Timeline for Regular sequence from prime ideal
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 22, 2018 at 20:35 | comment | added | Paul | Also, I would be happy with a nice reference (because, you know, each one is supposed to do their own research). I’m trying to follow pfister and de jong. But these seem more specific questions than the usual material covered in a textbook. I’m sorry if it feels that I asked without enough insight on the topic. | |
| Nov 22, 2018 at 20:30 | comment | added | Paul | I just require that there are exactly h of them and the condition on V(I) intersecting only at 0 the other irreducible components. | |
| Nov 22, 2018 at 20:27 | comment | added | Hailong Dao | Paul, if depends on specifics. Of course, if you don't require anything from the $g$s, then we can just take them to be generators of $I$. My answer actually says that as long as the quotient by the ideal $g$ has depth at least $2$, what you required forces $V(g) =V(I)$. | |
| Nov 22, 2018 at 20:08 | vote | accept | Paul | ||
| Nov 22, 2018 at 20:08 | comment | added | Paul | Right. I didn't see that assuming that $g_1, \ldots, g_h$ was a regular sequence was playing against me, rather than with me. Do you know if it is possible if we don't require $g_1, \ldots, g_h$ to be a regular sequence? or if we don't assume that we are in a cohen-macaulay ring? | |
| Nov 22, 2018 at 20:00 | comment | added | Hailong Dao | I don't understand what you are saying. The height of $I$ is $h$, so it is not an embedded component. | |
| Nov 22, 2018 at 19:56 | comment | added | Paul | I don't understand why this does not allow that the other irreducible components don't intersect in codim 1 sets. But on the other hand, your answer suggests that they have to intersect to $V(I)$ in something of codim > 2. Does this imply that $V(I)$ is neccesarily an embedded component in some other irreducible component? That $g_1, \ldots, g_h$ is a regular sequence just tells that $g_{i+1}$ does not vanish along a whole component of $V(g_1, \ldots, g_{i})$ but it may vanish along a part of $V(g_1, \ldots, g_{i})$. | |
| Nov 22, 2018 at 19:40 | comment | added | Hailong Dao | You wrote that $g_1,...,g_h$ is a regular sequence. | |
| Nov 22, 2018 at 19:39 | comment | added | Paul | I don't see why $\dim(V(I)) \geq 2$ implies that $g$ defines a complete intersection. This happens if the singular locus of $V(g)$ has codimension >1 by Hartshorne irreducibilty criterion, right? But the other irreducible components of $V(g)$ might intersect in codim 1 analytic sets. | |
| Nov 22, 2018 at 19:34 | history | answered | Hailong Dao | CC BY-SA 4.0 |