Bourbaki

Exercise 7, § 2

Commutative Algebra, X

Bourbaki marks this one as harder than the rest.

Let AA be a noetherian local ring, MM a finitely generated a-module. We say that a sequence (x1,…,xn)(x_1, \ldots, x_n) is weakly regular for MM if, for i=1,…,ni = 1, \ldots, n, the kernel of the homothety of ratio xi+1x_{i+1} in M/(x1M+…+xiM)M/(x_1 M + \ldots + x_i M) is annihilated by mAm_A.

a) If n⩽dim⁡(M)n \leqslant \dim(M), prove that a weakly regular sequence (x1,…,xn)(x_1, \ldots, x_n) for MM is sécant for MM (reason by induction on nn). Give an example of a weakly regular non-sécant sequence (one may take A=M=k[X]/(X2)A = M = k[X]/(X^2), where kk is a field).

b) We say that MM is a Buchsbaum module if every sécant sequence for MM is weakly regular for MM. A Cohen-Macaulay module is a Buchsbaum module; if MM is a Buchsbaum module, the ApA_p-module MpM_p is Cohen-Macaulay for every prime ideal p≠mAp \neq m_A.

c) Let MM be a Buchsbaum module, (x1,…,xn)(x_1, \ldots, x_n) a sécant sequence for MM; prove that M/(x1M+…+xnM)M/(x_1 M + \ldots + x_n M) is a Buchsbaum module.

d) For MM to be a Buchsbaum module, it is necessary and sufficient that, for every maximal sécant sequence (x1,…,xn)(x_1, \ldots, x_n) for MM, one has in Mn=M/(x1M+…+xn−1M)M_n = M/(x_1 M + \ldots + x_{n-1} M) the equality Ker(xn)Mn=Ker(xn2)Mn\mathrm{Ker}(x_n)_{M_n} = \mathrm{Ker}(x_n^2)_{M_n} (first treat the case n=1n = 1, observing that Ker(x1)M\mathrm{Ker}(x_1)_M is then of finite length and reasoning by induction on this length. In the general case, complete every sécant sequence (x1,…,xi)(x_1, \ldots, x_i) to a maximal sécant sequence (x1,…,xn)(x_1, \ldots, x_n), and consider the sequences (x1,…,xi−1,xi+1k,…,xnk,xi)(x_1, \ldots, x_{i-1}, x_{i+1}^k, \ldots, x_n^k, x_i) for k⩾1k \geqslant 1.

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.