By Friedrich Kasch, Adolf Mader

The publication generalizes the well known regularity of ring parts to regularity of homomorphisms in module different types, and additional to regularity of morphisms in any class. normal homomorphisms are characterised by way of decompositions of area and codomain, and diverse different effects are awarded. whereas the idea is easily constructed in module different types many questions stay approximately generalizations and extensions to different different types. The publication simply calls for the information of a easy path in smooth algebra. it's written essentially, with nice element, and is obtainable to scholars and researchers alike.

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Proof. Select isomorphisms σi : Ai → N and τj : Mj → N . 1) Suppose that Reg(N, N ) = 0 and let 0 = η ∈ Reg(N, N ). Then τ1−1 ησ1 ∈ HomR (A1 , M1 ) and τ1−1 ησ1 = 0. Let ⎡ ⎢ ⎢ ξ := ⎢ ⎣ τ1−1 ησ1 0 .. 0 0 .. 0 0 ⎤ ··· 0 ··· 0 ⎥ ⎥ .. ⎥ ∈ HomR (A, M ). ··· . 17, 0 = ξ ∈ Reg(A, M ) if and only if for all αi1 ∈ HomR (Ai , A1 ) and all μ1j it is true that μ1j (τ1−1 ησ1 )αi1 ∈ Reg(Ai , Mj ). But τj (μ1j τ1−1 ησ1 αi1 )σi−1 = (τj μ1j τ1−1 )η(σ1 αi1 σi−1 ) ∈ (End N )η(End N ) ⊆ Reg(N, N ) and it follows that μ1j (τ1−1 ησ1 )αi1 ∈ Reg(Ai , Mj ).

We conclude that μf hμf = μf , or, equivalently, μ∗ (f )hμ∗ (f ) = μ∗ (f ). 2. 2. 1) If φ ∈ HomR (A, A ) is surjective and HomR (A, M ) is regular, then HomR (A , M ) is regular. 2) If μ ∈ HomR (M, M ) is injective and HomR (A, M ) is regular, then HomR (A, M ) is regular. 2 is as follows. 3. 1) If B ⊆ A and HomR (A, M ) is regular, then also HomR (A/B, M ) is regular. 2) If N ⊆ M , and HomR (A, M ) is regular, then HomR (A, N ) is regular. 28 Chapter II. Regular Homomorphisms Proof. 2 with φ : A A/B and μ : N → M .

Suppose that ξ11 ξ12 ξ21 ξ22 ∈ Reg(A, M ). Then ξ21 = 0, ξ12 = 0, and for every μ21 ∈ Hom(M2 , M1 ), every α21 ∈ Hom(A2 , A1 ), every μ12 ∈ Hom(M1 , M2 ), and every α12 ∈ Hom(A1 , A2 ), μ21 ξ22 = 0, ξ11 α21 = 0, μ12 ξ11 = 0, ξ22 α12 = 0, 6. Inherited Regularity 37 and μ21 ξ22 α12 ∈ Reg(A1 , M1 ), μ12 ξ11 α21 ∈ Reg(A2 , M2 ). Proof. 4, Reg(A1 , M2 ) = 0 and Reg(A2 , M1 ) = 0. 14, ξ21 = 0 and ξ12 = 0. We also have that for all μ12 μ22 α11 α21 ∈ End(A), End(M ) and all α12 α22 μ11 ξ11 α11 + μ21 ξ22 α12 μ12 ξ11 α11 + μ22 ξ22 α12 = μ11 μ12 μ21 μ22 ξ11 0 μ11 ξ11 α21 + μ21 ξ22 α22 μ12 ξ11 α21 + μ22 ξ22 α22 0 ξ22 α11 α12 α21 α22 ∈ Reg(A, M ).