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I would like to ask about the inertia group of primes in the cyclotomic field extension. Indeed, let $n=p_{1}\cdots p_{k}$ for some distinct odd primes and let $K = \mathbb{Q}(\zeta_{n})$ so the Galois group of $K/\mathbb{Q}$ is isomorphic to $\mathbb{F}_{p_{1}}^{\times}\times\cdots\times\mathbb{F}_{p_{k}}^{\times}$ via CRT. I think the inertia group of a prime ideal $\mathfrak{p}$ over some prime $p_{i}$ is $1\times\cdots\times \mathbb{F}_{p_{i}}^{\times}\times\cdots\times 1$ under the above identification, but how can I show this rigorously?

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I would try the following. The field $K$ is the compositum of the fields $K_i=\Bbb{Q}(\zeta_{p_i})$, $i=1,2,\ldots,k$. The extensions $K_i/\Bbb{Q}$ are all Galois, and they are linearly disjoint (jointly, i.e. any two compositums without a common factor are linearly disjoint). BTW, this is one way of arriving at your CRT isomorphism. The rational prime $p_i$ is unramified in all the extensions $K_j/\Bbb{Q}$, $j\neq i$, so it is unramified in their compositum $$\hat K_i:=\prod_{j\neq i}K_j.$$ But $p_i$ is totally ramified in $K_i/\Bbb{Q}$, and as $[K:\hat K_i]=[K_i:\Bbb{Q}]=p_i-1=e(\mathfrak{p}|p_i)$, it follows that all the primes abov $p_i$ are totally ramified in $K/\hat K_i$. These facts means that $\hat K_i$ is the inertia field of all primes $\mathfrak{p}$ above $p_i$. Note that as $K$ is an abelian extension of the rationals, the conjugate prime ideals share the same inertia field.

The subgroup $1\times \cdots \times \Bbb{F}_{p_i}^{\times}\times\cdots\times 1$ is the group that Galois correspondence applied to $K/\Bbb{Q}$ associates with the field $\hat K_i$, so the claim follows from this.

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  • $\begingroup$ Very nice hat, comrade! $\endgroup$ – Bruno Joyal Dec 15 '14 at 22:02

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