$\mathbb{Q}(\theta) \neq \mathbb{Q}(\sqrt[3]{d})$.

Question

Let $\theta$ be a root of $X^3-2X-2$ show that $\mathbb{Q}(\theta) \neq \mathbb{Q}(\sqrt[3]{d})$.

A basis for $\mathbb{Q}(\sqrt[3]{d})$ would be of the form $1, \sqrt[3]{d}, \sqrt[\frac{2}{3}]{d}$

A basis for $\mathbb{Q}(\theta)$ would be of the form $1, \theta, \theta^2$

So is it correct in saying that for them to not be equal $\theta \neq \sqrt[3]{d}$ or is there a simpler way of doing this?

Answer 1

A vector space can have many different basis, so finding a base doesn't necessarily proves that the vector spaces (in our cases the fields) aren't same.

Taking the freedom to make an educated guess the problem wants you to prove that $d \not \in \mathbb{Q}(\theta)$. Assume that the opposite it's true. Then $\sqrt[3]{d} = a + b\theta + c\theta^2$, where $a,b,c \in \mathbb{Q}$. Now cube both side and conclude that the coeffivients in front of $\theta$ and $\theta^2$ are both $0$, which will give you a system of equations, which can't be solved. Hence hte proof.

On the other side if you are familiar with field automorphisms the easist way to prove it is to assume they are equal and note that $\mathbb{Q}(\sqrt[3]{d})$ has a non trivial automorphism, which sends $\sqrt[3]{d}$ to $\sqrt[3]{d^2}$ and fixes $\mathbb{Q}$. But such an automorphism sends roots of $x^3 - 2x -2 $ to roots of the same polynomials. So it fixes both $\mathbb{Q}$ and $\theta$, as $\theta$ is the only root of $x^3 - 2x - 2$ in $\mathbb{Q}(\theta)$. This is true, as it's not the splitting field of the polynomial, so has to not contain at least two of it's roots. This means that the automorphism fixes the whole field $\mathbb{Q}(\theta) = \mathbb{Q}(\sqrt[3]{d})$, so it's the identity automorphism, which is contradiction. Hence the proof.

Answer 2

Since bases are not unique, it's not correct to claim that $\theta \neq \sqrt[3]{d}$ implies the fields are not equal.

But here's one way to solve the problem . . .

If $\sqrt[3]{d}\in \mathbb{Q}$, then $\mathbb{Q}(\sqrt[3]{d})=\mathbb{Q}$, hence $\mathbb{Q}(\theta) \ne \mathbb{Q}(\sqrt[3]{d})$, since $\theta \notin \mathbb{Q}$.

Thus, we can assume $\sqrt[3]{d} \notin \mathbb{Q}$

Suppose the fields $\mathbb{Q}(\theta)$ and $\mathbb{Q}(\sqrt[3]{d})$ are equal.

Since the fields are equal, they must have the same splitting field, $F$ say, in $\bar{\mathbb{Q}}$,

The degree of $F$ over $\mathbb{Q}$ must be a multiple of $3$, and is at most $6$, so it's either $3$ or $6$. Actually, with a little more work, we can show that it must be exactly $6$, but we don't need to be that precise. All we need is that the degree of $F$ over $\mathbb{Q}$ is not a multiple of $4$.

Since the discriminant of the polynomial $x^3 - 2x - 2$ is $-76$, which must be a perfect square in $F$, it follows that $\sqrt{-19} \in F$.

Since the discriminant of the polynomial $x^3 - 2$ is $-108$, which must be a perfect square in $F$, it follows that $\sqrt{-3} \in F$.

Thus, we must have $\mathbb{Q}(\sqrt{-19},\sqrt{-3}) \subseteq F$, yielding the tower of fields $$ \mathbb{Q} \subset \mathbb{Q}(\sqrt{-19},\sqrt{-3}) \subseteq F $$ which is clearly impossible, since $[\mathbb{Q}(\sqrt{-19},\sqrt{-3}):\mathbb{Q}]=4$ and $[F:\mathbb{Q}]$ is not a multiple of $4$.