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".
Function field of integral scheme
algebraic-geometryschemes
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First, we note that $\mathcal{O}_X(X)$ is a field since $X$ is an integral projective scheme over a field. Next, for any locally ringed space $X$ and any affine scheme $Y$, there is a bijection between morphisms $X\to Y$ and ring homomorphisms $\mathcal{O}_Y(Y)\to \mathcal{O}_X(X)$ (see for instance Stacks 01I1, or EGA III Err 1 Prop 1.8.1 where it is attributed to Tate, or Hartshorne exercise II.2.4). The interesting portion of this bijection is that given a map on global sections $f$, we send a point $x\in X$ to the point $y\in Y$ corresponding to the preimage of $\mathfrak{m}_x\subset \mathcal{O}_{X,x}$ under the composite map $$\mathcal{O}_Y(Y)\stackrel{f}{\to}\mathcal{O}_X(X)\to\mathcal{O}_{X,x}.$$
So the morphisms $X\to Y$ are in bijection with maps from $\mathcal{O}_Y(Y)$ to a field. But such a map is given by a maximal ideal of $\mathcal{O}_Y(Y)$, and tracing through the bijection above, we see that this means that all points $x$ map to this maximal ideal.
Let me compile the discussion from the comments to mark this as answered.
If $X$ is a topological space, $x\in X$, $U\subset X$ an open subset containing $x$, and $\mathcal{F}$ a sheaf on $X$, then the stalk of $\mathcal{F}$ at $x$ is the same as the stalk of $\mathcal{F}|_U$ at $x$. Using your definition of the stalk as equivalence classes that agree on an open neighborhood of $x$, we can prove this by restricting a representative $(f,V)$ to $(f|_U,V\cap U)$. (With other definitions, there are other proofs - categorically, open neighborhoods of $x$ inside $U$ form a cofinal subset of the open neighborhoods of $x$, which means that calculating the limit along them gives the same value.)
Next, the generic point of $X$ is the generic point of the affine open subscheme $\operatorname{Spec} A\subset X$ as well (exercise: prove this, perhaps by assuming it's not and seeing what contradiction you can get, like disjoint open subsets or a decomposition of $X$ as a union of two proper closed subsets). Since $(0)$ is the generic point of $\operatorname{Spec} A$, we have that $\mathcal{O}_{X,\xi} \cong \mathcal{O}_{\operatorname{Spec} A, (0)} \cong A_{(0)} \cong K(A)$.
Best Answer
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.