Bourbaki

Exercise 6, § 2

Commutative Algebra, X

The hypotheses of the preceding exercise are retained; for every finitely generated A-module N such that N/xNN/xN is of finite length, we denote by χ(x,N)\chi(x, N) the integer χ(K∙(x,N))\chi(K_\bullet(x, N)).

a) For every submodule M′M' of MM, one has χ(x,M)=χ(x,M′)+χ(x,M/M′)\chi(x, M) = \chi(x, M') + \chi(x, M/M').

b) If x1M=0x_1 M = 0, prove that χ(x,M)\chi(x, M) is zero.

c) We denote by x′x' the sequence (x2,…,xn)(x_2, \ldots, x_n); prove the equality

χ(x,M)=χ(x′,M/x1M)−χ(x′,Ker(x1)M) \chi(x, M) = \chi(x', M/x_1 M) - \chi(x', \mathrm{Ker}(x_1)_M)

(use exercise 8 of A, X, p. 207).

d) For 1⩽i⩽n1 \leqslant i \leqslant n, we denote by KiK_i the kernel of the homothety of ratio xix_i in M/(x1M+…+xi−1M)M/(x_1 M + \ldots + x_{i-1} M). Prove the equality

lg⁡(M/xM)−χ(x,M)=∑i=1nχ((xi+1,…,xn),Ki). \lg(M/x M) - \chi(x, M) = \sum_{i=1}^n \chi((x_{i+1}, \ldots, x_n), K_i).

e) Let p1,…,pnp_1, \ldots, p_n be integers >1> 1, xPx^P the sequence (x1p1,…,xnpn)(x_1^{p_1}, \ldots, x_n^{p_n}). Prove the equality χ(xP,M)=p1…pnχ(x,M)\chi(x^P, M) = p_1 \ldots p_n \chi(x, M) (treat first the case n=1n = 1, then reduce to it with the aid of Cor. 2 of A, X, p. 157).

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.