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It is obvious that there is a parallel between the definition of structure sheaf of $\operatorname{Spec}(A)$ versus the sheafification of a pre-sheaf.

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

For any open set $U$, let $\mathscr F^+ (U)$ be the set of functions $s$ from $U$ to the union of stalks $\mathscr F_P$ of $\mathscr F$ over points $P$ of $U$ such that:

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

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

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

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

For each $\mathscr{p}$ in $U$, there is a neighborhood $V$ of $\mathscr{p}$, contained in $U$, and elements $a,f$ of $A$, such that for each $\mathscr{q}$ 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) that 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 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? $\endgroup$
    – S. Carnahan ♦
    Commented Nov 10, 2011 at 3:14
  • $\begingroup$ 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). $\endgroup$
    – urelement
    Commented Nov 10, 2011 at 3:48
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    $\begingroup$ 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... $\endgroup$
    – darij grinberg
    Commented Nov 10, 2011 at 5:06
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    $\begingroup$ 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. $\endgroup$
    – Qfwfq
    Commented Nov 10, 2011 at 19:55
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    $\begingroup$ 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 $f=0$). $\endgroup$
    – user2035
    Commented Nov 11, 2011 at 9:10

1 Answer 1

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For any open 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.

Claim: For open subsets of the form $U=\mathrm{Spec}(A_f)$ with $f\in A$ we have $\mathscr O'(U)=A_f$. (This shows that the associated sheaf of $\mathscr O'$ is indeed $\mathscr O_{\mathrm{Spec}(A)}$.)

Proof: Assume there is an $s\in S_U$ which does not 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 in an ideal $\mathfrak q$ which is maximal with respect to this property, but it is well-known that such an ideal $\mathfrak q$ is prime. By construction, $s\in\mathfrak q\in U$, contradicting $s\in S_U$.

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

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    $\begingroup$ Probably this is the best definition of the structure sheaf on the specrum ... and I wonder why I haven't seen it yet :). $\endgroup$
    – Martin Brandenburg
    Commented 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 constant sheaf $\underline{A}$. It is also called the "universal filter" or "generic filter" of $A$. The structure sheaf can then simply be constructed as the localization $\underline{A}[S^{-1}]$ (performed in the internal language of the topos of sheaves over $\operatorname{Spec}(A)$). $\endgroup$
    – Ingo Blechschmidt
    Commented May 13, 2017 at 21:45
  • $\begingroup$ $U=\text{Spec }(A_f)$ means the set of all prime ideals in $A$ which are correspondent with the prime ideals in $A_f$. That is, the set of all prime ideals in $A$ which are disjoint $\{1, f, f^2, ...\}$. $\endgroup$
    – bfhaha
    Commented Apr 13, 2019 at 6:07
  • $\begingroup$ Interesting! Hartshorne verified (or expected readers to verify) $\mathscr{O}_{\text{Spec }A}$ is a sheaf directly. Then proved that $\mathscr{O}(D(f))\cong A_f$. You did that in a reverse way. $\endgroup$
    – bfhaha
    Commented Apr 13, 2019 at 6:13
  • $\begingroup$ Why "This shows that 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 me? Thx! $\endgroup$
    – bfhaha
    Commented Apr 13, 2019 at 19:25

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