Can a zerodivisor reduce both the depth and the dimension? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:39:42Z http://mathoverflow.net/feeds/question/119111 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119111/can-a-zerodivisor-reduce-both-the-depth-and-the-dimension Can a zerodivisor reduce both the depth and the dimension? Mahdi Majidi-Zolbanin 2013-01-16T20:12:48Z 2013-01-17T07:46:53Z <p>In this question $R$ is a commutative noetherian local ring with unity. </p> <p>One can construct examples of rings $R$ and <em>zerodivisors</em> $z$ such that $\dim R/(z)=\dim R-1$, e.g., $S\colon=k[a,b,c],\ \mathfrak{m}\colon=(a,b,c),\ R\colon=S_\mathfrak{m}/(a^2,ab)S_\mathfrak{m},\ z\colon=b^2$.</p> <p>One can also construct examples of rings $R$ and <em>zerodivisors</em> $z$ such that $\mathrm{depth}\ R/(z)=\mathrm{depth}\ R-1$, e.g., $S\colon=k[a,b,c],\ \mathfrak{m}\colon=(a,b,c),\ R\colon=S_\mathfrak{m}/(a^2)S_\mathfrak{m},\ z\colon=ab.$</p> <p>What is an example of a zerodivisor that will reduce <em>both</em> the dimension <em>and</em> the depth by $1$, simultaneously? Is that possible?</p> http://mathoverflow.net/questions/119111/can-a-zerodivisor-reduce-both-the-depth-and-the-dimension/119122#119122 Answer by Allen Knutson for Can a zerodivisor reduce both the depth and the dimension? Allen Knutson 2013-01-16T23:59:01Z 2013-01-17T02:06:55Z <p>OOPS: As Sándor points out, I missed the assumption that the ring is local, in the following example of the wrong thing:</p> <p>"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 $2$.</p> <p>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 $1$.</p> <p>The resulting space is generically reduced, but not reduced, so I'm pretty sure its depth is $0$.</p> http://mathoverflow.net/questions/119111/can-a-zerodivisor-reduce-both-the-depth-and-the-dimension/119135#119135 Answer by Pham Hung Quy for Can a zerodivisor reduce both the depth and the dimension? Pham Hung Quy 2013-01-17T04:30:20Z 2013-01-17T06:50:52Z <p>$R = k[[a,b,c,d]]/(a,b,c)^2 \cap (c) \cap (c,d)^2$, $\dim R = 3$ 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 $\mathrm{depth}R/d = 0$.</p> <p>Edit: As Mahdi comment $\mathrm{depth}R/d = 2$. I repair as follows.</p> <p>I need the following interesting result (see, <a href="http://www.sciencedirect.com/science/article/pii/0021869379903065" rel="nofollow">http://www.sciencedirect.com/science/article/pii/0021869379903065</a> Proposition 9)</p> <p>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 $m \geq n$.</p> <p>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 $n\gg 0$.</p>