Bourbaki

Exercise 13, § 3

Commutative Algebra, X

Bourbaki marks this one as harder than the rest.

Let AA be a noetherian local ring. We suppose the following condition¹ is satisfied:

(GM) for every noetherian local AA-algebra BB, there exists a BB-module MM such that mBM≠Mm_B M \neq M and profB(M)=dim⁡(B)\mathrm{prof}_B(M) = \dim(B).

(a) Let PP be a complex of flat AA-modules, of finite length ℓ\ell, such that Supp(H(P))={mA}\mathrm{Supp}(H(P)) = \{ m_A \}. Prove that ℓ≥dim⁡(A)\ell \geq \dim(A) (use exerc. 6 of § 1)².

(b) Let BB be a noetherian ring, ρ:A→B\rho : A \to B a homomorphism, aρ:Spec(B)→Spec(A)^a\rho : \mathrm{Spec}(B) \to \mathrm{Spec}(A) the associated mapping, MM a finitely generated AA-module. Prove that the codimension of aρ−1(Supp(M))^a\rho^{-1}(\mathrm{Supp}(M)) in Spec(B)\mathrm{Spec}(B) is ≤dpA(M)\leq \mathrm{dp}_A(M) (apply a) to the complex L⊗ABqL \otimes_A B_q, where qq is the minimal prime ideal of a component of aρ−1(Supp(M))^a\rho^{-1}(\mathrm{Supp}(M)) and LL a free resolution of MM).

(c) Let M,NM, N be finitely generated AA-modules, with M≠0M \neq 0. Prove the inequality

dim⁡(N)≤dpA(M)+dim⁡(M⊗AN) \dim(N) \leq \mathrm{dp}_A(M) + \dim(M \otimes_A N)

(intersection theorem: reason by induction on the integer d=dim⁡(M⊗AN)d = \dim(M \otimes_A N). If d=0d = 0, apply b) to the ring B=A/Ann(M)B = A / \mathrm{Ann}(M); in the general case consider an element of AA which is secant for NN and for M⊗ANM \otimes_A N).

(d) For there to exist a finite length AA-module of finite projective dimension, it is necessary and sufficient that AA be a Macaulay ring.

Solution

No solution has been written yet.

Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.