---
tag: N36F259F1B
kind: definition
name: Decomposition group
msc: [11R32, 11S20]
defines: 7X92G8CQ33
uses: [MWBDFS2KJ5, 0QMZAQ6QQ4, 0XTY86BZKY, 00MHHZ9SNE, B7KNZCBFYT, 0X1JXT47GA, 0ZXZ82FTX2]
sources:
  - url: https://en.wikipedia.org/wiki/Splitting_of_prime_ideals_in_Galois_extensions
---

## Definition

Let $L/K$ be a finite [[Galois extension|0QMZAQ6QQ4]] of [[algebraic number fields|MWBDFS2KJ5]] with [[Galois group|0XTY86BZKY]] $G$, and let $\mathcal{O}_K \subseteq \mathcal{O}_L$ be the rings of integers.
Let $\mathfrak{P}$ be a nonzero [[prime ideal|00MHHZ9SNE]] of $\mathcal{O}_L$.
The *decomposition group* of $\mathfrak{P}$ is the [[subgroup|0X1JXT47GA]]
$$D_{\mathfrak{P}} = \set{\sigma \in G : \sigma(\mathfrak{P}) = \mathfrak{P}}.$$
The *inertia group* of $\mathfrak{P}$ is
$$I_{\mathfrak{P}} = \set{\sigma \in D_{\mathfrak{P}} : \sigma(x) \equiv x \pmod{\mathfrak{P}} \text{ for all } x \in \mathcal{O}_L}.$$

## Remarks

Let $\mathfrak{p} = \mathfrak{P} \cap \mathcal{O}_K$, with ramification index $e$, residue degree $f$ and $g$ primes over it, as in [[splitting of prime ideals|B7KNZCBFYT]].
Then $D_{\mathfrak{P}}$ has order $ef$ and index $g$ in $G$.
Each $\sigma \in D_{\mathfrak{P}}$ induces an automorphism of the [[finite field|0ZXZ82FTX2]] $\mathcal{O}_L/\mathfrak{P}$ over $\mathcal{O}_K/\mathfrak{p}$; this gives a surjective homomorphism onto the Galois group of the residue extension, with kernel $I_{\mathfrak{P}}$, so $\abs{I_{\mathfrak{P}}} = e$.
For $\tau \in G$, one has $D_{\tau(\mathfrak{P})} = \tau D_{\mathfrak{P}} \tau^{-1}$.

Example: in $\Q(i)/\Q$, with $G = \set{1, c}$ where $c$ is complex conjugation, take $\mathfrak{P} = (3)$.
Then $c(\mathfrak{P}) = \mathfrak{P}$, so $D_{\mathfrak{P}} = G$; and $c(i) = -i \not\equiv i \pmod 3$, so $I_{\mathfrak{P}}$ is trivial, matching $e = 1$, $f = 2$.

Non-example: for $\mathfrak{P} = (2 + i)$, the element $c$ is not in $D_{\mathfrak{P}}$, because $c(\mathfrak{P}) = (2 - i) \ne \mathfrak{P}$ (the quotient $(2+i)/(2-i)$ is not in $\Z[i]$).
So $D_{\mathfrak{P}}$ is trivial, matching $e = f = 1$ and $g = 2$.