HW11_soln - 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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 3

HW11_soln - 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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online