# 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.

The ebook has an answer handbook to be had. That makes is perfect for self-study.

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Download e-book for kindle: Foundations of Analysis: A Straightforward Introduction: by K. G. Binmore

In uncomplicated introductions to mathematical research, the remedy of the logical and algebraic foundations of the topic is unavoidably relatively skeletal. This booklet makes an attempt to flesh out the bones of such therapy by way of delivering a casual yet systematic account of the rules of mathematical research written at an easy 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 ].