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For every $ N \in \mathbb Z$ there exists an integer $n$ such that $ \sqrt N \in \mathbb Q(\zeta_n)$.

I am struggling where to start this question, please suggest me few hints.

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1 Answer 1

This follows from a very general result, called the Kronecker-Weber Theorem, which says that every finite abelian extension of $\mathbb Q$ is contained in a cyclotomic extension. The proof is rather involved, either using class field theory or deriving it from the corresponding theorem for local fields. The special case of quadratic extensions, however, can be proved directly.

Recall that $\mathbb Q(\zeta_n,\zeta_m) = \mathbb Q(\zeta_{\operatorname{lcm}(m,n)})$. So if $N = ab$ and we know that $\sqrt{a}$ and $\sqrt{b}$ are contained in a cyclotomic extension, then the same is true for $N$. Therefore, we can assume that $N$ is a prime $p$ or (since $\sqrt{-1} \in \mathbb Q(\zeta_4)$) the negative of a prime, $N=-p$. Thus, it suffices to show:

  1. $\sqrt{2} \in \mathbb Q(\zeta_8)$. Show that $\zeta_8 + \zeta_8^{-1}$ is a square root of $2$.
  2. If $p$ is a prime and $p \equiv 1 \pmod 4$ then $\sqrt{p} \in \mathbb Q(\zeta_p)$.
  3. If $p$ is a prime and $p \equiv 3 \pmod 4$ then $\sqrt{-p} \in \mathbb Q(\zeta_p)$.

The second and third part can be done by looking at the Gauss sum $\sum_{a =1}^{p-1} \left(\frac{a}{p}\right) \zeta_p^a$.

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thanks for the answer –  Ram Jan 20 '13 at 15:35
    
very nice answer! –  Bombyx mori Jan 20 '13 at 17:15

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