Suppose $m > 0$ is a square free integer and $m \equiv 1 $mod $4$. Then the $K = \mathbb{Q}(\sqrt{-m})$ is a subfield of $L = \mathbb{Q}(\theta)$, where $\theta$ is a primitive $4m$-th root of unity. For $j$ co-prime to $4m$, denote $\sigma_j \in$ Gal$(L/\mathbb{Q})$ such that $\sigma_j(\theta) = \theta^j$.
Now assume that $p>2$ is a prime such that $p \equiv 1$ mod $4m$. Then $p$ splits completely in $L/\mathbb{Q}$ and we write $p\mathcal{O}_K = \mathfrak{p}\bar{\mathfrak{p}}$, and $p\mathcal{O}_L = \mathcal{P}_1\dots\mathcal{P}_k\bar{\mathcal{P}}_1\dots\bar{\mathcal{P}}_k$, where $k = \phi(m)$ (Euler's totient function), and $\mathcal{P} \cap K = \mathfrak{p}$ and $\bar{\mathcal{P}_i} \cap K = \bar{\mathfrak{p}}$.
I would like to describe $j \in (\mathbb{Z}/4m\mathbb{Z})^{\times}$ such that $\sigma_j(\mathcal{P}_i) \cap K = \mathfrak{p}$. I know that conjugation $\sigma_{-1}$ sends $\mathfrak{p}$ to $\bar{\mathfrak{p}}$, and so there are $\phi(m) = \phi(4m)/2$ elements in Gal$(L/\mathbb{Q})$ that should then fix $K$.
From the value of certain Gauss sums, I can write $2\sqrt{-m} = \sum_{k = 0}^{4m-1}\theta^{k^2} - 2\sum_{k = 0}^{m-1} \theta^{(2k)^2}$ (maybe there is a better way to write this). So, $\sigma_j(\sqrt{-m})) = \sqrt{-m}$ if $j$ is a square mod $4m$. But this can't be all of the $\sigma_j$ that fix $K$, right?
Also, I can't seem to figure out how to write $\sqrt{-m}$ as a sum of $4m$-th roots of unity when $m \equiv 2$ mod $4$.
