April 16, 2014

http://coursework.tylerlogic.com/courses/upenn/math503/homework09
Assuming the equation

holds for f(T) = T^{3} + bT + c ∈ F[T] then we can determine the formula for a cubic polynomial
g(T) = T^{3} + AT^{2} + BT + C ∈ F[T] by substituting S - A∕3 for T [DF04]. So g(S - A∕3) yields a polynomial h(S) so
that

where P = 1∕3(3B - A^{2}) and Q = 1∕27(2A^{3} - 9AB + 27C). Now if α, β, and γ are roots of h(S), then
α - A∕3, β - A∕3, and γ - A∕3 are roots of g(T) since, for instance, g(α - A∕3) = h(α) = 0. Furthermore, since
[((α - A∕3) - (β - A∕3))((β - A∕3) - (γ - A∕3))((γ - A∕3) - (α - A∕3))]^{2} = [(α - β)(β - γ)(γ - α)]^{2}, then the
discriminents of g(T) and h(S) are the same, i.e. we can use the formula from the previous part of the problem (again
assuming it holds) for the discriminent of h(S). Hence

Exact values of cos(2π∕5) and sin(2π∕5) were computed with the help of Wolfram Alpha, [Wol14a], [Wol14b].

Since cos(2π∕5) = 1∕4( - 1) then 4cos(2π∕5) + 1 = , which implies (4cos(2π∕5) + 1)

| (3.1) |

since 16x^{2} + 8x - 4 = 4(4x^{2} + 2x - 1).

Since sin(2π∕5) = , then 8sin^{2}(2π∕5) - 5 = , which implies that 64sin^{4}(2π∕5) - 80sin^{2}(2π∕5) + 25 is 5.
Subtracting five yields 64sin^{4}(2π∕5) - 80sin^{2}(2π∕5) + 20 = 0, and thus that sin(2π∕5) is algebraic since it is the root
of

| (3.2) |

since 64x^{4} - 80x^{2} + 20 = 4(16x^{4} - 20x^{2} + 5).

The polynomial, 4x

The polynomial, 16x^{4} - 20x^{2} + 5, from equation 3.2 is irreducible over ℚ since it meets Eisenstein’s criteria, that is
5 ⁄|16, 5|20, 5|5, and 5^{2} ⁄|5. So ℚ(sin(2π∕5)) ℚ[x] ∕ (16x^{4} - 20x^{2} + 5), and therefore [ℚ(sin(2π∕5)) : ℚ] = 4.

Does one contain the other? Since sin(2π∕5) = , then

| (3.3) |

Therefore, because cos(2π∕5) = 1∕4( - 1) we have

Since the field ℚ(cos(2π∕5)) ℚ[x] ∕ (4x

Galois group of ℚ(sin(2π∕5))
Now attempting to determine the Galois group for ℚ(sin(2π∕5)) ℚ[x] ∕ (16x^{4} - 20x^{2} + 5) is slightly more tricky. Certainly
the roots ±sin(2π∕5) = ± of 16x^{4} - 20x^{2} + 5 are contained within, but it is not immediately clear whether or
not the roots ± are contained in the field. We note that

which implies that

Equation 3.3 informs us that is in ℚ(sin(2π∕5)), and being a field, (sin(2π∕5))^{-1} is in definitely in ℚ(sin(2π∕5)); hence
(and its negation) is also in the field. Hence all the roots of 16x^{4} - 20x^{2} + 5 are contained within the field. Thus
the field is therefore a splitting field of a separable polynomial, and hence Galois.

Now since ℚ(sin(2π∕5)) has degree four over ℚ, then its Galois group has size four. Automorphisms of the Galois group will just permute the roots ±, ±, because of which we can completely define two non-trivial automorphisms by

letting s = and t = . From this we see that σ^{2} and τ^{2} are both the identity automorphism and that the
the third non-trivial automorphism is

also with (στ)^{2} being the identity automorphism. Hence Gal(ℚ(sin(2π∕5))∕ ℚ) = {1,σ,τ,στ}, but futhermore, since each
non-trivial element has order two, then Gal(ℚ(sin(2π∕5))∕ ℚ) must be the Klein-four group.

[DF04] D.S. Dummit and R.M. Foote. Abstract Algebra. John Wiley & Sons Canada, Limited, 2004.

[Wol14a] Wolfram—Alpha. http://www.wolframalpha.com/input/?i=cos(2*pi/5), April 2014.

[Wol14b] Wolfram—Alpha. http://www.wolframalpha.com/input/?i=sin(2*pi/5), April 2014.