3
$\begingroup$

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?

$\endgroup$

2 Answers 2

8
$\begingroup$

$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 $n\gg 0$.

$\endgroup$
7
  • 1
    $\begingroup$ 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 $\mathrm{depth} R/d=0$? $\endgroup$ Jan 17, 2013 at 4:54
  • $\begingroup$ @ Mahdi: see my edit. $\endgroup$ Jan 17, 2013 at 5:56
  • 1
    $\begingroup$ 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 $2$. $\endgroup$ Jan 17, 2013 at 6:12
  • $\begingroup$ $(a, b, c)^2 \cap (c) = (ac, bc, c^2) = (a, b, c)(c)$ $\endgroup$ Jan 17, 2013 at 6:23
  • 2
    $\begingroup$ This is very good. Thank you! By the way, here we can take $n=2$. $\endgroup$ Jan 17, 2013 at 7:22
0
$\begingroup$

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 $0$.

$\endgroup$
1
  • 1
    $\begingroup$ @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. $\endgroup$ Jan 17, 2013 at 1:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.