Bourbaki

Exercise 17, § 2

Commutative Algebra, VIII, p. 86

Let AA be an integral domain. Put nk(A)=dim⁡(A[X1,...,Xk])n_k(A) = \dim(A[X_1, ..., X_k]) for every integer k⩾0k \geqslant 0. Let D\mathfrak{D} be the set of sequences of integers (ni)i⩾0(n_i)_{i \geqslant 0} such that there exists an integral domain AA with ni=ni(A)n_i = n_i(A) for every integer i⩾0i \geqslant 0. a) Prove the inequalities

sup⁡(n0(A)+i+1,ni(A)+1)⩽ni+1(A)⩽inf⁡(2ni(A)+1,2i+1(n0(A)+1)−1). \sup(n_0(A) + i + 1, n_i(A) + 1) \leqslant n_{i+1}(A) \leqslant \inf(2n_i(A) + 1, 2^{i+1}(n_0(A) + 1) - 1) .

b) Now prove that the only sequence of integers beginning with 1, 2 and belonging to D\mathfrak{D} is the sequence ni=i+1n_i = i + 1. (Let AA be such that n0(A)=1n_0(A) = 1, n1(A)=2n_1(A) = 2. By Exerc. 6, p. 84, the integral closure of AA is prüferian. Use Exerc. 6, cc) , p. 84 to conclude.) c) Prove the inequalities

n0(A)+i⩽ni(A)⩽(n0(A)+1)(i+1)−1. n_0(A) + i \leqslant n_i(A) \leqslant (n_0(A) + 1)(i + 1) - 1 .

Deduce that if n0(A)=0n_0(A) = 0, then ni(A)=in_i(A) = i for every i∈Ni \in \mathbf{N}. d) Prove that the sequence of differences ni+1(A)−ni(A)n_{i+1}(A) - n_i(A) is stationary. e) We use the notation of Exerc. 7, p. 84. Let δ\delta be the transcendence degree of L′L' over LL. Then: (i) If δ=0\delta = 0, we have ni(A)=ni(B)+1n_i(A) = n_i(B) + 1 for every ii. (ii) If 1⩽δ<∞1 \leqslant \delta < \infty, we have ni(A)=ni(B)+i+1n_i(A) = n_i(B) + i + 1 for 0⩽i⩽δ0 \leqslant i \leqslant \delta and ni(A)=ni(B)+δ+1n_i(A) = n_i(B) + \delta + 1 for i⩾δi \geqslant \delta. (iii) If δ=∞\delta = \infty, we have ni(A)=ni(B)+i+1n_i(A) = n_i(B) + i + 1 for every ii. (For a characterization of the elements of D\mathfrak{D}, one may consult T. Parker, Amer. J. Math., 97 (1975), p. 308-311.)

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Machine translation of the English, by gpt-5-6. Not checked by a person.