Timeline for Depth under localization over a Cohen-Macaulay ring
Current License: CC BY-SA 3.0
12 events
when toggle format | what | by | license | comment | |
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Dec 27, 2017 at 18:02 | vote | accept | Lisa S. | ||
Dec 27, 2017 at 18:00 | comment | added | Lisa S. | Thank you both for the counterexamples! For me, the default definition of (S$_2$) is the one from EGA. However, I am mostly interested in $M$ that are supported on the entire spectrum, so Hailong Dao's counterexample is more in the spirit of what I was looking for. | |
Dec 27, 2017 at 17:10 | comment | added | Hailong Dao | @JasonStarr: I have seen mistakes made in papers due to this confusion, which was my motivation for the question above. | |
Dec 27, 2017 at 17:08 | comment | added | Jason Starr | In the words of Shawn Spencer, "I've heard it both ways." :) I agree that Lisa S. should clarify the intended definition. | |
Dec 27, 2017 at 17:07 | comment | added | Hailong Dao | Perhaps the OP can clarify what s/he meant by $(S_2)$. | |
Dec 27, 2017 at 17:06 | comment | added | Hailong Dao | @JasonStarr: yes, indeed there are different notions of $(S_n)$ in the literature, which are radically different. Basically, whether a module should be $(S_n)$ on the whole spectrum of $R$ or just on it's support. I tried to alert people's attention to that issue here: mathoverflow.net/questions/22228/… | |
Dec 27, 2017 at 17:00 | comment | added | Jason Starr | @HailongDao. What definition of $S2$ are you using? I am using the definition from EGA IV_2, Definition 5.7.2, p. 103. According to the definition in EGA, the module in my comment is $S2$. | |
Dec 27, 2017 at 16:53 | comment | added | Hailong Dao | @JasonStarr: M in that case is not $(S_2)$. | |
Dec 27, 2017 at 16:52 | answer | added | Hailong Dao | timeline score: 9 | |
Dec 27, 2017 at 16:49 | comment | added | Jason Starr | The result is not true without further hypotheses. Here is a counterexample. Let $A$ be $k[x_1,\dots,x_n,y_1,\dots,y_n]_{\mathfrak{m}}$, for the maximal ideal $\mathfrak{m}=\langle x_1,\dots,x_n,y_1,\dots,y_n\rangle$. Let $\mathfrak{p}$ be $\langle x_1,\dots,x_n\rangle$. Let $M$ be $A/\mathfrak{p}.$ Then $M_{\mathfrak{p}}$ is the residue field of $A_\mathfrak{p}$. This has depth $0$. | |
Dec 27, 2017 at 16:42 | comment | added | Jason Starr | I do not immediately know the answer, but the numerical invariant that definitely behaves well with respect to localization is the "codepth" rather than the depth, cf. EGA IV_2, Proposition 6.11.2 of Auslander. | |
Dec 27, 2017 at 16:24 | history | asked | Lisa S. | CC BY-SA 3.0 |