Given a ring $A$ and his Spectrum $X=Spec(A)$ seen as a topological space with the Zariski topology, it's possible to build a sheaf on $X$ satisfying the conditions
- $O_X(X)=A$
- $O_X(X_f)=A_f$ where $X_f=X-V(f)$
- $O_{X,p}=A_p$
My question is: do these conditions determine the sheaf uniquely up to isomorphisms of sheaves, or there exists a ring with two non isomorphic sheafs satisfying these properties?
For a counterexample consider a field $k$ and define $A= k[x]_{(x)}$ which is a discrete valuation ring, so $X=\mathrm{Spec}(A) = \{0, (x)\}$ and $\{0\}$ is the only nontrivial open set. Now consider a sheaf $\cal{F}$ on $X$ with $\mathcal{F}(X) = A, \mathcal{ F}(\{0\})= k(x),$ and the restriction map $A=\mathcal{F}(X)\to k(x)=\mathcal{F}(\{0\})$ is given by sending $x$ to $x^2$ i.e. $k[x]_{(x)}\stackrel{\simeq}{\to} k[x^2]_{(x^2)}\to k(x).$ The quotient field of the image of this map is not equal to $\mathcal{F}(\{0\})$ and so this sheaf is not isomorphic to the structure sheaf of $\mathrm{Spec}(A)$.