# Decomposition and inerta fields in $\mathbb{Q}(\zeta_{12})$

Determine the decomposition of the primes $2, 3, 5$ and $7$ in $\mathbb{Q}(\zeta_{12})$. Find the decomposition fields and inertia fields for all prime ideals in $\mathbb{Z}[\zeta_{12}]$ in these decompositions. Show that $\mathbb{Q}(\zeta_{12})$ has trivial class group.

I've already done the following:

$$\Phi_{12}(x)=$$ \begin{cases} (x^2+x+1)^2 \text{ in }\mathbb{F}_2[x]\\ (x+1)^2(x+2)^2 \text{ in }\mathbb{F}_3[x]\\ (x^2+x+1)(x^2-x+1)\text{ in }\mathbb{F}_5[x]\\ (x+2)(x+3)(x+4)(x+5) \text{ in }\mathbb{F}_7[x]\\ \end{cases}

So that:

\begin{align*} (2)&=(2, (\zeta^2+\zeta+1))^2\\ (3)&=(3, (\zeta+1))^2\cdot(3, (\zeta+2))^2\\ (5)&=(5, (\zeta^2+\zeta+1))\cdot(3, (\zeta^2-\zeta+1))\\ (7)&=(7, (\zeta+2))\cdot(7, (\zeta+3))\cdot(7, (\zeta+4))\cdot(7, (\zeta+5))\\ \end{align*}

I also know that the Galois group of the extension is $\{\sigma_i:\zeta_{12}\mapsto \zeta_{12}^i\mid i=1, 5, 7, 11\}$. But I'm stuck because I can't find a way to calculate the decomposition and inertia groups, they seem really complicated. For example, if I want to check by brute force that $\sigma_i$ fixes the prime ideal $(7, (\zeta_{12}+2))$, I'd have to show that for every $z, w\in\mathbb{Z}[\zeta_{12}]$ we get $\sigma_i(7z+(\zeta_{12}+2)w)=7u+(\zeta_{12}+2)v$ for some $u, v\in\mathbb{Z}[\zeta_{12}]$, which by itself is pretty complicated. Is there a more adequate way to do this? And also, how can I relate the class group to these groups/fields?

• Notice that $7$ is totally split. This means that the inertia and decomposition groups are trivial, so the inertia and decomposition fields is $\Bbb Q(\zeta_{12})$. – Watson Feb 13 '17 at 16:57
• To expand on Watson's comment: it is a general fact that the Galois group acts transitively on the set of primes lying above a given prime in the ground field, and the decomposition groups are the stabilizers of these primes. ("Groups" is plural because generally there's one for each prime upstairs, all conjugate to each other but not necessarily equal. Here they're equal because the Galois group is abelian.) – Ravi Fernando Feb 14 '17 at 7:47
• @Watson, now I've noticed that some cases are really trivial. Like you said, for $7$ both groups are trivial. For $2$, since it factors in only one prime, we must have $[G:D]=1\Rightarrow D=G$. For $3$ we have from $[K:\mathbb{Q}]=efr$ that $4=4f\Rightarrow f=1$, so the inertia group is trivial. But now I'm stuck with the non trivial cases. How do I deal with them? – rmdmc89 Feb 14 '17 at 12:36
• @RaviFernando, theoreticaly I'm aware of the theorem about the transitive action, but I'm having trouble to make the action explicit in this concrete example. Trivial cases aside, it still looks complicated to me – rmdmc89 Feb 14 '17 at 12:38
• And by the way, how do the information about these fields allow me to conclude that the class group is trivial? – rmdmc89 Feb 14 '17 at 16:56