero divisor, then we know that the Krull dimension satisfies $\dim A/(x) = \dim A-1$, see e.g. Atiyah-MacDonald Corollary 11.18. But what if $x$ is a zero-divisor? By Theorem 13.6 in Matsumura (CRT) the dimension of $A/(x)$ can not drop below $\dim A -1$. Intuitively, it seems to me that there must exist cases where the dimension does drop to $\dim A-1$. Anyone knows any such example? I base this intuition on the fact that not all maximal chains of prime ideals of $A$ need to have the same length.
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Take $A=k[[x,y]]/(x^2,xy)$. Then both $x,y$ are zero divisors, $\dim A=1$, $\dim A/xA=1$, $\dim A/yA=0$.
Mohan
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Ok. I suppose you could have used $k[x,y]$ instead of $k[[x,y]]$ as well? – Manos Jul 08 '15 at 19:06
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1Except, $k[x,y]/(x^2,xy)$ is not local. – Mohan Jul 08 '15 at 20:01