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

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

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

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

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

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

so

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

Thus

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

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

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

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,

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

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

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 $e(L/\mathbf{Q})=f(L/\mathbf{Q}) = 1$. $\qedsymbol$