Timeline for Being Cohen-Macaulay open in Hilbert scheme?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2016 at 14:18 | vote | accept | Hans | ||
| Aug 4, 2016 at 14:18 | comment | added | Hans | okay, now I am more than happy! Thx! | |
| Aug 4, 2016 at 12:35 | comment | added | Jason Starr | I addressed your question in an edit above. Also, now I see that EGA does include the relative case of EGA $\textrm{IV}_2$ Section 6.11 later in EGA \textrm{IV}_3$ Section 12.1. | |
| Aug 4, 2016 at 12:34 | history | edited | Jason Starr | CC BY-SA 3.0 |
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| Aug 4, 2016 at 9:24 | comment | added | Hans | thanks for your answers Jason! I have a question regarding the second edit. For simplicity, assume that $R$ is a discrete valuation ring with residue field $k$. The following is not apparent to me from the proof: Why can it not be that the maximal open subscheme $U$ of $\mathbb{P}_R^d$, over which $f_*\mathcal{O}_Z$ is locally free, is empty, but $Z\times_R k \to \mathbb{P}_k^d$ is flat? (In that case the fiber over the closed point would be CM, but not the one over the generic point) | |
| Aug 3, 2016 at 20:09 | history | edited | Jason Starr | CC BY-SA 3.0 |
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| Aug 3, 2016 at 17:52 | history | edited | Jason Starr | CC BY-SA 3.0 |
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| Aug 3, 2016 at 17:50 | comment | added | Jason Starr | So, for a Noetherian ring $R$ and a closed subscheme $Z\subset \mathbb{P}^n_R$ such that the projection $p_R:Z \to \text{Spec}(R)$ is flat, the subset $CM(p_R,\mathcal{O}_Z)\subset Z$ where the geometric fibers are Cohen-Macaulay is an open subset. The complementary subset is a closed subset of $Z$, and this is then proper over $\text{Spec}(R)$. The projection of that closed subset to $\text{Spec}(R)$ is a closed subset of $\text{Spec}(R)$. The open complement is the maximal open subset of $\text{Spec}(R)$ over which all geometric fibers are Cohen-Macaulay. | |
| Aug 3, 2016 at 17:42 | comment | added | Jason Starr | You are correct that EGA states the results only in the absolute case, i.e., for a closed subscheme $Z$ of $\mathbb{P}^n$ over a field, they prove that the subset $CM(Z)\subset Z$ where $Z$ is (absolutely) CM is an open subset of $Z$. However, the same argument works verbatim in the relative case. Here is a link to an extract of a preprint of Chenyang Xu and I where we make the details explicit, math.stonybrook.edu/~jstarr/papers/RSCff_02_23_13a.pdf | |
| Aug 3, 2016 at 16:49 | comment | added | Hans | Wait, why is this the same as in my question? So Cor. 6.11.3 sais that the set of points $x$ such that $F_x$ is CM is open. But if the base is e.g. a field $K$ then being CM for a $K$-algebra is not the same as being CM as a $K$-module (which is always the case). | |
| Aug 3, 2016 at 16:40 | vote | accept | Hans | ||
| Aug 3, 2016 at 16:42 | |||||
| Aug 3, 2016 at 16:36 | comment | added | Jason Starr | Today is the 90th anniversary of Auslander's birth. | |
| S Aug 3, 2016 at 10:54 | history | answered | Jason Starr | CC BY-SA 3.0 | |
| S Aug 3, 2016 at 10:54 | history | made wiki | Post Made Community Wiki by Jason Starr |