# To calculate $[\mathbb Q(\sum_{j=1}^ka_j^{1/n_j}) : \mathbb Q]$ , when $a_i , n_i$ are positive integers with $a_1^{1/n_1}$ not an integer

Let $a_1 , n_1$ be positive integers such that $a_1^{1/n_1}$ is not an integer . Then I can show that for any positive integers $a_2,...,a_k,n_2,...,n_k$ , $\sum_{j=1}^ka_j^{1/n_j}$ is an irrational number .

My question is , can I determine the extension degree $[\mathbb Q(\sum_{j=1}^ka_j^{1/n_j}) : \mathbb Q]$ ? Or at least , is it possible to give a good upper and lower bounds ( I know it is bounded below by $2$) ?

• Obviously $\mathbb Q(\sum_{j=1}^ka_j^{1/n_j}) \subseteq \mathbb{Q}(a_1^{1/n_1},a_2^{1/n_2}, \dots , a_k^{1/n_k})$ and the later is an extension of a degree at most $n_1\cdot n_2 \cdot n_3 \cdots n_k$ – Stefan4024 Oct 21 '17 at 16:56
• @Stefan4024 : Yes that's true ... what about a lower bound (or explicit calculation ) ? – user Oct 21 '17 at 16:58
• Are you interested in the arbitrary case or the $\gcd(a_i,a_j)=1$ case? – Steven Stadnicki Oct 21 '17 at 17:39
• @StevenStadnicki : Arbitrary case , but if you have something for the co-prime case , I am eager to hear – user Oct 21 '17 at 17:41
• For the arbitrary case, I think Dionel's answer covers it - you need more info on the numbers to be able to say much more than '2'. – Steven Stadnicki Oct 21 '17 at 20:58

## 1 Answer

You cannot be guaranteed a lower bound greater than 2. For instance let $a_i = 2^{i}$ and $n_i = 2i$

Then each $a_i^{\frac{1}{n_i}} = (2^i)^{\frac{1}{2i}} = 2^{\frac{1}{2}}= \sqrt2$