Prove that $[\mathbb{Q}(\sqrt[4]{2}, \sqrt{3}):\mathbb{Q}] = 8$

Question

Prove that $[\mathbb{Q}(\sqrt[4]{2}, \sqrt{3}):\mathbb{Q}] = 8$.

I proved that

-(i) $$[F[\alpha_{1},...,\alpha_{n}]:F] \leq \prod_{1}^{n}[F(\alpha_{i}):F]$$ with $K$ a field extension of $F$ and $\alpha_{i} \in K$ algebraic over $F$;

-(ii) If $K$ is a field extension of $F$ and $\alpha \in K$ is algebraic over $F$, then $F[\alpha] = F(\alpha)$;

-(iii) Let $F \subseteq L \subseteq K$ be a fields, so $$[K:F]=[K:L][L:F].$$

I can to use this for solve the question? I cannot see how. I don't know if I can claim that $[\mathbb{Q}(\sqrt[4]{2},\sqrt{3}):\mathbb{Q}] = [\mathbb{Q}[\sqrt[4]{2},\sqrt{3}]:\mathbb{Q}]$.

Answer 1

Hint:

$$[\Bbb Q(\sqrt[4]2,\,\sqrt3):\Bbb Q]=[\Bbb Q(\sqrt[4]2,\,\sqrt3):\Bbb Q(\sqrt[4]2)]\cdot[\Bbb Q(\sqrt[4]2):\Bbb Q]$$

Now check that

$$\begin{cases}x^4-2\in\Bbb Q[x]\\{}\\ x^2-3\in\Bbb Q(\sqrt[4]2)[x]\end{cases}\;\;\;\;\text{are irreducible}$$

Answer 2

HINT:

Assume that $\sqrt{3} \in \mathbb{Q}(\sqrt[4]{2})$. The morphism of fields from $\mathbb{Q}(\sqrt[4]{2})$ to $\mathbb{Q}(i\sqrt[4]{2})$ maps $\sqrt{3}$ to a root of $x^2 -3$, so to a real number. Therefore we must have $\sqrt{3}\in \mathbb{Q}(\sqrt{2})$, and we get a contradiction.