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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?

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2 Answers 2

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$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 $n\gg 0$.

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    $\begingroup$ Dear Pham: If I am not mistaken $\mathrm{depth} R/d=2$. In fact, I think $\{a,b\}$ forms a regular sequence in $R/d$. Can you explain why you think $\mathrm{depth} R/d=0$? $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Jan 17, 2013 at 4:54
  • $\begingroup$ @ Mahdi: see my edit. $\endgroup$
    – Pham Hung Quy
    Commented Jan 17, 2013 at 5:56
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    $\begingroup$ I don't think this is correct. The reason is every element in $(a,b,c)^2\cap(c)\cap(c,d)^2$ must be a multiple of $c$. Therefore, $a^2,b^2\not\in(a,b,c)^2\cap(c)\cap(c,d)^2$. Therefore the isomorphism you wrote is not correct. I compute $(a,b,c)^2\cap(c)\cap(c,d)^2=(c^2,bcd,acd)$. Therefore, $R/(d)\cong k[[a,b,c,d]]/(c^2,d)$, which has dimension and depth both equal to $2$. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Jan 17, 2013 at 6:12
  • $\begingroup$ $(a, b, c)^2 \cap (c) = (ac, bc, c^2) = (a, b, c)(c)$ $\endgroup$
    – Pham Hung Quy
    Commented Jan 17, 2013 at 6:23
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    $\begingroup$ This is very good. Thank you! By the way, here we can take $n=2$. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Jan 17, 2013 at 7:22
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OOPS: As Sándor points out, I missed the assumption that the ring is local, in the following example of the wrong thing:

"Inside 3-space, glue together a plane $y=0$ transversely with a parabola $z=0, x=y^2$ and a line $z=1, x=0$ meeting it in a separate point. This is reduced, and I'm pretty sure its depth is $1$. Because of the plane, its dimension is $2$.

Now cut it with $x=0$, which cuts the plane to a line and the parabola to a double point leaving the line alone. Hence $x$ is a zero divisor in $k[x,y,z]/(\langle y\rangle \cap \langle z,x-y^2\rangle \cap \langle z-1,x\rangle)$, and cutting with it drops the dimension from $2$ to $1$.

The resulting space is generically reduced, but not reduced, so I'm pretty sure its depth is $0$.

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    $\begingroup$ @Allen: $\mathrm{depth}$ is usually considered over a local ring and in fact, the question includes that assumption. The $x$ in your example seems like a red herring. It's only a zero-divisor away from where the interesting thing is happening. In particular, I don't see why it would be a zero-divisor after localizing at the point where you are reducing the depth. $\endgroup$
    – Sándor Kovács
    Commented Jan 17, 2013 at 1:28

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