Proposition 8

Let k be a field and A a k-algebra essentially of finite type. Let I denote the kernel of the homomorphism of k-algebras $\mu : A \otimes_k A \to A$ such that $\mu(x \otimes y) = xy$ for x and y in A. The following conditions are equivalent:

(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.