Automorphism-invariant non-singular rings and modules

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Proof. Consider the map g:A/I → C , a+I 7→ f (a). It is well defined: a+I = a′ +I implies a− a′ ∈ I implies f (a) = f (a′). The element a + I belongs to the kernel of g iff g(a + I) = f (a) = 0, i.e. a ∈ I , i.e. a + I = I is the zero element of A/I . Thus, ker(g) = 0. The image of g is g(A/I) = {f (a) : a ∈ A} = C . Thus, g is an isomorphism. The inverse morphism to g is given by f (a) 7→ a + I .

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ژورنال

عنوان ژورنال: Journal of Algebra

سال: 2017

ISSN: 0021-8693

DOI: 10.1016/j.jalgebra.2017.05.013