This follows from III, § 2, No. 10, Cor. 3 of Th. 2 and II, § 2, No. 4, Cor. 2 of Prop. 10.

The following properties are deduced from the definition:

b) Every ring of fractions of a $k$-algebra essentially of finite type is a $k$-algebra essentially of finite type.

c) The product $k$-algebra of a finite family of $k$-algebras essentially of finite type is essentially of finite type.

d) Let $k \to k'$ be a homomorphism of rings; for every $k$-algebra $A$ essentially of finite type, the $k'$-algebra $A_{(k')} = k' \otimes_k A$ is essentially of finite type.

In fact, it is a $B$-algebra essentially of finite type (prop. 2, d)), and one applies Proposition 1.

Let us denote by $\rho : A \to B$ the canonical mapping. Let $x = (x_i)_{i \in I}$ be a finite essentially generating family of the $k$-algebra $A$, and let $A'$ be the subalgebra generated by it; let $y = (y_j)_{j \in J}$ be a finite essentially generating family of the $A$-algebra $B$. Let us denote by $B'$ the sub-$k$-algebra of $B$ generated by the $\rho(x_i)$ and the $y_j$. Let $b \in B$; by hypothesis there exist polynomials $P$ and $Q$ in $A[(Y_j)_{j \in J}]$ such that $Q(y)$ is invertible in $B$ and such that one has $Q(y) b = P(y)$. The non-zero coefficients of $P$ and $Q$ are finite in number; there exists a polynomial $R \in k[(X_i)_{i \in I}]$ such that $R(x)$ is invertible in $A$ and that $R(x) P$ and $R(x) Q$ belong to $A'[(Y_j)_{j \in J}]$. Then $\rho(R(x)) Q(y)$ is invertible in $B$, and one has

Thus the elements $\rho(x_i)$ for $i \in I$ and $y_j$ for $j \in J$ form a finite essentially generating family of the $k$-algebra $B$.

For let $A$ and $B$ be two $k$-algebras essentially of finite type. Then $A \otimes_k B$ is essentially of finite type over $A$ (Prop. 2, d)), hence over $k$ (Prop. 3).

a) If $A$ and $B$ are Macaulay rings, the same is true of $A \otimes_k B$.

b) If $A$ and $B$ are Gorenstein rings, the same is true of $A \otimes_k B$.

Suppose that $A$ and $B$ are Macaulay rings (resp. Gorenstein rings), and let us prove that the same is true of $A \otimes_k B$. The ring $A \otimes_k B$ is noetherian (No. 1, cor. of Prop. 2). The $A$-module $A \otimes_k B$ is free, hence flat. By Prop. 10 of § 2, No. 7 (resp. Cor. 1 of Prop. 12 of § 3, No. 8), it is enough for us to prove that $\kappa(p) \otimes_k B$ is a Macaulay ring (resp. a Gorenstein ring) for every prime ideal $p$ of $A$. The extension $\kappa(p)$ of $k$ is of finite type (No. 1, Prop. 2 and Example 1); we are therefore reduced to proving the statement in the case where the $k$-algebra $A$ is a finite-type extension $K$ of $k$.

The ring $K \otimes_k B$ is then a free $B$-module of finite rank, hence a Macaulay $B$-module. By § 2, No. 6, Cor. 1 of Prop. 8, it is a Macaulay ring, which proves a). Suppose now that $B$ is a Gorenstein ring. There exist intermediate extensions $K_i,\ 0 \leq i \leq m$, of $K$ such that

and such that $K_i$ is a monogenic $K_{i-1}$-algebra for $i = 1, \ldots, m$; this enables us to reduce to the case where the extension $K$ of $k$ is monogenic (of finite degree). The canonical homomorphism $B \to K \otimes_k B$ makes $K \otimes_k B$ into a flat $B$-algebra and, for every $q \in \mathrm{Spec}(B)$, the ring $(K \otimes_k B) \otimes_B \kappa(q)$, isomorphic to $K \otimes_k \kappa(q)$, is a monogenic $\kappa(q)$-algebra, hence a Gorenstein ring ($§ 3$, n° 9, example 5). Thus $K \otimes_k B$ is a Gorenstein ring ($§ 3$, n° 8, cor. 1 of Prop. 12), which completes the proof of b).

If $A$ is a Macaulay ring (resp. a Gorenstein ring), the same is true of $A_{(K)}$ by Prop. 4. Since $A_{(K)}$ is a faithfully flat $A$-module, the converse follows from Prop. 10 of § 2, No. 7 (resp. Cor. 1 of Prop. 12 of § 3, No. 8).

Suppose first that $B$ is a field; let $\varphi$ denote the canonical homomorphism from $k$ into $B$, and $r$ its kernel. The homomorphism $\varphi$ induces a homomorphism from the field of fractions $\kappa(r)$ of $k/r$ into $B$. Then $A \otimes_k B$ is identified with $(A \otimes_k \kappa(r)) \otimes_{\kappa(r)} B$; since $\varphi^{-1}(0) = r$, the ring $A \otimes_k \kappa(r)$ is a Macaulay ring (resp. a Gorenstein ring) by Prop. 10 of § 2, No. 7 (resp. Cor. 1 of Prop. 12 of § 3, No. 8). The assertion then follows in this case from Cor. 1.

Let us pass to the general case. The $B$-algebra $A \otimes_k B$ is flat, and noetherian by the corollary to Prop. 2 of No. 1. For each prime ideal $q$ of $B$, the ring $(A \otimes_k B) \otimes_B \kappa(q)$, which is identified with $A \otimes_k \kappa(q)$, is a Macaulay ring (resp. a Gorenstein ring) by what precedes. We conclude by applying Prop. 10 of § 2, No. 7 (resp. Cor. 1 of Prop. 12 of § 3, No. 8).

a) If the rings $A_\alpha$ are regular and if $A$ is a flat $A_\alpha$-module for every $\alpha \in I$, the ring $A$ is regular.

b) If the rings $A_\alpha$ are normal ($§ 1$, No. 8), $A$ is normal.

a) Let $m$ be a maximal ideal of $A$; for each $\alpha \in I$, let us denote by $m_\alpha$ the ideal $m \cap A_\alpha$ of $A_\alpha$. Since $A$ is noetherian, $m$ is of finite type and there exists an element $\alpha$ of $I$ such that $m = A m_\alpha$, so that the $A$-module $(A_\alpha / m_\alpha) \otimes_{A_\alpha} A$ is isomorphic to $A / m$. Since $A$ is flat over $A_\alpha$, we have $\mathrm{dp}_A(A/m) \leq \mathrm{dp}_{A_\alpha}(A_\alpha / m_\alpha)$ (A, X, p. 141, Lemma 2). Since the rings $A_\alpha$ are regular, it follows that $A$ is regular ($§ 4$, No. 2, Prop. 4).

b) Since the $A_\alpha$ are reduced, $A$ is reduced. Let $a$ and $b$ be elements of $A$ such that $b$ is not a zero divisor and that the element $a/b$ of the total ring of fractions of $A$ is integral over $A$. There exists a monic polynomial $P \in A[X]$ such that $P(a/b) = 0$. Let $\alpha$ be an element of $I$ such that the ring $A_\alpha$ contains $a$, $b$ and the coefficients of $P$. Since $A_\alpha$ is normal, there exists $c \in A_\alpha$ such that $a = bc$. Thus $a/b = c$ belongs to $A$ and $A$ is normal.

Let $A$ denote the ring $K \otimes_k L$; it is noetherian by cor. of prop. 2 (No. 2). Suppose first that the extension $K$ is separable. By A, V, p. 130, corollary, there exists a transcendence basis $t = (t_1, \ldots, t_n)$ of $K$ such that $K$ is a finite separable extension of the pure extension $k(t)$. The ring $E = k(t) \otimes_k L$, isomorphic to a ring of fractions of a polynomial ring over $L$, is regular ($§ 4$, No. 2, cor. 5 of prop. 4); for every prime ideal $p$ of $E$, the ring $\kappa(p) \otimes_E A$ is isomorphic to $\kappa(p) \otimes_{k(t)} K$, hence to a finite product of fields (A, V, p. 35, def. 1, p. 34, th. 4, and p. 33, prop. 5). Since $A$ is a free $E$-module, it is a regular ring by cor. of prop. 9 of $§ 4$, No. 5.

Suppose now that $L$ is separable over $k$. By the first part of the proof, the ring $K \otimes_k L'$ is regular for every finite-type subextension $L'$ of $L$. One concludes by applying lemma 1.

¹ In this number, an algebra over a field $k$ is said to be regular if it is a regular ring. In particular, every extension of $k$ is a regular algebra. Care will be taken not to confuse this notion with that of regular extension, introduced in A, V, p. 135, def. 2, which will not occur here.

a) If the ring $A_{(K)}$ is regular (resp. normal), the ring $A$ is regular (resp. normal).

b) If the ring $A$ is regular (resp. normal) and if the extension $K$ of $k$ is separable, the ring $A_{(K)}$ is regular (resp. normal).

The $A$-module $A_{(K)}$ is free, hence faithfully flat; assertion a) therefore follows from cor. of prop. 8 of I, § 3, No. 5 and from prop. 8 of § 4, No. 5 (resp. cor. 2 of th. 4 of § 1, No. 10). Under the hypotheses of b), for every prime ideal $\mathfrak{p}$ of $A$, the ring $\kappa(\mathfrak{p}) \otimes_k K$ is regular (lemma 2), and $a fortiori$ normal (§ 4, No. 2, cor. 2 of prop. 4). Assertion b) therefore follows from cor. of prop. 9 of § 4, No. 5 (resp. cor. 3 of th. 4 of § 1, No. 10).

a) If A is absolutely regular (resp. absolutely normal, resp. separable), the same is true of S$^{-1}$A for every multiplicative subset S of A.

b) If $A_m$ is absolutely regular (resp. absolutely normal, resp. separable), for every maximal ideal m of A, then A is absolutely regular (resp. absolutely normal, resp. separable).

a) This follows from the fact that the ring $(S^{-1}A)_{(k')}$ is isomorphic to a ring of fractions of the ring $A_{(k')}$ for every extension $k'$ of k.

b) Suppose that $A_m$ is absolutely regular (resp. absolutely normal, resp. separable) for every maximal ideal m of A. Let $k'$ be a radicial extension of k of finite degree and let $m'$ be a maximal ideal of $A_{(k')}$. It remains to verify that the local ring $(A_{(k')})_{m'}$ is regular (resp. normal, resp. reduced).

The canonical homomorphism $A \to A_{(k')}$ makes $A_{(k')}$ into a finite A-algebra. The ideal $m'$ lies over a maximal ideal m of A (V, § 2, No. 1, prop. 1), and $(A_{(k')})_{m'}$ is isomorphic to a ring of fractions of the regular (resp. normal, resp. reduced) ring $(A_m)_{(k')}$, and is therefore regular (resp. normal, resp. reduced).

Let us choose a composite extension E of K and of a perfect closure $\hat{k}$ of k, and elements $(t_1, \ldots, t_n)$ of E such that E is a separable algebraic extension of $\hat{k}(t_1, \ldots, t_n)$ (A, V, p. 130, cor.). Let I be a finite generating subset of K over k. The field $\hat{k}(t_1, \ldots, t_n)$ (resp. $\hat{k}K$) is the union of the subfields $k'(t_1, \ldots, t_n)$ (resp. $k'K$), where $k'$ runs through the set of subextensions of $\hat{k}$ of finite degree over k; hence one can find such a subextension $k'$ such that the elements $t_1, \ldots, t_n$ of E = $\hat{k}K$ belong to $k'K$ and that each element of I is algebraic and separable over $k'(t_1, \ldots, t_n)$. Then L = $k'K$ is a separable extension of $k'$, of finite degree over K, and $k'$ is a finite radicial extension of k.

Let K be an extension of finite type of k; let us prove that the ring $A_{(K)}$ is regular (resp. normal). Consider indeed extensions L and $k'$ having the properties of lemma 3. Then the ring $A_{(k')}$ is regular (resp. normal) by definition, and the ring $A_{(L)}$, which identifies with $A_{(k')} \otimes_{k'} L$, is regular (resp. normal) by Prop. 5, b) of No. 3. Consequently, $A_{(K)}$ is regular (resp. normal) by Prop. 5, a).

Suppose the ring B regular (resp. normal), and prove the proposition. The canonical homomorphism $B \to A \otimes_k B$ makes $A \otimes_k B$ into a free B-module. For every prime ideal $p$ of B, the ring $(A \otimes_k B) \otimes_B \kappa(p)$ identifies with $A \otimes_k \kappa(p)$; by the corollary of Prop. 9 of § 4, No. 5 (resp. Cor. 3 of Th. 4 of § 1, No. 10), it is enough to prove that $A \otimes_k \kappa(p)$ is regular (resp. normal) for every prime ideal $p$ of B.

If the $k$-algebra B is essentially of finite type, the extension $\kappa(p)$ of $k$ is of finite type and the ring $A \otimes_k \kappa(p)$ is regular (resp. normal) by what we have seen above. Suppose now the $k$-algebra A essentially of finite type; the ring $A \otimes_k \kappa(p)$ is noetherian and is the union of the increasing filtered family of noetherian subrings $A \otimes_k K$, where $K$ runs through the subextensions of finite type of $\kappa(p)$. The latter are regular (resp. normal), and one applies lemma 1 of No. 3.

Let $A$ and $B$ be two $k$-algebras satisfying the hypotheses of the corollary. Let $k'$ be a finite radicial extension of $k$. The ring $B_{(k')}$ is regular (resp. normal); the ring $A \otimes_k B_{(k')}$ is regular (resp. normal) by Prop. 7, as is the ring $(A \otimes_k B) \otimes_k k'$ which is isomorphic to it.

a) The ring $A_{(K)}$ is regular (resp. normal).

b) If the extension $K$ of $k$ is separable, the $k$-algebra $A_{(K)}$ is absolutely regular (resp. absolutely normal).

Assertion a) results from proposition 7; assertion b) results from cor. 1 and example 2.

Suppose $A$ absolutely regular (resp. absolutely normal) and let $K'$ be a purely inseparable extension of $K$ of finite degree. The ring $K' \otimes_K A_{(K)}$, isomorphic to $K' \otimes_k A$, is regular (resp. normal) by cor. 2.

Suppose conversely that the $K$-algebra $A_{(K)}$ is absolutely regular (resp. absolutely normal), and let $k'$ be a purely inseparable extension of $k$ of finite degree. Let $L$ be a composite extension of $k'$ and $K$; then the ring $A_{(L)}$ may be identified with $L \otimes_K A_{(K)}$, hence is regular (resp. normal); consequently, the ring $A_{(k')}$ is regular (resp. normal) by prop. 5, a) of No. 3.

This results from cor. 3 and example 1.

(i) the k-algebra A is absolutely regular; (ii) for every regular k-algebra C, the ring $A \otimes_k C$ is regular; (iii) the ring $A \otimes_k A$ is regular; (iv) the ideal I of $A \otimes_k A$ is completely secant ($§ 5$, No. 1, Definition 1).

Let B denote the ring $A \otimes_k A$ and provide it with the structure of an A-algebra deduced from the homomorphism $\rho : A \to A \otimes_k A$ such that $\rho(x) = x \otimes 1$; then $\mu$ is a homomorphism of A-algebras, and induces by passing to the quotient an isomorphism of $B/I$ onto A.

(i) $\Rightarrow$ (ii): this results from Prop. 7. (ii) $\Rightarrow$ (iii): it is enough to apply (ii) with $C = k$, then with $C = A$. (iii) $\Rightarrow$ (iv): the A-module B is free, hence faithfully flat. If the ring B is regular, A is regular ($§ 4$, No. 5, Prop. 8); the ideal I is then completely secant ($§ 5$, No. 3, Prop. 2). (iv) $\Rightarrow$ (i): let us suppose the ideal I completely secant and first prove that A is regular. Let m be a maximal ideal of A and let $\nu : (A/m) \otimes_k A \to A/m$ be the homomorphism deduced from $\mu$. The maximal ideal $n = \mathrm{Ker}\,\nu$ is equal to $1((A/m) \otimes_k A)$; applying Prop. 6 of $§ 5$, No. 6 to the A-algebra $A' = A/m$, one sees that the ideal n is completely secant in $(A/m) \otimes_k A$. Consequently ($§ 5$, No. 3, Prop. 3) the local ring $((A/m) \otimes_k A)_n$ is regular. Let us denote by $j : A \to (A/m) \otimes_k A$ the homomorphism $x \mapsto 1 \otimes x$; since $\nu \circ j$ is the canonical homomorphism of A into $A/m$, one has $j^{-1}(n) = m$. Thus j extends to a local homomorphism of local rings from $A_m$ into $((A/m) \otimes_k A)_n$, which makes the latter into a faithfully flat $A_m$-module. By Prop. 8 of $§ 4$, No. 5, the ring $A_m$ is therefore regular. We have thus proved that A is regular.

Now let $k'$ be an extension of k. The kernel of the mapping $\mu' : A_{(k')} \otimes_{k'} A_{(k')} \to A_{(k')}$ deduced from the multiplication of $A_{(k')}$ is none other than $IA_{(k')}$; it is therefore completely secant in $A_{(k')}$ ($§ 5$, No. 6, Prop. 6). The $k'$-algebra $A_{(k')}$ therefore satisfies condition (iv); by what has just been seen, it is regular, and this proves (i).

Recall (A, III, p. 133-134) that the quotient $I/I^2$ endowed with the A-module structure deduced from $\rho$ is denoted by $\Omega_k(A)$ and called the module of k-differentials of A. When the k-algebra A is essentially of finite type, the ring $A \otimes_k A$ is noetherian, so that the A-module $\Omega_k(A)$ is finitely generated.

One denotes by $d_{A/k}$, or simply by d, the k-linear mapping from A into $\Omega_k(A)$ which associates with an element x of A the class of $x \otimes 1 - 1 \otimes x$ in $\Omega_k(A)$. The mapping d is a k-derivation; for every A-module M and every k-derivation D : A $\to$ M, there exists a unique A-linear mapping $g : \Omega_k(A) \to M$ such that $D = g \circ d$ (loc. cit., Prop. 18).

If S is a multiplicative subset of A, the canonical $S^{-1}A$-linear mapping (loc. cit., p. 136)

is bijective: in fact, it is enough to verify that, for every $S^{-1}A$-module M, every k-derivation D : A \to M extends in a unique way to a k-derivation D : S^{-1}A \to M; but this results from the reasoning of loc. cit., p. 123, Prop. 5.

Let k be a field and A a local k-algebra essentially of finite type. Put n = \dim(A) + \deg.\mathrm{tr}_k(\kappa_A). Let B be a k-algebra of finite type and q a prime ideal of B such that the k-algebra A is isomorphic to $B_q$ (cf. No. 1, Example 2); one has n = \dim_q(B) (VIII, § 2, No. 4, Cor. 5 of Th. 3). By loc. cit., Th. 3, c) and Cor. 5, one also has

where \mathfrak{p} runs through the (finite) family of minimal prime ideals of A. If A is an integral domain, n is therefore the transcendence degree of the field of fractions of A.

a) One has [\kappaA \otimes_A \Omega_k(A) : \kappa_A] \geq n.*

b) The following conditions are equivalent: (i) the k-algebra A is absolutely regular; (ii) the A-module \Omegak(A) is free of rank n;* (iii) one has [\kappaA \otimes_A \Omega_k(A) : \kappa_A] = n.*

Consider the assertion (iii') one has [\kappaA \otimes_A \Omega_k(A) : \kappa_A] \leq n;* it is clear that (ii) implies (iii) and that (iii) implies (iii'). To prove the theorem it is enough to prove the implications (i) \Rightarrow (ii) and (iii') \Rightarrow (i).

Choose a k-algebra of finite type B and a prime ideal q of B such that A is isomorphic to $B_q$ (No. 1, Example 2). One has \dim_q(B) = n; replacing B by a ring $B_f$ , with $f \in B - q$, one may impose that the spectrum of B is connected and of dimension n, which we shall henceforth suppose.

(i) \Rightarrow (ii): Suppose the k-algebra A absolutely regular. As in the proof of Prop. 8, let us denote by \mu : A \otimes_k A \to A and \nu : \kappa_A \otimes_k A \to \kappa_A the homomorphisms deduced from the multiplication of A; put I = \mathrm{Ker}(\mu) and n = \mathrm{Ker}(\nu). The ideal I is completely secant; the ideal n is maximal and the local ring (\kappa_A \otimes_k A)_n is regular (loc. cit.). By Theorem 1 of § 5, No. 2, the finitely generated A-module I/I^2, which is none other than \Omega_k(A), is projective, hence free. But by the remark of § 5, No. 6, the ideal n identifies with \kappa_A \otimes_A 1, so that n/n^2 is isomorphic to \kappa_A \otimes_A \Omega_k(A). One therefore has

Let us denote by B' the ring \kappa_A \otimes_k B, and by S the image in B' of B - q. The ring \kappa_A \otimes_k A identifies with $B' \otimes_B B_q$, that is to say with $S^{-1}B'$. There therefore exists a maximal ideal q' of B' not meeting S such that one has n = $S^{-1}q'$; the ring (\kappa_A \otimes_k A)_n is then isomorphic to $B'_{q'}$ , and one has

Since n contains \kappa_A \otimes_k qB_q , q' contains the ideal \kappa_A \otimes q of B' , hence its inverse image in B contains q ; it is equal to q since q' does not meet S . By VIII, § 2, No. 4, cor. to prop. 5, one has

which proves (ii).

(iii') \Rightarrow (i) : Suppose that one has [\kappa_A \otimes_A \Omega_k(A) : \kappa_A] \leq n , that is to say [\kappa(q) \otimes_B \Omega_k(B) : \kappa(q)] \leq n , and let us prove that the k-algebra B is absolutely regular. Let (x_1, \ldots , x_n) be a sequence of elements of B such that 1 \otimes dx_1, \ldots , 1 \otimes dx_n generate the \kappa(q)-vector space \kappa(q) \otimes_B \Omega_k(B) . Replacing B by B_f , for a suitable element f of B - q , one may suppose that dx_1, \ldots , dx_n generate the B-module \Omega_k(B) (Nakayama lemma and II, § 5, No. 1, prop. 2). Let \bar{k} be an algebraically closed extension of k . It is enough to prove that the \bar{k}-algebra B_{(\bar{k})} is regular, since this will imply that B is absolutely regular (cor. 4 to prop. 7 of No. 4). For every canonical factor C of B_{(k)} , the differentials d(1 \otimes x_i) generate the C-module \Omega_{\bar{k}}(C) (A, III, § 10, No. 12, prop. 20). The theorem therefore follows from the two following lemmas:

Let us first treat the case where the extension K is of finite degree over k . The B-module C is a direct factor of a free module, hence finitely generated projective. Since the function p \mapsto \mathrm{rg}_p C is constant on Spec(B) (II, § 5, No. 3), the support of the B-module C is equal to Spec(B) , so that the B-module C is faithfully flat. This implies \dim(C) \geq \dim(B) (VIII, § 2, No. 1, prop. 2), hence the required equality since one has \dim(C) \leq \dim(B_{(K)}) = \dim(B) (VIII, § 2, No. 4, cor. to prop. 5).

Let us pass to the general case. Let e be the idempotent element of B_{(K)} such that C = B_{(K)}/B_{(K)}e . There exists a subextension K' of K of finite degree such that e is the image of an idempotent element e' of B_{(K')} . Put C' = B_{(K')}/B_{(K')}e' . The ring C' \otimes_{K'} K is identified with C , and C' is a canonical factor of B_{(K')} . One has \dim(C') = \dim(B) by the case already treated, and \dim(C') = \dim(C) by loc. cit. , whence the lemma.

Let m be a maximal ideal of A such that \dim(A_m) = n . One has [A/m : k] = 1 (V, § 3, No. 3, prop. 1 (iii)), hence A = m \oplus k1_A . Let p and q denote the corresponding projections. For $a$ and $b$ in $A$, one has

whence $p(ab) \equiv p(a)q(b) + q(a)p(b) \pmod{\mathfrak{m}^2}$. Consequently the mapping $\delta : A \to \mathfrak{m}/\mathfrak{m}^2$ which associates to each element $x$ of $A$ the class of $p(x)$ modulo $\mathfrak{m}^2$ is a $k$-derivation of $A$ into the $k$-vector space $\mathfrak{m}/\mathfrak{m}^2$. There therefore exists a $A$-linear mapping $\phi : \Omega_k(A) \to \mathfrak{m}/\mathfrak{m}^2$ such that $\delta(x) = \phi(dx)$ for every $x \in A$. Since $\delta$ is surjective, the $\phi(dx_i)$ generate the $A/\mathfrak{m}$-vector space $\mathfrak{m}/\mathfrak{m}^2$, and one has $[\mathfrak{m}/\mathfrak{m}^2 : A/\mathfrak{m}] \leq n = \dim(A_{\mathfrak{m}})$. It follows that $A_{\mathfrak{m}}$ is regular and that the images of the $dx_i$ form a basis of the $A/\mathfrak{m}$-vector space $A/\mathfrak{m} \otimes_A \Omega_k(A)$.

Let us now prove that the ring $A$ is an integral domain and regular. There exists a minimal prime ideal $\mathfrak{q}$ of $A$ such that $\dim(A/\mathfrak{q}) = n$. For every maximal ideal $\mathfrak{m}$ of $A$ containing $\mathfrak{q}$, one has $\dim(A_{\mathfrak{m}}) = n$ (VIII, § 2, No. 4, cor. 2 to th. 3), so that $A_{\mathfrak{m}}$ is regular by what we have just seen. In particular $A_{\mathfrak{m}}$ is an integral domain, which implies $\mathfrak{q}A_{\mathfrak{m}} = 0$. Since this holds for all the maximal ideals $\mathfrak{m}$ of $V(\mathfrak{q})$, one deduces that one has $\operatorname{Supp}(\mathfrak{q}) \cap V(\mathfrak{q}) = \varnothing$. But $\operatorname{Spec}(A)$ is connected, $V(\mathfrak{q})$ is nonempty and one has $\operatorname{Supp}(\mathfrak{q}) \cup V(\mathfrak{q}) = \operatorname{Spec}(A)$ (II, § 4, No. 4, prop. 16). One deduces that $\operatorname{Supp}(\mathfrak{q}) = \varnothing$, whence $\mathfrak{q} = 0$, which means that $A$ is an integral domain. One then has $\dim(A_{\mathfrak{m}}) = n$ for every maximal ideal $\mathfrak{m}$ of $A$; applying the first part of the proof, one deduces that $A$ is regular.

Finally, let $\sum_{i=1}^n a_i dx_i = 0$ be a linear relation between the $dx_i$ with coefficients in $A$. If the $a_i$ are not all zero, there exists an index $i$ and a maximal ideal $\mathfrak{m}$ of $A$ such that $a_i$ does not belong to $\mathfrak{m}$ (V, § 3, No. 3, Prop. 1, (iii) and (iv)); but this contradicts the fact proved above that the classes of $dx_i$ in $(A/\mathfrak{m}) \otimes_A \Omega_k(A)$ are linearly independent.

We may suppose that the $k$-algebra $A$ is of finite type. The set considered is then formed of the prime ideals $\mathfrak{p}$ such that $[\kappa(\mathfrak{p}) \otimes_k \Omega_k(A) : \kappa(\mathfrak{p})] \leq \dim_{\mathfrak{p}}(A)$. Now the mapping $\mathfrak{p} \mapsto \dim_{\mathfrak{p}}(A)$ is lower semi-continuous by definition, and the mapping $\mathfrak{p} \mapsto [\kappa(\mathfrak{p}) \otimes_k \Omega_k(A) : \kappa(\mathfrak{p})]$ is upper semi-continuous (Nakayama's lemma and II, § 5, No. 1, Prop. 2).

We shall see later (§ 7, No. 9, Cor. 4 of Th. 3) that under the hypotheses of Cor. 1, the set of prime ideals $\mathfrak{p}$ of $A$ such that the ring $A_{\mathfrak{p}}$ is regular is open in $\operatorname{Spec}(A)$.

Conversely, suppose that the $A$-module $\Omega_k(A)$ is projective and that the $k$-algebra $A_q$ is separable for every minimal prime ideal $q$ of $A$. Let $p$ be a prime ideal of $A$, and $q$ a minimal prime ideal of $A$ contained in $p$. Since the $A_p$-module $\Omega_k(A)_p$ is free (II, § 3, No. 2, Cor. 2 of Prop. 5), we have

The $k$-algebra $A_q$ is Artinian and separable, hence absolutely regular (No. 4, Example 2). Th. 1 implies

It then follows from Th. 1 that the $k$-algebra $A_p$ is absolutely regular, whence the corollary.

By Corollary 2, it is enough to prove that an artinian local $k$-algebra $A$ such that $\Omega_k(A)$ is a free $A$-module is a field. Now, by IX, § 3, No. 3, Theorem 1, there exists a subfield $K$ of $A$ such that $A = K \oplus m_A$. Let $\delta : A \to m_A/m_A^2$ be the composite mapping of the projection of $A$ onto $m_A$ and the canonical mapping $m_A \to m_A/m_A^2$. One verifies as in the proof of Lemma 4 that $\delta$ is a $k$-derivation. But every $k$-derivation of $A$ into an $A$-module annihilates $m_A$: let in fact $x \in m_A$, and $n$ an integer $\geq 1$ such that one has $x^{n-1} \neq 0,\ x^n = 0$. This implies in the $A$-module $\Omega_k(A)$ the relation $nx^{n-1} dx = 0$, hence $dx = 0$ since $nx^{n-1} \neq 0$. Thus $\delta(m_A) = 0$, hence $m_A = m_A^2$, and finally $m_A = \{0\}$.