Bourbaki

Exercise 9, § 3

Commutative Algebra, X

Bourbaki marks this one as harder than the rest.

Let AA be a ring, (L,d)(L, d) a bounded complex of free AA-modules of finite type, zero in degrees < 0. We propose to prove that the following conditions are equivalent: (i) One has Hi(L)=0H_i(L) = 0 for i>0i > 0 ; (ii) One has \operatorname{prof}_A(\partial(d_i); A) \geq i and rg(di)+rg(di+1)=rgA(Li)rg(d_i) + rg(d_{i+1}) = rg_A(L_i) for all i≥1i \geq 1. a) Show that it suffices to prove this statement when the ring AA is noetherian (consider the subring of AA generated by the coefficients of the matrices did_i for i≥1i \geq 1 ). We henceforth suppose this condition satisfied. b) If AA is a local ring of dimension 0, prove that (ii) implies (i) (use exerc. 8, b)).

c) Suppose that the ring AA is local, that condition (ii) is satisfied, and that Supp(Hi(L))={mA}\mathrm{Supp}(H_i(L)) = \{ m_A \} for i≥1i \geq 1. Prove that Hi(L)=0H_i(L) = 0 for i≥0i \geq 0 (let pp be the largest integer such that d(dp)≠A\mathfrak{d}(d_p) \neq A; deduce from exerc. 8, b) that Hi(Li)=0H_i(L_i) = 0 for i>pi > p, then from exerc. 8, a) and cor. 2 of prop. 3 (No. 2) that the complex 0→Lp/Bp(L)→Lp−1→…→L00 \to L_p/B_p(L) \to L_{p-1} \to \ldots \to L_0 is acyclic in degrees >0> 0.

d) Prove the implication (ii) ⇒\Rightarrow (i) (reduce to the case where AA is local, and reason by induction on dim⁡(A)\dim(A)).

e) Suppose that the complex LL is acyclic in degrees >0> 0. Let pp be a prime ideal associated with AA. Show that for every i≥1i \geq 1 (Coker di\mathrm{Coker}\ d_i)p_p is a free ApA_p-module, of rank ∑p≥i−1(−1)prgA(Lp)\sum_{p \geq i-1} (-1)^p \mathrm{rg}_A(L_p); moreover one has rg((di)p)=∑p≥i(−1)p−irgA(Lp)\mathrm{rg}((d_i)_p) = \sum_{p \geq i} (-1)^{p-i} \mathrm{rg}_A(L_p) and d((di)p)=Ap\mathfrak{d}((d_i)_p) = A_p (exerc. 8).

f) Deduce that one has rg(di)=∑p≥i(−1)p−irgA(Lp)\mathrm{rg}(d_i) = \sum_{p \geq i} (-1)^{p-i} \mathrm{rg}_A(L_p) and consequently rg(di)+rg(di+1)=rgA(Li)\mathrm{rg}(d_i) + \mathrm{rg}(d_{i+1}) = \mathrm{rg}_A(L_i) for all i≥1i \geq 1.

g) Let ii be an integer ≥1\geq 1, and pp a prime ideal of AA. Show that the condition dpAp(H0(Lp))≥i\mathrm{dp}_{A_p}(H_0(L_p)) \geq i is equivalent to d(di)Ap≠Ap\mathfrak{d}(d_i)A_p \neq A_p, that is to say d(di)⊂p\mathfrak{d}(d_i) \subset p (use exerc. 8, a)). With the aid of prop. 8 of § 1, No. 5 and th. 1, deduce that one has profA(d(di);A)≥i\mathrm{prof}_A(\mathfrak{d}(d_i); A) \geq i, which proves the implication (i) ⇒\Rightarrow (ii).

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.