Is $\{1,3^{1/2},3^{1/3},3^{2/3}\}$ a basis of $\Bbb Q(3^{1/2},3^{1/3})$

Question

Consider the extension $\Bbb Q(3^{1/2},3^{1/3})\supset\Bbb Q$

I think it is an extension of degree $4$ and a basis is $\{1,3^{1/2},3^{1/3},3^{2/3}\}$ but I'm not sure exactly how to prove it. Maybe it's a degree $2$ extension?

I know that $\Bbb Q(2^{1/3})$ is a degree $3$ extension of $\Bbb Q$ because the minimal polynomial is $f(x)=x^3-2$

Answer 1

Here's a useful lemma to prove:

Lemma: Let $F$ be a field, $E/F$ a finite field extension, and $S \subseteq E$ be a set of elements. Then $S$ is an $F$-linear spanning set for $E$ if and only if:

• $S$ is a set of generators for the extension $E/F$
• For every $a,b \in S$, the product $ab$ is in the $F$-linear span of $S$

$\{1,3^{1/2},3^{1/3},3^{2/3}\}$ is a $\mathbb{Q}$-linear basis for $\mathbb{Q}(3^{1/3}, 3^{1/2})$ if and only if it is a spanning set and linearly independent.

Once you've proven this lemma, you can use it to check if this set is a spanning set. And if its not, the calculation you do for the lemma will suggest more elements to take.

The main idea of the proof is to relate the second bullet of the lemma to the assertion that the span of $S$ is closed under multiplication.

Answer 2

Your field extension $K:=\mathbb{Q}\left(3^{\frac12},3^{\frac13}\right)$ of $\mathbb{Q}$ contains intermediate fields $\mathbb{Q}\left(3^{\frac12}\right)$ and $\mathbb{Q}\left(3^{\frac13}\right)$ of degrees $2$ and $3$, respectively. Thus, $[K:\mathbb{Q}]$ is divisible by both $2$ and $3$, so $$[K:\mathbb{Q}]\geq 6\,.$$ Therefore, you are missing at least two elements in your "basis." (In fact, $[K:\mathbb{Q}]=6$, so you need exactly two more basis elements.)