nothing. the degree of an extension matches the degree of a polynomial only if the polynomial is IRREDUCIBLE.

let's look at x^{6}- 1:

this isn't irreducible, it factors as:

x^{6}- 1 = (x^{3}- 1)(x^{3}+ 1) = (x - 1)(x + 1)(x^{2}+ x + 1)(x^{2}- x +1).

so one thing is clear from this factorization, if we adjoin the roots of x^{2}- x + 1 to Q, we get at least 4 of the roots of x^{6}- 1.

the 6-th roots of 1 form a cyclic subgroup of the multiplicative group of Q(u)*, where u is a root of x^{2}- x + 1 (the other root is u^{5}= 1/u.

it turns out that u + 1/u = 1. you can use trigonometry to prove this: u = 1/2 + (√3/2)i)

we obviously can't have a subgroup of a group of order 6 of order 4: if we have at least 4 elements of it, we have them all.

also, if u is a 6-th root of 1, clearly u^{2}is a 3-rd root of 1, so u^{2}is a root of x^{3}- 1, so we can write:

x^{6}- 1 = (x - 1)(x + 1)(x - u^{2})(x - 1/u^{2})(x - u)(x - 1/u)

(it turns out that u^{2}+ 1/u^{2}= -1. this can be proved using some complex number algebra, since u^{2}= -1/2 + (√3/2)i).

another way to look at it:

suppose u is a non-rational root of x^{3}+ 1. this means that u^{3}= -1.

but -1 is a SQUARE root of 1 (remember every positive number has TWO square roots: the positive one, and the negative one. we often forget about the negative one).

so if u^{3}= -1, then:

u^{6}= (u^{3})^{2}= (-1)^{2}= 1, that is: u is a 6-th root of unity.

6 = 3*2, so it is NOT surprising that including the "3 part" (the 6-th roots that come from the cubic x^{3}+ 1), we get everything, because Q already contains the "2-part"

(that is: the factor x^{2}- 1 of x^{6}- 1 splits completely over Q).

what IS sort of surprising is this, we have:

Q < Q(u^{2}) ≤ Q(u)

but since [Q(u):Q] = 2, we must have either Q(u^{2}) = Q, or Q(u^{2}) = Q(u).

it turns out that Q(u^{2}) = Q(u). the easiest way to prove this is to show we can write u in terms of u^{2}:

so convince yourself that u = -1/u^{2}, which will prove Q(u) ≤ Q(u^{2}).