Corollary 4

Let $A$ be an integrally closed domain, $K$ its field of fractions, $K$ a Galois extension of $K$, $A'$ the integral closure of $A$ in $K'$, $\mathfrak{p}'$ a prime ideal of $A$, $\mathfrak{p} = A \cap \mathfrak{p}'$ and $k$ and $k'$ the fields of fractions of $A/\mathfrak{p}$ and $A'/\mathfrak{p}'$ respectively. Then:

(i) The field of fractions of $A^{\mathbf{Z}}/(\mathfrak{p}' \cap A^{\mathbf{Z}})$ is equal to $k$ and the maximal ideal of the local ring of $A^{\mathbf{Z}}$ relative to $\mathfrak{p}' \cap A^{\mathbf{Z}}$ is generated by $\mathfrak{p}$.

(ii) The field of fractions $k^T$ of $A^T/(\mathfrak{p}' \cap A^T)$ is the greatest separable extension of $k$ contained in $k'$.

The ring $A$ is the ring of invariants in $A'$ of the Galois group of $K'$ over $K$; if $K'$ is of finite degree over $K$, the corollary then follows from Propositions 4 and 5 of no. 2. Consider now the general case, $K'$ therefore being the union of a right directed set $(K_i)$ of finite Galois extensions of $K$. Then:

(i) If $x, y$ are two elements of $A^{\mathbf{Z}}$, where $y \notin \mathfrak{p}'$, there is an index $\alpha$ such that $x$ and $y$ belong to $A^{\mathbf{Z}}(\mathfrak{p}' \cap K_{\alpha})$ (Remark 2); by Proposition 4 of no. 2, there are $x_0, y_0$ in $A$ with $y_0 \notin \mathfrak{p}'$ such that $xy_0 - x_0y \in \mathfrak{p}'$, which proves the first assertion of (i); if further $x \in \mathfrak{p}'$, we may assume that $y_0$ satisfies

which proves the second assertion of (i).

(ii) Suppose now that $x \in A^T$; there exists $\alpha$ such that $x \in A^T(p' \cap K_\alpha)$ (Remark 2) and Proposition 5 of no. 2 shows that the class $\bar{x}$ of $x \mod.(p' \cap K_\alpha \cap A^T)$ is algebraic and separable over $k$; a fortiori the class $\mod.(p' \cap A^T)$ of $x$ is separable over $k$; to complete the proof of the corollary, it is sufficient to show that $k'$ is a radicial extension of $k^T$. Now, $k'$ is the union of the right directed family of fields of fractions $k_\alpha$ of the rings $(A' \cap K_\alpha)/(p' \cap K_\alpha)$. It follows therefore from Proposition 5 that, if an element of $k'$ belongs to $k$, it is radicial over the field of fractions of

and a fortiori over $k^T$ (by virtue of Remark 2).