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Unformatted text preview: Homework #11 Solutions p 348, #12 Since π 3 ∈ F , π is a root of the polynomial x 3 π 3 ∈ F [ x ]. This polynomial is irreducible over F since the only possible root in F would be π itself, and it is easy to show that π 6∈ F (if it were, π would be algebraic over Q ). Therefore a basis for F ( π ) over F is { 1 , π, π 2 } . p 348, #14 Let F = Q ( 3 √ 5) = { a + b 3 √ 5 + c ( 3 √ 5) 2  a, b, c ∈ Q } and φ : F → F be an automorphism. Arguing as we have several times before, we can show that φ ( r ) = r for all r ∈ Q . It follows that 5 = φ (5) = φ (( 3 √ 5) 3 ) = φ ( 3 √ 5) 3 . Since φ ( 3 √ 5) ∈ F ⊂ R , we can therefore conclude that φ ( 3 √ 5) = 3 √ 5. Finally, this means that φ ( a + b 3 √ 5 + c ( 3 √ 5) 2 ) = φ ( a ) + φ ( b ) φ ( 3 √ 5) + φ ( c ) φ ( 3 √ 5) 2 = a + b 3 √ 5 + c ( 3 √ 5) 2 for any rational a, b, c . But every element of F can be written in the form above, and so it must be that φ ( α ) = α for every α ∈ F . That is, the only automorphism of F is the identity....
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This note was uploaded on 02/29/2012 for the course MATH 4363 taught by Professor Ryandaileda during the Spring '07 term at Trinity University.
 Spring '07
 RyanDaileda
 Algebra

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