Theorem 2

Let A be a finitely generated integral k-algebra, K its field of fractions and $A'$ the integral closure of A in a field K' which is a finite algebraic extension of K. Then $A'$ is a finitely generated A-module and a finitely generated k-algebra.

By Theorem 1 there exists a subalgebra C of A isomorphic to a polynomial algebra $k[X_1, \ldots, X_n]$ and such that A is integral over C; $A'$ is obviously the integral closure of C in K' (§ 1, no. 1, Proposition 6); we may therefore confine our attention to the case where $A = k[X_1, \ldots, X_n]$. Let $N$ be the quasi-Galois extension of $K$ (in an algebraic closure of $K$) generated by $K'$, which is a finite algebraic extension of $K$ (Algebra, Chapter V, § 6, no. 3, Corollary 1 to Proposition 9). It will suffice to prove that the integral closure $B$ of $A$ in $N$ is a finitely generated $A$-module, for $A'$ is a sub-$A$-module of $B$ and $A$ is a Noetherian ring (Chapter III, § 2, no. 10, Corollary 2 to Theorem 2). We may therefore confine our attention to the case where $K'$ is a quasi-Galois extension of $K$. Then we know (Algebra, Chapter V, § 10, no. 9, Proposition 14) that $K'$ is a (finite) Galois extension of a (finite) radicial extension $K''$ of $K$. If $A''$ is the integral closure of $A$ in $K''$, $A'$ is the integral closure of $A''$ in $K'$ and it will suffice to prove that $A''$ is a finitely generated $A$-module and $A'$ is a finitely generated $A''$-module. Now, if it has been proved that $A''$ is a finitely generated $A$-module, it is a Noetherian domain, integrally closed by definition; the fact that $A$ is a finitely generated $A''$-module will follow from § 1, no. 6, Corollary 1 to Proposition 18.

We see therefore that we may confine our attention to the case where $A = k[X_1, \ldots, X_n]$ and where $K'$ is a finite radicial extension of $K = k(X_1, \ldots, X_r)$. Then $K'$ is generated by a finite family of elements $(y_i)_{1 \leq i \leq m}$ and there exists a power $q$ of the characteristic exponent of $k$ such that $y_i^q \in k(X_1, \ldots, X_r)$. Let $c_j$ ($1 \leq j \leq r$) be the coefficients of the numerators and denominators of the rational functions in $X_1, \ldots, X_r$, equal to $y_i^q$ ($1 \leq i \leq m$). Then $K'$ is contained in the extension $L = k'(X_1^{q^{-1}}, \ldots, X_n^{q^{-1}})$, where $k' = k(c_1^{q^{-1}}, \ldots, c_r^{q^{-1}})$ (we are in an algebraic closure of $K'$) and $A$ is contained in the algebraic closure $B'$ of $A$ in $L'$. Now, $k'$ is algebraic over $k$ and hence $C' = k'[X_1, \ldots, X_n]$ is integral over $A$ ($§ 1$, no. 1, Proposition 5); as $k'[X_1^{q^{-1}}, \ldots, X_n^{q^{-1}}]$ is integrally closed ($§ 1$, no. 3, Corollary 2 to Proposition 13), it is seen that this ring is the integral closure of $C'$ in $L'$ and hence also that of $A$ ($§ 1$, no. 1, Proposition 6), in other words $B' = k'[X_1^{q^{-1}}, \ldots, X_n^{q^{-1}}]$. Now clearly $B'$ is a finitely generated $C'$-module ($§ 1$, no. 1, Proposition 4) and, as $k'$ is a finite extension of $k$, $C'$ is a finitely generated $A$-module and hence $B'$ is a finitely generated $A$-module; since $A$ is Noetherian and $A \subset B'$, $A'$ is a finitely generated $A$-module.