Clarifications about extension fields, its basis

Question

I am quite confused about this topic and so I want some clarification by answering this question. Find the degree of $\mathbb{Q}$$(\sqrt[3]{2} ,\sqrt{3} )$ and its basis over the field $\mathbb{Q}$. Note that $\mathbb{Q}$$(\sqrt[3]{2} ,\sqrt{3} )$ is a simple extension field over $\mathbb{Q}$

Answer 1

The basis could be (there are many) $\{1,\sqrt[3]{2},\sqrt[3]{4},\sqrt3,\sqrt3\cdot\sqrt[3]2,\sqrt3\cdot\sqrt[3]4 \}$.

It is not hard to prove that this is a basis as they are linearly independent and span the whole space, $\mathbb{Q}(\sqrt[3]2,\sqrt3)$.

One way you can find the basis is that: first, pick any random vector in it, say $1$. Then, pick another that is linearly independent to the previous ones. $\sqrt[3]2$ is in the space and linearly independent to $1$, so it is a good pick.

Since $\mathbb{Q}(\sqrt[3]2,\sqrt3)$ is a field, $(\sqrt[3]2)^2=\sqrt[3]4$ must be in it, and independent to the previous two, so add it in your basis. You can keep doing this until you get a basis of $6$ vectors, which shows it is a field extension of degree $6$.

Answer 2

Let me answer your question in a more general context.

Consider an extension field $k\subset K$ and two subextensions $k\subset E\subset K, k\subset F\subset K$.
If these extensions have finite degrees $[E:k]=n, [F:k]=r$ and if these degrees are relatively prime, i.e. (n,r)=1, then the composite extension has degree $[E\cdot F:k]=nr$.

The proof is short: if $[E\cdot F:k]=x$, multiplicativity in towers of fields implies the divisibility relations $n\mid x, r\mid x$, which in turn imply $$nr\mid x$$ by relative primeness of $n,r$.
On the other hand, we have $x=[E\cdot F:k]=[E\cdot F:F][ F:k]$by multiplicativity in towers again and thus $$x\leq nr$$ once you realize that $[E\cdot F:F]\leq [E :k]$
The displayed relations together show what we want, namely$$x=[E\cdot F:k]=nr$$

In slightly more sophisticated language this shows that the canonical morphism of $k$-algebras $E\otimes F\to E\cdot F:x\otimes y \mapsto xy$ is bijective.
This makes it crystal clear that given $k$-vector space bases $(e_i)$ of $E$ and $(f_j)$ of $F$ the products $e_i\cdot f_j$ form a base of $E\cdot F$