abstract algebra - Finding degree of a finite field extension

Let $x=\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}, n\geq 2$. I want to show that $[\mathbb{Q}(x):\mathbb{Q}]=2^{\phi(n)}$, where $\phi$ is Euler's totient function.

I know that if $p_1,\ldots,p_n$ are pairwise relatively prime then $[\mathbb{Q}(\sqrt{p_1}+\ldots+\sqrt{p_n}):\mathbb{Q}]=2^n$. But how to proceed in the above case? I could not apply induction also. Any help is appreciated.

The assertion is false. Actually $[\mathbb{Q}(x):\mathbb{Q}]=2^{\pi(n)}$, where $\pi(n)$ is the number of prime numbers less than or equal to $n$.

Answer

Let $L= \mathbb Q ( \sum _{j=1} ^n \sqrt j )$ , $k= \mathbb Q$ and $N = \mathbb Q ( \sqrt 2, \sqrt 3 ,... , \sqrt n )$ .

Clearly $N|_k$ is Galois and the Galois group is of the form $\mathbb Z_2 ^m$ for some $m$ since every $k$ automorphism of $N$ has order at most $2$. Note that each element of $Gal (N|_k)$ is completely specified by it's action on $\{ \sqrt p : \ p \ prime \ \ p \leq n \}$ by the fundamental theorem of arithmetic. So this gives $$m \leq \pi (n)$$

Now if the Galois group is $\mathbb Z_2 ^m$ then it will have $2^m -1$ subgroups of index $2$ and hence there exist $2^m -1$ subfields $F$ of $N$ containing $k$ such that $F:k=2$ . But we already have $2^ {\pi (n)} -1$ many such subfields by taking product of a nonempty subset of $\{ \sqrt p : \ p \ prime \ \ p \leq n \}$ and hence we get $$2^ {\pi (n)} -1 \leq 2^ m -1$$

$$\implies \pi (n) \leq m$$

And hence $$Gal ( N|_k) = \mathbb Z_2 ^ {\pi(n)}$$

Now we just observe that the orbit of $\sum _{j=1} ^n \sqrt j$ under the action of $Gal(N|_k)$ contains $2^ {\pi (n)}$ distinct elements by linear independence of $\{ \sqrt {p_i }, \sqrt {p_ip_j},... \}$ and hence $N= L$

So $$Gal \left ( \mathbb Q ( \sum _{j=1} ^n \sqrt j ) |_ {\mathbb Q} \right ) \cong \mathbb Z _2 ^ {\pi (n)}$$