Adv Alegbra HW Solutions 99 - Solution. Let A B C be ring...

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99 a generator of the cyclic group F × . It is now straightforward to show that F has the addition and multiplication tables displayed in the solution to Exercise 3.19, and this is enough to show that any two f elds with four elements are isomorphic, for being isomorphic means having the same multiplication and addition tables. 3.44 (i) Let ϕ : A R be an isomorphism, and let ψ : R A be its inverse. Show that ψ is an isomorphism. Solution. Let r , s R . Since ϕ is surjective, there are elements a , b A with ϕ( a ) = r and ϕ( b ) = s ; since ϕ is injective, these elements a and b are unique. Now ϕ an isomorphism gives a + b the unique element of A with ϕ( a + b ) = r + s and ab the unique element of A with ϕ( ab ) = rs . Therefore, ψ( r + s ) = a + b = ψ( r ) + ψ( s ) and ψ( rs ) = ab = ψ( r )ψ( b ) . Finally, since 1 is the unique element of A with ϕ( 1 ) = 1, we have ψ( 1 ) = 1. We conclude that ψ is a ring isomorphism. (ii) Show that the composite of two homomorphisms (or two isomor- phisms) is again a homomorphism (or an isomorphism).
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Unformatted text preview: Solution. Let A B C be ring homomorphisms. Now : 1 7 ( 1 ) = 1 7 ( 1 ) = 1 , so that the composite preserves 1. Let a , a A . Then : a + a 7 ( a + a ) = ( a ) + ( a ) 7 (v ar phi ( a ) + ( a )) = (v ar phi ( a )) + (( a )). One shows that the composite preserves multiplication in the same way, and one concludes that the composite of ring homomorphisms is again a ring homomorphism. Since the composite of bijections is always a bijection, it follows that the composite of ring isomor-phisms is again an isomorphism. (iii) Show that A = R de f nes an equivalence relation on any family of commutative rings. Solution. (i) Re F exive: If A is a ring, then the identity map 1 A : A A is an isomorphism....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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