Question about determining $[\Bbb Q(\sqrt[3]2, \sqrt[4]3) : \Bbb Q]$

Question

I'm trying to show that $[\Bbb Q(\sqrt[3]2, \sqrt[4]3) : \Bbb Q] = 12$.

Part of the proof specifies that $[\Bbb Q(\sqrt[3]2, \sqrt[4]3) : \Bbb Q(\sqrt[3]2)]$ is at most $4$ because $\sqrt[4]3$ is a zero of $x^4 - 3 \in \Bbb Q(\sqrt[3]2)[x]$.

Why does the fact that $\sqrt[4]3$ is a zero of $x^4 - 3 \in \Bbb Q(\sqrt[3]2)[x]$ imply that it is at most $4$?

Answer 1

By definition we have that $[F(k) : F] = n$, where $n$ is the degree of the minimal polynomial $P(x) \in F[x]$ s.t. $P(k) = 0$. Now we have that:

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

As you've mentioned we have that $\sqrt[4]{3}$ is a zero of $x^4 - 3 \in \mathbb{Q}(\sqrt[3]{2})[x]$. Therefore we have that the minimal polynomial of $\sqrt[4]{3}$ divides $x^4-3$. This is true as any principal ideal generated by an irreducible polynomial in $F[x]$ where $F$ is a field is a prime ideal too. Now show that $x^4 -3$ is irreducible over $\mathbb{Q}(\sqrt[3]{2})$, meaning that it's infact the minimal polynomial itself. Therefore $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) : \mathbb{Q}(\sqrt[3]{2}) ] = 4$.

Similiar reasoning leads us to concluding that $[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q} ] = 3$. Therefore: $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) : \mathbb{Q} ] = 12$

Answer 2

$3$ divides $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) : \mathbb{Q} ]$ because $ \mathbb{Q}(\sqrt[3]{2}) \subseteq \mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3})$ and $ [\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q} ] = 3 $.

$4$ divides $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) : \mathbb{Q} ]$ because $ \mathbb{Q}(\sqrt[4]{3}) \subseteq \mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3})$ and $ [\mathbb{Q}(\sqrt[4]{3}) : \mathbb{Q} ] = 4 $.

Therefore, $12$ divides $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) : \mathbb{Q} ]$. In particular, $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) : \mathbb{Q} ] \ge 12$

Now $\sqrt[4]{3}$ is a zero of $X^4-3 \in \mathbb{Q}[X] \subseteq \mathbb{Q}(\sqrt[3]{2})[X]$ and so $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) : \mathbb{Q}(\sqrt[3]{2}) ] \le 4$. Hence, $$[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) : \mathbb{Q} ] = [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) : \mathbb{Q}(\sqrt[3]{2}) ] \ [\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q} ] \le 4 \cdot 3 = 12$$

Therefore $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) : \mathbb{Q} ] = 12$.