Degree of Extensions Compute $|\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}|$
My attempt so far :
$|\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}|$
$=|\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})|$.$|\mathbb Q(\sqrt[5] {3}):\mathbb{Q}|$
where:
$|\mathbb Q(\sqrt[5] {3}):\mathbb{Q}|=5$
Now i need to calculate:
$|\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})|$
Which i think is $2$ but i am unsure to calculate . Is this the best method and if so how do i finish the problem?
Also how do i find a bases of $\mathbb Q(\sqrt {2},\sqrt[5] {3})$ over $\mathbb Q   $ ?
 A: $[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}]$ is a multiple of both $2$ and $5$ because
$$
[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}]
= [\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt {2})]
\,[\mathbb Q(\sqrt {2}):\mathbb{Q}]
= 2[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt {2})]
$$
$$
[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}]
= [\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})]
\, [\mathbb Q(\sqrt[5] {3}):\mathbb{Q}]
=5[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})]
$$
On the other hand,
$$
[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})]
=
[\mathbb Q(\sqrt[5] {3})(\sqrt {2}):\mathbb Q(\sqrt[5] {3})]\le 2
$$
because $\sqrt {2}$ is a root of quadratic polynomial with coefficients in $\mathbb Q \subseteq \mathbb Q(\sqrt[5] {3})$,
and so
$$
[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}]
= [\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})]
\, [\mathbb Q(\sqrt[5] {3}):\mathbb{Q}]
\le 2 [\mathbb Q(\sqrt[5] {3}):\mathbb{Q}] = 2 \cdot 5 = 10
$$
Therefore, $[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}]=10$.
