For a nonzero prime ideal $\mathfrak{p}$ of $\mathbb{Z}[\sqrt{-6}]$ which does not divide $2$, does $\mathfrak{p}$ decompose completely in the extension $\mathbb{Q}(\zeta\_{24})/\mathbb{Q}(\sqrt{-6})$ of degree $3$ if and only if $\mathfrak{p} = (\alpha)$ for some $\alpha \in \mathbb{Z}[\sqrt{-6}]$ such that $a \equiv 1 \text{ mod }2\mathbb{Z}[\sqrt{-6}]$?


I suspect this is a true based off working out the following example. $5 + 6\sqrt{-6}$ is a prime divisor of the prime number $241$ in $\mathbb{Z}[\sqrt{-6}]$ and $5 + 6\sqrt{-6} \equiv 1 \text{ mod }2\mathbb{Z}[\sqrt{-6}]$. We have$$241 = \prod\_{a = 1, 5, 7, , 11, 13, 17, 19, 23} (2 - \zeta\_{24}^a), \quad 5 + 6\sqrt{-6} = -\prod\_{a = 1, 5, 7, 11}(2 - \zeta\_{24}^a).$$



A prime ideal $\mathfrak{p}$ decomposes in $\mathbb{Q}(\zeta_{24})/\mathbb{Q}(\sqrt{-6})$ iff it is generated by $\alpha\in1+2\Bbb{Z}[\sqrt{-6}]$
OK, so here are the details. The decomposition law in cyclotomic extensions tells you that a prime $p$ coprime to $m$ splits completely in ${\mathbb Q}(\zeta\_{m})$ if and only if $p \equiv 1 \bmod m$. Since ${\mathbb Q}(\zeta\_{24}) = {\mathbb Q}(\sqrt{-1},\sqrt{2},\sqrt{-3})$ this is equivalent to the condition $(-1/p) = (2/p) = (-3/p) = +1$.


Prime ideals in the subfield $k = {\mathbb Q}(\sqrt{-6}\,)$ split completely in $K$ if and only if they split in $k/{\mathbb Q}$ and in $K/k$. Primes $p$ split in $k$ iff $(-6/p) = +1$, i.e. iff $(2/p) = (-3/p)$. By genus theory, the prime ideals above $p$ are principal if both Legendre symbols are $+1$, and nonprincipal otherwise. 


For finding out which principal primes split completely in $K/k$ we look at their norms: clearly $p = x^2 + 6y^2 \equiv 1 \bmod 4$ is equivalent to $2 \mid y$, which means that exactly those principal primes split completely whose norm is a prime $p \equiv 1 \bmod 4$. 


On a more advanced level, the extension $K/k$ consists of the Hilbert class field and genus field $F = {\mathbb Q}(\sqrt{2},\sqrt{-3})$ and a quadratic extension $K/F$ unramified outside $2$ that happens to live inside the ray class field modulo $(2)$ over $k$. The Hilbert class field step explains why only principal primes split, the ray class part explains the congruence condition modulo $2$.


