Proposition 5

Let $K$ be a field, $S$ a subring of $K, h$ a homomorphism from $S$ to a field and $p$ the kernel of $h.$ For an element $x$ of $K$ to be integral over the local ring $S,$ it is necessary and sufficient that every place of $K$ extending $h$ be finite at $x.$

If $f$ is a place of $K$ extending $h, f$ is finite on $S_p$ and zero on $pS_p$ and hence the ring of the place $f$ dominates $S_p.$ Conversely, if $V$ is a valuation ring of $K$ which dominates $S_p,$ $V$ is the ring of a placef whose restriction to $S$ is a homomorphism with the same kernel as $h;$ replacing $f$ by an equivalent place, it is seen that $V$ is the ring of a place of $K$ which extends $h.$ To say that every place of $K$ extending $h$ is finite at $x$ is equivalent to saying that $x$ belongs to all the valuation rings of $K$ which dominate $S_p.$ The proposition then follows from Theorem 3 of § 1, no. 3.