Why is $L=\mathbb{Q}(\sqrt[1]{2})\cup\mathbb{Q}(\sqrt[2]{2})\cup\mathbb{Q}(\sqrt[3]{2})\cup\cdots$ a field?

Question

The title sais it already:

Why is $L=\mathbb{Q}(\sqrt[1]{2})\cup\mathbb{Q}(\sqrt[2]{2})\cup\mathbb{Q}(\sqrt[3]{2})\cup\cdots$ a field?

The hint provided in my textbook is: $\mathbb{Q}(\sqrt[n]{2})\cup\mathbb{Q}(\sqrt[m]{2})\subset\mathbb{Q}(\sqrt[mn]{2})$, but this doesn't really get me anywhere. Actually, I have no idea what to do whatsoever. Could anyone clarify or give some hint please?

Answer 1

The smallest field containing $K(a^{1/n})$ and $K(a^{1/m})$ is $K(a^{1/n},a^{1/m}) = K(a^{1/lcm(n,m)})$

Answer 2

Suppose you take any two elements $x, y\in L.$ This means that there are some positive integers $m$ and $n$ such that $x\in \mathbb{Q}(\sqrt[m]{2})$ and $y\in \mathbb{Q}(\sqrt[n]{2})$.

From the hint in your textbook (that you wrote in your question), this implies that $x,y\in \mathbb{Q}(\sqrt[mn]{2}).$

Now, your end goal is to show that $x+y$ and $xy$ are both elements of $L$ as well (and that $L$ is closed under inverses for nonzero elements).

Is there any information about $\mathbb{Q}(\sqrt[mn]{2})$ that you can use which might help here (since $x$ and $y$ are elements of this set)?