Proving an extension is simple

[PsychonautQQ]

Homework Statement

Let c be a primitive 3rd root of unity in the complex numbers and b be the real root of x^4-2=0. If a = c*b, show that Q(b,c) = Q(a)

Homework Equations

The Attempt at a Solution

So [Q(a):Q(c)]=3 and [Q(a):Q(b)]=4, and c and b contain no 'overlapping material', so [Q(a):Q)=12.

The usual way I prove things are a simple extension is by starting off by taking (a+b)^2 all the way up to (a+b)^n-1 where n i the deg(a)*deb(b) and then playing around with these expressions trying to come up with a way to get either a or b by itself as to show that a and b are elements of Q(a+b).

Since the degree here is 12, things could get quite messy if I use that method. Is there another way to look at this?

 

[fresh_42]

Homework Statement

Let c be a primitive 3rd root of unity in the complex numbers and b be the real root of x^4-2=0. If a = c*b, show that Q(b,c) = Q(a)

Homework Equations

The Attempt at a Solution

So [Q(a):Q(c)]=3 and [Q(a):Q(b)]=4, and c and b contain no 'overlapping material', so [Q(a):Q)=12.

The usual way I prove things are a simple extension is by starting off by taking (a+b)^2 all the way up to (a+b)^n-1 where n i the deg(a)*deb(b) and then playing around with these expressions trying to come up with a way to get either a or b by itself as to show that a and b are elements of Q(a+b).

Since the degree here is 12, things could get quite messy if I use that method. Is there another way to look at this?

I would not use a statement like "no overlapping material". This actually shouts for an explanation and contains every flaw, if there is one. Furthermore, I think $[ \mathbb{Q}(a) \, : \, \mathbb{Q}(c) ] = 4$ and $[ \mathbb{Q}(a) \, : \, \mathbb{Q}(b) ] = 3$.

Since $\mathbb{Q}(a) = \mathbb{Q}(b \cdot c) \subseteq \mathbb{Q}(b,c)$ it only has to be shown, that $\, b\, , \,c \in \mathbb{Q}(a)$.
A little basic algebra on the powers of $a=b \cdot c$ should do.