Corollary 4

Let $A$ be a Zariski ring such that $\hat{A}$ is an integral domain, $L$ the field of fractions of $\hat{A}$ and $K \subset L$ the field of fractions of $A$; then $\hat{A} \cap K = A$.

Clearly $A \subset \hat{A} \cap K$; on the other hand, if $x \in \hat{A} \cap K$, then $\hat{A}x \subset \hat{A}$ and hence, as $\hat{A}x = \hat{A} \otimes_A (Ax)$ (no. 4, Theorem 3), $\hat{A} \otimes_A ((Ax + A)/A) = 0$. As $\hat{A}$ is a faithfully flat $A$-module (Proposition 9), we deduce that $Ax \subset A$, whence $x \in A$.