Lemma 2

Suppose the ring A local and noetherian. Then $A_n$ is local and noetherian. The following conditions are equivalent:

(i) $A_n$ is regular; (ii) A is regular and the elements $1 \otimes df_i$ of the $\kappa_A$-vector space $\kappa_A \otimes_A \Omega(A)$ are linearly independent.

A) Let us first treat the case $n = 1$, setting $T_1 = T$, $f_1 = f$. Let $a$ denote the class of $f$ in $\kappa_A$, and distinguish two cases according as $a$ does or does not belong to $\kappa_A^p$. If $a \notin \kappa_A^p$, then the polynomial $T^p - a$ is irreducible in $\kappa_A$ (A, V, p. 24, lemma 1), and $\kappa_A \otimes_A A_1$ is isomorphic to the field $\kappa_A[T]/(T^p - a)$. The ideal $m_A A_1$ of $A_1$ is therefore maximal, so that the ring $A_1$ is local (V, § 2, No. 1, prop. 1). If $A$ is regular, $A_1$ is regular (VIII, § 5, No. 1, prop. 1). By A, V, p. 99, prop. 6, the element $da$ of $\Omega(\kappa_A)$ is not zero; since it is the image under the canonical mapping $\kappa_A \otimes_A \Omega(A) \to \Omega(\kappa_A)$ of $1 \otimes df$, the latter is not zero. This proves the lemma in this case.

Suppose now that $a$ belongs to $\kappa_A^p$. There therefore exists an element $g$ of $A$ such that $f - g^p \in m_A$. Put $h = f - g^p$. Since $T^p - f = (T - g)^p - h$, the $A$-algebra $A_1$ is isomorphic to $A[T]/(T^p - h)$. By VIII, § 5, No. 4, prop. 4, the ring $A_1$ is local and, in order that it be regular, it is necessary and sufficient that $A$ be regular and that $h$ should not belong to $m_A^2$. Now, since $\kappa_A$ is formally smooth over the prime field (No. 3, th. 1), the canonical mapping

is injective (No. 2, remark 1); but the image under $\bar{d}$ of the class of $h$ modulo $m_A^2$ is equal to $1 \otimes dh = 1 \otimes d(f - g^p) = 1 \otimes df$. This proves the lemma in this second case and completes the proof of the case $n = 1$.

B) Suppose $n > 1$. The ring $A_1$ is local and noetherian by the case already treated. The $A_1$-algebra $A_n$ is identified with the quotient of $A_1[T_2, \ldots, T_n]$ by the ideal generated by the $T_i^p - f_i$, $i \geq 2$; by the induction hypothesis, it is a local ring and condition (i) is equivalent to the conjunction of the following two:

(i') $A_1$ is regular; (ii') the elements $1 \otimes df_2, \ldots, 1 \otimes df_n$ of the $\kappa_{A_1}$-vector space $\kappa_{A_1} \otimes_{A_1} \Omega(A_1)$ are linearly independent.

But (i') is equivalent, as we have just seen, to (ii'') $A$ is regular and the element $1 \otimes df_1$ of the $\kappa_A$-vector space $\kappa_A \otimes_A \Omega(A)$ is not zero.

By lemma 1, the canonical homomorphism $A_1 \otimes_A \Omega(A) \to \Omega(A_1)$ induces an isomorphism of $((A_1 \otimes_A \Omega(A))/A_1(1 \otimes df_1)) \oplus A_1$ onto $\Omega(A_1)$, and consequently an injective homomorphism of $(\kappa_{A_1} \otimes_A \Omega(A))/\kappa_{A_1}(1 \otimes df_1)$ into $\kappa_{A_1} \otimes_{A_1} \Omega(A_1)$. Since $\kappa_{A_1} \otimes_A \Omega(A)$ is identified with $\kappa_{A_1} \otimes_{\kappa_A} (\kappa_A \otimes_A \Omega(A))$, assertion (ii'') is therefore equivalent to: (ii''') the elements $1 \otimes df_2, \ldots, 1 \otimes df_n$ are linearly independent in $(\kappa_A \otimes_A \Omega(A))/\kappa_A(1 \otimes df_1)$.

But the conjunction of (ii') and (ii'') is equivalent to (ii), which proves the lemma.