Adv Alegbra HW Solutions 93

# Adv Alegbra HW Solutions 93 - r f = r f if r ∈ R f g = f...

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93 3.34 (i) If p is a prime and m , n N , prove that ( pm pn ) ( m n ) mod p . Solution. Absent. (ii) Prove that ( p r m p r n ) ( m n ) mod p for all r 0. Solution. Absent. (iii) Give another proof of Exercise 1.72: if p is a prime not dividing an integer m 1, then p - ( p r m p r ) . Solution. Absent. 3.35 Let α C , and let Z [ α ] be the smallest subring of C containing α ; that is, Z [ α ]= T S , where S ranges over all those subrings of C containing α . Prove that Z [ α ]={ f (α) : f ( x ) Z [ x ]} . Solution. Absent. 3.36 Prove that the usual rules of calculus hold for derivatives of polynomials in R [ x ] , where R is a commutative ring; that is, ( f + g ) 0 = f 0 + g
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Unformatted text preview: ; ( r f ) = r f if r ∈ R ; ( f g ) = f g + f g ; ( f n ) = n f n − 1 f for all n ≥ 1 . Solution. Let f ( x ) = ∑ a n x n and g ( x ) = ∑ b n x n . (i) ( f + g ) = ³ X a n x n + X b n x n ´ = ³ X ( a n + b n ) x n ´ = X n ( a n + b n ) x n − 1 = X na n x n − 1 + X nb n x n − 1 = f + g . (ii) ( r f ) = ³ r X a n x n ´ = ³ X ra n x n ´ = X rna n x n − 1 = r X na n x n − 1 = r f ....
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