Corollary 1

Let $A$ be an integral domain and $B$ a finitely generated $A$-algebra containing $A$ as a subring. Then there exist an element $s \neq 0$ of $A$ and a subalgebra $B'$ of $B$ isomorphic to a polynomial algebra $A[Y_1, \ldots, Y_n]$ such that $B[s^{-1}]$ (Chapter II, § 2, no. 1) is integral over $B'[s^{-1}]$.

We write $S = A - \{0\}$ and let $k = S^{-1}A$ the field of fractions of $A$; clearly $S^{-1}B$ is a finitely generated $k$-algebra and, as it contains $k$ by hypothesis (Chapter II, § 2, no. 4, Theorem 1), it is not reduced to 0. By Theorem 1 (applied top = 1 and $a_i = 0$) there exists therefore a finite sequence $(x_i)_{1 \leq i \leq n}$ of elements of $S^{-1}B$ which are algebraically independent over $k$ and such that $S^{-1}B$ is integral over $k[x_1, \ldots, x_n]$. Let $(z_j)_{1 \leq j \leq m}$ be a system of generators of the $A$-algebra $B$; in $S^{-1}B$ each of the $z_j/1$ satisfies an equation of integral dependence

where the $P_{hj}$ are polynomials in the $x_i$ with coefficients in $k$. There exists an element $s \neq 0$ of $A$ such that we may write $x_i = y_i/s$ whereby, $\in B$ for $1 \leq i \leq n$ and all the coefficients of the $P_{hj}$ are of the form $c/s$ where $c \in A$; finally, replacing if need be, $s$ by a product of elements of $S$, we may assume that in $B$

where the $Q_{hj}$ are polynomials in $y_1, \ldots, y_n$ with coefficients in $A$; if we write $z'_j = sz_j$, it is seen, multiplying (6) by $s^{a_j-1}$, that $z'_j$ is integral over $B' = A[y_1, \ldots, y_n]$. We show that the $y_i$ are algebraically independent over $A$; if there is a relation of the form $\sum_p a_p y_1^{p_1} \ldots y_n^{p_n} = 0$ where $a_p \in A$ for all $p$, we deduce that $\sum_p a'_p x_1^{p_1} \ldots x_n^{p_n} = 0$ in $S^{-1}B$, where $a'_p = a_p s^{p_1 + \cdots + p_n}$ in $k$; by hypothesis therefore $a'_p = 0$ for all $p$, whence $a_p = 0$ for all $p$. Moreover, in the ring $B[s^{-1}]$ each of the $z'_j/1$ is integral over $B'[s^{-1}]$ (§ 1, no. 1, Proposition 2) and, as $z_j/1 = (z'_j/1)(1/s)$ in $B[s^{-1}]$, it is seen that the $z_j/1$ are integral over $B'[s^{-1}]$, which completes the proof (§ 1, no. 1, Proposition 4).