I'm studying chapter 7.2 Algebraic extensions in Abstract Algebra by S. Lovett, and I'm stuck with an exercise problem.
Let $S = \{ \sqrt[n]{2} : n \in \mathbb{Z}$ with $n \geq 2 \}$. Prove that $\sqrt{3} \notin \mathbb{Q}[S]$.
Here is my argument.
Suppose that $\sqrt{3} \in \mathbb{Q}[S]$.
Since $\mathbb{Q}[S] = \cup \mathbb{Q}[\sqrt[n]{2}]$, there exists an $n$ such that $\sqrt{3} \in \mathbb{Q}[\sqrt[n]{2}]$.
If $n$ is odd, $[\mathbb{Q}[\sqrt[n]{2}]:\mathbb{Q}]=[\mathbb{Q}[\sqrt[n]{2}]:\mathbb{Q}[\sqrt{3}]][\mathbb{Q}[\sqrt{3}]:\mathbb{Q}]$ and $[\mathbb{Q}[\sqrt{3}]:\mathbb{Q}]=2$, so we have a contradiction.
Now I'm stuck at proving $\sqrt{3} \notin \mathbb{Q}[\sqrt[n]{2}]$ for $n$ even.
I appreciate any help on this part or suggestion of another approach on whole problem.
If you can show the statement for $n$ a power of $2$, then you can argue with degrees to take care of all other $n$. Proof by induction works here. Let $\beta_k = \sqrt[2^k]{2}$. Suppose $\sqrt{3} \not\in \mathbb{Q}(\beta_k)$ where $k \leq m$. Now suppose $\sqrt{3} \in \mathbb{Q}(\beta_{m+1})$. Then there exist $a_0,a_1 \in \mathbb{Q}(\beta_m)$ such that $$\sqrt{3} = a_0 + a_1\beta_{m+1}.$$ Square both sides and try to get a contradiction.