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Let $A$ be a Noetherian ring and $\mathfrak{p} \in Spec(A).$ If $x \in \mathfrak{p}$ then it is well-known that $ht_{A/(x)}(\mathfrak{p}/(x)) \leq ht_A(\mathfrak{p}) \leq ht_{A/(x)}(\mathfrak{p}/(x))+1$ where $ht$ stands for height.

Moreover, if $x$ is not a zero-divisor, then $ht_A(\mathfrak{p})= ht_{A/(x)}(\mathfrak{p}/(x))+1.$

Does the converse hold? that is, if the equality holds, is it possible to deduce that $x$ is not a zero-divisor.

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up vote 5 down vote accepted

No, the converse does not hold. The point is that $x$ may not be a zero-divisor locally at $\mathfrak p$, but could still be a zero divisor elsewhere on Spec $A$.

As an extreme example, imagine that $A = B \times C$ with $B$ and $C$ non-zero, and take $\mathfrak p = \mathfrak q \times C$ for some prime ideal $\mathfrak q$ of $B$. Let $b \in \mathfrak q$ be a non-zero divisor, and set $x = (b,0)$. Then $x$ is a zero-divisor in $A$, but $$ht_A(\mathfrak p) = ht_B(\mathfrak q) = ht_{B/(b)}(\mathfrak q/(b))+1 = ht_{B/(b) \times C}(\mathfrak q \times C) + 1 = ht_{A/(x)}(\mathfrak p/(x)) + 1.$$ You can also find examples in which $A$ is not a product, though; I'll leave this as an exercise.

The general moral here is that such questions are best considered from the point of view of the geometry/topology of Spec $A$.

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