Infinite algebraic extension of $\mathbb{Q}$

Solution 1:

As has been mentioned, the result is a direct consequence of the fact that $[\mathbf{Q}(2^{1/n}):\mathbf{Q}]$ = n. This is because we then have, for every $n$, the inequality $$n = [\mathbf{Q}(2^{1/n}):\mathbf{Q}] \leq [K:\mathbf{Q}].$$ Since $[K:\mathbf{Q}]$ is larger than every natural number, it must be infinite.

The trickiest point is why $2^{1/n}$ has degree $n$ over $\mathbf{Q}$. For this it is enough to establish that the polynomial $X^n - 2$ is irreducible. This follows immediately from Eisenstein's criterion for the prime $p = 2$.

There is an error in your proof that every element of $K$ is algebraic over $\mathbf{Q}$. It is not clear at all why you claim any $\alpha \in K$ can be written in the form of a linear combination of various roots of $2$. The only thing that is obvious is that $\alpha$ can be written as a polynomial expression of these numbers, not a linear combination.

The proof that $K$ is algebraic is as follows. The field $K$ is generated by the set $S = \mathbf{Q} \cup \{2^{1/2}, 2^{1/3}, \dots\} \subseteq \overline{\mathbf{Q}}$, where $\overline{\mathbf{Q}}$ is the field of algebraic numbers over $\mathbf{Q}$. It follows that $K$ itself is contained in $\overline{\mathbf{Q}}$. Thus $K$ is algebraic over $\mathbf{Q}$.