8 1. THE KRULL-REMAK-SCHMIDT THEOREM 1.11. Lemma. Every pair (M, N) of modules over a local ring (R, m) has a lifting number. Proof. Choose exact sequences F1 α −→ F0 −→ M −→ 0 , G1 β −→ G0 −→ N −→ 0 , where Fi and Gi are finite-rank free R-modules. Define an R-homo- morphism Φ: HomR(F0,G0) × HomR(F1,G1) −→ HomR(F1,G0) by Φ(μ, ν) = μα−βν. Applying the Artin-Rees Lemma to the submodule im(Φ) of HomR(F1,G0), we get a positive integer e such that (1.2) im(Φ) me+f HomR(F1,G0) mf im(Φ) for each f 0 . Suppose now that f 0 and ξ : M/me+fM −→ N/me+fN is an R-homomorphism. We can lift ξ to homomorphisms μ0 and ν0 making the following diagram commute. (1.3) F1/me+fF1 α ν0 F0/me+fF0 μ0 M/me+fM ξ 0 G1/me+fG 1 β G0/me+fG 0 N/me+fN 0 Now lift μ0 and ν0 to homomorphisms μ0 HomR(F0,G0) and ν0 HomR(F1,G1). The commutativity of (1.3) implies that the im- age of Φ(μ0,ν0): F1 −→ G0 lies in me+fG0. Choosing bases for F1 and G0, we see that the matrix representing Φ(μ0,ν0) has entries in me+f, so that Φ(μ0,ν0) me+f HomR(F1,G0). By (1.2), Φ(μ0,ν0) mf im(Φ) = Φ(mf(HomR(F0,G0) × HomR(F1,G1))). Choose a pair of maps (μ1,ν1) mf(HomR(F0,G0) × HomR(F1,G1)) with Φ(μ1,ν1) = Φ(μ0,ν0), and set (μ, ν) = (μ0,ν0) (μ1,ν1). Then Φ(μ, ν) = 0, so μ induces an R-homomorphism σ : M −→ N. Since μ and μ0 agree modulo mf, it follows that σ and ξ induce the same map M/mfM −→ N/mfN. 1.12. Lemma. If e is a lifting number for (M, N) and e e, then e is also a lifting number for (M, N). Proof. Let f be a positive integer, and let ξ : M/me +f M −→ N/me +f N be an R-homomorphism. Put f = f + e e. Since e + f = e + f and e is a lifting number, there is a homomorphism σ : M −→ N such that σ and ξ induce the same homomorphism M/mfM −→ N/mfN. Now
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