Finding degree and basis for a field extension I want to find the degree and basis for $\mathbb{Q}(\sqrt{2},\sqrt[3]{3})$ over $\mathbb{Q}(\sqrt{2})$. I can show that the degree is $3$, since $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}] = 6$ and $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2$. Intuitively, I think $\{1,\sqrt[3]{3},\sqrt[3]{9}\}$ should be a basis, which means I should show that $x^3 - 3$ is irreducible over $\mathbb{Q}(\sqrt{2})$.
How do I do that?
 A: Your intuition is correct.
A degree three polynomial is reducible if and only if it has a root. Suppose $x^3-3$ has a root $a+b\sqrt{2}$ in $\mathbb{Q}(\sqrt{2})$; then
$$
a^3+3a^2b\sqrt{2}+6ab^2+2b^3\sqrt{2}-3=0
$$
implies
$$
\begin{cases}
a^3+6ab^2=3 \\[4px]
3a^2b+2b^3=0
\end{cases}
$$
The second equation yields $b=0$ or $3a^2+2b^2=0$ and it's easy to conclude the root does not exist.

A proof based on dimensions is possible, too. We have
$$
n=[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]=
[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt{2})]
[\mathbb{Q}(\sqrt{2}),\mathbb{Q}]\tag{*}
$$
so $2\mid n$; similarly
$$
n=[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]=
[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt[3]{3})]
[\mathbb{Q}(\sqrt[3]{3}),\mathbb{Q}]\tag{**}
$$
so $3\mid n$. Hence $6\mid n$. On the other hand, the two relation above show also $n\le 6$, hence $n=6$. Then (*) proves
$$
[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt{2})]=3
$$
and, since $x^3-3$ has $\sqrt[3]{3}$ as root, it must be irreducible over $\mathbb{Q}(\sqrt{2})$: indeed it divides the minimal polynomial of $\sqrt[3]{3}$, which must have degree $3$.
A: Consider a typical diamond of fields. First determine $\mathbb{Q}(\sqrt{2}, \sqrt[3]{3})$ over the rationals. Verify that $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]=6$. This is since $x^3-3$ is irreducible by Eisenstein with $p=3$ and $x^2-2$ irreducible over the rationals as well thus,
$$\text{gcd}([\mathbb{Q}(\sqrt{2}):\mathbb{Q}],[\mathbb{Q}(\sqrt[3]{3}):\mathbb{Q}])=1 \Rightarrow [\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]=6$$
 Thus if you consider $\mathbb{Q}\subset \mathbb{Q}(\sqrt{2}) \subset\mathbb{Q}(\sqrt{2},\sqrt[3]{3})$ then we must have $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}:\mathbb{Q}(\sqrt{2})]=3$ since we have that $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$; hence just apply the tower theorem. 

A vector space basis is $\{1,\sqrt[3]{3},\sqrt[3]{9}\}$ so your intitution was correct. 
