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Let $\zeta_n $ be a primitive root of unity generating the cyclotomic field $\Bbb Q(\zeta_n)$. Is/are there quick and/or "standard" techniques" to check if a given real quadratic roots $\sqrt{d}, d >0$ is contained in $\Bbb Q(\zeta_n)$?

A "canonical" quadratic root contained in $\Bbb Q(\zeta_n)$ is given by square root of discriminant given (up to sign issues) by $\prod_{i,j}(\alpha_i-\alpha_j)$ where $\alpha_i$ are pairwise disjoint roots of (over $\Bbb Q$ irreducible) cyclotomic polynomials $\Phi_n$.

Also, one could try to take sum of $\zeta_n^k +\overline{\zeta_n^k}$ as a "candidate" and try to analyze when is even a square root.

But are there other say "standard" methods to approach the posed proplem above? Conceptional approsches?

user267839
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  • Yes, see the posts on the Galois group of $X^n-d$. For example $\sqrt{2}\in \Bbb Q(\zeta_n)$ if and only if $8\mid n$. – Dietrich Burde May 26 '24 at 14:49
  • see https://math.stackexchange.com/questions/4881966/when-is-sqrt-p-contained-in-bbb-q-omega-n/4882068#4882068 for the case of a prime $d$; for an arbitrary square-free $d$ the ramification argument implies that every prime dividing $d$, except possibly $2$, must be in the cyclotomic field too, but the case o $2$ needs to be clarified – user8268 May 26 '24 at 14:57
  • I meant "the square root of every prime dividing ..." – user8268 May 26 '24 at 15:08
  • @DietrichBurde: But does it happen "accidentally" with $\sqrt{2}$ (eg with interplay of $ 8 | n $ and $\sqrt{2}\in \Bbb Q(\zeta_n)$ or can from this more general techniques be extracted to decide about general $\sqrt{d}, d$ positive integer? Eg something like a "congruence criterion" $\sqrt{d}\in \Bbb Q(\zeta_n)$ iff $m(d) | n$ with mysterious function $m(-)$? Or something more complicated but "in same vein"? – user267839 May 26 '24 at 15:24
  • Basically, $n$ has to be a multiple of the absolutely minimal discriminant among quadratic equations over the integers whose roots contain $\sqrt{d}$. Thus a multiple of $5$ for $d=5$ because of $x^2-x-1=0$, but a multiple of $24$ for $d=6$ where we can't do better than $x^2-6=0$. – Oscar Lanzi Jun 20 '25 at 15:00

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The Galois group of $\mathbb Q(\zeta_n)/\mathbb Q$ is $G_n=(\mathbb Z/n\mathbb Z)^\times$. If $n=\prod_{i=1}^k p_i^{m_i}$ is the factorization of $n$, this gives us $G_n\cong \prod_{i=1}^k G_{p_i^{m_i}}$.

We want to know all the fields $\mathbb Q\subset K\subset \mathbb Q(\zeta_n)$ s.t. $[K:\mathbb Q]=2$, i.e. all the surjective homomorphisms $G_n\to C_2$. Such a homomorphism is then the product of homomorphisms $G_{p_i^{m_i}}\to C_2$, of which at least one is surjective.

For $p\neq 2$ the group $G_{p^{m}}$ ($m\geq1$) is cyclic, with a unique surjective homomorphism to $C_2$, the fixed field is $\mathbb Q(\sqrt{p^*})$, where $p^*=(-1)^{(p-1)/2}p$ (obtained via the Gauss sum).

For $p=2$ it is a bit more complicated: $G_2$ is trivial, $G_4\cong C_2$, the fixed field is $\mathbb Q(\sqrt{-1})=\mathbb Q(\zeta_4)$, and for $m\geq3$ there are 3 surjective homomorpisms $G_{2^m}\to C_2$ and the corresponding fixed fields are $\mathbb Q(\sqrt{-1})$, $\mathbb Q(\sqrt{2})$, $\mathbb Q(\sqrt{-2})$.

To sum up which square roots of integers are in $\mathbb Q(\zeta_n)$: If $S$ is the set of all the odd primes dividing $n$ and if $T$ is a subset of $S$ then $$\sqrt{d_T}\in \mathbb Q(\zeta_n)\quad\text{ where }\quad d_T=\prod_{p\in T}p^*$$ ($d_T$ may or may not be positive), if $4|n$ then it contains also $\sqrt{-d_T}$, and if $8|n$ then also $\sqrt{\pm 2 d_T}$. And these are all the square-free integers $d$ s.t. $\sqrt d$ is in the field.

user8268
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