Does the valuative criterion of separatedness guarantee uniqueness of maps from normal schemes? Suppose you have a normal integral scheme $S$, and a separated scheme $X$. Let $\eta = \text{Spec }K(S)$ be its generic point, and suppose we have a morphism $\eta\rightarrow X$.
(How) can we deduce that there is at most one morphism $S\rightarrow X$ factoring $\eta\rightarrow X$?
 A: Recall that if $S \rightrightarrows X$ are pair of continuous maps of topological spaces which coincide on a dense subset of $S$, and if $X$ is Hausdorff, then these two maps coincide on all of $S$.
One way to see this is as follows: the pair of morphisms can be encoded into a single morphism $S \to X \times X$, and the assumption is that on a dense subset of $S$, this map lands in the diagonal $\Delta(X) \subset X\times X$.  If $X$ is Hausdorff, then $\Delta(X)$ is closed in $X\times X$, and so it follows that all of $S$ lands in $\Delta(X)$.
Now we can imitate this argument for schemes.  The pair of morphisms $S \rightrightarrows X$ can be encoded as a single morphism
$S \to X\times_{\operatorname{Spec} \mathbb Z} X.$   When composed with the natural morphism $\operatorname{Spec} K(S) \to S$, the resulting morphism factors through the closed subscheme $\Delta(X)$ of $X\times_{\operatorname{Spec} \mathbb Z} X$.  (Here we use the assumption that $X$ is separated, to know that $\Delta(X)$ is a closed subscheme.)
Since $S$ is integral, the morphism $\operatorname{Spec} K(S) \to S$ is scheme-theoretically dominant, which is to say that its image is dense in $S$, even in the scheme-theoretic sense, i.e. if any section $f \in
\mathcal O_S(U)$, for an non-empty open subset $U$ of $S$, vanishes when pulled back over $\eta$ to yield an element of $K(S)$, then it already vanishes in 
$\mathcal O_S(U)$.   This means that any section (over some open subset $V$) of the ideal sheaf cutting out $\Delta(X)$ in $X\times_{\operatorname{Spec} \mathbb Z}X$ vanishes when pulled back to (the preimage of $V$ in) $S$ (since by assumption it vanishes when pulled back to $K(S)$), and so in fact the morphism $S \to X\times_{\operatorname{Spec} \mathbb Z} X$ factors through $\Delta(X)$.

One could try to argue via the valuative criterion: if $s \in S$, then we can think of $\mathcal O_{S,s}$ as a subring of $K(S)$, with the latter being the field of fractions of the former.  We can then find a valuation ring lying between $\mathcal O_{S,s}$ and $K(S)$ whose maximal ideal intersects $\mathcal O_{S,s}$ in its maximal ideal.  We could then apply the valuative criterion.  But we would still need to use a Zariski density/scheme-theoretic dominance result to pass information from the Spec of the valuation ring back to the Spec of $\mathcal O_{S,s}$, and so it doesn't really simplify anything.  (The arguments I just sketched are related to how you prove the valuative criterion in the first place.)
