Timeline for Can $\mathcal O_X$ be recognized abstract-nonsensically?
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Dec 13, 2016 at 1:53 | comment | added | nfdc23 | @მამუკაჯიბლაძე: I was just referring to Ben Webster's assertion about a criterion for invertibility of a coherent sheaf, nothing to do with your additional conditions (i.e., I was pointing out that Ben Webster's argument had some issues to be fixed up). | |
| Dec 12, 2016 at 13:21 | comment | added | Will Sawin | In particular, the definition of a spectral object in that paper shows how to deal with non-closed points. | |
| Dec 12, 2016 at 13:20 | comment | added | Will Sawin | This is essentially done for quasicoherent sheaves on quasiseparated schemes by Martin Brandenburg in the link HeinrichD posted in the comments of my answer arxiv.org/pdf/1310.5978v3.pdf - one can use Martin's construction to define the internal Hom as a bifunctor from the abstract category to a category of sheaves, and a sheaf is invertible if and only if internal homs from it are an equivalence. | |
| Dec 12, 2016 at 7:58 | comment | added | მამუკა ჯიბლაძე | No this does not work even over $\mathbf P^1$ since $\operatorname{Hom}({\mathcal O}(n),{\mathcal O}(m))$ is zero for $n>m$ :( | |
| Dec 12, 2016 at 7:39 | comment | added | მამუკა ჯიბლაძე | @nfdc23 What about the stronger requirement I mentioned in a comment above? Or even so: that $\operatorname{Hom}(L,X)$ is an invertible $\operatorname{End}(X)$-module for any $X$ with $\operatorname{End}(X)$ commutative? Does not it rule out your counterexample? | |
| Dec 12, 2016 at 1:23 | history | edited | Ben Webster♦ | CC BY-SA 3.0 |
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| Dec 11, 2016 at 22:25 | comment | added | nfdc23 | @BenWebster It is not true that a coherent sheaf is invertible if it has residual rank 1 at all points: consider non-reduced $X$ and $O_{X_{\rm{red}}}$ as an $O_X$-module. Also, what does it mean that a scheme is "of finite type"? Over what? It seems you assume $X$ is a Jacobson scheme (the class for which "closed point" is well-behaved such as being of local nature; schemes of finite type over a dvr are generally not Jacobson). For Jacobson affine schemes one has a "Nullstellensatz" with maximal ideals, so for reduced Jacobson schemes the residual criterion with closed points holds. | |
| Dec 11, 2016 at 20:40 | vote | accept | მამუკა ჯიბლაძე | ||
| Dec 11, 2016 at 20:40 | comment | added | მამუკა ჯიბლაძე | Anyway, although I still don't understand some essential moments I understood enough to be sure it is the answer to accept :D | |
| Dec 11, 2016 at 20:07 | history | edited | Ben Webster♦ | CC BY-SA 3.0 |
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| Dec 11, 2016 at 20:00 | comment | added | მამუკა ჯიბლაძე | Hmm I guess it should be that simple objects with commutative endomorphism fields cogenerate but I am not sure. In any case what I want is a condition on objects which is formulable for arbitrary abelian categories and gives exactly line bundles wherever they make sense. I realize this sounds ambitious, but after your answer I have growing confidence that it is doable... | |
| Dec 11, 2016 at 19:40 | comment | added | Ben Webster♦ | @მამუკაჯიბლაძე You should know in exactly what sense you need enough, since it's covered in the answer: the closed points need to be dense. I'm sure one can modify things to deal with non-closed points as well, but at the cost of slightly more song and dance (since those give quasi-coherent, not coherent sheaves). | |
| Dec 11, 2016 at 18:36 | comment | added | მამუკა ჯიბლაძე | (I deleted one comment here, it was rubbish; here is another attempt, sorry for too many questions.) Would it make sense in the general case (not necessarily finite type, and not for only coherent sheaves) to require something like $\operatorname{Hom}(L,X)$ being an indecomposable projective generator in $\operatorname{End}(X)$-modules for every (indecomposable? injective?) object $X$? | |
| Dec 11, 2016 at 18:21 | comment | added | მამუკა ჯიბლაძე | One more question - have you seen this concept somewhere or it is new?? | |
| Dec 11, 2016 at 18:19 | comment | added | მამუკა ჯიბლაძე | Wait you also need to have enough simples in I am not sure what sense... | |
| Dec 11, 2016 at 18:14 | comment | added | მამუკა ჯიბლაძე | Oh I see now, over the residue field - so let me make sure. You say $L$ is a "line object" if for every simple $S$ the vector space $\operatorname{Hom}(L,S)$ over $\operatorname{End}(S)$ is one-dimensional, right? This includes the requirement that all $\operatorname{End}(S)$ are commutative, yes? | |
| Dec 11, 2016 at 18:14 | comment | added | Ben Webster♦ | @მამუკაჯიბლაძე You're thinking about dimension wrong; I don't mean over a fixed base field (that makes no sense for a scheme like $\mathrm{Spec}(\mathbb{Z})$ anyways). Each closed point has a residue field, and you have to look at the dimension at that point over that field. That will be 1 for the structure sheaf by definition (and line bundles are indistinguishable from the structure sheaf for local purposes). | |
| Dec 11, 2016 at 18:11 | history | edited | Ben Webster♦ | CC BY-SA 3.0 |
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| Dec 11, 2016 at 18:08 | comment | added | მამუკა ჯიბლაძე | Thanks for the edit, it is clearer now. But I am still in doubt about the necessity part - just in the affine case, algebras (of finite type) over a non-closed field may possess simple modules of dimension greater than one, no? Say, just the affine line - its simple modules are in 1-1 with irreducible polynomials, their dimensions being degrees of these polynomials. | |
| Dec 11, 2016 at 18:03 | history | edited | Ben Webster♦ | CC BY-SA 3.0 |
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| Dec 11, 2016 at 17:56 | comment | added | მამუკა ჯიბლაძე | And what exactly do you require on the base? I believe you need to be over an algebraically closed field for simples to automatically become one-dimensional? | |
| Dec 11, 2016 at 16:08 | comment | added | Ben Webster♦ | @YosemiteSam Smooth and proper sounds way too restrictive. I might need something like finite type (since I need the closed points to be Zariski dense). | |
| Dec 11, 2016 at 3:42 | comment | added | Yosemite Sam | for the criterion for invertible objects in Coh(X) you give, you need some assumptions on X, like smooth and proper. right? | |
| Dec 11, 2016 at 2:53 | history | answered | Ben Webster♦ | CC BY-SA 3.0 |