Timeline for In a noetherian commutative ring with only one associated prime, is the nilradical locally free?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 8, 2015 at 9:40 | vote | accept | Matthieu Romagny | ||
| Oct 8, 2015 at 9:39 | comment | added | Matthieu Romagny | All right, I see I'm a bit messed up. I have to think about it a bit more. | |
| Oct 7, 2015 at 21:21 | comment | added | Eric Wofsey | Well, as you're defining it, $d(x)$ is usually not finite (if $A=B\otimes C$ where $B$ is a domain and $C$ is artinian local, then $N_x$ is finite rank over a localization of $B$, not over $k$). I have a sense of the intuition you're going for with $d(x)$, but I don't know the right way to define it in general. Maybe you want to say $d(x)$ is the length of a maximal chain of ideals in $\mathcal{O}_{X,x}$ which are all annihilators of submodules of the nilradical, or something similar? | |
| Oct 7, 2015 at 21:03 | comment | added | Matthieu Romagny | you are absolutely right that my question is nonsense. I guess that what I have in mind is something along the following lines : if X is a scheme of finite type over a field, and N is the sheaf of nilpotent regular functions, then the dimension d(x)=dim_k(N_x)=_dim_k(Nil(O_{X,x})) is finite, and in case X has only one associated point, the function d should be constant. Does that make sense to you ? | |
| Oct 7, 2015 at 20:46 | answer | added | Eric Wofsey | timeline score: 6 | |
| Oct 7, 2015 at 20:37 | comment | added | Eric Wofsey | Isn't $k[x]/(x^2)$ (or more generally, any artinian local ring) a counterexample? Actually, I'm having more trouble coming up with examples (where the nilradical is nonzero) than counterexamples. | |
| Oct 7, 2015 at 20:15 | history | asked | Matthieu Romagny | CC BY-SA 3.0 |