MATHS
pma427_0607_sols

# 2 consider the polynomial f x x 4 2 x 2 4 x 2 i as f

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(2) Consider the polynomial f ( x ) = x 4 + 2 x 2 + 4 x + 2. (i) As f ( x ) is monic with f ( x ) = x 4 (mod 2) and f (0) = 2 not divisible by 2 2 , Eisenstein’s criterion tells us that f ( x ) is irreducible over Q . It is a standard formula that for a quartic f ( x ) = x 4 + bx 2 + cx + d , the resolvent cubic is g ( x ) = x 3 + 2 bx 2 + b 2 x - 4 dx - c 2 . In our case we have b = d = 2 and c = 4, so g ( x ) = x 3 + 4 x 2 - 4 x - 16. We are asked to show that this has only integer roots. Any integer roots must divide the constant term, which is - 16, so the only possible roots are ± 1, ± 2, ± 8 and ± 16. After a little experimentation we find that the roots are ± 2 and - 4, and that g ( x ) = ( x - 2)( x + 2)( x + 4). (ii) I suspect that the examiner had some more cunning approach in mind, which he probably covered in lectures. This is the only method that I can see at the moment. We would like to write f ( x ) as ( x 2 + αx + γ )( x 2 + βx + δ ), so x 4 + 2 x 2 + 4 x + 2 = x 4 + ( α + β ) x 3 + ( γ + δ + αβ ) x 2 + ( αδ + βγ ) x + γδ. By comparing coefficients of x 3 and x 0 , we see that β = - α and δ = 2 . After substituting these, the remaining two equations reduce to 2 α/γ - αγ = 4 2 - α 2 + γ = 2 . 1

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We multiply the second of these by α , and then consider the sum and difference of the two equations and rearrange to get γ - 1 = α 2 / 4 + 1 / 2 + 1 γ = α 2 / 2 + 1 - 2 /α. We then multiply these and rearrange to get α 4 / 8 + α 2 / 2 - 1 / 2 - 2 2 = 0 . This can be factorised as ( α 2 + 4)( α 2 + 2)( α 2 - 2) / (8 α 2 ) = 0. The possible values of α are therefore ± 2 i , ± 2 i and ± 2. (iii) Now take α = 2 i . The above formulae then give γ = α 2 / 2 - 1 - 2 = i - 1 and β = - α = - 2 i and δ = 2 = - i - 1, so the factorisation is f ( x ) = ( x 2 + 2 ix + i - 1)( x 2 - 2 ix - i - 1) = (( x + i ) 2 + i )(( x - i ) 2 - i ) . The roots are thus α 1 = i + i α 2 = i - i α 3 = - i + - i α 4 = - i - - i. Here i = e iπ/ 2 = e iπ/ 4 = (1 + i ) / 2, and similarly - i = (1 - i ) / 2.
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• Fall '13
• 427
• Nth root, Galois theory, Cyclic group, Galois group, Root of unity

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