Adv Alegbra HW Solutions 148

# Adv Alegbra HW Solutions 148 - 148(ix The alternating group...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 148 (ix) The alternating group A 5 is a solvable group. Solution. False. 5.16 Let ϕ : A → H be a group homomorphism. If B C A and B ≤ ker ϕ , prove that the induced map ϕ ∗ : A / B → H , given by aB 7→ ϕ( a ) , is a well-de fi ned homomorphism with im ϕ ∗ = im ϕ . Solution. If aB = cB , then c − 1 a ∈ B ≤ ker ϕ , and so ϕ( c − 1 a ) = 1. Therefore, ϕ( c ) = ϕ( a ) , and ϕ ∗ is well-de fi ned. 5.17 If z ∈ C is a constructible number, prove that Q ( i , z )/ Q is a radical exten- sion. Solution. Absent. 5.18 Let k be a fi eld and let f ( x ) ∈ k [ x ] . Prove that if E and E are splitting fi elds of f ( x ) over k , then Gal ( E / k ) ∼ = Gal ( E / k ) . Solution. Absent. 5.19 Prove that F 3 [ x ] /( x 3 − x 2 − 1 ) ∼ = F 3 [ x ] /( x 3 − x 2 + x − 1 ) . Solution. We saw, in Chapter 3, that both cubics are irreducible in F 3 [ x ] , and so the quotient rings are fi elds of order 3 3 = 27. But any two fi nite fi elds of the same order are isomorphic.elds of the same order are isomorphic....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online