# 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 $$n$$.

• Is this a conjecture or a thm? – Wuestenfux Apr 13 at 15:18
• This is a theorem. – Anupam Apr 13 at 15:23
• Why not $2^{\pi(n)}$? – Lord Shark the Unknown Apr 13 at 15:40

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

• Yeah I am wrong. Thus I am changing the question. – Anupam Apr 13 at 16:45
• @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. – nguyen quang do Apr 14 at 7:28
• OP already knows $\mathbb Q(\sqrt { p_i } : 1=1,...,n) :\mathbb Q$ is of degree $2^n$ . Then linear independence is a triviality. – Ignorant Mathematician Apr 14 at 7:37

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 $$n\ge 2$$, 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.