Bourbaki

Lemma 1

Tag 0D99, permanent. Cite it as [Bourbaki, Tag 0D99].

Commutative Algebra, chapter X, § 7. Algèbres lisses

Suppose that the AA-algebra BB admits a generating family t=(ti)i∈It = (t_i)_{i \in I}, with defining relations fλ∈A[T]f_\lambda \in A[T] (λ∈A\lambda \in A). The BB-linear homomorphism

ψ:(B⊗AΩk(A))⊕B(I)⟶Ωk(B) \psi : (B \otimes_A \Omega_k(A)) \oplus B^{(I)} \longrightarrow \Omega_k(B)
defined by ψ(α,(bi))=Ω0(ρ)(α)+∑i∈Ibidti\psi(\alpha, (b_i)) = \Omega_0(\rho)(\alpha) + \sum_{i \in I} b_i dt_i, is surjective; its kernel is generated by the elements rλ=(dAfλ(t),(∂fλ∂Ti(t))i∈I)r_\lambda = \left( d^A f_\lambda(t), \left( \frac{\partial f_\lambda}{\partial T_i}(t) \right)_{i \in I} \right) for λ∈A\lambda \in A.

Consider the sequence of B-modules and B-linear mappings

B(A)→φ(B⊗AΩk(A))⊕B(I)→ψΩk(B)⟶0, B^{(A)} \xrightarrow{\varphi} (B \otimes_{A} \Omega_k(A)) \oplus B^{(I)} \xrightarrow{\psi} \Omega_k(B) \longrightarrow 0,

where φ\varphi is the homomorphism such that φ(eλ)=nλ\varphi(e_\lambda) = n_\lambda; it is a matter of proving that this sequence is exact. By A, II, p. 36, th. 1, it is enough to prove that, for every B-module M, the sequence

0→HomB(Ωk(B),M)→Hom(ψ,1)HomB((B⊗AΩk(A))⊕B(I),M)→Hom(φ,1)HomB(B(A),M) 0 \to \mathrm{Hom}_B(\Omega_k(B), M) \xrightarrow{\mathrm{Hom}(\psi, 1)} \mathrm{Hom}_B((B \otimes_A \Omega_k(A)) \oplus B^{(I)}, M) \xrightarrow{\mathrm{Hom}(\varphi, 1)} \mathrm{Hom}_B(B^{(A)}, M)

is exact. In view of the universal property of the module of differentials (A, III, p. 134), this sequence is identified with

0→Dk(B,M)→ψ′Dk(A,M)⊕M1→φ′MA 0 \to D_k(B, M) \xrightarrow{\psi'} D_k(A, M) \oplus M^1 \xrightarrow{\varphi'} M^A

where ψ′(D)=(D∘ρ,(D(ti)))\psi'(D) = (D \circ \rho, (D(t_i))) and φ′(Δ,(mi))=(fλΔ(t)+∑i∂fλ∂Ti(t)mi)λ∈A\varphi'(\Delta, (m_i)) = (f^\Delta_\lambda(t) + \sum_i \frac{\partial f_\lambda}{\partial T_i}(t) m_i)_{\lambda \in A} (in accordance with A, V, p. 121, for every polynomial f=∑α∈N(I)cαTαf = \sum_{\alpha \in \mathbf{N}^{(I)}} c_\alpha T^\alpha of A[T]A[T], fΔ(t)f^\Delta(t) denotes the element ∑αtαΔ(cα)\sum_\alpha t^\alpha \Delta(c_\alpha)). But the exactness of this sequence follows from loc. cit., prop. 1, in view of the fact that a derivation D:B→MD : B \to M is kk-linear if and only if this is so for D∘ρD \circ \rho.

Let A be a ring. There exists a unique structure of Z\mathbf{Z}-algebra on A; one writes simply Ω(A)\Omega(A) for the A-module ΩZ(A)\Omega_{\mathbf{Z}}(A). If ρ:k→A\rho : k \to A is a homomorphism of rings, there is a canonical exact sequence of A-modules (A, III, p. 136, prop. 21)

A⊗kΩ(k)→Ω(A)→Ωk(A)→0. A \otimes_k \Omega(k) \to \Omega(A) \to \Omega_k(A) \to 0.

Suppose that A contains a subfield, and let P be the prime subfield of A; then Ω(P)\Omega(P) is zero and the canonical homomorphism of A-modules Ω(A)→ΩP(A)\Omega(A) \to \Omega_P(A) is bijective. If moreover A is of characteristic p≠0p \neq 0 (which means by definition that pp is a prime number, that p1A=0p1_A = 0 and 1A≠01_A \neq 0), then P is identified with Fp\mathbf{F}_p. Moreover, every derivation of A vanishes on the subring ApA^p; for every subring kk of A contained in ApA^p (and, in particular, for every perfect subfield kk of A), the canonical mapping Ω(A)→Ωk(A)\Omega(A) \to \Omega_k(A) is bijective.

Let A be a ring of characteristic p≠0p \neq 0 and (fi)1≤i≤n(f_i)_{1 \leq i \leq n} a finite sequence of elements of A. Let AnA_n denote the quotient ring of the polynomial ring A[T1,…,Tn]A[T_1, \ldots, T_n] by the ideal generated by the polynomials Tip−fiT_i^p - f_i, for 1≤i≤n1 \leq i \leq n.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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