Let $S=\big\{\sqrt[n]{3}\colon n\in \mathbb{N}\big\}$. Is the extension $\mathbb{Q}[S]\colon\mathbb{Q}$ algebraic? A field extension $L\colon K$ is algebraic if every element in $\alpha \in L$ is algebraic over $K$. An elemenet $\alpha \in L$ is algebraic over $K$ if there exists a polynomial $f \in K[x]$ such that $f(\alpha)=0$.
$\mathbb{Q}[S]=\{g(s)\colon g \in \mathbb{Q}[x], s \in S\}$.
I'm not sure how to go about proving if an element of $\mathbb{Q}[S]$ is algebraic over $\mathbb{Q}$? Any help would be appreciated, thank you.
 A: The key fact you need to show is that if $ \alpha, \beta$ algebraic over some field $k$, then $\alpha+\beta,\alpha/\beta, \alpha\beta$ are all algebraic as well (this tells you that algebraic combinations of your $S$ are all algebraic over $\Bbb Q$). 
To show this, you can use the following equivalence. $\alpha$ is algebraic over $k$ if and only if $k(\alpha)$ is a finite dimensional $k$-vector space (obviously you need to show this if you haven't done so before). 
A: Note that $[{\Bbb Q}[\sqrt[n]{3}]:\Bbb{Q}]=n$. Can you go on with the theorem that any finite extension must be algebraic?

${\Bbb Q}[\sqrt{3},\sqrt[3]{3},\cdots,\sqrt[n]{3},\cdots]$ is not a finite extension. But ${\Bbb Q}[\sqrt{3},\sqrt[3]{3},\cdots,\sqrt[n]{3}]$ is a finite extension for any $n$ and hence algebraic. Let $b\in$${\Bbb Q}[\sqrt{3},\sqrt[3]{3},\cdots,\sqrt[n]{3},\cdots]$, then $b$ is a finite ${\Bbb Q}$-linear combination of some elements in $S$. It follows that
$$
b\in{\Bbb Q}[\sqrt{3},\sqrt[3]{3},\cdots,\sqrt[k]{3}]
$$
for some $k\in{\Bbb N}$. Therefore $b$ is algebraic.
A: Here is an example of a useful finiteness principle: let $S_k = \{ \sqrt[n]{3} \mid n \in \mathbb{N}, n \leq k \} $. Then, we have
Lemma: If $x \in \mathbb{Q}[S]$, then there exists a $k$ such that $x \in \mathbb{Q}(S_k)$
A: Another approach, closely related to that of @Hurkyl, could consist in noting that
$$
\mathbf{Q}[S] = \bigcup_{k \in \mathbf{N}} \mathbf{Q}[\sqrt[k!]{3}],
$$
where the union is ascending.
