Adv Alegbra HW Solutions 85

# Adv Alegbra HW Solutions 85 - = 1 f 1 = 0 and f r = 0...

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85 Solution. Choose f to be any nonconstant function all of whose values are 0 or 1. 3.9 Find all the units in the commutative ring F ( R ) de f ned in Example 3.11(i). Solution. We claim that f is a unit if and only if f ( r ) ±= 0 for every r R . If f is a unit, there is g F ( R ) with fg = 1; that is, f ( r ) g ( r ) = 1 for all r R , and so f ( r ) ±= 0 for all r R . Conversely, if f ( r ) ±= 0 for all r R ,de f ne g F ( R ) by g ( r ) = 1 / f ( r ) ; then fg = 1 and f is a unit. 3.10 Generalize the construction of F ( R ) to a set X and an arbitrary commu- tative ring R : let F ( X , R ) be the set of all functions from X to R , with pointwise addition f + g : x 7→ f ( x ) + g ( x ) and pointwise multiplication fg : x 7→ f ( x ) g ( x ) for x X . (i) Show that F ( X , R ) is a commutative ring. Solution. One proves that F ( X , R ) is a commutative ring just as one proves that F ( R ) is a commutative ring, for the only aspect if R used in the proof is that it is a commutative ring. (ii) Show that if X has at least two elements, then F ( X , R ) is not a domain. Solution. Denote two elements in X by 0 and 1. De f ne f : R R by f ( 0 )
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Unformatted text preview: = 1, f ( 1 ) = 0, and f ( r ) = 0 otherwise; de f ne g : R → R by g ( ) = 0, g ( 1 ) = 1, and g ( r ) = 0 otherwise. Then neither f = 0 nor g = 0, but f g = 0. (iii) If R is a commutative ring, denote F ( R , R ) by F ( R ) : F ( R ) = { all functions R → R } . Show that F ( I 2 ) has exactly four elements, and that f + f = for every f ∈ F ( I 2 ) . Solution. There are exactly four functions from a 2-point set to itself. If f ∈ F ( I 2 ) , then ( f + f )( r ) = f ( r ) + f ( r ) = for all r ∈ I 2 . 3.11 (i) Prove that the commutative ring C is a domain. Solution. If z w = 0 and z = a + ib ±= 0, then z ¯ z = a 2 + b 2 ±= 0, and µ ¯ z ¯ zz ¶ z = 1 . (ii) Prove that Z , Q , and R are domains. Solution. Every subring of a domain is a domain....
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