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 14.1.1 (Decomposition group)   The of $\mathfrak{p}$ is the subgroup

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

(Note: The decomposition group is called the ``splitting group'' in Swinnerton-Dyer. Everybody I know calls it the decomposition group, so we will too.)

Let $\mathbf{F}_{\mathfrak{p}} = \O _K/\mathfrak{p}$ denote the residue class field of $\mathfrak{p}$. In this section we will prove that there is a natural exact sequence

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

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

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

Recall that $G$ acts on the set of primes $\mathfrak{p}$ lying over $p$. Thus the decomposition group is the stabilizer in $G$ of $\mathfrak{p}$. The orbit-stabilizer theorem implies that $[G:D_\mathfrak{p}]$ equals the orbit of $\mathfrak{p}$, which by Theorem 13.2.2 equals the number $g$ of primes lying over $p$, so $[G:D_\mathfrak{p}]=g$.

Lemma 14.1.2   The decomposition subgroups $D_\mathfrak{p}$ corresponding to primes $\mathfrak{p}$ lying over a given $p$ are all conjugate in $G$.

Proof. We have $\tau(\sigma(\tau^{-1}(\mathfrak{p}))) = \mathfrak{p}$ if and only if $\sigma(\tau^{-1}(\mathfrak{p})) = \tau^{-1}\mathfrak{p}$. Thus $\tau\sigma\tau^{-1}\in D_p$ if and only if $\sigma\in D_{\tau^{-1}\mathfrak{p}}$, so $\tau^{-1}D_p\tau = D_{\tau^{-1}\mathfrak{p}}$. The lemma now follows because, by Theorem 13.2.2, $G$ acts transitively on the set of $\mathfrak{p}$ lying over $p$. $\qedsymbol$

The decomposition group is extremely useful because it allows us to see the extension $K/\mathbf{Q}$ as a tower of extensions, such that at each step in the tower we understand well the splitting behavior of the primes lying over $p$. Now might be a good time to glance ahead at Figure 14.1.2 on page .

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

Proposition 14.1.3   The fixed field $K^D$ of $D$

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

is the smallest subfield $L\subset K$ such that $\mathfrak{p}\cap L$ does not split in $K$ (i.e., $g(K/L)=1$).

Proof. First suppose $L=K^D$, and note that by Galois theory $\Gal (K/L)\cong D$, and by Theorem 13.2.2, the group $D$ acts transitively on the primes of $K$ lying over $\mathfrak{p}\cap L$. One of these primes is $\mathfrak{p}$, and $D$ fixes $\mathfrak{p}$ by definition, so there is only one prime of $K$ lying over $\mathfrak{p}\cap L$, i.e., $\mathfrak{p}\cap L$ does not split in $K$. Conversely, if $L\subset K$ is such that $\mathfrak{p}\cap L$ does not split in $K$, then $\Gal (K/L)$ fixes $\mathfrak{p}$ (since it is the only prime over $\mathfrak{p}\cap L$), 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 14.1.4   Let $L=K^D$ for our fixed prime $p$ and Galois extension $K/\mathbf{Q}$. Let $e=e(L/\mathbf{Q}),f=f(L/\mathbf{Q}),g=g(L/\mathbf{Q})$ be for $L/\mathbf{Q}$ and $p$. Then $e=f=1$ and $g=[L:\mathbf{Q}]$, i.e., $p$ does not ramify and splits completely in $L$. Also $f(K/\mathbf{Q})=f(K/L)$ and $e(K/\mathbf{Q})=e(K/L)$.

Proof. As mentioned right after Definition 14.1.1, the orbit-stabilizer theorem implies that $g(K/\mathbf{Q})=[G:D]$, and by Galois theory $[G:D]=[L:\mathbf{Q}]$. Thus

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

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

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 $f(K/\mathbf{Q})=f(K/L)\cdot f(L/\mathbf{Q})$, the proposition follows. $\qedsymbol$