Lemma 3

Let $A$ be a $k$-algebra of finite type and $p_0 \subset ... \subset p_m$ a maximal chain of prime ideals of $A$. There exists an integer $n \geq m$, a sequence $(x_1, ..., x_n)$ of elements of $A$, algebraically free over $k$ (A, IV, p. 4), and such that : a) $A$ is entire over the ring $B = k[x_1, ..., x_n]$; b) for every $j$ such that $0 \leq j \leq m$, the ideal $p_j \cap B$ is generated by the $x_k$ with $1 \leq k \leq n - m + j$.

According to the normalization lemma (V, § 3, No. 1, th. 1), there exists an integer $n \geq 0$, a finite sequence $(x_1, ..., x_n)$ of elements of $A$ algebraically free and an increasing sequence $(h(j))_{0 \leq j \leq m}$ of integers $\leq n$ such that the ideal $p_j \cap B$ is equal to the prime ideal $q_j$ of $B$ generated by the $x_k$ with $1 \leq k \leq h(j)$, and that $A$ is entire over the ring $B$. Let $j$ be an integer such that $0 \leq j < m$. By passing to quotients, one deduces from the canonical injection of $B$ into $A$ an injective homomorphism of $B/q_j$ into $A/p_j$ which makes $A/p_j$ a finite $(B/q_j)$-algebra. Since the ring $B/q_j$ is isomorphic to an algebra of polynomials in $n - h(j)$ indeterminates over $k$, it is integrally closed (V, § 1, No. 3, Cor. 3 of Prop. 13). According to th. 2 of No. 3, one therefore has

It follows that one has $h(j+1) \leq h(j) + 1$ and $q_{j+1} \neq q_j$, whence $h(j+1) = h(j) + 1$. But one has $h(m) = n$ since $q_m$ is maximal (V, § 2, No. 1, Prop. 1), whence finally $h(j) = n - m + j$.