This is Exercise 4.5 of Hartshorne Chapter II
Let $X$ be an integral scheme of finite type over a field $k$, having function field $K$. We say a valuation of $K/k$ has center $x$ on $X$ if its valuation ring dominates some local ring $\mathcal{O}_{x,X}$. Show that if every valuations of $v$ has at most one (resp: a unique) center, then $X$ is separated (resp: proper).
Let us say an integral scheme finite type $X$ over $k$ has (I) (resp:(II)) if every valuation of $k(X)/k$ has almost one center(resp: a unique) on $X$. I think I need to show that if $X$ has (I) or (II), then so does any of its integral closed subscheme.
For (I): Let $Z$ be an integral closed subscheme of $X$, and $\mu$ be a valuation of $K(Z)/k$, and suppose $p,q\in Z$ s.t. $\mathcal{O}_{p,Z}$ and $\mathcal{O}_{q,Z}$ are both centers of $\mu$, I need to show that $p=q$.
Let $F$ be the algebraic closure of $K(Z)$, and let us denote also by $\mu$ some valuation of $F/k$ that extends the original $\mu$. We can extend the maps $$\mathcal{O}_{p,X}\to \mathcal{O}_{p,Z}\to R_\mu$$ and $$\mathcal{O}_{q,X}\to \mathcal{O}_{q,Z}\to R_\mu$$ to maps $$A\to R\mu$$ and $$B\to R\mu$$ for some valuation rings $A,B$ dominating $\mathcal{O}_{p,X}$ and $\mathcal{O}_{q,X}$ respectively. I guess I have to prove that $A=B$, or to show the following statement(but I am not sure whether this statement is true or not):
Let $k$ be a field, and $K,L$ be field extensions of $k$ with $L$ algebraically closed. Let $R$ be a valuation ring of $L/k$, and $(S,\mathfrak{p})\subset K$ be a local ring s.t. we have an injective map $S/\mathfrak{p} \to L$. Suppose $A,B$ are two valuation rings of $K/k$ and we have maps $A\to R$ and $B\to R$ which agree with the map $S\to L$ on $A\cap S$ and $B\cap L$ respectively. Then $A=B$.
Any help or hints are appreciated!