Abstract Algebra: need help with next step finding $[\mathbb{Q}(\sqrt{5},\sqrt[3]{2} ): \mathbb{Q}(\sqrt{5})]$

Question

Question asks: Compute $[\mathbb{Q}(\sqrt{5},\sqrt[3]{2} ): \mathbb{Q}] $

I've worked it to

=$[\mathbb{Q}(\sqrt{5},\sqrt[3]{2} ): \mathbb{Q}(\sqrt{5})][\mathbb{Q}(\sqrt{5}): \mathbb{Q}] $

= $[\mathbb{Q}(\sqrt{5},\sqrt[3]{2} ): \mathbb{Q}(\sqrt{5})] * 2 $

so what is the degree $[\mathbb{Q}(\sqrt{5},\sqrt[3]{2} ): \mathbb{Q}(\sqrt{5})]$

Answer 1

The degree of $\mathbb{Q}(\sqrt{5}, \sqrt[3]{2})$ over $\mathbb{Q}(\sqrt{5})$ is the degree of the minimal polynomial $p$ for $\sqrt[3]{2}$ over $\mathbb{Q}(\sqrt{5})$. An obvious candidate is the usual $p(x) = x^3 - 2$. Prove that this is indeed the minimal polynomial by showing that no polynomial of lower degree can annihilate $\sqrt[3]{2}$. For instance, clearly no linear polynomial will. Why can no quadratic?

Answer 2

The polynomial $x^3-2$ in $\mathbb{Q}(\sqrt{5})$ is reducible if and only if it has a root. So suppose $(a+b\sqrt{5})^3=2$. This becomes $$ a^3+3a^2b\sqrt{5}+15ab^2+5b^3\sqrt{5}=2 $$ and so $$ \begin{cases} a^3+15ab^2=2\\[4px] 3a^2b+5b^3=0 \end{cases} $$ Can you finish?

Answer 3

You have the following diagram of field extensions and degrees:

So, $[\mathbb Q(\sqrt 5,\sqrt[3]{2}):\mathbb Q]$ is a multiple of $2$, a multiple of $3$ and at most $2 \cdot 3$, so it is $6$.

Note that it follows that $x^3 - 2$ is the minimum polynomial of $\sqrt[3]{2}$ over ${\mathbb Q}(\sqrt 5)$ and that $x^2 - 5$ is the minimum polynomial of $\sqrt5$ over ${\mathbb Q}(\sqrt[3]{2})$.