Infinite union of fields. Let $L=\cup_{n}\mathbb{Q}(\sqrt[n]{2})$, prove that L is a field.
and L is not finite over Q, but L is algebraic over Q.
My first step is to show that $\mathbb{Q}(\sqrt[n]{2})\cup \mathbb{Q}(\sqrt[m]{2})$ is contained in $\mathbb{Q}(\sqrt[mn]{2})$. 
Can you help me?
Thanks!
 A: Your first step of showing that $\mathbb Q (\sqrt[n]{2}) \cup \mathbb Q(\sqrt[m]{2}) \subset \mathbb Q(\sqrt[nm]{2})$ is very useful! This fact will help you show that if you add or multiply any two elements in $L$, the answer will also be in $L$. This will then enable you to show that $L$ is a field... but you probably already knew that!
Next, you would like to show that $L$ is not finite over $\mathbb Q$. Perhaps you could ask yourself: what is $[\mathbb Q(\sqrt[n]{2}) : \mathbb Q]$? 
(Remember, if $\alpha \in \mathbb C$ has a minimal polynomial of degree $N$ over $\mathbb Q$, then $[\mathbb Q(\alpha) : \mathbb Q] = N$. So what is the minimal polynomial of $\sqrt[n]{2}$ over $\mathbb Q$? And therefore, what is $[\mathbb Q(\sqrt[n]{2}) : \mathbb Q]$?)
Once you have done this, note that $L$ has each $\mathbb Q (\sqrt[n]{2})$ as a subfield, so $[L:\mathbb Q] \geq [\mathbb Q(\sqrt[n]{2}) : \mathbb Q]$ for each $n$. Can you see where this is going?
Finally, you would like to show that every element of $L$ is algebraic over $\mathbb Q$. But since $L$ is the union of the $\mathbb Q(\sqrt[n]{2})$'s, you just need to show that for each $n$, it is true that every element of $\mathbb Q(\sqrt[n]{2})$ is algebraic over $\mathbb Q$. Can you do this?
(Remember, if $\alpha$ is NOT algebraic over $\mathbb Q$, then $[\mathbb Q(\alpha) : \mathbb Q] = \infty$.)
