Splitting Field Proof for f(x) = x^3-5: Vector Space Basis B for K over Q

[AlexChandler]

Homework Statement

Consider $$f(x) = x^3-5$$
and its splitting field $$K = Q(5^{1/3}, \omega)$$
where $$\omega = e^{2 \pi i/3}$$
Show that $$B = \{1, 5^{1/3}, 5^{2/3}, \omega, \omega 5^{1/3} , \omega 5^{2/3} \}$$
is a vector space basis for K over Q.

The Attempt at a Solution

I am just a bit confused. Since $$5^{1/3}$$ and $$\omega$$ are in K, and K is a field, then $$B'= \{ \omega ^2, \omega ^2 5^{1/3}, \omega ^2 5^{2/3} \} \subseteq K$$
But how can we get any of these elements using only the shown basis B with scalars in Q? I would think that B+B' would be the vector space basis.

 

[Deveno]

try building up K in 2 steps:

first adjoin 5^1/3. what is [Q(5^1/3):Q]?

what does a basis of E = Q(5^1/3) over Q look like?

now K = E(ω). what is [K:E]? what is a basis for K over E?

if dim_Q(E) = m, and dim_E(K) = n,

what must dim_Q(K) be?

this tells you how many basis elements you have to have.

then it's just a matter of proving linear independence over Q.

*******

as for how we get the elements of B', i'll show you for ω²:

ω satisfies the polynomial x² + x + 1 = (x³ - 1)/(x - 1).

so ω² = (-1)1 + (-1)ω, see?

 

[AlexChandler]

Yes, thanks so much I was confusing myself.