Lemma 2

Let $B$ be an integral domain and $A$ a subring of $B$ such that $B$ is integral over $A$. For $B$ to be a field, it is necessary and sufficient that $A$ be a field.

If $A$ is a field, then, for ally $\neq 0$ in $B$, $A[y]$ is by hypothesis ($§ 1$, Theorem 1) a finitely generated $A$-module; as $A[y]$ is an integral domain, it is a field (Algebra, Chapter V, § 2, no. 1, Proposition 1) and a fortiori $y$ is invertible in $B$ and hence $B$ is a field. Conversely, suppose that $B$ is a field and let $z \neq 0$ in $A$; as $z^{-1} \in B$, $z^{-1}$ is integral over $A$, in other words there is an equation of integral dependence

where the $a_i \in A$; now this relation shows that

hence $A$ is certainly a field.