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.

  • 1
    $\begingroup$ As mentioned in the answer below, you may assume that $N$ is either $-1$, or a prime $p$, or $-p$. If $N=p>2$, then $2$ divides the degree of the cyclic extension $\Bbb Q(\zeta_p) / \Bbb Q$, which has then a unique quadratic subextension $K/\Bbb Q$. Since only $p$ ramifies in $K$, and has odd discriminant, one finds $$K = \Bbb Q(\sqrt{(-1)^{(p-1)/2} p})$$ (as mentioned here). $\endgroup$
    – Watson
    Nov 22, 2018 at 13:05
  • $\begingroup$ See also math.stackexchange.com/questions/30111 $\endgroup$
    – Watson
    Nov 22, 2018 at 17:45

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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    $\begingroup$ very nice answer! $\endgroup$ Jan 20, 2013 at 17:15

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