Bourbaki

Exercise 5, § 3

Commutative Algebra, IX, p. 77

Let A be a noetherian local ring containing a field k, and let m be the maximal ideal of A. We suppose that A is a formally smooth k-algebra (the field k being discrete). (a) Let K be a field of representatives of the ring A/m2A/m^2 (exerc. 4, b)), and let x1,...,xdx_1, ..., x_d be elements of m whose images form a basis of the K-vector space m/m2m/m^2. Let K

X1,...,Xd X_1, ..., X_d
be the ring of polynomials in d variables X1,...,XdX_1, ..., X_d, n its ideal generated by X1,...,XdX_1, ..., X_d, and φ the homomorphism from K
X1,...,Xd X_1, ..., X_d
into A/m2A/m^2 which associates xix_i to XiX_i for 1≤i≤d1 \leq i \leq d and induces the identity on K. Then φ induces an isomorphism φ‾\overline{\varphi} of K
X1,...,Xd X_1, ..., X_d
/n2n^2 onto A/m2A/m^2. (b) For every integer n≥1n \geq 1, there exists a homomorphism ψn:A→K[X1,...,Xd]/nn+1\psi_n : A \to K[X_1, ..., X_d]/n^{n+1} which, by passing to the quotients, induces φ‾−1\overline{\varphi}^{-1}. Such a homomorphism ψn\psi_n is surjective. (c) For every integer n≥1n \geq 1, the length of the A-module A/mn+1A/m^{n+1} is at least (d+nd)\binom{d+n}{d} and one has dim⁡(A)≥d\dim(A) \geq d. (d) For every extension k′k' of finite degree of k, and every maximal ideal m′m' of the semi-local ring A⊗kk′A \otimes_k k', the local ring (A⊗kk′)m′(A \otimes_k k')_{m'} is regular.

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.