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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?

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    Notice that $7$ is totally split. This means that the inertia and decomposition groups are trivial, so the inertia and decomposition fields is $\Bbb Q(\zeta_{12})$. – 2017-02-13
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    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.) – 2017-02-14
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    @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? – 2017-02-14
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    @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 – 2017-02-14
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    And by the way, how do the information about these fields allow me to conclude that the class group is trivial? – 2017-02-14
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    7 does not split completely, as it is inert in ${\mathbb Q}(i)$. The splitting field of $7$ obviously is ${\mathbb Q}(\zeta_3)$. – 2017-02-14
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    Good catch @franzlemmermeyer. It looks like OP may be using the polynomial $x^4 + x^2 + 1$ (roots are sixth roots of unity, except $\pm 1$) instead of $x^4 - x^2 + 1$; the factorizations should change accordingly. – 2017-02-16
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    As far as the class group, I would just compute it with the Minkowski bound. I believe for this field the Minkowski bound is less than 2, so the class group is automatically trivial. – 2017-02-16

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