Let $(S,m)$ be a commutative Gorenstein local ring, $I$ an ideal of $S$ such that $\operatorname{ht} I=t$, and $R=S/I$. Let $a \in m$ be an $R$-regular element such that for any prime ideal $p\in\operatorname{Ass}_{S}(R)$, $\operatorname{ht}p=t$. Then, for any prime ideal $p \in \operatorname{Ass}_{S}(R/aR)$, $\operatorname{ht}p=t+1$?
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On the line line, you probably wanted to say ht$p = t + 1$. This is more or less a duplication of the question http://math.stackexchange.com/q/724772/84157. The answer I wrote shows that it is not true. – Youngsu Mar 27 '14 at 17:27
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In your counter example, R is Gorenstein? – user138400 Mar 28 '14 at 15:35
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@user138400: No. It is not Cohen-Macaulay. If it were what you are asking is true even for regular sequences. – Youngsu Mar 30 '14 at 19:10
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All these questions about height unmixed ideals are solved by a (counter)example given here. – user26857 Apr 10 '14 at 21:18