In this question $R$ is a commutative noetherian local ring with unity. 


One can construct examples of rings $R$ and *zerodivisors* $z$ such that $\dim R/(z)=\dim R-1$, e.g., $S\colon=k[a,b,c],\ \mathfrak{m}\colon=(a,b,c),\ R\colon=S\_\mathfrak{m}/(a^2,ab)S\_\mathfrak{m},\ z\colon=b^2$.


One can also construct examples of rings $R$ and *zerodivisors* $z$ such that $\mathrm{depth}\ R/(z)=\mathrm{depth}\ R-1$, e.g., $S\colon=k[a,b,c],\ \mathfrak{m}\colon=(a,b,c),\ R\colon=S\_\mathfrak{m}/(a^2)S\_\mathfrak{m},\ z\colon=ab.$


What is an example of a zerodivisor that will reduce *both* the dimension *and* the depth by $1$, simultaneously? Is that possible?



Can a zerodivisor reduce both the depth and the dimension?
$R = k[[a,b,c,d]]/(a,b,c)^2 \cap (c) \cap (c,d)^2 $, $\dim R = 3$ and $\mathrm{depth}R = 1$. We have $d$ is a zerodivisor. Because $R/d \cong k[[a,b,c]]/(a,b,c)^2 \cap (c)$.
So $\dim R/d = 2$ and $\mathrm{depth}R/d = 0$.


Edit: As Mahdi comment $\mathrm{depth}R/d = 2$. I repair as follows.


I need the following interesting result (see, <http://www.sciencedirect.com/science/article/pii/0021869379903065> Proposition 9)


Lemma: Let $\mathfrak{q} \in \mathrm{Ass}R$, $\mathfrak{p}$ is minimal over $\mathfrak{q}+I$. Then there exists $n$ such that $\mathfrak{p} \in \mathrm{Ass}R/I^n$ for all $m \geq n$.


Applying for our ring we have $(a, b, c) \in \mathrm{Ass}R$ hence $(a,b,c,d) \in \mathrm{Ass}R/d^n$ for $n \gg 0$. So $\mathrm{depth}R/d^n = 0$ for all $n\gg 0$.


