# Adv Alegbra HW Solutions 120 - f elds k and k with k ∼ =...

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120 If a ( x ) k [ x ] , denote its congruence class in k [ x ] /( f ( x )) by [ a ] f . It is easy to check that ϕ : k [ x ]→ k [ x ] /( f ( x )) × k [ x ] /( g ( x )), given by a 7→ ( [ a ] f , [ a ] g ), is a ring homomorphism. We claim that ker ϕ = ( fg ) . Clearly, ( fg ) ker ϕ . For the reverse inclusion, if a ker ϕ , then [ a ] f = [ 0 ] f and [ a ] g =[ 0 ] g ; that is, a ( f ) and a ( g ) ; that is, f | a and g | a . Since f and g are relatively prime, Exercise 3.94 gives fg | a , and so a ( fg ) , that is, ker ϕ ( fg ) and ker ϕ = ( fg ) . We now show that ϕ is surjective. If ( [ a ] f , [ b ] g ) k [ x ] /( f ( x )) × k [ x ] /( g ( x )), is there h ( x ) k [ x ] with ϕ( h ) = ( [ h ] f , [ h ] g ) = ( [ a ] f , [ b ] g ) ; that is, is there h k [ x ] with h a ( f ) and h b ( g ) ? Since f and g are relatively prime, part (i) provides a solution h . The f rst isomorphism theorem now gives k [ x ] /( f ( x ) g ( x )) = = k [ x ] /( f ( x )) × k [ x ] /( g ( x )). 3.95 Generalize Exercise 3.84 by proving that if k is a f eld of characteristic 0 and if p ( x ) k [ x ] is an irreducible polynomial, then p ( x ) has no repeated roots. Solution. Absent. 3.96 (i) Prove that a f eld K cannot have sub
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Unformatted text preview: f elds k and k with k ∼ = Q and k ∼ = F p for some prime p . Solution. Suppose such sub f elds k and k exist, and consider their intersection k = k ∩ k . The one element 1 lies in k ; since k ≤ k , we have 0 = p · 1 = 1 + 1 + ··· + 1, where there are p summands equal to 1. On the other hand, in k , we have p · 1 ±= 0. This is a contradiction. (ii) Prove that a f eld K cannot have sub f elds k and k with k ∼ = F p and k ∼ = F q , where p ±= q . Solution. The argument in part (i) also works here: if p < q , then p · 1 = 0 in k = k ∩ k , because k ≤ k , but p · 1 ±= 0 in k because k ≤ k ....
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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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