7. (a) We have α ∈ C× , so x − α is the minimum polynomial of α over C.
(b) Since α ∈ C we have R(α) ⊆ C. Since α 6∈ R we have [R(α) : R] > 1. Now [C : R] = 2, so
2 = [C : R] = [C : R(α)] · [R(α) : R] > [C : R(α)], so [C : R(α)] < 2, so [C : R(α)] = 1, so
R(α) = C.
√
√
√
We have α = cos(π/6) + i sin(π/6) = 21 ( 3 + i). So α − 21 3 = 21 i, so (α − 21 3)2 + 14 = 0, so
(x −
1
2
√
3)2 +
1
4
= x2 −
√
3x + 1
is a monic polynomial in R[x] of degree 2, which is equal to [R(α) : R], with α as a root. So this
is the minimum polynomial of α over R.
(c) Observe that α2 − α4 = α2 + α2 = 2<(α2 ) = 2 cos(2π/6) = 1. [The way to come up with this
observation is to plot 1, α, α2 , α3 , α4 in the complex plane, and notice that −α4 is the reflection
of α2 in the real axis, so α2 = −α4 .] So
x4 − x2 + 1
is a monic polynomial in Q[x] with α as a root. So [Q(α) : Q] ≤ 4. If we can show that
[Q(α) : Q] = 4 then this will show that x4 − x2 + 1 is the minimum polynomial of α over Q.
√
√
√
√
We claim that Q(α) = Q( 3, i). Indeed, α = 12 ( 3 + i) ∈ Q( 3, i), so Q(α) ⊆ Q( 3, i), and
√
√
i = α3 ∈ Q(α), so also 3 = 2α − i ∈ Q(α), so Q( 3, i) ⊆ Q(α). Hence these fields are equal.
Now
√
√
√
√
[Q(α) : Q] = [Q( 3, i) : Q] = [Q( 3, i) : Q( 3)] · [Q( 3) : Q].
√
√
√
√
Since √3 6∈ Q we√have Q( 3)
√ 6= Q, so [Q(
√ 3) : Q] ≥ 2. Similarly, since Q( 3) ⊆ R we have
i 6∈ Q( 3), so Q( 3, i) 6= Q( 3), so [Q( 3, i) : Q] ≥ 2. So [Q(α) : Q] ≥ 4. But we’ve already
seen that [Q(α) : Q] ≤ 4. So [Q(α) : Q] = 4, and x4 − x2 + 1 is the minimum polynomial of α
over Q.
8. (a) Let α = eiπ/6 and let f = x12 − 1. If 0 ≤ j < 12 then f (αj ) = e12ijπ/6 − 1 = e2πij − 1 = 0.
So x − αj |f for 0 ≤ j < 12. Since no two of α0 , α1 , α2 , . . . , α11 are equal, this shows that
(x − α0 )(x − α1 )(x − α2 ) . . . (x − α11 ) is a monic polynomial which divides the monic polynomial f ,
and since it has degree 12, which is the same as the degree of f , it is equal to f . Each of the
factors x − αj is monic and linear, so is irreducible in C[x]. So this is the required factorisation.
(b) Observe that αj = α12−j . So (x − αj )(x − α12−j ) = x2 − 2<(αj ) + 1 ∈ R[x]. If αj 6∈ R then (by
the uniqueness of factorisation in C[x]) this polynomial is irreducible in R[x]. Hence, taking the
factorisation from (a) and pairing off the factors which are not in R[x] in this manner, we obtain
the following factorisation of f into irreducible polynomials in R[x]:
f = (x − 1)(x2 − 2<(α)x + 1)(x2 − 2<(α2 )x + 1)(x2 − 2<(α3 )x + 1)(x2 − 2<(α4 )x + 1)(x2 − 2<(α5 )x + 1)(x + 1)
√
√
= (x − 1)(x2 − 3x + 1)(x2 − x + 1)(x2 + 1)(x2 + x + 1)(x2 + 3x + 1)(x + 1)
√
√
The only factors not in Q[x] are x2 − 3x + 1 and x2 + 3x + 1. Their product is (x2 + 1)2 − 3x2 =
x4 − x2 + 1, which is necessarily irreducible in Q[x], by uniqueness of factorisation in R[x]. (And
we’ve already proven, independently, that this is irreducible in Q[x] in 7(c)). So the factorisation
of f into irreducibles in Q[x] is
f = (x − 1)(x4 − x2 + 1)(x2 − x + 1)(x2 + 1)(x2 + x + 1)(x + 1).