Bourbaki

Theorem 4

Tag 0D9Q, permanent. Cite it as [Bourbaki, Tag 0D9Q].

Commutative Algebra, chapter X, § 7. Algèbres lisses

Let A be a noetherian ring and B an A-algebra essentially of finite type. The following conditions are equivalent:

(i) the A-algebra B is formally smooth ; (ii) for every q∈Spec(B)q \in \mathrm{Spec}(B), the A-algebra BqB_q is formally smooth (resp. formally smooth for the qBqqB_q-adic topology) ; (iii) the A-module B is flat and, for every p∈Spec(A)p \in \mathrm{Spec}(A), the κ(p)-algebra κ(p)⊗AB\kappa(p)\text{-algebra} \ \kappa(p) \otimes_A B is absolutely regular ; (iv) the A-module B is flat and, for every regular A-algebra R, the ring R⊗ABR \otimes_A B is regular ; (v) the A-module B is flat and the kernel of the homomorphism μ:B⊗AB→B\mu : B \otimes_A B \to B such that μ(x⊗y)=xy\mu(x \otimes y) = xy is a completely secant ideal.

The equivalence of (i) and (ii) follows from cor. 2 of th. 3 (No. 9).

(i)⇒(v) : suppose the A-algebra B formally smooth. Let q be a prime ideal of B, and p its inverse image in A. The ApA_p-algebra BqB_q is formally smooth (prop. 4, a) of No. 2), hence flat (lemma 5) ; consequently the A-module B is flat (II, § 3, No. 4, prop. 15). On the other hand, the ring B⊗ABB \otimes_A B is noetherian (§ 6, No. 1, cor. of prop. 2), hence the ideal Ker μ\mathrm{Ker}\,\mu is completely secant by cor. 3 of th. 2 (No. 5).

(v)⇒(iii) : suppose condition (v) satisfied. Put I=Ker(μ)I = \mathrm{Ker}(\mu). Let p∈Spec(A)p \in \mathrm{Spec}(A). The mapping

1⊗μ:κ(p)⊗A(B⊗AB)→κ(p)⊗AB 1 \otimes \mu : \kappa(p) \otimes_A (B \otimes_A B) \to \kappa(p) \otimes_A B

is identified with the mapping

μp:(κ(p)⊗AB)⊗κ(p)(κ(p)⊗AB)→κ(p)⊗AB \mu_p : (\kappa(p) \otimes_A B) \otimes_{\kappa(p)} (\kappa(p) \otimes_A B) \to \kappa(p) \otimes_A B

deduced from the multiplication of the κ(p)-algebra κ(p)⊗AB\kappa(p)\text{-algebra} \ \kappa(p) \otimes_A B. The ideal Ker(μp)\mathrm{Ker}(\mu_p) is identified with I(κ(p)⊗A(B⊗AB))I(\kappa(p) \otimes_A (B \otimes_A B)). It is completely secant since the A-module B is flat (§ 5, No. 6, prop. 6). Assertion (iii) then follows from prop. 8 of § 6, No. 5.

(iii)⇒(ii) : let q be a prime ideal of B, and p its inverse image in A. Under the hypotheses of (iii), the ApA_p-module BqB_q is flat, and the κ(p)-algebra κ(p)⊗ApBq\kappa(p)\text{-algebra} \ \kappa(p) \otimes_{A_p} B_q, which is identified with a ring of fractions of κ(p)⊗AB\kappa(p) \otimes_A B, is absolutely regular (§ 6, No. 4, prop. 6). It follows from lemma 5 that BqB_q is formally smooth over ApA_p, hence over A (No. 2, prop. 3 and 4).

(iii)⇒(iv) : let us place ourselves under the hypotheses of (iii). Let R be a regular A-algebra. The R-module R⊗ABR \otimes_A B is flat (I, § 2, No. 7, cor. 2 to prop. 8). Let r be a prime ideal of R and p its inverse image in A ; the ring κ(r)⊗R(R⊗AB)\kappa(r) \otimes_R (R \otimes_A B), which is identified with κ(r)⊗κ(p)(κ(p)⊗AB)\kappa(r) \otimes_{\kappa(p)} (\kappa(p) \otimes_A B), is regular (§ 6, No. 4, cor. 2 of prop. 7). The ring R⊗ABR \otimes_A B is therefore regular (§ 4, No. 5, cor. of prop. 9).

(iv) ⇒ (iii) : let pp be a prime ideal of AA and let kk be an extension of κ(p)\kappa(p); under the hypotheses of (iv), the ring k⊗κ(p)(κ(p)⊗AB)k \otimes_{\kappa(p)} (\kappa(p) \otimes_A B), which is identified with k⊗ABk \otimes_A B, is regular, whence (iii).

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

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