i think for every prime $p$ we have $\sqrt p \in \mathbb Q(\zeta_{4p})$ when $\zeta_{4p}$ is a primitive 4p-th root of unity.but i have no idea to prove it.

is it true? can any one help me with a proof?



2 Answers 2


Yes this is true. For $p=2$ this is clear because $\zeta_8 = \frac{1+i}{\sqrt{2}}$. Otherwise, let $p$ be an odd prime. Then by evaluating the quadratic Gauss sum $\sum_{i=1}^{p-1} \left(\frac{i}{p} \right) \zeta_p^{i} = \sqrt{p^{*}}$, where $p^* = p$ if $p \equiv 1 \pmod{4}$, and $-p$ otherwise, we see that $\sqrt{p^*} \in \mathbb Q(\zeta_p)$. So if we further adjoin $i= \zeta_4$, we get that $\sqrt{p} \in \mathbb Q(\zeta_{4p})$.


Another way to see that $\sqrt{n}$ is in the field $\mathbb{Q}( \zeta_n, i)$ (for $n$ odd) is to consider the matrix of the discrete Fourier transform of order $n$ $$ \mathcal{F}_n\colon =\frac{1}{\sqrt{n}} \cdot (e^{\frac{2\pi i k l} {n}})_{0\le k,l \le n-1}$$ $\mathcal{F}_n$ is a unitary matrix of order $4$ ($\mathcal{F}_n^4 = I_n$) so its eigenvalues are in the set $\{1, i, -1, -i\}$. Therefore, its trace $$\frac{1}{\sqrt{n}} \sum_{k=0}^{n-1} e^{\frac{2\pi i k^2}{n}}$$ is an integer combination of $1$ and $i$ (one can determine precisely the value). Moreover, if $n$ is odd then the trace is not $0$.

Obs: for $n=p$ odd prime the quadratic Gauss sums $\sum_{k=0}^{p-1} e^{\frac{2\pi i k^2}{p}}$ and $\sum \left( \frac{a}{p}\right ) e^{2\pi i a /p}$ coincide.


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