证明
Theorem 4.1 ([The fixed-\(K\) missing-vertex implication]). Fix the ring data…Corollary 8 ([Ring-class lower bound]). Every simple-zero ring instance in the…观点
Flat modules over an Artin local ring are free, including modules of infinite rank. Here is a useful version of the argument. Lift a basis of M/ mM to a flat module M , and map the corresponding free module to M . Its cokernel is zero because m is nilpotent. Flatness shows that the kernel has zero reduction, and nilpotence again makes the kernel zero. Split the residual complex as its cohomology, with zero differential, plus contractible two-term disks. Lift all the graded bases. These lifts are graded isomorphisms: an inverse modulo m lifts to an inverse by a finite geometric series. Let d_0 be the lifted disk differential and write d=d_0+ε . All coefficients of ε belong to m . For the disk contraction (i_0,p_0,h_0) , arrange h_0^2=h_0i_0=p_0h_0=0 and 1-i_0p_0=d_0h_0+h_0d_0 . The operators 1+h_0ε and 1+εh_0 have finite geometric inverses. The elementary perturbation identities give the differential d_C=p_0ε(1+h_0ε)^ -1 i_0 on the lifted residual cohomology, and give i=(1+h_0ε)^ -1 i_0, p=p_0(1+εh_0)^ -1 , h=h_0(1+εh_0)^ -1 . Substitution, using (d_0+ε)^2=0 , proves the asserted contraction identities. These calculations are valid for infinite disk sums: each operator is a module homomorphism and every geometric series terminates at the nilpotence exponent of m . For a map between original complexes use p'fi ; insert the contraction homotopies when composing maps. A quasi-isomorphism between bounded free complexes has a contractible cone and hence a homotopy inverse. This gives the stated finite-diagram assertion, including the homotopies expressing triangles and perfect duality.