Find the minimal polynomial of $2^{1/5}$ over $\mathbb{Q}(\sqrt{3})$ What I have been trying to do: we have $\mathbb{Q}\left(2^{1/5}3^{1/2}\right) = \mathbb{Q}\left(3^{1/2},2^{1/5}\right)$, $\left[\mathbb{Q}\left(2^{1/5}3^{1/2}\right): \mathbb{Q}\right] = \left[\mathbb{Q}\left(2^{1/5}3^{1/2}\right):\mathbb{Q}\left(3^{1/2}\right)\right]\cdot \left[\mathbb{Q}\left(3^{1/2}\right):\mathbb{Q}\right]$, and $\left[\mathbb{Q}\left(3^{1/2}\right):\mathbb{Q}\right] = 2$. I would like to finish by showing that $x^{10} - 2^23^5$ is the minimal polynomial of $2^{1/5}3^{1/2}$ over $\mathbb{Q}$ (maybe it isn't?). Thanks in advance,
 A: Hint
As you observed
$$\left[\mathbb{Q}\left(2^{1/5}3^{1/2}\right): \mathbb{Q}\right] = \left[\mathbb{Q}\left(2^{1/5}3^{1/2}\right):\mathbb{Q}\left(3^{1/2}\right)\right]\cdot \left[\mathbb{Q}\left(3^{1/2}\right):\mathbb{Q}\right]$$
and $\left[\mathbb{Q}\left(3^{1/2}\right):\mathbb{Q}\right]=2$. Therefore $\left[\mathbb{Q}\left(2^{1/5}3^{1/2}\right): \mathbb{Q}\right] $ is divisible by $2$.
Now, switch 2 and 3:
$$\left[\mathbb{Q}\left(2^{1/5}3^{1/2}\right): \mathbb{Q}\right] = \left[\mathbb{Q}\left(2^{1/5}3^{1/2}\right):\mathbb{Q}\left(2^{1/5}\right)\right]\cdot \left[\mathbb{Q}\left(2^{1/5}\right):\mathbb{Q}\right]$$
and $\left[\mathbb{Q}\left(2^{1/5}\right):\mathbb{Q}\right]=5$. Therefore $\left[\mathbb{Q}\left(2^{1/5}3^{1/2}\right): \mathbb{Q}\right] $ is divisible by $5$.
Since $2$ and $5$ are relatively prime, $\left[\mathbb{Q}\left(2^{1/5}3^{1/2}\right): \mathbb{Q}\right] $ is divisible by $10$. 
Therefore, the minimal polynomial of $2^{1/5}3^{1/2}$ has degree $10$. Since your polynomial is divisible by the minimal polynomial and had degree 10 you can conclude that....
Note Your also need to explain why $2^{1/5}, 3^{1/2}\in \mathbb{Q}\left(2^{1/5}3^{1/2}\right)$. This is easy, but you need to explain it as you use it in the towering lemma.
A: The minimal polynomial of $2^{1/5}$ over $\mathbb{Q}$ is $x^5-2$ (use Einsenstein Criterion to prove it!). So $[\mathbb{Q}(2^{1/5}):\mathbb{Q}]=5$. Now $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]=2$. Since $2$ and $5$ are relatively prime, we have that $[\mathbb{Q}(2^{1/5},\sqrt{3}):\mathbb{Q}]=2 \times 5=10$. Now, you can easily show that $[\mathbb{Q}(\sqrt{3})(2^{1/5}):\mathbb{Q}(\sqrt{3})]=5$. Hence the minimal polynomial of $2^{1/5}$ over $\mathbb{Q}(\sqrt{3})$ has degree $5$ and hence is $x^5-2$.
