pma427_0607_sols

# 2 this is a monic polynomial of degree 8 with ? as a

• Notes
• 3

This preview shows pages 2–3. Sign up to view the full content.

2

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

This is a monic polynomial of degree 8 with ζ as a root. (iv) The Galois group G ( Q ( ζ ) / Q ) is isomorphic to ( Z / 20) × . In this group, the elements - 1 , 9 and - 9 all have order precisely 2, but in a cyclic group there can be at most one element of order precisely 2. It follows that this Galois group is not cyclic. (4) Put α = p 6 + 11 and β = p 6 - 11 and K = Q ( α ). We note that 6 + 11 and 6 - 11 are positive, and we take α and β to be the positive square roots as usual. (i) We will take it as given that α 6∈ Q ( 11), although to be really rigorous we should prove that. This means that K is a proper quadratic extension of Q ( 11), so [ K : Q ] = [ K : Q ( 11)][ Q ( 11) : Q ] = 2 × 2 = 4, so the minimal polynomial of α must be quartic. We have α 2 - 6 = 11, so ( α 2 - 6) 2 - 11 = 0, so α is a root of the polynomial f ( x ) = ( x 2 - 6) 2 - 11 = x 4 - 12 x 2 + 25. As the degree of f ( x ) is the same as that of the minimal polynomial, we see that f ( x ) must be the minimal polynomial. It is easy to see that - α, β and - β are also roots of f ( x ), and they are all distinct, so f ( x ) = ( x - α )( x + α )( x - β )( x + β ). (ii) Next note that αβ = 6 2 - 11 = 25 = 5, so β = 5 Q ( α ) = K . It follows that all roots of f ( x ) lie in K , so K is a splitting field for f ( x ), and so is a Galois extension of Q . This means that | G ( K/ Q ) | = [ K : Q ] = 4. (iii) As f ( x ) is irreducible, we know that the Galois group G ( K/ Q ) acts transitively on the roots, so there is an automorphism σ with σ ( α ) = β . It follows that σ ( β ) = σ (5 ) = 5 ( α ) = 5 = α . Similarly, there is another automorphism τ with τ ( α ) = - β and thus τ ( β ) = - α . The composite στ is the same as τσ ; it sends α to - α and β to - β . It is also clear that σ 2 = τ 2 = id . We conclude that G ( K/ Q ) = { id, σ, τ, στ } ’ C 2 × C 2 . (iv) We now have the following table, in which 11 = α 2 - 6 = - ( β 2 - 6): α β α + β α - β 11 σ β α α + β - ( α - β ) - 11 τ - β - α - ( α + β ) α - β - 11 στ - α - β - ( α + β ) - ( α - β ) 11 It follows that K h σ i = Q ( α + β ), and K h τ i = Q ( α - β ), and K h στ i = Q
This is the end of the preview. Sign up to access the rest of the document.
• Fall '13
• 427
• Nth root, Galois theory, Cyclic group, Galois group, Root of unity

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business â€˜17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. Itâ€™s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania â€˜17, Course Hero Intern

• The ability to access any universityâ€™s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLAâ€™s materials to help me move forward and get everything together on time.

Jill Tulane University â€˜16, Course Hero Intern