Bourbaki

Exercise 11, § 5

Commutative Algebra, X, p. 168

Bourbaki marks this one as harder than the rest.

We retain the notations of the preceding exercise; we suppose that the rings AA and BB are local and complete.

a) Now prove that BB is isomorphic to the quotient of an algebra of formal series S=A[[X1,…,Xn]]S = A[[X_1, \ldots, X_n]] by an ideal aa contained in (X1,…,Xn)(X_1, \ldots, X_n); one can take n=dim⁡κA(n/n2)n = \dim_{\kappa_A}(\mathfrak{n}/\mathfrak{n}^2), where n\mathfrak{n} is the image of mBm_B in B/mABB/m_A B.

b) Prove that the following conditions are equivalent: (i) BB is a finite and flat AA-algebra and a complete intersection ring ; (ii) the ideal aa is generated by a sequence (P1,…,Pn)(P_1, \ldots, P_n) completely transverse for κA⊗AS\kappa_A \otimes_A S. (cf. exerc. 5.)

c) If these conditions are satisfied, prove that one has η(B)=γ(B)=v(ε(det⁡(∂Pi∂Xj)))\eta(B) = \gamma(B) = v(\varepsilon(\det(\frac{\partial P_i}{\partial X_j}))) (use exerc. 9 and 10, d)).

d) Now prove that one can find a sequence (f1,…,fn)(f_1, \ldots, f_n) of elements of aa completely transverse for κA⊗AS\kappa_A \otimes_A S such that the AA-algebra B′=S/(f1,…,fn)B' = S/(f_1, \ldots, f_n) satisfies γ(B′)=γ(B)\gamma(B') = \gamma(B).

(Deduce from the exact sequence a/a2→ΩA(S)⊗SB→ΩA(B)→0a/a^2 \to \Omega_A(S) \otimes_S B \to \Omega_A(B) \to 0 (A, III, p. 137) an exact sequence of AA-modules a/a2⊗SA→ΩA(S)⊗SA→I/I2→0a/a^2 \otimes_S A \to \Omega_A(S) \otimes_S A \to I/I^2 \to 0 ; take elements f1,…,fnf_1, \ldots, f_n of aa whose images in ΩA(S)⊗SA\Omega_A(S) \otimes_S A form a basis over AA of the image of a/a2⊗SAa/a^2 \otimes_S A.)

e) Deduce using exerc. 10, e) that one has γ(B)≥η(B)\gamma(B) \geq \eta(B) and that equality holds if and only if BB is a complete intersection ring.

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.