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HW11_soln

# HW11_soln - Homework#11 Solutions p 348#12 Since 3 F is a...

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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 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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