Let $X$ be a separated integral scheme with function field $K\left(X\right)$. If $R$ is a valuation ring of $K\left(X\right)$, we say that $x\in X$ is a center of $R$ if $\mathcal{O}_{X,x}$ is dominated by $R$.
Exercise 2.4.5 of Hartshorne says that if the center exists, then it is unique.
To get started on this exercise, we should use the valuative criterion of separatedness. We see that if $\operatorname{Spec} K \left(X\right)\to X$ maps to $x\in X$, then $x$ being a center is equivalent to the commutativity of the following diagram given by inclusions $\mathcal{O}_{X,x} \subseteq R \subseteq K\left(X\right)$:
Now since $X$ is separated, the composition $\operatorname{Spec}R \to \operatorname{Spec} \mathcal{O}_{X,x}\to X$ is unique. From this, how can we conclude that $x\in X$ is unique? If there exists another center $y\in X$ for $R$ then we must consider a different morphism $\operatorname{Spec} K \left(X\right)\to X$, and by the same reasoning we get another unique morphism $\operatorname{Spec}R\to X$. I don't see how we can conclude that $x=y$ since we have two different morphisms $\operatorname{Spec} K\left(X\right)\to X$.
I looked at the answer given here but it does not seem to answer my concern.
Best Answer
First of all, it's important to specify that $X$ is separated over $\operatorname{Spec} k$, and that $R$ is a valuation ring for the extension $K/k$, where $K = K(X)$. That way we have $k \subset R \subset K$.
In this setup we have the following diagram. $\require{AMScd}$ \begin{CD} \operatorname{Spec} K @>{}>> X\\ @VVV @VVV\\ \operatorname{Spec} R @>{}>> \operatorname{Spec} k \end{CD}
where the top map is the inclusion of the generic point, the bottom map is induced by $k \subset R$, the left map is localization, and the right map is the separated structure map of $X$.
Now let $x$ be a center of this valuation. This means that we have an inclusion $\mathcal{O}_{X, x} \subset R$ as subrings of $K(X)$, and that $\mathfrak{m}_R \cap \mathcal{O}_{X, x} = \mathfrak{m}_x$.
Now, let's try and understand the induced map $\operatorname{Spec} R \to \operatorname{Spec} \mathcal{O}_{X, x}$. By definition, this sends $\mathfrak{p} \mapsto \mathfrak{p} \cap \mathcal{O}_{X, x}.$ Hence, the special point of $\operatorname{Spec} R$ goes to the special point of $\mathcal{O}_{X, x}$ (which then gets sent to $x$ under $\operatorname{Spec} \mathcal{O}_{X, x} \to X$) and the generic point of $\operatorname{Spec} R$ goes to the generic point of $\operatorname{Spec} \mathcal{O}_{X, x}$ (which then goes to the generic point of $X$). Hence a choice of center gives a map fitting diagonally into the diagram above, so the uniqueness of a center follows from the valuative criterion of separatedness.
Edit: To elaborate on my comment, your diagram is not commutative, even as a map of sets, unless the top map sends $\operatorname{Spec} K(X)$ to the generic point of $X$. That's because the bottom path is a composition of three dominant morphisms.