Basis of $\mathbb Q (3^{1/3}, \xi )$

Question

$\xi = e^{\frac{\pi}3 i}$

$K=\mathbb Q (3^{1/3}, \xi ) = \mathbb Q (3^{1/3})(\xi)$

So every element can be written as $a+b \xi$ where $a,b \in \mathbb Q (3^{1/3})$

And we can rewrite these as $a=c+d (3^{1/3})$ and $b=e+f (3^{1/3})$ with $c,d,e,f \in \mathbb Q$

Then $a+b \xi = c+d (3^{1/3}) + be \xi+bf \xi(3^{1/3})$

So all elements can be written as $a+b (3^{1/3}) + c \xi+d \xi(3^{1/3})$

So claim that a basis is $\{1, 3^{1/3}, \xi, \xi(3^{1/3}) \}$

So this spans $K$ as we can see. Lets prove it is linearly independent.

$a+b (3^{1/3}) + c \xi+d \xi(3^{1/3})=0$

Compare $\xi$ terms: $c+d (3^{1/3})=0$. Suppose $d \neq 0$ so $3^{1/3}= c/d$, contradiction. So $d=0$ so $c=0$.

Compare non $\xi$ terms: $a+b (3^{1/3})=0$, similarly we get $a,b=0$.

So it is linearly independent. So it is a basis.

Is this OK? I was unsure if i could compare the $\xi$ terms and "non $\xi$" terms.

Answer 1

You seem to be saying that the basis consists of four elements, namely $\{1,\xi,\lambda,\xi\lambda)$ where $\lambda^3=3$. But $\lambda^2$ can not be written as a $\Bbb Q$-linear combination of these four. If you’re not saying that the basis consists of four elements, then I would like you to make your notation more explicit.

EDIT: The basic fact never to forget is that field extension degree (the dimension of the big field as a vector space over the little field) is multiplicative. It should be clear that the dimension of $\Bbb Q(\sqrt[3]3\,)$ over $\Bbb Q$ is three — the degree of a simple extension is also the degree of the minimal polynomial for the generator. Our biggest field contains an extension of degree two, and an extension of degree three, so must have degree at least six.

Maybe the sermon I should be delivering is for you to do examples, examples, examples. If our insight comes only from quadratic extensions, then we will have the wrong mental image and expectations when it comes to bigger extensions. What’s the reciprocal of $2+\sqrt[3]3$? Apply the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$, which I hope you learned in high school, with $a=2$ and $b=\sqrt[3]3$, so that $8+3=(2+\sqrt[3]3)(4+2\sqrt[3]3+\sqrt[3]9)$. This is trick that isn’t always useful, but here it gives you an answer.