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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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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? –  S. Carnahan Nov 10 '11 at 3:14
    
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). –  urelement Nov 10 '11 at 3:48
    
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... –  David Roberts Nov 10 '11 at 4:35
    
Why should it be a presheaf of local ring? You probably want the stalks to be local... –  darij grinberg Nov 10 '11 at 5:04
    
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... –  darij grinberg Nov 10 '11 at 5:06

1 Answer 1

up vote 24 down vote accepted

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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Probably this is the best definition of the structure sheaf on the specrum ... and I wonder why I haven't seen it yet :). –  Martin Brandenburg Nov 10 '11 at 17:58

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