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 decomposition group 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

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

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

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