Consider the field extension $K = \mathbb { Q } ( \sqrt { 2 } , \sqrt [ 3 ] { 2 } )$ over $\mathbb { Q }$.
By looking at the intermediate extensions $\mathbb{Q}(\sqrt 2)$ of degree 2 and $\mathbb{Q}(\sqrt [ 3 ] { 2 })$ of degree 3 and using the fact that $\mathrm{gcd}(2,3) = 1$, one can show using degree and divisibility arguments that $K$ is a $\mathbb{Q}-$vector space of dimension 6 (i.e. $[ \mathbb { Q } ( \sqrt { 2 } , \sqrt [ 3 ] { 2 } ) : \mathbb { Q } ] = 6$).
One can then find bases for the intermediate extensions: $$ \mathbb{Q}(\sqrt 2) = \mathrm{span} \{ 1, \sqrt 2\}\\ \mathbb{Q}(\sqrt [ 3 ] { 2 }) = \mathrm{span} \{ 1, \sqrt[3]{2}, \sqrt[3]{2}^{~2}\} $$
and thus obtain a basis for $K$ by taking the 6 distinct products of basis elements from above: $$ \mathbb { Q } ( \sqrt { 2 } , \sqrt [ 3 ] { 2 } ) = \mathrm{span} \left\{ 1 , \sqrt { 2 } , \sqrt [ 3 ] { 2 } , \sqrt [ 3 ] { 2 } ^ { ~2 } , \sqrt { 2 } \sqrt [ 3 ] { 2 } , \sqrt { 2 } \sqrt [ 3 ] { 2 } ^ { ~2 } \right\} $$
Motivated by the last step of taking all products of basis elements, I am wondering: is it true that $$ \mathbb { Q } ( \sqrt { 2 } , \sqrt [ 3 ] { 2 } ) = \mathbb { Q } ( \sqrt { 2 }) \otimes_{\mathbb Q} \mathbb { Q } (\sqrt [ 3 ] { 2 } ) $$ where the tensor product is taken in the category of $\mathbb{Q}-$modules?
In cases when there are perhaps more intermediate extensions, are there any criteria for when combining bases in this way will yield the correct basis for the total extension?
(For example, it seems to work here because neither intermediate extension is contained in the other - so do you perhaps need intermediate extensions that only intersect trivially?)