# Series of Extension Fields

Let $$(a_n)_{n \in \mathbb{Z\geq0}}$$ with $$a_0 = 2$$ and $$a_{n+1}= \sqrt{a_n}$$ with $$a_{n+1}>0$$. We need to show that $$[\mathbb{Q}(a_n):\mathbb{Q}]=2^n \forall n \in \mathbb{Q}$$.

To prove this we need to show that $$\mathbb{Q}(a_n) \subset \mathbb{Q}(a_{n+1})$$ and $$[\mathbb{Q}(a_{n+1}):\mathbb{Q}(a_n)]=2$$ trough induction.

Now the easiest part is to show that $$\mathbb{Q}(a_n) \subseteq \mathbb{Q}(a_{n+1})$$ and I managed to do this. The tough part is to show that $$\mathbb{Q}(a_n) \subset \mathbb{Q}(a_{n+1})$$. If anyone can help me out here I would very much appreciate it.

• You can draw all conclusions you wish from the explicit solution of the recursion, which is $a_{n}=(a_{0})^\frac{1}{2^n}$ – Dr. Wolfgang Hintze Mar 12 at 14:17

I think it's easier than you think. $$a_n$$ is a root of $$x^{2^n} - 2$$ which, by Eisenstein's criterion, is irreducible over the rational numbers.