Let $X$ be an integral scheme. For any affine open $U = \operatorname{Spec}(A)$ the field of fractions $K(A)$ of $A$ is the stalk of $X$ at its generic point. This is called the function field of $X$ and denoted by $K(X)$. Now let $U$ be some open set of $X$, not necessarily affine. Then the field of fractions $K(\mathcal{O}_X(U))$ of the ring of sections on $U$ is a subfield of $K(X)$ because the restriction maps of the structure sheaf of $X$ are injective. But is it true that $K(\mathcal{O}_X(U)) = K(A)$? I have the feeling that this is not true, while if we take the sheaf associated to the presheaf $U \mapsto K(\mathcal{O}_X(U))$, then we do get the constant sheaf associated to $K(X)$. On the other hand, the wikipedia article https://en.wikipedia.org/wiki/Function_field_(scheme_theory) states ''the fraction fields of the rings of regular functions on any open set will be the same".
2026-04-02 17:36:42.1775151402
Function field of integral scheme
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You're correct - $\Bbb P^1_k$ is a counterexample. The open subset $\Bbb P^1_k$ has functions (and thus fraction field of those functions) $k$, while any other nonempty open subset has fraction field of it's regular functions $k(x)$.
As for the wikipedia page, one potential way to fix the inaccuracy of the claim is to include an "affine" in the sentence you quote:
(quoted directly from Wikipedia at the time of posting, bold denotes my addition.) The thing that's going on here is that the wikipedia page is secretly sheafifying in this sentence and not explaining it - starting from the presheaf $U\mapsto \operatorname{Frac}(\mathcal{O}_X(U))$ on an integral scheme $X$, when we sheafify, we get the constant sheaf $K_X$: every open can be covered by affine opens, and every affine open has the same sections, so via the sheaf property we can glue those sections together to get the right sections on the open set we started with.