# Adv Alegbra HW Solutions 103 - nition of ideal or note that...

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103 Solution. By Exercise 3.49, we may assume that there is an injec- tive ring homomorphism 8 : Frac ( R [ x ] ) Frac ( F ( x )) = F ( x ) with [ f ( x ), g ( x ) ] 7→ f ( x ) g ( x ) 1 . It suf f ces to show that 8 is a surjection. If α( x ) F [ x ] , then clearing denominators of coef- f cients gives a factorization α( x ) = r 1 f ( x ) , where r R and f ( x ) R [ x ] . Therefore, if α( x )β( x ) 1 F ( x ) , then there are r , s R and f ( x ), g ( x ) R [ x ] with α( x )β( x ) 1 = sf ( x ) [ rg ( x ) ] 1 = 8( [ sf ( x ), rg ( x ) ] ). (ii) Prove that Frac ( R [ x 1 , x 2 ,..., x n ] ) = F ( x 1 , x 2 ,..., x n ) . Solution. The proof is by induction on n 1, the base step being part (i). For the inductive step, set S = R [ x 1 , x 2 ,..., x n ] and the base step gives Frac ( R [ x 1 , x 2 ,..., x n + 1 ] ) = Frac ( S [ x n + 1 ] ) = Frac ( S )( x n + 1 ) = F ( x 1 , x 2 ,..., x n )( x n + 1 ) = F ( x 1 , x 2 ,..., x n , x n + 1 ). 3.53 (i) If R and S are commutative rings, show that their direct product R × S is also a commutative ring. Solution. Straightforward. (ii) Show that R ×{ 0 } is an ideal in R × S . Solution. Either do directly, checking each part of the de f
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Unformatted text preview: nition of ideal, or note that R × { } is the kernel of the homomorphism R × S → S given by ( r , s ) 7→ s . (iii) Show that R × S is not a domain if neither R nor S is the zero ring. Solution. ( 1 , ) · ( , 1 ) = ( , ) . 3.54 (i) If R and S are commutative rings, prove that U ( R × S ) = U ( R ) × U ( S ), where U ( R ) is the group of units of R . Solution. We f rst prove that ( r , s ) is a unit in R × S if and only if r is a unit in R and s is a unit in S . If r , s ) is a unit in R × S , then there is ( a , b ) ∈ R × S with ( r , s )( a , b ) = ( 1 , 1 ) . It follows that ra = 1 and sb = 1; that is, r is a unit in R and s is a unit in S ....
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