A prime ideal $\mathfrak{p}$ decomposes in $\mathbb{Q}(\zeta_{12})/\mathbb{Q}(i)$ iff it is generated by $\alpha\in1+3\Bbb{Z}[i]$ 
Prove that for a nonzero prime ideal $\mathfrak{p}$ of $\mathbb{Z}[i]$ which does not divide $3$, $\mathfrak{p}$ decomposes completely in the quadratic extension $\mathbb{Q}(\zeta_{12})/\mathbb{Q}(i)$ if and only if $\mathfrak{p} = (\alpha)$ for some $\alpha \in \mathbb{Z}[i]$ such that $a \equiv 1 \text{ mod }3\mathbb{Z}[i]$.

What I have attempted so far. I've worked out an example of this. The prime divisor $(2 + 3i)$ of $13$ is generated by $-2 - 3i = 1 - (3 + 3i) \equiv 1 \text{ mod }3\mathbb{Z}[i]$. We have$$13 = \prod_{a = 1, 5, 7, 11} (2 - \zeta_{12}^a),\text{ }3 + 2i = i(2 - \zeta_{12})(2 - \zeta_{12}^5)$$in $\mathbb{Z}[\zeta_{12}]$.
I know that for a prime number $p \neq 2$, $3$, $p = x^2 + 9y^2$ for some $x$, $y \in \mathbb{Z}$ if and only if $p \equiv 1 \text{ mod }12$.
I don't know what do from here with the original statement. Could anybody help?
 A: Since everything in the tower $\Bbb{Q}(\zeta_{12})/\Bbb{Q}(i)/\Bbb{Q}$ is abelian Galois, and what happens in the extension $\Bbb{Q}(i)/\Bbb{Q}$ is well known, we can as well look at the picture all the way down to the level of rational primes. So assume that $\mathfrak{p}$ is above the rational prime $p$.
Excluding the ramified cases $p=2$ and $p=3$ for now, we can use Dedekind's theorem telling us that the splitting behavior of $p$ in $\Bbb{Q}(\zeta_{12})$ (resp. $\Bbb{Q}(i)$) is accurately reproduced in the factorization of the minimal polynomial $x^4-x^2+1$ of $\zeta_{12}$ (resp. the minimal polynomial $x^2+1$ of $i$) modulo $p$. This depends on the residue class of $p$ modulo $12$ as follows. The fact that whenever $p>3$ we have $p^2\equiv1\pmod{12}$ comes to the fore.


*

*If $p\equiv1\pmod{12}$, then there are twelfth roots of unity in the prime field $\Bbb{F}_p$. Consequently $x^4-x^2+1$ splits into linear factors over $\Bbb{F}_p$, and $p$ decomposes totally in $\Bbb{Q}(\zeta_{12})/\Bbb{Q}$. So in this case $\mathfrak{p}$ also decomposes.

*If $p\equiv5\pmod{12}$, then we need to go to the extension $\Bbb{F}_{p^2}$ to find twelfth roots of unity. Therefore $x^4-x^2+1$ is a product of two irreducible quadratic factors modulo $p$, and $p$ is a product of two prime ideals with inertia degree $f=2$. But as $p$ decomposes in $\Bbb{Q}(i)$, this means that $\mathfrak{p}$ is inert.

*If $p\equiv7\pmod{12}$ or $p\equiv11\pmod{12}$, then, again $x^4-x^2+1$ splits into a product of two irreducible quadratic factors modulo $p$. However, this time $p$ is inert in $\Bbb{Q}(i)$, so $\mathfrak{p}=(p)$ decomposes in the extension $\Bbb{Q}(\zeta_{12})/\Bbb{Q}(i)$.


Relating this to residue class of generator of $\mathfrak{p}$ is not difficult. Observe that we have the liberty to replace $\alpha$ with $i^k\alpha$ if the necessity arises. The multiplicative group of the field $\Bbb{Z}[i]/3\Bbb{Z}[i]$
consists of the subgroup $H=\langle i\rangle$ and its coset $(1+i)H$. Therefore we only need to know whether $\alpha+3\Bbb{Z}[i]\in H$ or not. A useful way of distinguishing the cosets is the observation that $\alpha+3\Bbb{Z}[i]\in H$, iff $N(\alpha)=\alpha\overline{\alpha}\equiv1\pmod3$. This can be verified case-by-case: $N(i^k)=1$, $N((1+i)i^k)=2$.


*

*If $p\equiv1\pmod{12}$, then there are two prime ideals $\mathfrak{p}$, $\mathfrak{p'}$ above $p$, generated by some Gaussian integers $\alpha$ and $\overline{\alpha}$ respectively. In this case $p=N(\alpha)=\alpha\overline{\alpha}$, so $\alpha\overline{\alpha}=p\equiv1\pmod3$, and we can conclude that both $\alpha$ and $\overline{\alpha}$ have an associate in $1+3\Bbb{Z}[i]$.

*If $p\equiv5\pmod{12}$, then a similar argument shows that the generators $\alpha$ and $\overline{\alpha}$ of the two prime ideals above $p$ satisfy
$N(\alpha)=N(\overline{\alpha})=p\equiv2\pmod3$, so in this case $\mathfrak{p}$ hasn't got a generator in $1+3\Bbb{Z}[i]$. This is just as well as we earlier saw that $\mathfrak{p}$ and $\mathfrak{p'}$ are both inert.

*In the case $p\equiv7\pmod{12}$ we see that $\mathfrak{p}=(p)$. Here $p\equiv1\pmod3\Bbb{Z}[i]$. As we saw that $\mathfrak{p}$ splits, this is what was to be verified.

*In the case $p\equiv11\pmod{12}$ we again have $\mathfrak{p}=(p)$, and this prime ideal splits in $\Bbb{Q}(\zeta_{12})/\Bbb{Q}(i)$. This time we can use $\alpha=-p\in1+3\Bbb{Z}[i]$ as a generator.


This takes care of all the cases except the prime ideal $\mathfrak{p}=(1+i)$. Leaving that to you.
