1

(a)

(b)

holds for f(T) = T³ + bT + c ∈ F[T] then we can determine the formula for a cubic polynomial g(T) = T³ + AT² + 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 - A²) and Q = 1∕27(2A³ - 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

(c)

(d)

2

(a)

(b)

(c)

3

(a) Show that cos(2π∕5) and sin(2π∕5) are algebraic over ℚ

(3.1)

since 16x² + 8x - 4 = 4(4x² + 2x - 1).

Since sin(2π∕5) = , then 8sin²(2π∕5) - 5 = , which implies that 64sin⁴(2π∕5) - 80sin²(2π∕5) + 25 is 5. Subtracting five yields 64sin⁴(2π∕5) - 80sin²(2π∕5) + 20 = 0, and thus that sin(2π∕5) is algebraic since it is the root of

(3.2)

since 64x⁴ - 80x² + 20 = 4(16x⁴ - 20x² + 5).

(b) Determine the degrees [ℚ(cos(2π∕5) : ℚ] and [ℚ(sin(2π∕5)) : ℚ]. Does one contain the other?

The polynomial, 16x⁴ - 20x² + 5, from equation 3.2 is irreducible over ℚ since it meets Eisenstein’s criteria, that is 5 ⁄|16, 5|20, 5|5, and 5² ⁄|5. So ℚ(sin(2π∕5)) ℚ[x] ∕ (16x⁴ - 20x² + 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

(c) Determine the Galois groups of ℚ(cos(2π∕5)) and ℚ(sin(2π∕5))

Galois group of ℚ(sin(2π∕5)) Now attempting to determine the Galois group for ℚ(sin(2π∕5)) ℚ[x] ∕ (16x⁴ - 20x² + 5) is slightly more tricky. Certainly the roots ±sin(2π∕5) = ± of 16x⁴ - 20x² + 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 16x⁴ - 20x² + 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 σ² and τ² 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.

4 Extra Credit

(a)

(b)

(c)

References