Degree of field extension of $ [\mathbb Q( 2^{1/2} + 2^{1/3} +2^{1/4}+2^{1/5}+\cdots 2^{1/n}) : \mathbb Q(2^{1/2}) ]$

Question

The title says the question.

How to find the degree of the extension $ [\mathbb Q( 2^{1/2} + 2^{1/3} +2^{1/4}+2^{1/5}+\cdots 2^{1/n}) : \mathbb Q(2^{1/2} )]$ ?

Here $n$ is any fixed natural number greater than $1$.

I was trying to prove that $ \mathbb Q( 2^{1/2} + 2^{1/3} +2^{1/4}+2^{1/5}+\cdots 2^{1/n})$ = $\mathbb Q( 2^{1/2}, 2^{1/3}, \cdots 2^{1/n} ) $, but failed to show.

Any insight.

Any help will be appreciated. Thank you.

Answer 1

Questions like this are the subject of something called coGalois theory. There may be a simpler proof in this specific case of roots of $2$, but in general questions like this are hard and (the last I've heard) coGalois theory is essentially the only way to treat them. A nice survey of this theory can be found here. In particular, we have the following theorem (see section 5.2 in the survey):

Let $F$ be a subfield of $\mathbb R$, $r,n_1,\dots,n_r\in\mathbb N$ and $a_1,\dots,a_r\in F^*$ positive, then the extension $F(\sqrt[n_1]{a_1}+\dots+\sqrt[n_r]{a_r})$ coincides with the extension $F(\sqrt[n_1]{a_1},\dots,\sqrt[n_r]{a_r})$.

For the problem at hand, this confirms your conjecture that $\mathbb Q( 2^{1/2} + 2^{1/3} \cdots +2^{1/n})$ coincides with $\mathbb Q(2^{1/2},2^{1/3}\cdots,2^{1/n})$. To find the degree of this extension over $\mathbb Q(2^{1/2})$ (note, addressing Dietrich's doubt in the comment above, we now know this extension indeed contains $2^{1/2}$), it is enough to find its degree over $\mathbb Q$ and divide by $2$.

Let $p_1,\dots,p_k$ be the primes up to $n$, and for each $p_i$ let $p_i^{e_i}$ be its largest power which is at most $n$. Then $\mathbb Q(2^{1/p_1^{e_1}},\dots,2^{1/p_k^{e_k}})$ is clearly contained in our extension, and in fact is equal to it - I will leave out the proof, which amounts to showing that $1/m$ for $m\leq n$ can be written as an integer linear combination of $p_1^{e_1},\dots,p_k^{e_k}$.

It remains to find the degree of $\mathbb Q(2^{1/p_1^{e_1}},\dots,2^{1/p_k^{e_k}})/\mathbb Q$. But this extension is the compositum of the $k$ extensions $\mathbb Q(2^{1/p_i^{e_i}})/\mathbb Q$. These extensions have degree $p_i^{e_i}$, which are relatively prime, so this compositum has degree equal to the product $p_1^{e_1}\dots p_k^{e_k}$. This number can be more simply written as $\mathrm{lcm}(1,2,3,\dots,n)$.

Altogether, we find that $[\mathbb Q( 2^{1/2} + 2^{1/3} \cdots +2^{1/n}):\mathbb Q(2^{1/2})]=\frac{1}{2}\mathrm{lcm}(1,2,3,\dots,n)$.