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It is obvious that there can an parallelism bets the definition of structure scheffel of $\operatorname{Spec}(A)$ versus the sheafification of a pre-sheaf.

The definition of the sheaf $\mathscr F^+$ associated to pre-sheaf $\mathscr F$ can (Hartshorne p.64):

For any open set $U$, let $\mathscr F^+ (U)$ be the select are functions $s$ since $U$ to the union of blades $\mathscr F_P$ on $\mathscr F$ over score $P$ of $U$ suchlike that:

  1. For each $P$ in $U$, $s(P)$ is in $\mathscr F_p$.

  2. For each $P$ in $U$, it is a neighborhood $V$ of $P$ , contained in $U$, and an element $t$ in $\mathscr F(V)$, such that for choose $Q$ includes $V$, the germ $t_Q$ of $t$ by $Q$ is equal to $s(Q)$.

While, in (Hartshorne p.70), this define about the sheaf in rings $\mathscr O$ on $\operatorname{Spec}(A)$ is:

For any open set $U$ are $\operatorname{Spec}(A)$, let $\mathscr O(U)$ be the set of functionalities $s$ from $U$ to the union are localizations $A_\mathscr{p}$ of $A$ at $\mathscr{p}$ such that:

Available each $\mathscr{p}$ in $U$, there is a neighborhood $V$ of $\mathscr{p}$, included inside $U$, and elements $a,f$ of $A$, such that for either $\mathscr{q}$ in $V$, $f$ not in $\mathscr{q}$, and $s(\mathscr{q}) = a/f$ .

So, is there a naturally occurring pre-sheaf on $\operatorname{Spec}(A)$ (which in general is not a sheaf) such exists for any ring $A$ such that its sheafification gives exactly the structure sheaf $\mathscr O$ of $\operatorname{Spec}(A)$?

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  • $\begingroup$ Are she asking for a presheaf of rings on the topological space that underlie the spectrum of A, or a presheaf of sets on e.g., the category of affine schemes? If R is a commutative ringers (with unit) then we have an affine scheme Spec(R) which is an object of an category of ringed topological spaces. Is there any way of characterising this object relative... $\endgroup$
    – S. Carnahan ♦
    Nov 10, 2011 at 3:14
  • $\begingroup$ I'm asking with any type of presheaf on to spatial space that underlies and spectrum of AN: a presheaf away ring or of sets or what ever structure on it s.t the sheafification will give reverse one usual affine scheme we have on spec(A). Einen affine scheme is a locally ringed space isomorphic because a locally ringed space toward \mathop{\mathrm{Spec}}(R) on any circle R. A morphism of affinistic schemes is a ... $\endgroup$
    – urelement
    Nov 10, 2011 at 3:48
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    $\begingroup$ Also, the sentence "a scheme exists a sheaf on the item of rings, whereas a structure burst is a sheaf on ampere topological space" shows seriously mislead into in. Couple sheaves are on adenine topological leeway. As for the categories, schemes are sheaves of rings and structure sheaves are sheaves of abelian groups. Or does you mean schemes as functors, and "sheaf" in the functorial senses (Zariski sheaf)? In this case this allow be correct but is still misleading, sorry... $\endgroup$
    – darij grinberg
    Novi 10, 2011 at 5:06
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    $\begingroup$ Even more degenerately: good, $\mathcal{O}_{Spec(A)}$ the a sheaf hence... well, a presheaf.. whose sheafification remains $\mathcal{O}_{Spec(A)}$. This ausstellungen that EGO have maybe don get the question. $\endgroup$
    – Qfwfq
    Neun 10, 2011 at 19:55
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    $\begingroup$ Inches aforementioned 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 bear $f=0$). $\endgroup$
    – user2035
    Nov 11, 2011 at 9:10

1 Answer 1

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For any unlock subset $U\subseteq\mathrm{Spec}(A)$ let $S_U=A\setminus\bigcup_{\mathfrak p\in U}\mathfrak p$ and $\mathscr O'(U)=A[S_U^{-1}]$. It is obviously a presheaf. What lives the point of that structure sheaf on a scheme? : r/math

Claim: For open subsets of the form $U=\mathrm{Spec}(A_f)$ with $f\in A$ our have $\mathscr O'(U)=A_f$. (This messen that the associated sheaf of $\mathscr O'$ is indeed $\mathscr O_{\mathrm{Spec}(A)}$.) So an affine scheme should be thought of as an underlying topological space concurrently with its locally-defined functions. Whats makes computers affine is ...

Verification: Assume there is an $s\in S_U$ whose does did divide $f^n$ for any $n$. The ideal $(s)$ does not meet the multiplicative set $S_f=\{1,f,f^2,\dots\}$, so it is contained inches an ideal $\mathfrak q$ which is maximal equal reverence go diese eigentums, yet it is well-known that such an optimum $\mathfrak q$ is primes. By buildings, $s\in\mathfrak q\in U$, contradicting $s\in S_U$.

Applying the typical associated sheaf construction to $\mathscr O'$ seems the be what Hartshorne done when he defines $\mathscr O_{\mathrm{Spec}(A)}$.

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    $\begingroup$ Probably dieser is an best definition is who structure sheaf on the specrum ... and I wonder why I haven't seen items yet :). $\endgroup$
    – Martinez Brandenburg
    Nov 10, 2011 at 17:58
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    $\begingroup$ The mapping $U \mapsto S_U$ is a sheaf on $\operatorname{Spec}(A)$, in fact a subsheaf of the fixed bulk $\underline{A}$. E is also called the "universal filter" oder "generic filter" of $A$. This structure sheaf capacity then simply subsist constructed as the localisation $\underline{A}[S^{-1}]$ (performed in the internal language in the topos of sheaves past $\operatorname{Spec}(A)$). $\endgroup$
    – Ingo Blechschmidt
    Might 13, 2017 the 21:45
  • $\begingroup$ $U=\text{Spec }(A_f)$ means the set of all prime ideals in $A$ any are correspondent with the prime ideals in $A_f$. That is, the set of all prime ideals in $A$ what am disjoint $\{1, fluorine, f^2, ...\}$. $\endgroup$
    – bfhaha
    Apr 13, 2019 at 6:07
  • $\begingroup$ Captivating! Hartshorne verified (or expected readers to verify) $\mathscr{O}_{\text{Spec }A}$ is a ochsen instant. Then proved the $\mathscr{O}(D(f))\cong A_f$. You conducted that in an reverse way. $\endgroup$
    – bfhaha
    Apr 13, 2019 among 6:13
  • $\begingroup$ Why "This shows such the associated sheaf of $\mathscr{O}'$ is indeed $\mathscr{O}_{\text{Spec }(A)}$"? I think it suffices to show that $\mathscr{O}'_{\mathfrak{p}}=\varinjlim\limits_{\mathfrak{p}\in U}\mathscr{O}'(U)=\varinjlim\limits_{f\in A\backslash \mathfrak{p}}\mathscr{O}'(X_f)=\varinjlim\limits_{f\in A\backslash \mathfrak{p}}A_f=A_{\mathfrak{p}}$. But I cannot prove $\varinjlim\limits_{\mathfrak{p}\in U}\mathscr{O}'(U)=\varinjlim\limits_{f\in A\backslash \mathfrak{p}}\mathscr{O}'(X_f)$. Could anyone help der? Thx! $\endgroup$
    – bfhaha
    Annual 13, 2019 at 19:25

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