Sheaf of meromorphic functions on an integral scheme

Question

It is a theorem that the sheaf of meromorphic functions on an integral scheme is equal to the constant sheaf where each open set is assigned the function field of the scheme. See, for example, the Stacks Project. I would like to know why this is true.

Recall that the sheaf of meromorphic functions is the sheafification of the presheaf that assigns to each open set $U$ the set $\mathcal R_X(U)^{-1}\mathcal O_X(U)$, where $\mathcal R(U)$ is the set of regular sections. In the case of an integral scheme, $\mathcal R(U)$ is simply all nonzero sections.

Answer 1

Let $\mathcal{F}$ be the presheaf $U\mapsto K(\mathcal{O}_X(U))$, where $K(A)$ is the quotient field of an integral domain $A$. Then $\mathcal{K}$, the sheaf of total quotient rings of $\mathcal{O}$ (i.e. sheaf of meromorphic functions on $X$), is by definition the sheafification of $\mathcal{F}$. Let $K$ be the constant sheaf $\mathcal{O}_\xi$ on $X$, $\xi$ is the generic point of $X$.

Actually, there is a natural morphism of presheaves $\rho:\mathcal{F}\rightarrow K$ defined below:

For any $f=g/h\in K(\mathcal{O}_X(U))$, $f_\xi=g_\xi/h_\xi\in K(\mathcal{O}_\xi)=\mathcal{O}_\xi$. Define $\rho_U(f)=f_\xi$, then $\rho$ is the restriction of rational functions to their germs at $\xi$, so $\rho$ is automatically a morphism of presheaves. By the universal property of sheafification, there is a morphism of sheaves $\bar\rho: \mathcal{K}\rightarrow K$ induced by $\rho$.

Claim: $\bar\rho$ is an isomorphism of sheaves.

Proof: For any affine open subset $U=\operatorname{Spec}A\subset X$, $\mathcal{F}(U)=K(\mathcal{ O}_X(U))=K(A)\cong \mathcal{O}_\xi$, so $\rho_U:\mathcal{F}(U)\rightarrow K(U)=\mathcal{O}_\xi$ is an isomorphism. Since affine open subsets form a topological basis of the Zariski topology of $X$, $\rho_x: \mathcal{F}_x\rightarrow K_x$ is an isomorphism for any $x\in X$.

Since $\mathcal{K}_x$ is isomorphic to $\mathcal{F}_x$ via the natural morphism of presheaves $\mathcal{F}\rightarrow \mathcal{K}$, $\bar\rho_x:\mathcal{K}_x\rightarrow K_x$ is an isomorphism for any $x\in X$, hence $\bar\rho$ is an isomorphism of sheaves. QED