Finding a square root in a cyclotomic field

Question

Let $x\in\mathbb{Q}[\zeta_{p}]$. Suppose that $$x=a_0+a_1\zeta_p+...+a_{p-2}\zeta_p^{p-2},\quad a_i\in\mathbb{Q}$$ and that $x$ has a square root in $\mathbb{Q}[\zeta_{p}]$, i.e. there exists an element $$\sqrt{x}=b_0+b_1\zeta_p+...+b_{p-2}\zeta_p^{p-2}\in\mathbb{Q}[\zeta_{p}],\quad b_i\in\mathbb{Q}$$ such that $(\sqrt{x})^2=x$. The question is, given the existence of the square root, do we have any concrete and efficient algorithm (preferably in a practical sense) to find $b_i$ apart from brute-force solving equations? Any related result or comment is welcome.

Answer 1

First you can catch all your $\sqrt x$ just with Galois theory. You are surely aware of this, but I recall the details that we need. For simplicity, given an odd prime $p$, let $K=\mathbb Q (\zeta)$, where $\zeta$ is a primitive $p$-th root of $1$. It is well known that the extension $K/\mathbb Q$ is normal, with cyclic Galois group $G$ of order $p-1$. This cyclic $G$ contains a unique automorphism of order $2$, the so called "complex conjugation" sending $\zeta$ to its inverse. Noting $K^+$ its fixed field, the extension $K/K^+$ is quadratic, $K=K^+(\sqrt x)$, with $x$ in $K^+$, and this quadratic extension is uniquely determined by the class of $x$ modulo the squares of $K^+$. One canonical $x$ is $\delta=\zeta+ \zeta^{-1}$ and all the $x$ 's we want will have the form $x= \delta. y^2$, with a non zero $y \in K^+$. But $K^+=\mathbb Q (\delta)$ is explicitly known, and this explicit knowledge is as effective as any algorithm, I think.