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Let $x=sqrt{2}+sqrt{3}+ldots+sqrt{n}, ngeq 2$. 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 $n$.

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

This is false...take n=5 for instance

I think that Kummer gives the neatest proof, using only the multiplicative structure of $mathbf Q^*$. For a fixed $nge 2$, let $K=mathbf Q (sqrt 2, sqrt 3,...,sqrt n)$ and $mu_2=(pm 1)$. Kummer theory tells us that $K/mathbf Q$ is an abelian extension, with Galois group $Gcong Hom (V,mu_2)$, where $V$ is the subgroup of $mathbf Q^*/{mathbf Q^*}^2$ generated by the classes $bar 2,...,bar n$ mod ${mathbf Q^*}^2$. Although $V$ is a multiplicative group, it will be convenient to view it as a vector space over $mathbf F_2$, a linear combination of $bar 2,...,bar n$ mod ${mathbf Q^*}^2$ being just a product ${bar 2}^{epsilon_2}...{bar n}^{epsilon_n}$, with $epsilon_i=0$ or $1$. We aim to show that the $mathbf F_2$-dimension of $V$ is $pi (n)$, the number of rational primes $le n$. Let $W$ be the $mathbf F_2$-subspace generated by the classes of these primes. For for any $mle n$, the prime factorization of $m$ in $mathbf Z$ immediately shows that $bar m$ is a linear combination of the classes of the primes $le m$, which implies that $V=W$. It remains only to show that $W$ has $mathbf F_2$-dimension $pi (n)$, e.g. that the classes $bar p_i$ of the primes $le n$ form a basis. But a relation of linear dependence among them would mean that some finite product $prod p_i$ is a rational square, which contradicts the fact that $mathbf Z$ is a UFD.