Consider a scheme $X$; for every $x\in X$ with residue field $k(x)$, we have the canonical surjection $\mathcal O_{X,x}\longrightarrow k(x)$ that induces the morphism of affine schemes $\operatorname{Spec}(k(x))\longrightarrow\operatorname{Spec}(\mathcal O_{X,x})$. Now if $\operatorname{Spec} A=U\subseteq X$ is an affine open containing $x$, then we have a morphism
$$A=\mathcal O_X(U)\longrightarrow \mathcal O_{U,x}=\mathcal O_{X,x} $$
that induces a morphism of schemes $\operatorname{Spec}(\mathcal O_{X,x})\longrightarrow U$.
Reassuming, by composing with the immersion of $U$ in $X$ we obtain a morphism of schemes $\operatorname{Spec}(k(x))\longrightarrow X$. Many texts say that this morphism is independent from the choice of the open affine $U$, but I don't understand why.
Best Answer
There are multiple ways of attacking this:
Method 1:
Show that if $V\subseteq U$ is another affine containing $x$, then the maps $\text{Spec}(\mathcal{O}_{X,x})\to V$ and $\text{Spec}(\mathcal{O}_{X,x})\to U$ are the same (this isn't that hard) (EDIT: As Asal Beag Dubh points out below, this means that $\text{Spec}(\mathcal{O}_{X,x})\to U$ factors as $\text{Spec}(\mathcal{O}_{X,x})\to V\to U$). Then, for any two affines $W,U$ pass to some affine open $V\subseteq W\cap U$.
Method 2:
Let $Z\subseteq X$ be the subset of $X$ consisting of points which generalize $x$. Consider the topological map $i:Z\hookrightarrow X$. Define $\mathcal{O}_Z:=i^{-1}\mathcal{O}_X$. Shown then that $(Z,\mathcal{O}_Z)$ is a scheme, and that for any choice of affine $x\in U$ one has that $Z\to X$ is isomorphic to $\mathcal{O}_{X,x}$ (i.e. that $Z\cong \mathcal{O}_{X,x}$ in a way compatible with these mappings).