# How can we prove $\mathbb{Q}(\sqrt 2, \sqrt 3, … , \sqrt n ) = \mathbb{Q}(\sqrt 2 + \sqrt 3 + … + \sqrt n )$ [duplicate]

I want to prove this statement.

$$\mathbb{Q}(\sqrt 2, \sqrt 3, ..... , \sqrt n ) = \mathbb{Q}(\sqrt 2 + \sqrt 3 + .... + \sqrt n )$$ for any $n >1$.

It looks like a very hard problem.

How can I approach this one?

• – Martin R Dec 14 '15 at 18:58
• An answer to your question is buried in the comments to this answer by yours truly. – Jyrki Lahtonen Dec 14 '15 at 18:59
• MartinR, Jyrki Thanks! – nicksohn Dec 14 '15 at 19:45
• It does look hard, but some aspects are easy. Try to break it into an "easy" half and a "difficult" half, and then develop an approach to the difficult direction by considering the situation for small $n$. – hardmath Dec 15 '15 at 0:50

Let us show that $$\mathbb{Q}(\sqrt[d_1]{a_1}, \ldots, \sqrt[d_n]{a_n}) = \mathbb{Q}( \sqrt[d_1]{a_1}+ \cdots + \sqrt[d_n]{a_n})$$ ( $a_l$, $d_l$ positive integers).
By Galois theory, if $K$ be a field, $L\supset K$ a Galois extension, $\alpha$, $\beta$ in $L$ so that every $\sigma \in \text{Gal}(L/K)$ that fixes $\alpha$ also fixes $\beta$, then $\beta \in K(\alpha)$.
Let now $\alpha = \sqrt[d_1]{a_1}+ \cdots + \sqrt[d_n]{a_n}$ and $\beta_l = \sqrt[d_l]{a_l}$, $1\le l \le n$. Consider a Galois extension $L$ containing all the $\beta_l$ (and so $\alpha$, too). Let $\sigma \in \text{Gal}(L/\mathbb{Q})$. We have $\sigma(\beta_l) = \omega_l \beta_l$ where $\omega_l$ is a $d_l$-th root of $1$. Assume that $\sigma(\alpha) = \alpha$, that is $$\sum \omega_l \beta_l = \sum \beta_l.$$ Now the $\beta_l$'s are real and positive and we have $$| \sum \omega_l \beta_l| \le \sum | \omega_l \beta_l | = \sum \beta_l$$ and if we have equality then all the modulus $1$ numbers $\omega_l$ have to be equal, and from the above we conclude $\omega_l=1$ for all $l$, and so $\sigma(\beta_l)= \beta_l$ for all $l$.
We conclude $\sqrt[d_l]{a_l} \in \mathbb{Q}( \sqrt[d_1]{a_1}+ \cdots + \sqrt[d_n]{a_n})$ for all $l$.