Definition 5

With the hypotheses and notation of Proposition 6, the field of invariants $K^Z(p')$ (resp. $K^T(p')$) of the group $G^Z(p')$ (resp. $G^T(p')$) in the field $K'$ is called the decomposition field (resp. inertia field) of $p'$ with respect to $K$.

We also write $K^Z$ (resp. $K^T$) in place of $K^Z(p')$ (resp. $K^T(p')$). It follows from § 1, no. 9, Proposition 23 that $K^Z$ (resp. $K^T$) is the field of fractions of the ring $A^Z$ (resp. $A^T$); $A^Z$ (resp. $A^T$) is the integral closure of $A$ in $K^Z$ (resp. $K^T$).

Remarks

(3) Under the conditions of Corollary 4 of Proposition 6 and assuming that $[K':K]$ is finite, the number of distinct prime ideals lying above $p$ is $[K^Z:K]$, this degree being equal to the index ('3:*G*Z*) by Galois theory; moreover, it follows from Galois theory that

(4) Let $A$ be an integrally closed domain, $K$ its field of fractions, $K'$ a finite algebraic extension of $K$ and $A'$ the integral closure of $A$ in $K'$. Then, for every prime ideal $p$ of $A$, the number $\&$ prime ideals $\&$ of $A'$ lying above $p$ is at most $[K':K]_s$ (the separable factor of the degree of $K$ over $K$). We may first restrict our attention to the case where $K$ is a separable extension of $K$, for in general $K$ is a radicial extension of the greatest separable extension $K_0$ of $K$ contained in $K'$, $[K':K]_s = [K_0:K]$ by definition and, if $A$, is the integral closure of $A$ in $K_0$, the prime ideals of $A$, and $A'$ are in one-to-one correspondence (Lemma 4). Suppose therefore that $K$ is separable over $K$ and let $N$ be the Galois extension of $K$ generated by $K$ in an algebraic closure of $K$, $G$ its Galois group, $B$ the integral closure of $A$ in $N$ and $\mathfrak{P}$ a prime ideal of $B$ lying above $p$. Let $\mathscr{H}$ be the Galois group of $N$ over $K$ and $G^Z$ the decomposition group of $\mathfrak{P}$; the prime ideals of $B$ lying above $p$ are the $s.\mathfrak{P}$ where $s \in G$ (no. 2, Theorem 2) and the relation $s.\mathfrak{P} = s'.\mathfrak{P}$ means that $s' = sg$ where $g \in G^Z$. On the other hand in order that $s.\mathfrak{P} \cap K' = s'.\mathfrak{P} \cap K'$, it is necessary and sufficient that $s'.\mathfrak{P} = ts.\mathfrak{P}$, where $t \in \mathscr{H}$ (no. 2, Theorem 2), whence finally $s' = tsg$ where $t \in \mathscr{H}$ and $g \in G^Z$. The number of prime ideals of $A'$ lying above $p$ is therefore equal to the number of classes of $G$ under the equivalence relation "there exists $t \in \mathscr{H}$ and $g \in G^{\mathbf{Z}}$ such that $s' = tsg$" between $s$ and $s'$; clearly this number is at most equal to the index $(G : \mathscr{H})$, the number of right cosets of $\mathscr{H}$ in $G$, and $(G : \mathscr{H}) = [K' : K]$ by Galois theory.