Show $ \mathbb{Q}(\sqrt[4]{2},\sqrt[4]{18}) = \mathbb{Q}(\sqrt[4]{2},\sqrt{3}) $

Question

The problem :

Note $a = \sqrt[4]{2}$ and $b = \sqrt[4]{18}$. Show $\mathbb{Q}(a,b) = \mathbb{Q}(a,\sqrt{3})$

My works :

I found $[ \mathbb{Q}(\sqrt[4]{2}) : \mathbb{Q} ] = [ \mathbb{Q}(\sqrt[4]{18}) : \mathbb{Q} ] = 4$.

In addition, since $b = \sqrt[4]{2} \sqrt{3} = a \sqrt{3}$, so $\mathbb{Q}(a,b) = \mathbb{Q}(a,a\sqrt{3})$ and here I don't know what to do. Should I show the following equality to finish the proof?

$$ \mathbb{Q}(a,a\sqrt{3}) = \mathbb{Q}(a,\sqrt{3}) $$

I am a beginner in number theory. Thanks you

Answer 1

$\mathbb{Q}(a,a\sqrt{3})\subseteq \mathbb{Q}(a,\sqrt{3})$ since $a\in \mathbb{Q}(a,\sqrt{3})$ and $a\sqrt{3}\in \mathbb{Q}(a,\sqrt{3})$

$\mathbb{Q}(a,a\sqrt{3})\supseteq \mathbb{Q}(a,\sqrt{3})$ since $a\in \mathbb{Q}(a,a\sqrt{3})$ and $\sqrt{3}=\frac{a\sqrt{3}}{a}\in \mathbb{Q}(a,a\sqrt{3})$

Answer 2

$$\sqrt[4]{18}=\sqrt[4]2\cdot\sqrt3,$$ which says $$\mathbb{Q}(\sqrt[4]{2},\sqrt[4]{18})\subset \mathbb{Q}(\sqrt[4]{2},\sqrt{3})$$ The second is similar: $$\sqrt3=\frac{\sqrt[4]{18}}{\sqrt[4]2},$$ which says $$ \mathbb{Q}(\sqrt[4]{2},\sqrt{3})\subset\mathbb{Q}(\sqrt[4]{2},\sqrt[4]{18}).$$