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pma427_0607_sols

# Here √ i = √ e iπ 2 = e iπ 4 =(1 i √ 2 and

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Unformatted text preview: Here √ i = √ e iπ/ 2 = e iπ/ 4 = (1 + i ) / √ 2, and similarly √- i = (1- i ) / √ 2. (iv) The splitting field K contains i + √ i and i- √ i . By adding and subtracting these, we see that it contains i and √ i . It therefore also contains (1 + i ) / √ i = √ 2. All roots can be expressed in terms of i and √ 2, so K = Q ( i, √ 2), so [ K : Q ] = 4. Splitting fields are always Galois extensions, so | G ( K/ Q ) | = [ K : Q ] = 4. (v) There is an automorphism σ with σ ( i ) =- i and σ ( √ 2) = √ 2, and another automorphism τ with τ ( i ) = i and τ ( √ 2) =- √ 2. Moreover, we have G ( K/ Q ) = { 1 ,σ,τ,στ } ’ C 2 × C 2 . Using the above formulae for √ ± i , we see that σ ( √ i ) = √- i σ ( √- i ) = √ i τ ( √ i ) =- √ i τ ( √- i ) =- √- i. Given this, we can check that the corresponding permutations are as follows: σ = (1 3)(2 4) τ = (1 2)(3 4) στ = (1 4)(2 3) . (3) (i) The cyclotomic polynomial λ n ( x ) (denoted by ϕ n ( x ) in the 2009/10 course) is the product of the terms x- ζ as ζ runs over all primitive n th roots of unity. Exquivalently, ζ runs over the numbers e 2 kπi/n where 0 ≤ k < n and k is coprime with n . The degree is the number of such k , which is denoted by φ ( n ); if the distinct primes dividing n are p 1 ,...,p r , then φ ( n ) = n Q i (1- 1 /p i ). (ii) Put f n ( x ) = x n- 1, so f n ( x ) is monic, and the roots are just the n ’th roots of unity. Note that f n ( x ) = nx n- 1 , which has no roots in common with f n ( x ), so f n ( x ) has no repeated roots. It follows that f n ( x ) = Q ζ ( x- ζ ), with ζ running over the n th roots of unity. Each such ζ is a primitive d ’th root of unity for some divisor d of n (namely, the order of ζ ). If we group the roots according to their orders, we obtain the factorisation x n- 1 = Y d | n λ d ( x ) . (iii) Now take ζ = e πi/ 10 , which is a primitive 20th root of unity, and thus a root of λ 20 ( x ). We can divide the two equations x 20- 1 = λ 20 ( x ) λ 10 ( x ) λ 5 ( x ) λ 4 ( x ) λ 2 ( x ) λ 1 ( x ) x 10- 1 = λ 10 ( x ) λ 5 ( x ) λ 2 ( x ) λ 1 ( x ) to obtain x 20 + 1 = λ 20 ( x ) λ 4 ( x ). The primitive 4th roots of 1 are ± i , so λ 4 ( x ) = x 2 + 1. We thus have λ 20 ( x ) = x 10 + 1 x 2 + 1 = x 8- x 6 + x 4- x 2 + 1 ....
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Here √ i = √ e iπ 2 = e iπ 4 =(1 i √ 2 and...

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