Bourbaki

Exercise 20, § 7

Commutative Algebra, VIII, p. 106

Let AA be a noetherian ring, MM a finitely generated AA-module, and q1,...,qkq_1, ..., q_k ideals of AA such that M/qi⋅MM/q_i \cdot M is of finite length for every i. Suppose that M≠0M \neq 0 and put d=dim⁡A(M)d = \dim_A(M). (a) Show that M/q1n1...qknk⋅MM/q_1^{n_1} ... q_k^{n_k} \cdot M is of finite length for every k-tuple of positive integers (n1,...,nk)(n_1, ..., n_k). (b) Consider the graded ring H of type Nk\mathbf{N}^k such that

Hn1,...,nk=q1n1q2n2...qknk/q1n1+1q2n2...qknk, H_{n_1, ..., n_k} = q_1^{n_1} q_2^{n_2} ... q_k^{n_k}/q_1^{n_1+1} q_2^{n_2} ... q_k^{n_k} ,
and the graded H-module N such that
Nn1,...,nk=q1n1q2n2...qknk⋅M/q1n1+1q2n2...qknk⋅M; N_{n_1, ..., n_k} = q_1^{n_1} q_2^{n_2} ... q_k^{n_k} \cdot M/q_1^{n_1+1} q_2^{n_2} ... q_k^{n_k} \cdot M ;
show that N is a finitely generated graded H-module generated by N0,...,0N_{0,...,0} and a finite number of elements whose degrees are vectors of the canonical basis of Zk\mathbf{Z}^k. Deduce that there exists a polynomial Q(T)∈Z[T1,...,Tk]Q(T) \in \mathbf{Z}[T_1, ..., T_k] and integers s1⩾1,...,sk⩾1s_1 \geqslant 1, ..., s_k \geqslant 1 such that one has
∑n∈NrlongA(Hn)⋅Tn=Q(T)(1−T1)s1...(1−Tk)sk. \sum_{n \in \mathbf{N}^r} \mathrm{long}_A(H_n) \cdot T^n = \frac{Q(T)}{(1 - T_1)^{s_1} ... (1 - T_k)^{s_k}} .
(Proceed by induction on d.)

(c) Deduce, by extending to Z((T1,...,Tk))Z((T_1, ..., T_k)) the results of § 6, No. 1, the existence of integers c(α1,...,αk)c(\alpha_1, ..., \alpha_k) for the sequences (α1,...,αk)(\alpha_1, ..., \alpha_k) such that ∣α∣=∑i=1kαi=d|\alpha| = \sum_{i=1}^k \alpha_i = d, such that one has

eq1v1...qkvk(M)=∑∣α∣=dd!α1!...αk!c(α1,...,αk)v1α1...vkαk e_{q_1^{v_1} ... q_k^{v_k}}(M) = \sum_{|\alpha|=d} \frac{d!}{\alpha_1! ... \alpha_k!} c(\alpha_1, ..., \alpha_k) v_1^{\alpha_1} ... v_k^{\alpha_k}

for (v1,...,vk)(v_1, ..., v_k) in Nk\mathbf{N}^k.

(d) Suppose now that A is local, with infinite residue field κ\kappa. Show, using the existence of superficial elements for the graded H-module N, that one has

c(α1,...,αk)=inf⁡qα(eqα(M)) c(\alpha_1, ..., \alpha_k) = \inf_{q_\alpha} (e_{q_\alpha}(M))

where qαq_\alpha runs through the set of ideals of A generated by α1\alpha_1 elements of q1q_1, α2\alpha_2 elements of q2q_2, ..., αk\alpha_k elements of qkq_k. One sometimes denotes the integer c(α1,...,αk)c(\alpha_1, ..., \alpha_k) by e(q1[α1]+⋯+qk[αk];M)e(q_1^{[\alpha_1]} + \cdots + q_k^{[\alpha_k]}; M), which is abbreviated to e(q1[α1]+⋯+qk[αk])e(q_1^{[\alpha_1]} + \cdots + q_k^{[\alpha_k]}) when M=AM = A.

(e) Prove the equality

eq1.q2(M)=∑i=1d(di)e(q1[i]+q2[d−i];M). e_{q_1.q_2}(M) = \sum_{i=1}^d \binom{d}{i} e(q_1^{[i]} + q_2^{[d-i]}; M).

(f) Show that one has :

e(q1‾[α1]+⋯+qk‾[αk];M)=e(q1[α1]+⋯+qk[αk];M) e(\overline{q_1}^{[\alpha_1]} + \cdots + \overline{q_k}^{[\alpha_k]}; M) = e(q_1^{[\alpha_1]} + \cdots + q_k^{[\alpha_k]}; M)

where qi‾\overline{q_i} denotes the integral closure in A of the ideal qiq_i (cf. exerc. 19).

Solution

No solution has been written yet.

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