Corollary 3

Let $A'$ be an integral domain, $\mathscr{G}$ a finite commutative group operating on $A'$, $A$ the ring of invariants of $\mathscr{G}$, $\mathfrak{p}$ a prime ideal of $A$ and $K$ and $K'$ the fields of fractions of $A$ and $A'$ respectively. Then the prime ideals of $A'$ lying above $\mathfrak{p}$ all have the same decomposition ring $A^{\mathbf{Z}}$ and the field of fractions $K^{\mathbf{Z}}$ of $A^{\mathbf{Z}}$ is the greatest intermediate field between $K$ and $K'$ in which $\mathfrak{p}$ decomposes completely.

If $\mathfrak{p}'$ is a prime ideal of $A'$ lying above $\mathfrak{p}$, then $\mathscr{G}^{\mathbf{Z}}(\sigma.\mathfrak{p}') = \mathscr{G}^{\mathbf{Z}}(\mathfrak{p}')$ since $\mathscr{G}$ is commutative (formula (2)), hence (Theorem 2 (i)) all the prime ideals of $A'$ lying above $\mathfrak{p}$ have the same decomposition group $\mathscr{G}^{\mathbf{Z}}$ and therefore the same decomposition ring $A^{\mathbf{Z}}$; their number is $(\mathscr{G}:\mathscr{G}^{\mathbf{Z}})$. Let L be an intermediate field between K and $K'$ and let $\mathscr{H}$ be the subgroup of $\mathscr{G}$ leaving invariant the elements of L; the decomposition group of $\mathfrak{p}'$ with respect to $\mathscr{H}$ is $\mathscr{G}^{\mathbf{Z}} \cap \mathscr{H}$; as $A' \cap L$ is the ring of invariants of $\mathscr{H}$ in $A'$, the number of prime ideals of $A'$ lying above $\mathfrak{p}' \cap L$ is $(\mathscr{H}:(\mathscr{G}^{\mathbf{Z}} \cap \mathscr{H})) = (\mathscr{H}\mathscr{G}^{\mathbf{Z}}:\mathscr{G}^{\mathbf{Z}})$ (since $\mathscr{G}$ is commutative). The number of prime ideals of $A' \cap L$ lying above $\mathfrak{p}$ is therefore $(\mathscr{G}:\mathscr{H}\mathscr{G}^{\mathbf{Z}})$. For $\mathfrak{p}$ to decompose completely in L, it is necessary and sufficient therefore that $(\mathscr{G}:\mathscr{H}\mathscr{G}^{\mathbf{Z}}) = [L:K] = (\mathscr{G}:\mathscr{H})$ and, as $\mathscr{H} \subset \mathscr{H}\mathscr{G}^{\mathbf{Z}}$, this is equivalent to $\mathscr{H}\mathscr{G}^{\mathbf{Z}} = \mathscr{H}$ or also to $\mathscr{G}^{\mathbf{Z}} \subset \mathscr{H}$ and finally to $L \subset K^{\mathbf{Z}}$.