Let $a_1,\ldots, a_n$ be integers such that $[\mathbb{Q}(\sqrt{a_1},\ldots, \sqrt{a_n}):\mathbb{Q}]=2^n$. I want to show that $\sqrt{a_1}+\cdots+\sqrt{a_n}$ is a primitive element of $\mathbb{Q}(\sqrt{a_1},\ldots, \sqrt{a_n})$ over $\mathbb{Q}$.
I tried like this.
Clearly $\sqrt{a_1}+\ldots+\sqrt{a_n}\in \mathbb{Q}(\sqrt{a_1},\ldots, \sqrt{a_n})$ and so $\mathbb{Q}(\sqrt{a_1}+\ldots+\sqrt{a_n})\subset \mathbb{Q}(\sqrt{a_1},\ldots, \sqrt{a_n})$.
Now how to show $[\mathbb{Q}(\sqrt{a_1},\ldots, \sqrt{a_n}):\mathbb{Q}(\sqrt{a_1}+\ldots+\sqrt{a_n})]=1$? Any hint will be appreciated.
Why not show that there are $2^n$ isomorphisms of $\mathbb{Q}(\sqrt{a_1}+\ldots+\sqrt{a_n})$ into a subfield of $\overline{\mathbb{Q}},$ hence $$[\mathbb{Q}(\sqrt{a_1}+\ldots+\sqrt{a_n}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{a_1},\ldots,\sqrt{a_n}):\mathbb{Q}],$$ so by the Tower theorem $$[\mathbb{Q}(\sqrt{a_1},\ldots,\sqrt{a_n}):\mathbb{Q}(\sqrt{a_1}+\ldots+\sqrt{a_n})]=1.$$ Why are there $2^n$ isomorphisms? Well consider the map $$\sigma:\mathbb{Q}(\sqrt{a_1}+\ldots+\sqrt{a_n})\to\overline{\mathbb{Q}}$$ given by $$\sigma(\sqrt{a_1}+\ldots+\sqrt{a_n})=(-1)^{\beta_1}\sqrt{a_1}+\ldots+(-1)^{\beta_n}\sqrt{a_n}.$$ To show this map is an isomorphism consider the 'extended map' $$\sigma':\mathbb{Q}(\sqrt{a_1},\ldots,\sqrt{a_n})\to\mathbb{Q}(\sqrt{a_1},\ldots,\sqrt{a_n})$$ by $$\sigma'(\sqrt{a_i})=(-1)^{\beta_i}\sqrt{a_i}.$$ This is an isomorphism, hence since $\sigma'$ extends $\sigma$ it follows that $\sigma$ is additive, multiplicative, and has a trivial kernel, therefore $\sigma$ is an isomorphism into $\overline{\mathbb{Q}}.$ Since $\beta_1,\ldots,\beta_n$ were arbitrary and $\sigma$ is completely determined by their least residues modulo $2$ it follows that there are at least $2^n$ isomorphisms of $\mathbb{Q}(\sqrt{a_1}+\ldots+\sqrt{a_n})$ into a subfield of $\overline{\mathbb{Q}}$ as desired.