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 $$Gal \left ( \mathbb Q ( \sum _{j=1} ^n \sqrt j ) |_ {\mathbb Q} \right ) \cong \mathbb Z _2 ^ {\pi (n)} $$
No comments:
Post a Comment