Let $K=\mathbb{Q}(\sqrt{5},\sqrt[3]{7})\subset \mathbb{C}$. To compute $[K:\mathbb{Q}]$, I know that
\begin{align*} [\mathbb{Q}(\sqrt{5}):\mathbb{Q}]&=2\\ [\mathbb{Q}(\sqrt[3]{7}):\mathbb{Q}]&=3. \end{align*}
Can I say that $$2<[K:\mathbb{Q}]\leq 6$$ and because $2\not |$ $3,5$ and $3\not | $ $4$, then $[K:\mathbb{Q}]=6$?
And, if $\gamma=\sqrt{5}\sqrt[3]{7}$, the minimal polynomial of $\gamma$ over $\mathbb{Q}(\gamma)$ would be $p(x)=x^6-5^37^2?$
I'm a bit insecure about this.
Thanks in advance.
If $K=\Bbb Q(\sqrt5,\sqrt[3]7)$ then $|K:\Bbb Q|=|K:\Bbb Q(\sqrt5)| |\Bbb Q(\sqrt5):\Bbb Q|$ so $|K:\Bbb Q|$ is divisible by $2$. Similarly $|K:\Bbb Q|$ is divisible by $3$ (why?). So it is divisible by $6$ (as $2$ and $3$ are coprime). But $|K:\Bbb Q|\le 6$ (why?).