I have this problem in a exercise list:
"Prove that $K=\mathbb{Q}(2^{\frac{1}{2}},2^{\frac{1}{3}}, 2^{\frac{1}{4}}, \ldots)$ is an algebraic extension, but not a finite extension of $\mathbb{Q}$."
What I did:
Let $\alpha \in K$, then $\alpha = a_0 + a_12^{\frac{1}{n_1}}+a_22^{\frac{1}{n_2}}+\cdots+a_m2^{\frac{1}{n_m}}$.
Now, since each factor of the sum above is algebraic over $\mathbb{Q}$, it follows that $\alpha$ is indeed algebraic over $\mathbb{Q}$ (because the set of algebraic numbers is a field).
Suppose now that $K$ is a finite extension of $\mathbb{Q}$. Then, by Steinitz's theorem, there is $u \in K$ such that $K=\mathbb{Q}(u)$. Let $p(x)$ be the minimal polynomial of $u$, and let $n$ be $p$'s degree. Then, K is an extension os degree $n$. However, the minimal polynomial of $2^{\frac{1}{n+1}}$ has degree $n+1$, so K is an extension of degree at least $n+1$, contradicting the hypothesis that $K$ is an extension of degree $n$. So, $K$ is an infinite extension of $\mathbb{Q}$.
My doubts: Can $\alpha$ be an infinite sum? If yes, can I use the same argument?