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Adv Alegbra HW Solutions 92

# Adv Alegbra HW Solutions 92 - But degrees are nonnegative...

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92 (vi) If R is a domain and f ( x ), g ( x ) R [ x ] are nonzero, then either f ( x ) + g ( x ) = 0ordeg ( f + g ) max { deg ( f ), deg ( g ) } . Solution. True. (vii) If k is a f eld, then k [ x ]= k ( x ) . Solution. False. 3.30 Show that if R is a nonzero commutative ring, then R [ x ] is never a f eld. Solution. If f ( x ) = x 1 , then xf ( x ) = 1. But the degree of the left side is at least 1, while the degree of the right side is 0. 3.31 Let k be a f eld and let A be an n × n matrix with entries in k (so that the powers A i are de f ned). If f ( x ) = c 0 + c 1 x +···+ c m x m k [ x ] ,de f ne f ( A ) = c 0 I + c 1 A +···+ c m A m . (i) Prove that k [ A ] ,de f ned by k [ A ]={ f ( A ) : f ( x ) k [ x ]} ,isa commutative ring under matrix addition and matrix multiplication. Solution. Absent. (ii) If f ( x ) = p ( x ) q ( x ) k [ x ] and if A is an n × n matrix over k , prove that f ( A ) = p ( A ) q ( A ) . Solution. Absent. (iii) Give examples of n × n matrices A and B such that k [ A ] is a domain and k [ B ] is not a domain. Solution. Absent. 3.32 (i) Let R be a domain. Prove that a polynomial f ( x ) is a unit in R [ x ] if and only if f ( x ) is a nonzero constant which is a unit in R . Solution. If f ( x ) is a unit, then there is g ( x ) R [ x ] with f ( x ) g ( x ) = 1. Since R is a domain, deg ( f ) + deg ( g ) = 0.
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Unformatted text preview: But degrees are nonnegative, so that deg ( f ) = 0 and f ( x ) is a nonzero constant. If R is a f eld, then every nonzero element is a unit in R ; that is, there is v ∈ R with u v = 1. Since R ⊆ R [ x ] , we have v ∈ R [ x ] , and so u is a unit in R [ x ] . (ii) Show that ( [ 2 ] x +[ 1 ] ) 2 = [ 1 ] in I 4 [ x ] . Conclude that the statement in part (i) may be false for commutative rings that are not domains. Solution. ( [ 2 ] x + [ 1 ] ) 2 = [ 4 ] x 2 + [ 4 ] x + [ 1 ] = [ 1 ] in I 4 [ x ] . 3.33 Show that if f ( x ) = x p − x ∈ F p [ x ] , then its polynomial function f [ : F p → F p is identically zero. Solution. Let f ( x ) = x p − x ∈ F p [ x ] . If a ∈ F p , Fermat ’ s theorem gives a p = a , and so f ( a ) = a p − a = 0....
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