Corollary 2

Let $A$ be an integral domain, $K$ its field of fractions, $R$ an algebra over $K$ and $B$ the integral closure of $A$ in $R$. The elements of $R$ which are algebraic over $K$ (no. 1, Example 1) are the elements of the form $a^{-1} b$ where $b \in B$ and $a \in A, a \neq 0$; if $L$ is the algebraic closure of $K$ in $R$, there exists a basis of $L$ over $K$ contained in $B$.

The first assertion follows from Proposition 16 applied in the case $S = A - \{0\}$. If $(x_i)_{i \in I}$ is a basis of $L$ over $K$, then there exists for all $i \in I$ an element $a_i \neq 0$ of $A$ such that $a_i x_i \in B$; then $(a_i x_i)_{i \in I}$ is also a basis of $L$ over $K$.