Field extension $\mathbb{Q} \subset \mathbb{Q}(\sqrt[4]{5} i, \sqrt[3]{2} )$

Question

I was studying for my algebra exam when I came across this problem. It asked to find $[\mathbb{Q}(\sqrt[4]{5} i, \sqrt[3]{2} ): \mathbb{Q}]$ which was not a problem using towers.

The next two bits of the question I struggled on. It asked to show that $\sqrt[3]{2}\sqrt[4]{5} i $ is a primitive element of $\mathbb{Q}(\sqrt[4]{5} i, \sqrt[3]{2} )$. I suppose that I have to show that $\mathbb{Q}(\sqrt[4]{5} i, \sqrt[3]{2} ) = \mathbb{Q}(\sqrt[3]{2}\sqrt[4]{5} i)$. I'm not sure on how to start this.

The third part asks to find a minimal polynomial of $\sqrt[3]{2}\sqrt[4]{5} i$ over $\mathbb{Q}$ where again I am lost.

Any help is appreciated.

Answer 1

To show that $\mathbb{Q}(\sqrt[4]{5} i, \sqrt[3]{2} ) = \mathbb{Q}(\sqrt[3]{2}\sqrt[4]{5} i)$, you only need to show that both $\sqrt[4]{5} i$ and $\sqrt[3]{2}$ are in the field $\mathbb{Q}(\sqrt[3]{2}\sqrt[4]{5} i)$.

But this is easy: we have $(\sqrt[3]{2}\sqrt[4]{5} i)^4 = 10\sqrt[3]{2}$ and $(\sqrt[3]{2}\sqrt[4]{5} i)^9 = 200\sqrt[4]{5} i$, hence $\sqrt[3]{2} = \frac 1 {10}(\sqrt[3]{2}\sqrt[4]{5} i)^4$ and $\sqrt[4]{5} i = \frac 1{200} (\sqrt[3]{2}\sqrt[4]{5} i)^9$ are both in $\mathbb{Q}(\sqrt[3]{2}\sqrt[4]{5} i)$.

Finally, by the first two questions, it should be clear that the degree of $\mathbb{Q}(\sqrt[3]{2}\sqrt[4]{5} i)$ over $\Bbb Q$ is $12$. But the element $\sqrt[3]{2}\sqrt[4]{5} i$ is already root of a degree $12$ polynomial over $\Bbb Q$, namely $x^{12} - 2000$. Hence this is the minimal polynomial of $\mathbb{Q}(\sqrt[3]{2}\sqrt[4]{5} i)$ over $\Bbb Q$.

Answer 2

Hint

Obviously, $$\sqrt[4]{5}\sqrt[3]2i\in \mathbb Q(\sqrt[4]{5}i,\sqrt[3]{2}).$$ Moreover, its minimal polynomial is given by $$x^{12}-5^32^4.$$ Therefore $$\mathbb Q(\sqrt[4]{5}i,\sqrt[3]{2})=\mathbb Q(\sqrt[4]{5}\sqrt[3]2i),$$ because both extension has the same degree.