Find basis of $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$ Find a basis of $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$ over $\mathbb{Q}$.
I think the basis should be $1, \sqrt{2}, \sqrt[3]{4}, \sqrt[3]{4}\cdot\sqrt{2},\sqrt[3]{4}^2,\sqrt[3]{4}^2\cdot\sqrt{2}$. So I want to show that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})=\mathbb{Q}(\sqrt{2},\sqrt[3]{4})$.
It is clear that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{4})$, but how can I show that $\mathbb{Q}(\sqrt{2},\sqrt[3]{4})\subseteq \mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$?
 A: Say $\Bbb Q(\sqrt{2},\sqrt[3]{4})\ne\Bbb Q(\sqrt{2}+\sqrt[3]{4})$, so $\sqrt{2},\sqrt[3]{4}\not\in\Bbb Q(\sqrt{2}+\sqrt[3]{4})$. Let $n$ be the index of the two fields.
What does $\Bbb Q(\sqrt{2},\sqrt[3]{4})=\Bbb Q(\sqrt{2}+\sqrt[3]{4})(\sqrt{2})$ and $\sqrt{2}\not\in\Bbb Q(\sqrt{2}+\sqrt[3]{4})$ tell us about $n$?
On the other hand, as $\sqrt[3]{4}\not\in\Bbb Q(\sqrt{2}+\sqrt[3]{4})$ and the other two roots of $X^3-4$ are complex, what do we know about $X^3-4$? Then what does $\Bbb Q(\sqrt{2},\sqrt[3]{4})=\Bbb Q(\sqrt{2}+\sqrt[3]{4})(\sqrt[3]{4})$ say about $n$?

As for finding a basis. You know that if $K(\alpha)/K$ has degree $n$ then $\{1,\alpha,\cdots,\alpha^{n-1}\}$ is a $K$-basis for $L$. Furthermore if $L/M/K$ and $\cal A$ is an $M$-basis for $L$ and $\cal B$ is a $K$-basis for $M$, then
$${\cal AB}=\{ab:a\in{\cal A},b\in{\cal B}\}$$
is a $K$-basis for $L$. Can you prove this? This is a nice fact to have at hand.
Use the above to find a basis for $\Bbb Q(\sqrt{2})/\Bbb Q$ and $\Bbb Q(\sqrt{2})(\sqrt[3]{4})/\Bbb Q(\sqrt{2})$ and then construct a basis for the extension $\Bbb Q(\sqrt{2},\sqrt[3]{4})/\Bbb Q$ out of these two bases. This is exactly what you have, but the above discussion provides formal justification for why it is a basis.
A: Let’s call $\sqrt2=\alpha$, $\root3\of4=\beta$ and $\alpha+\beta=\gamma$. It’s enough to show that $\sqrt2\in\mathbb Q(\gamma)$. Since the minimal polynomial for $\beta$ over $\mathbb Q(\alpha)$ is $f(X)=X^3-4$, the minimal polynomial for $\gamma$ over the same field is $g(X)=f(X-\alpha)=X^3-3\alpha X^2+3\alpha^2X-\alpha^3-4
=X^3-3\alpha X^2+6X-2\alpha-4$. Thus we have $0=g(\gamma)$, which we can solve for $\alpha=(\gamma^3+6\gamma-4)/(3\gamma^2+2)$, if my calculations involve no error.
