Bourbaki

Exercise 13, § 1

Commutative Algebra, X, p. 152

Bourbaki marks this one as harder than the rest.

Let AA be a ring, JJ an ideal of AA, MM an AA-module. For every integer n≥0n \geq 0, denote by profn(J;M)\mathrm{prof}_n(J;M) the upper bound (in N‾\overline{\mathbf{N}}) of the lengths of the M[X1,…,Xn]M[X_1,\ldots,X_n]-regular sequences of elements of JA[X1,…,Xn]JA[X_1,\ldots,X_n]. a) Now prove that the sequence (profn(J;M))n≥0(\mathrm{prof}_n(J;M))_{n \geq 0} is increasing; denote by prof∞(J;M)\mathrm{prof}_\infty(J;M) its limit (in N‾\overline{\mathbf{N}}). Prove that one has prof∞(J;M)≤profA(J;M)\mathrm{prof}_\infty(J;M) \leq \mathrm{prof}_A(J;M).

EXERCISES

b) When the ideal JJ is of finite type, prove the equality prof∞(J;M)=profA(J;M)\mathrm{prof}_{\infty}(J; M) = \mathrm{prof}_A(J; M) (reason as in the proof of th. 2 of no. 4, using exerc. 12).

c) In the general case, show that prof∞(J;M)\mathrm{prof}_{\infty}(J; M) is the upper bound of the numbers profA(J′;M)\mathrm{prof}_A(J'; M), where J′J' runs through the set of ideals of finite type contained in JJ.

d) Let J′J' be an ideal of AA such that V(J′)=V(J)V(J') = V(J). Prove that one has prof∞(J′;M)=prof∞(J;M)\mathrm{prof}_{\infty}(J'; M) = \mathrm{prof}_{\infty}(J; M).

e) Let AA be a valuation ring of height 1, non noetherian, mm its maximal ideal. Prove that one has prof∞(m;A)=1\mathrm{prof}_{\infty}(m; A) = 1 and profA(m;A)⩾2\mathrm{prof}_A(m; A) \geqslant 2 (cf. exerc. 2).

Solution

No solution has been written yet.

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