Timeline for Is the structure sheaf of an integral stack torsion-free?
Current License: CC BY-SA 3.0
15 events
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| Dec 8, 2016 at 15:51 | comment | added | Sam | @Jason Thanks for your reply. I understand that the rings $\mathcal O_U(U)$ are reduced. However, I am trying to define torsion free sheaves in general by setting torsion submodule to zero. If $M$ is an $A$-module, then to define the torsion submodule of $M$, we need the ring $A$ to be an integral domain, not just reduced. It is in this sense that I wanted to ask how we can define torsion for sheaves on an integral stack. | |
| Dec 8, 2016 at 13:30 | comment | added | Jason Starr | @Sam: I do not understand your comment. For a stack that is integral, the sections of the structure sheaf over any 'etale open is a reduced ring in the sense that every nilpotent element equals zero. This can be checked on stalks. | |
| Dec 7, 2016 at 21:01 | comment | added | Sam | @Jason: I tried but couldn't really understand why the sections of the structure sheaf being a reduced ring is enough to define torsion free sheaf on $ X_{et}$ (or stack) ? Since we don't have irreducibility, are you asking me to consider a cover by the irreducible components in some sense?Just to avoid confusion, let me mention the definition I am using for torsion free sheaf for the usual Zariski case. For $X$ integral, a sheaf $F$ on $X$ is torsion free, if $F(U)$ is a torsion free $O_U(U)$ module for all $U$. | |
| Dec 4, 2016 at 15:13 | comment | added | Qfwfq | @JasonStarr: Oh I see, so it's only the $\mathcal{O}_{\mathcal{X}}(U)$'s that may not be integral domains, just because the $U$'s may not be connected. So, maybe, for what the OP is after one can look at the stalks (that don't see the possible disconnectedness of étale "opens")? | |
| Dec 4, 2016 at 11:37 | comment | added | Jason Starr | @Sam. For a stack that is integral, the sections of the structure sheaf over any 'etale open is a reduced ring. | |
| Dec 4, 2016 at 9:10 | comment | added | Sam | @Jason...thanks! I got it. But now how to define a torsion free sheaf on the small etale site of a scheme ? As you have shown $\mathcal O_U(U)$ need not be a domain in case of $X_{etale}$, we can't use the same definition of torsion-free as in case of Zariski site. | |
| Dec 4, 2016 at 1:10 | comment | added | Jason Starr | @Qfwfq. For every scheme $\mathcal{X}$, for the structure sheaf $\mathcal{O}_{\mathcal{X},\text{et}}$ on the small 'etale site, the global sections of $\mathcal{O}_{\mathcal{X},\text{et}}$ is canonically isomorphic to the global sections of the usual Zariski sheaf $\mathcal{O}_{\mathcal{X}}$. | |
| Dec 3, 2016 at 22:38 | comment | added | Qfwfq | @Jasonstarr: is there a way to explicitely describe the ring $\Gamma(\mathcal{X}_{\mathrm{ét}},\mathcal{O}_{\mathcal{X}})$? (maybe, for $\mathcal{X}:=\mathrm{Spec}(\mathbb{Z})$) | |
| Dec 3, 2016 at 18:41 | comment | added | Jason Starr | Okay: Let $V$ be $\text{Spec}\ \mathbb{Z}[e,f]/\langle e^2-e,f^2-f,e+f-1\rangle$, i.e., a disjoint union of two copies of $\text{Spec}\ \mathbb{Z}$. The unique morphism $V\to \text{Spec}\ \mathbb{Z}$ is 'etale. Let $U$ be the fiber product $\mathcal{X}\times_{\text{Spec}\ \mathbb{Z}} V$ with its projection $U\to \mathcal{X}$. This morphism is 'etale. Yet $U$ is not connected. Therefore $\mathcal{O}_U(U)$ is not an integral domain. This has nothing to do with stacks -- the same issue arises for schemes. | |
| Dec 3, 2016 at 14:54 | comment | added | Sam | @Jason...Sorry but I didn't understand what you meant by that case of disjoint union of copies of $X$. Could you kindly explain a little ? | |
| Dec 3, 2016 at 14:49 | comment | added | Sam | Now if $\mathcal {O_X}(U \xrightarrow{etale} \mathcal X)$ is not an integral domain in general, then by the usual definition of a torsion-free sheaf, the structure sheaf wouldn't be torsion-free. But it seems fair enough to ask if we have the structure sheaf torsion-free while considering an integral stack...because this happens in case of an integral scheme. | |
| Dec 3, 2016 at 14:45 | comment | added | Jason Starr | Consider the case that $U\xrightarrow{\text{etale}} X$ is a disjoint union of two copies of $X$. | |
| Dec 3, 2016 at 14:43 | comment | added | Sam | I am asking about those 1-morphisms that are represented by etale maps which are not open immersions. The ring of global sections $\Gamma(\mathcal X_{etale}, \mathcal {O_X})$ of the small etale site of $\mathcal X$ is shown to be an integral domain in the Stack Project link I have shared. But I don't know if $\mathcal O_X(U \xrightarrow{etale} \mathcal X):= \mathcal O_U(U)$ would also be an integral domain since being integral is not an etale local property of schemes. | |
| Dec 3, 2016 at 11:00 | comment | added | Jason Starr | Sections over what? For every $1$-morphism that is represented by open immersions, the domain of that stack is again integral, so the sections are again an integral domain. Even for schemes, for 'etale morphisms, the domain can be disconnected, and thus the global sections would not be a domain. Are you asking about nilpotent elements in the ring of global sections? | |
| Dec 3, 2016 at 9:46 | history | asked | Sam | CC BY-SA 3.0 |