The Decomposition Group

Suppose $ K$ is a number field that is Galois over $ \mathbf {Q}$ with group $ G=\Gal (K/\mathbf{Q})$. Fix a prime $ \mathfrak{p}\subset \O_K$ lying over $ p\in\mathbf{Z}$.

Definition 9.3.1 (Decomposition group)   The decomposition group of $ \mathfrak{p}$ is the subgroup

$\displaystyle D_\mathfrak{p}= \{\sigma \in G : \sigma(\mathfrak{p})=\mathfrak{p}\} \subset G.

Note that $ D_\mathfrak{p}$ is the stabilizer of $ \mathfrak{p}$ for the action of $ G$ on the set of primes lying over $ p$.

It also makes sense to define decomposition groups for relative extensions $ K/L$, but for simplicity and to fix ideas in this section we only define decomposition groups for a Galois extension $ K/\mathbf{Q}$.

Let $ k_\mathfrak{p}= \O_K/\mathfrak{p}$ denote the residue class field of $ \mathfrak{p}$. In this section we will prove that there is an exact sequence

$\displaystyle 1\to I_\mathfrak{p}\to D_\mathfrak{p}\to \Gal (k_{\mathfrak{p}}/\mathbf{F}_p)\to 1,

where $ I_\mathfrak{p}$ is the inertia subgroup of $ D_\mathfrak{p}$, and $ \char93 I_\mathfrak{p}=e$, where $ e$ is the exponent of $ \mathfrak{p}$ in the factorization of $ p\O_K$. The most interesting part of the proof is showing that the natural map $ D_\mathfrak{p}\to \Gal (k_{\mathfrak{p}}/\mathbf{F}_p)$ is surjective.

We will also discuss the structure of $ D_\mathfrak{p}$ and introduce Frobenius elements, which play a crucial role in understanding Galois representations.

Recall from Theorem 9.2.2 that $ G$ acts transitively on the set of primes  $ \mathfrak{p}$ lying over $ p$. The orbit-stabilizer theorem implies that $ [G:D_\mathfrak{p}]$ equals the cardinality of the orbit of  $ \mathfrak{p}$, which by Theorem 9.2.2 equals the number $ g$ of primes lying over $ p$, so $ [G:D_\mathfrak{p}]=g$.

Lemma 9.3.2   The decomposition subgroups $ D_\mathfrak{p}$ corresponding to primes $ \mathfrak{p}$ lying over a given $ p$ are all conjugate as subgroups of $ G$.

Proof. We have for each $ \sigma, \tau \in G$, that

$\displaystyle \tau^{-1}\sigma \tau\mathfrak{p}= \mathfrak{p}
\sigma\tau \mathfrak{p}= \tau \mathfrak{p},


$\displaystyle \sigma \in D_{\tau\mathfrak{p}} \iff \tau^{-1}\sigma\tau\in D_\mathfrak{p}.


$\displaystyle \sigma \in D_\mathfrak{p}\iff \tau \sigma \tau^{-1} \in D_{\tau \mathfrak{p}}.

Thus $ \tau D_{\mathfrak{p}}\tau^{-1} = D_{\tau \mathfrak{p}}$. $ \qedsymbol$

The decomposition group is useful because it allows us to refine the extension $ K/\mathbf{Q}$ into a tower of extensions, such that at each step in the tower we understand well the splitting behavior of the primes lying over $ p$.

We characterize the fixed field of $ D=D_\mathfrak{p}$ as follows.

Proposition 9.3.3   The fixed field

$\displaystyle K^D=\{a \in K : \sigma(a) = a$ for all $\displaystyle \sigma \in D\}$

of $ D$ is the smallest subfield $ L\subset K$ such that the prime ideal $ \mathfrak{p}\cap\O_L$ has $ g(K/L)=1$, i.e., there is a unique prime of $ \O_K$ over $ \mathfrak{p}\cap\O_L$.

Proof. First suppose $ L=K^D$, and note that by Galois theory $ \Gal (K/L)\cong
D$, and by Theorem 9.2.2, the group $ D$ acts transitively on the primes of $ K$ lying over $ \mathfrak{p}\cap\O_L$. One of these primes is $ \mathfrak{p}$, and $ D$ fixes $ \mathfrak{p}$ by definition, so there is only one prime of $ K$ lying over $ \mathfrak{p}\cap\O_L$, i.e., $ g=1$. Conversely, if $ L\subset K$ is such that $ \mathfrak{p}\cap\O_L$ has $ g=1$, then $ \Gal (K/L)$ fixes $ \mathfrak{p}$ (since it is the only prime over $ \mathfrak{p}\cap\O_L$), so $ \Gal (K/L)\subset D$, hence $ K^D\subset L$. $ \qedsymbol$

Thus $ p$ does not split in going from $ K^D$ to $ K$--it does some combination of ramifying and staying inert. To fill in more of the picture, the following proposition asserts that $ p$ splits completely and does not ramify in $ K^D/\mathbf{Q}$.

Proposition 9.3.4   Fix a finite Galois extension $ K$ of  $ \mathbf {Q}$, let  $ \mathfrak{p}$ be a prime lying over $ p$ with decomposition group $ D$, and set $ L=K^D$. Let $ e=e(L/\mathbf{Q}),f=f(L/\mathbf{Q}),g=g(L/\mathbf{Q})$ be for $ L/\mathbf{Q}$ and $ p$. Then $ e=f=1$, $ g=[L:\mathbf{Q}]$, $ e(K/\mathbf{Q})=e(K/L)$ and $ f(K/\mathbf{Q})=f(K/L)$.

Proof. As mentioned right after Definition 9.3.1, the orbit-stabilizer theorem implies that $ g(K/\mathbf{Q})=[G:D]$, and by Galois theory $ [G:D]=[L:\mathbf{Q}]$, so $ g(K/\mathbf{Q}) = [L:\mathbf{Q}]$. Proposition 9.3.3,, $ g(K/L)=1$ so by Theorem 9.2.2,

$\displaystyle e(K/L)\cdot f(K/L)$ $\displaystyle = [K:L] =[K:\mathbf{Q}]/[L:\mathbf{Q}]$    
  $\displaystyle = \frac{e(K/\mathbf{Q})\cdot f(K/\mathbf{Q}) \cdot g(K/\mathbf{Q})}{[L:\mathbf{Q}]} = e(K/\mathbf{Q})\cdot f(K/\mathbf{Q}).$    

Now $ e(K/L)\leq e(K/\mathbf{Q})$ and $ f(K/L)\leq f(K/\mathbf{Q})$, so we must have $ e(K/L)=e(K/\mathbf{Q})$ and $ f(K/L)=f(K/\mathbf{Q})$. Since $ e(K/\mathbf{Q})=e(K/L)\cdot e(L/\mathbf{Q})$ and $ f(K/\mathbf{Q})=f(K/L)\cdot f(L/\mathbf{Q})$, it follows that $ e(L/\mathbf{Q})=f(L/\mathbf{Q}) = 1$. $ \qedsymbol$

William Stein 2012-09-24