Decomposition group

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. 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 $\mathcal{O}_L/\mathfrak{P}$ 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 $g = 2$.