Theorem 1 (Normalization Lemma)

Let A be a finitely generated k-algebra and let $a_1 \subset a_2 \subset \ldots \subset a_p$ be an increasing finite sequence of ideals of A such that $p \geq 1$ and $a_p \neq A$. There exists a finite sequence $(x_i)_{1 \leq i \leq n}$ of elements of A which are algebraically independent over k (Chapter III, § 1, no. 1) and such that: (a) A is integral over the ring $B = k[x_1, \ldots, x_n]$. (b) For all j such that $1 \leq j \leq p$, there exists an increasing sequence $(h(j))_{j \leq j \leq p}$ of integers such that for all j the ideal $a_j \cap B$ of B is generated $x_1, \ldots, x_{h(j)}$.

Note first that it is sufficient to prove the theorem when A is a polynomial algebra $k[Y_1, \ldots, Y_m]$. For in the general case, A is isomorphic to a quotient of such an algebra A' by an ideal $a'_0$; let $a'_j$ denote the inverse image of $a_j$ in A' and let $x'_i$ ($1 \leq i \leq r$) be elements of A' satisfying the conditions of the statement for the ring A' and the increasing sequence of ideals $a'_0 \subset a'_1 \subset \ldots \subset a'_p$. Then the images $x_i$ of the $x'_i$ in A for $i > h(0)$ satisfy the desired conditions; this is obvious for condition (b) and for condition (a) this follows from § 1, no. 1, Proposition 2; finally, if the $x_i(h(0) + 1 \leq i \leq r)$ were not algebraically independent over k, there would be a non-zero polynomial

where the $Q_j$ are polynomials in $k[Y_1, \ldots, Y_m]$; as $a_p \neq 0$ is invertible in $k$, (4) is certainly an equation of integral dependence with coefficients in B, whence our assertion.

The field of fractions $k(Y_1, \ldots, Y_r)$ of A is therefore algebraic over the field of fractions $k(x_1, \ldots, x_m)$ of B, which proves (Algebra, Chapter V, § 5, no. 3, Theorem 4) that the $x_i$ ($1 \leq i \leq m$) are algebraically independent. Moreover, $a \cap B = Bx_1$; for every element $z \in a, \cap B$ may be written $z = x_1 z'$ where $z' \in A \cap k(x_1, \ldots, x_r)$; but $A \cap k(x_1, \ldots, x_r) = k[x_1, \ldots, x_m] = B$ since $B$ is integrally closed ($\S$ 1, no. 3, Corollary 2 to Proposition 13); therefore $z' \in B$, which completes the proof of properties (a) and (b) in this case.

(A2) General case ($p = 1$). We argue by induction on $m$, the case $m = 0$ being trivial. We may obviously suppose that $a_i \neq 0$ (otherwise we may take $x_i = Y$, for $1 \leq i \leq m$ and $h(1) = 0$). Let $x_1$ be a non-zero element of $a_i$; by (A1) there exist $t_2, \ldots, t_m$ such that $x_1, t_2, \ldots, t$, are algebraically independent over $k$, $A$ is integral over $C = k[x_1, t_2, \ldots, t_m]$ and $x_1 A \cap C = x_1 C$. By the induction hypothesis there exist elements $x_2, \ldots, x_m$ of $k[t_2, \ldots, t_m]$ and an integer $h$ such that $k[t_2, \ldots, t_m]$ is integral over $B' = k[x_2, \ldots, x_m]$, $x_2, \ldots, x_m$ are algebraically independent over $k$ and the ideal $a_i \cap B'$ is generated by $x_2, \ldots, x_h$. Then $C$ is integral over $B = k[x_1, x_2, \ldots, x_m]$ ($\S$ 1, no. 1, Corollary to Proposition 5 ) and hence so is $A$ ($\S$ 1, no. 1, Proposition 6 ); the same argument as in the case (A1) shows that $x_1, \ldots, x_m$ are algebraically independent over $k$; finally, as $x_1 \in a_i$, and $B = B'[x_1], a_i \cap B = Bx, + (a_i \cap B')$ and, as $a_i \cap B'$ is generated (in $B'$) by $x_2, \ldots, x_h, a_i \cap B$ is generated (in $B$) by $x_1, x_2, \ldots, x_h$.

(B) Passage from $p - 1$ top. Let $t_1, \ldots, t_r$, be elements of $A$ satisfying the conditions of the theorem for the increasing sequence of ideals $a_1 \subset \ldots \subset a_{p-1}$ and let us write $r = h(p - 1)$. By (A2) there exist elements $x_{r+1}, \ldots, x_s$ of $k[t_{r+1}, \ldots, t_m]$ and an integer s such that

is integral over $B' = k[x_{r+1}, \ldots, x_m]$, $x_{r+1}, \ldots, x_s$ are algebraically independent over $k$ and the ideal $a_p \cap B'$ is generated by $x_{r+1}, \ldots, x_s$. Writing $x_i = t_i$ for $i \leq r$ the family $(x_i)_{1 \leq i \leq m}$ obtained solves the problem with $h(p) = s$. For $A$ is integral over $C[t_1, \ldots, t_r] = C[x_1, \ldots, x_r]$ and hence also over $B = k[x_1, \ldots, x_m] = B'[x_1, \ldots, x_r]$ since $C$ is integral over $B'$ ($\S$ 1, no. 1, Corollary to Proposition 5 and Proposition 6); it can be shown as in the case (A1) that the $x_i$ are algebraically independent over $k$. On the other hand, for $j \leq p - 1$, the ideal is by hypothesis the set of polynomials in $x_1, \ldots, x_r, t_{r+1}, \ldots, t_m$ all of whose monomials contain one of the elements $x_1, \ldots, x_{h(j)}$; as $x_{r+1}, \ldots, x_m$ are polynomials in $t_{r+1}, \ldots, t$, with coefficients in $k$, it is seen immediately that a polynomial in $x_1, \ldots, x_r, x_{r+1}, \ldots, x_s$ (with coefficients in $k$) can belong to $a_j$ only if all its monomials contain one of the elements $x_1, \ldots, x_{h(j)}$. Finally, as $x_1, \ldots, x_r$ belong to $a_{p-1}$, and hence also to $a_p$, $a_p \cap B'[x_1, \ldots, x_r]$ consists of the polynomials in $x_1, \ldots, x_r$ with coefficients in $B'$ whose constant term belongs to $a_p \cap B'$; this ideal is therefore generated by $x_1, \ldots, x_r, x_{r+1}, \ldots, x_t$.