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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 $n$.

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  • $\begingroup$ Is this a conjecture or a thm? $\endgroup$ – Wuestenfux Apr 13 at 15:18
  • $\begingroup$ This is a theorem. $\endgroup$ – Anupam Apr 13 at 15:23
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    $\begingroup$ Why not $2^{\pi(n)}$? $\endgroup$ – Lord Shark the Unknown Apr 13 at 15:40
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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 $$Gal \left ( \mathbb Q ( \sum _{j=1} ^n \sqrt j ) |_ {\mathbb Q} \right ) \cong \mathbb Z _2 ^ {\pi (n)} $$

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  • $\begingroup$ Yeah I am wrong. Thus I am changing the question. $\endgroup$ – Anupam Apr 13 at 16:45
  • $\begingroup$ @Soumik Ghosh You should have insisted on the "linear independence of {$\sqrt {p_i},\sqrt {p_ip_j},...$}", which I think is the cumbersome point. $\endgroup$ – nguyen quang do Apr 14 at 7:28
  • $\begingroup$ OP already knows $\mathbb Q(\sqrt { p_i } : 1=1,...,n) :\mathbb Q$ is of degree $2^n$ . Then linear independence is a triviality. $\endgroup$ – Ignorant Mathematician Apr 14 at 7:37
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This is false...take n=5 for instance

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I think that Kummer gives the neatest proof, using only the multiplicative structure of $\mathbf Q^*$. For a fixed $n\ge 2$, 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 $G\cong Hom (V,\mu_2)$, 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 $m\le n$, 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.

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