$\mathbb{Q}(\omega2^{1/3}) \neq \mathbb{Q}(\omega^{2}2^{1/3})$

Question

I am reading An Introduction to The Theory of Numbers by Niven et al.. In the section on Algebraic Number Fields (section 9.3 in the 5th edition) towards the end the authors mention that $\mathbb{Q}(\omega2^{1/3}) \neq \mathbb{Q}(\omega^{2}2^{1/3})$. I am able to reduce the problem to showing that if $\omega^{2}2^{1/3}\notin\mathbb{Q}(\omega 2^{1/3})$ or $\omega 2^{1/3}\notin\mathbb{Q}(\omega^{2}2^{1/3})$ holds then the claim is true. Now I am stuck and don't know how to proceed. Thanks in advance.

Note: If possible, use the theorems given in the book as I am not too comfortable with the methods from an algebra course (will take one soon :) ).

Answer 1

Suppose $\omega^22^{1/3}\in\mathbb{Q}\left(\omega2^{1/3}\right)$. Then $\omega\in\mathbb{Q}\left(\omega2^{1/3}\right)$. Thus we have a tower of fields: $$\mathbb{Q}\subseteq\mathbb{Q}\left(\omega\right)\subseteq\mathbb{Q}\left(\omega2^{1/3}\right)$$ The minimal polynomial of $\omega$ over $\mathbb{Q}$ has degree 2, so we see that the first field extension above has degree 2. So by multiplicativity, we'd have to have that the total extension had degree divisible by 2. But the minimal polynomial of $\omega2^{1/3}$ over $\mathbb{Q}$ is the same as that of $2^{1/3}$, which has degree 3. Thus we cannot have $\omega^22^{1/3}\in\mathbb{Q}\left(\omega2^{1/3}\right)$.

(Make sure you can prove all the steps I take for granted—e.g., the minimal polynomials of the relevant elements, field extension degree multiplicativity).

Answer 2

I am posting a solution based on the theorems given in the book for future reference.

First note that $\omega 2^{1/3}$ is an algebraic number of degree 3. By theorem 9.14, if $\omega^{2}2^{1/3}\in \mathbb{Q}(\omega 2^{1/3})$ then,

$$\omega^{2}2^{1/3} = a_{0}+a_{1}\omega 2^{1/3} + a_{2}(\omega 2^{1/3})^{2}$$

where $a_{0}, a_{1}, a_{2}\in \mathbb{Q}$.

Multiplying both sides by $\omega$ and comparing the real and imaginary parts, we get

$$a_{0} = a_{1}2^{1/3}$$ and

$$2^{1/3} = \frac{a_{0}+a_{1}2^{1/3}}{2}+a_{2}4^{1/3}$$

$$\implies 2^{1/3} = a_{1}2^{1/3}+a_{2}4^{1/3}$$

$$\implies 2^{1/3} = \frac{1 - a_{1}}{a_{2}} \in \mathbb{Q}$$

A contradiction.

Answer 3

Observe that $a=2^{1/3},b=2^{1/3}\omega,c=2^{1/3}\omega^2$ are the distinct roots of $f(x) =x^3-2\in\mathbb {Q} [x] $.

If $\mathbb {Q} (b) =\mathbb{Q} (c)=K $ then both the roots $b, c$ are in $K$ and hence $a=-(b+c) \in K$. Hence $K$ contains all the roots $a, b, c$.

Next note that $f(x) $ is irreducible of degree $3$ thus each of $\mathbb {Q} (a), \mathbb {Q} (b) =K, \mathbb{Q} (c) =K$ is of degree $3$ over $\mathbb {Q} $. Since $\mathbb{Q} \subseteq \mathbb{Q} (a) \subseteq K$ it follows that $\mathbb{Q} (a) =K$.

But this is a contradiction as $\mathbb{Q} (a) \subseteq \mathbb {R} $ and $K$ contains complex numbers $b, c$.