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35 Let Λ1 , Λ2 be generalized Euclidean rings; then Λ1 × Λ2 is also generalized Euclidean. To consider an example, we denote by Zn the ring of residues mod n; that is, Zn = Z/n. If n is square free we may write it as a product of distinct primes n = p1 p2 · · · pk . Then Zn ∼ = Fp1 × · · · × Fpk where Fp is the field with p elements and so Zn [t, t −1 ] ∼ = Fp1 [t, t −1 ] × · · · × Fpk [t, t −1 ]. 35 we see that: If n is square free then Zn [t, t −1 ] is generalized Euclidean. 37 Let Λ1 be a generalized Euclidean ring; if ϕ : Λ1 → Λ2 is a surjective ring homomorphism then Λ2 is also generalized Euclidean ring.

3 Matrices with a Smith Normal Form 17 We shall say that X ∈ Mn (Λ) has a Smith Normal Form when X = E+ Δ(α1 , . . , αn )E− for some E+ , E− ∈ En (Λ) and some α1 , . . , αn ∈ Λ. More generally, if X, Y ∈ Mn (Λ) we write X ∼ Y when X = E+ Y E− for some E+ , E− ∈ En (Λ). 24 Let X ∼ Y ∈ Mn (Λ); if Λ is commutative then det(X) = det(Y ). The identities of Sect. 1 allow us to move units around; let α1 , . . , αn ∈ Λ and u1 , . . , un ∈ Λ∗ . Writing D(r) = Δ(1, ur )Δ(r, u−1 r ), D = D(2)D(3) · · · D(n) and u = u1 · · · un we see that Δ(α1 u1 , .

18) In this context one has Eilenberg’s trick. 19 If P ∈ Mod∞ is projective then P ⊕ Λ∞ ∼ = Λ∞ . Proof As P is countably generated choose a surjective Λ-homomorphism η : Λ∞ → P . Putting Q = Ker(η) gives an exact sequence 0 → Q → Λ∞ → P → 0 which splits since P is projective. 18), (P ⊕ Q) ⊕ (P ⊕ Q) ⊕ · · · ⊕ (P ⊕ Q) ⊕ (P ⊕ Q) ⊕ · · · ∼ = Λ∞ which we may re-bracket as P ⊕ (Q ⊕ P ) ⊕ (Q ⊕ P ) ⊕ · · · ⊕ (Q ⊕ P ) ⊕ (Q ⊕ P ) ⊕ · · · ∼ = Λ∞ . 18), P ⊕ Λ∞ ∼ = Λ∞ as stated. (II) 12 1 Preliminaries We shall say that a module M ∈ Mod∞ is hyperstable when M ∼ = M0 ⊕ Λ∞ for ∞ ∞ ∞ ∼ ∼ some module M0 ∈ Mod∞ .