Exercise 8, § 2

A local ring $A$ is called unibranche if its integral closure is a local ring. (a) Let $A$ be a local integral domain and $K$ its field of fractions. For $A$ to be unibranche, it is necessary and sufficient that every sub-$A$-algebra of $K$ which is a finitely generated $A$-module be a local ring. (b) Show that, if the completion of a Noetherian local ring $A$ is an integral domain, $A$ is unibranche (if $B$ is a finite $A$-algebra contained in $K$, show that the subring of the field of fractions of $\hat{A}$ generated by $B$ and $\hat{A}$ is isomorphic to $B \otimes_A \hat{A}$, using the flatness of the $A$-module $\hat{A}$ and the fact that $B$ is contained in a finitely generated free $A$-module contained in $K$; then use (a) and Chapter III, § 2, no. 13, Corollary to Proposition 19).