# decomposition group

## 1 Decomposition Group

Let $A$ be a Noetherian   integrally closed  integral domain  with field of fractions  $K$. Let $L$ be a Galois extension  of $K$ and denote by $B$ the integral closure  of $A$ in $L$. Then, for any prime ideal   ${\mathfrak{p}}\subset A$, the Galois group  $G:=\operatorname{Gal}(L/K)$ acts transitively on the set of all prime ideals ${\mathfrak{P}}\subset B$ containing ${\mathfrak{p}}$. If we fix a particular prime ideal ${\mathfrak{P}}\subset B$ lying over ${\mathfrak{p}}$, then the stabilizer  of ${\mathfrak{P}}$ under this group action  is a subgroup   of $G$, called the at ${\mathfrak{P}}$ and denoted $D({\mathfrak{P}}/{\mathfrak{p}})$. In other words,

 $D({\mathfrak{P}}/{\mathfrak{p}}):=\{\sigma\in G\mid\sigma({\mathfrak{P}})=({% \mathfrak{P}})\}.$

If ${\mathfrak{P}}^{\prime}\subset B$ is another prime ideal of $B$ lying over ${\mathfrak{p}}$, then the decomposition groups $D({\mathfrak{P}}/{\mathfrak{p}})$ and $D({\mathfrak{P}}^{\prime}/{\mathfrak{p}})$ are conjugate   in $G$ via any Galois automorphism     mapping ${\mathfrak{P}}$ to ${\mathfrak{P}}^{\prime}$.

## 2 Inertia Group

Write $l$ for the residue field  $B/{\mathfrak{P}}$ and $k$ for the residue field $A/{\mathfrak{p}}$. Assume that the extension    $l/k$ is separable  (if it is not, then this development is still possible, but considerably more complicated; see [serre, p. 20]). Any element $\sigma\in D({\mathfrak{P}}/{\mathfrak{p}})$, by definition, fixes ${\mathfrak{P}}$ and hence descends to a well defined automorphism of the field $l$. Since $\sigma$ also fixes $A$ by virtue of being in $G$, it induces an automorphism of the extension $l/k$ fixing $k$. We therefore have a group homomorphism

 $D({\mathfrak{P}}/{\mathfrak{p}})\longrightarrow\operatorname{Gal}(l/k),$
 (1)