Timeline for How to think about CM rings?
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 25, 2009 at 14:27 | comment | added | user709 | Oh yes I forgot about the embedded primes.. thanks! | |
| Nov 25, 2009 at 14:19 | comment | added | Graham Leuschke | That ring is isomorphic to $k[X,Y,Z]/(XZ,YZ)$. Setting $Y+Z=0$ gives the quotient ring $k[X,Z]/(XZ,Z^2)$, in which $X+Z$ is not a nonzerodivisor. The big difference here is that it's not enough that the hypersurface "intersects $V(I)$ but not in a component". That corresponds to avoiding the minimal primes of the coordinate ring. What you need for a nonzerodivisor is to avoid the <em>associated</em> primes of the ring. For the cut-down ring $k[X,Z]/(XZ,Z^2)$, there is a unique minimal primes $(Z)$, but $(X,Z)$ is also an associated prime. | |
| Nov 25, 2009 at 4:11 | comment | added | user709 | I guess I made some silly mistakes somewhere but can someone help me out? Using Justin's idea I tried to work with $Spec k[X,Y,Z]/(X,Y) \cap Z$. This shouldn't be CM at the origin because it is not locally equidimensional. Its Krull dimension should be 2. But when I compute the depth, I try to first use Y+Z = 0 to cut it, so that only the X-axis remains, and then use X+Z = 0 to cut it, so that the origin remains. This should then give me a regular sequence of length 2. But then this local ring can't be CM... | |
| Nov 24, 2009 at 19:25 | history | answered | Justin DeVries | CC BY-SA 2.5 |