Download PDF by John B. Fraleigh: A First Course in Abstract Algebra, 7th Edition

By John B. Fraleigh

A well-known e-book in introductory summary algebra at undergraduate point.

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Use the surjectivity of ϕ to choose, for each x ∈ X, an element mx ∈ M such that ϕ(mx ) = α(fx ). Since every f ∈ F is a (finite) unique sum f = x∈X rx fx with all rx ∈ R, we can define β by β(f ) = x∈X rx mx . We then get the desired assertion: ϕ(β(f )) = ϕ( r x mx ) = rx ϕ(mx ) = x∈X x∈X rx α(fx ) = α(f ) . 9) Projective but not free. Consider the commutative group R = R2 ( = R × R); the addition is (r1 , r2 )+(r1 , r2 ) =def (r1 +r1 , r2 +r2 ). The group R becomes a commutative ring with the multiplication (r1 , r2 )(r1 , r2 ) =def (r1 r1 , r2 r2 ); the indentity element is (1, 1).

If F is a free R–module, then F is projective. 8) Proof. Assume that F is free with basis (fx )x∈X , and let the solid part of the next projectivity diagram with exact row, cf. 2), be given. F β M α { ϕ  / / N O. Use the surjectivity of ϕ to choose, for each x ∈ X, an element mx ∈ M such that ϕ(mx ) = α(fx ). Since every f ∈ F is a (finite) unique sum f = x∈X rx fx with all rx ∈ R, we can define β by β(f ) = x∈X rx mx . We then get the desired assertion: ϕ(β(f )) = ϕ( r x mx ) = rx ϕ(mx ) = x∈X x∈X rx α(fx ) = α(f ) .

Let ϕ : M → N be any R–homomorphism, and consider the induced commutative diagram: O HomR (Q, M ) / HomR (π,M ) HomR (S, M ) / HomR (Q,ϕ) O / HomR (P, M ) / HomR (S,ϕ)  HomR (Q, N ) HomR (ι,M ) HomR (π,N ) /  HomR (S, N ) HomR (ι,N ) / O / HomR (P,ϕ)  HomR (P, N ) / O The rows are exact as they result from application of an additive functor to a split-exact sequence, cf. 48). Proof of “⇒”. Assume that S is projective. By symmetry it suffices to prove that P is projective. Let ϕ : M → N be surjective [ and we want to prove that HomR (P, ϕ) is surjective ].

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A First Course in Abstract Algebra, 7th Edition by John B. Fraleigh


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