most recent 30 from http://mathoverflow.net 2013-05-19T13:18:18Z http://mathoverflow.net/feeds/question/93428 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93428/a-problem-on-cohen-macaulay-ring mickeyivy 2012-04-07T16:50:06Z 2012-04-08T06:33:09Z

<p>If $A$ is a Cohen Macaulay local ring, and $B$ is a quotient ring of $A$ and $B$ is also Cohen Macaulay, Then is $B$ always a quotient by a regular sequence of $A$?</p>

http://mathoverflow.net/questions/93428/a-problem-on-cohen-macaulay-ring/93429#93429 Liran Shaul 2012-04-07T17:00:41Z 2012-04-07T17:00:41Z

<p>The answer is no. Let $k$ be a field. Let $A=k[x]/x^2$, and let $B=k$.</p> <p>$A$ is Artinian, so it is Cohen Macaulay. $B$ is clearly Cohen Macaulay. But the only non-trivial ideal in $A$ is $(x)$ which is not generated by a regular sequence, because every element of it is a zero-divisor.</p>

http://mathoverflow.net/questions/93428/a-problem-on-cohen-macaulay-ring/93430#93430 Mahdi Majidi-Zolbanin 2012-04-07T17:11:30Z 2012-04-07T17:11:30Z

<p>As another example, say $\mathfrak{m}$ is the maximal ideal of $A$. Then $B:=A/\mathfrak{m}$ is Cohen-Macaulay (field), but $\mathfrak{m}$ is not generated by a regular sequence, unless $A$ is regular.</p>

http://mathoverflow.net/questions/93428/a-problem-on-cohen-macaulay-ring/93452#93452 Sándor Kovács 2012-04-07T20:07:40Z 2012-04-07T20:40:36Z

<p>A more general series of examples (when this fails) is given by choosing $B$ to be Cohen-Macaulay, but not Gorenstein and $A$ to be Gorenstein.</p> <p>An explicit example is when <code>$B=(k[x,y,z]/(xy,yz,zx))_{\mathfrak m}$</code> is the affine coordinate ring of three lines in $\mathbb A^3$ meeting in one point, but not contained in a plane and $A=k[x,y,z]_{\mathfrak m}$ where ${\mathfrak m}=(x,y,z)$.</p>

http://mathoverflow.net/questions/93428/a-problem-on-cohen-macaulay-ring/93460#93460 Sándor Kovács 2012-04-07T20:39:43Z 2012-04-08T06:33:09Z

<p>Here is an even better example:</p> <p><code>$A=(k[x,y,z]/(xy-z^2))_{\mathfrak m}$</code> with ${\mathfrak m}=(x,y,z)$, $B=k[x]_{(x)}$. Here $A$ is Gorenstein and $B$ is regular. The map is given by $x\mapsto x$ and $y,z\mapsto 0$.</p> <p>Geometrically, $B$ corresponds to a line going through the vertex of a quadric cone corresponding to $A$. It is not a quotient by a regular sequence, because that would have to be just a regular element for dimension reasons, but the line is not a Cartier divisor, so it cannot be defined by a single equation. To see that it is not a Cartier divisor, for example one can compute its self-intersection number which is $\frac 12$.</p>