# 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^{} $\U0001d52d\subset A$, the Galois group^{} $G:=\mathrm{Gal}(L/K)$ acts transitively on the
set of all prime ideals $\U0001d513\subset B$ containing $\U0001d52d$. If we fix a
particular prime ideal $\U0001d513\subset B$ lying over $\U0001d52d$, then the
stabilizer^{} of $\U0001d513$ under this group action^{} is a subgroup^{} of
$G$, called the decomposition group^{} at $\U0001d513$ and denoted
$D(\U0001d513/\U0001d52d)$. In other words,

$$D(\U0001d513/\U0001d52d):=\{\sigma \in G\mid \sigma (\U0001d513)=(\U0001d513)\}.$$ |

If ${\U0001d513}^{\prime}\subset B$ is another prime ideal of $B$ lying over $\U0001d52d$, then
the decomposition groups $D(\U0001d513/\U0001d52d)$ and $D({\U0001d513}^{\prime}/\U0001d52d)$ are conjugate^{} in
$G$ via any Galois automorphism^{} mapping $\U0001d513$ to ${\U0001d513}^{\prime}$.

## 2 Inertia Group

Write $l$ for the residue field^{} $B/\U0001d513$ and $k$ for the residue field
$A/\U0001d52d$. 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 $\sigma \in D(\U0001d513/\U0001d52d)$, by definition, fixes $\U0001d513$ 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

$$D(\U0001d513/\U0001d52d)\u27f6\mathrm{Gal}(l/k),$$ |

and the kernel (http://planetmath.org/KernelOfAGroupHomomorphism) of this homomorphism^{} is called the inertia group of
$\U0001d513$, and written $T(\U0001d513/\U0001d52d)$. It turns out that this homomorphism is
actually surjective^{}, so there is an exact sequence^{}

(1) |