Skip to main content
12 events
when toggle format what by license comment
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