Bourbaki

Exercise 15, § 3

Algebra, X, p. 182

Let AA be a noetherian local ring, mm its maximal ideal, k=A/mk = A/m. Let PP be a minimal projective resolution of the a-module kk; for n⩾0n \geqslant 0, denote by bnb_n the rank of the free a-module PnP_n. (a) Show that one has b0=1b_0 = 1 and that b1=dim⁡k(m/m2)b_1 = \dim_k(m/m^2) is the minimal number of generators of mm. (b) If AA is commutative, one has b2⩾12b1(b1−1)b_2 \geqslant \frac{1}{2} b_1(b_1 - 1) (consider the exact sequence P2→mP1→m2→0P_2 \to mP_1 \to m^2 \to 0). (c) Suppose that the ring AA is artinian (not necessarily commutative). Prove the inequality b2⩾14b12b_2 \geqslant \frac{1}{4} b_1^2 (let MtM_t be the submodule of P2P_2 formed by the elements x such that d2(x)∈mtP1d_2(x) \in m^t P_1; show that one has

mtP2⊂Mt+1 m^t P_2 \subset M_{t+1}
and that there exists an exact sequence 0→Mt/Mt+1→mtP1/mt+1P1→mt+1/mt+2→00 \to M_t/M_{t+1} \to m^t P_1/m^{t+1} P_1 \to m^{t+1}/m^{t+2} \to 0. Deduce that the polynomial p(t)=∑l(dim⁡k(ml/ml+1))tlp(t) = \sum_l (\dim_k(m^l/m^{l+1})) t^l satisfies (b2t2−b1t+1)p(t)⩾0(b_2 t^2 - b_1 t + 1) p(t) \geqslant 0 for 0⩽t⩽10 \leqslant t \leqslant 1. Conclude by examining separately the cases b1=1,b1⩾2b_1 = 1, b_1 \geqslant 2.

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.