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Let $L/K$ be a finite Galois extension. Let $\text{Gal}(L/K)$ be its Galois group.

For every cyclic subgroup $H$ of $\text{Gal}(L/K)$, why does there exists a prime $v$ of ring of integers of $K$ and a prime $w$ of $L$ above $v$ such that $H\cong \text{Gal}(L_w/K_v)$ , in other words, every cyclic sub extension appears as decomposition group for certain prime $v$ ?

If $H=\text{Gal}(L/K)$, from Chebotarev density theorem, there exists $v$ such that it splits completely in $L$, so it is ok.

If $\#H=n$ and $\#\text{Gal}(L/K)=m$, we should find $v$ such that it splits into $m/n$ primes in $L/K$.

But if $H$ is proper subgroup of $\text{Gal}(L/K)$, how Chebotarev density ensures the existence of $v$ ?

Martin Sleziak
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Poitou-Tate
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1 Answers1

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Let $G = {\rm Gal}(L/K)$. Chebotarev says each conjugacy class in $G$ is a conjugacy class of Frobenius elements at infinitely many primes in $L$. Once one element in a conjugacy class in $G$ is a Frobenius element at some prime ideal $\mathfrak P$ in $L$, every other element in its conjugacy class is a Frobenius element at some prime in $L$ lying over the same prime in $K$ as $\mathfrak P$ does, since $g{\rm Frob}_{\mathfrak P}(L/K)g^{-1} = {\rm Frob}_{g(\mathfrak P)}(L/K)$.

So if you pick a cyclic subgroup $H$ in $G$ and a generator $h$ of $H$, Chebotarev says $h$ is a Frobenius element at infinitely many primes in $L$. The subgroup generated by a Frobenius element at an unramified prime is the decomposition group at that prime, so $H$ is the decomposition group at infinitely many primes in $L$.

KCd
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