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Jun 15, 2020 at 7:27 history edited CommunityBot
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Apr 14, 2019 at 12:01 comment added user137767 @darijgrinberg I am not accusing anyone of anything but the phrase "a scheme is a sheaf on the category of rings" was clear to me (and when I first read your comment, "both sheaves are on a topological space...schemes are sheaves of rings", I at first thought you were the one making misleading claims here). I guess the way you think of schemes depends on the type of work you do. However, David Roberts' comment is definitely not misleadging for everyone.
Nov 13, 2011 at 0:05 history edited urelement CC BY-SA 3.0
improved formatting: added a line break
Nov 12, 2011 at 14:34 comment added urelement @a-fortiori I have fixed the formula as per your suggestion. Thanks for the answer!
Nov 12, 2011 at 14:30 history edited urelement CC BY-SA 3.0
Fixed formula as per comment from a-fortiori
Nov 11, 2011 at 12:20 vote accept urelement
Nov 11, 2011 at 9:10 comment added user2035 In the definition of $\mathscr O_{\mathrm{Spec}(A)}$, it should be $(\exists a,f\in A)(\forall\mathfrak q\in V)\ f\notin\mathfrak q\land s(\mathfrak q)=a/f$, not $(\exists a,f\in A)(\forall\mathfrak q\in V)\ f\notin\mathfrak q\to s(\mathfrak q)=a/f$ (you could always take $f=0$).
Nov 11, 2011 at 5:51 history edited S. Carnahan CC BY-SA 3.0
Spelling, formatting
Nov 11, 2011 at 5:04 comment added David Roberts @darij #1 - hmm, possibly. I'm not an algebraic geometer, and I haven't thought about how the stalks of a sheaf of rings end up as local rings #2 - I'm thinking of a scheme as its functor of points. This is a sheaf on $Ring$, if I'm not mistaken.
Nov 11, 2011 at 4:27 history edited urelement CC BY-SA 3.0
Spelling and improved formatting
Nov 10, 2011 at 19:55 comment added Qfwfq Even more degenerately: well, $\mathcal{O}_{Spec(A)}$ is a sheaf hence... well, a presheaf.. whose sheafification remains $\mathcal{O}_{Spec(A)}$. This shows that I have probably not understood the question.
Nov 10, 2011 at 6:17 answer added user2035 timeline score: 45
Nov 10, 2011 at 6:00 comment added S. Carnahan The answer to your question is "yes" for degenerate reasons. For example, if Spec(A) is covered by two affine opens not equal to Spec(A), then it is easy to change the presheaf so that the ring of global sections is altered but nothing else, and the associated sheaf remains unchanged. Your question would be more clear if you spelled out the "obvious" parallel in more detail.
Nov 10, 2011 at 5:06 comment added darij grinberg Also, the sentence "a scheme is a sheaf on the category of rings, whereas a structure sheaf is a sheaf on a topological space" looks seriously misleading to me. Both sheaves are on a topological space. As for the categories, schemes are sheaves of rings and structure sheaves are sheaves of abelian groups. Or do you mean schemes as functors, and "sheaf" in the functorial sense (Zariski sheaf)? In this case this may be correct but is still misleading, sorry...
Nov 10, 2011 at 5:04 comment added darij grinberg Why should it be a presheaf of local ring? You probably want the stalks to be local...
Nov 10, 2011 at 4:35 comment added David Roberts Well it will have to be a presheaf of local rings, otherwise it won't sheafify to the structure sheaf of anything. And please make a distinction between the scheme and the structure sheaf; a scheme is a sheaf on the category of rings, whereas a structure sheaf is a sheaf on a topological space. There seems to be a misunderstanding here...
Nov 10, 2011 at 3:48 comment added urelement I'm asking for any type of presheaf on the topological space that underlies the spectrum of A: a presheaf of rings or of sets or what ever structure on it s.t the sheafification will give back the usual affine scheme we have on spec(A).
Nov 10, 2011 at 3:14 comment added S. Carnahan Are you asking for a presheaf of rings on the topological space that underlies the spectrum of A, or a presheaf of sets on e.g., the category of affine schemes?
Nov 10, 2011 at 2:58 history asked urelement CC BY-SA 3.0