Proposition 5

With the hypotheses and notation of Theorem 2, the field of fractions $k^T$ of $A^T/(\mathfrak{p}' \cap A^T)$ is equal to the greatest separable extension $k'_s$ of $k$ contained in $k'$.

As in Proposition 4 this may be reduced to the case where $\mathfrak{p}$ is a maximal ideal of $A$, which implies that $\mathfrak{p}'$, $\mathfrak{p}' \cap A^{\mathbf{Z}}$ and $\mathfrak{p}' \cap A^T$ are maximal in $A'$, $A^{\mathbf{Z}}$ and $A^T$ respectively (no. 1, Proposition 1).

For all $x \in A'$, the polynomial $P(X) = \prod_{\sigma \in \mathscr{G}^T} (X - \sigma.x)$ has its coefficients in the inertia ring $A^T$ and, by definition of $\mathscr{G}^T$, all its roots in $A'$ are congruent mod. $\mathfrak{p}'$; the polynomial $\pi'(P)$ over $A^T/(\mathfrak{p}' \cap A^T)$ whose coefficients are the canonical images of those of P under the homomorphism $\pi':A' \to A'/\mathfrak{p}'$ therefore has all its roots in $A'/\mathfrak{p}'$ equal to the image of $x$, which shows that $k'$ is a radicial extension of $k^T$; whence $k'_s \subset k^T$, since every element of $k'_s$ is separable over k and a fortiori over $k^T$.

We know that $k'_s$ is a Galois extension of k (Algebra, Chapter V, § 10, no. 9, Proposition 14) and it follows from Theorem 2 that its Galois group is isomorphic to $\mathscr{G}' = \mathscr{G}^z / \mathscr{G}^T$. As $k^T$ is a radicial extension of $k'_s$, $k^T$ is a quasi-Galois extension of k and the separable factor of the degree of $k^T$ over k is $q = (\mathscr{G}^z : \mathscr{G}^T )$. It remains to see that $k^T$ is a separable extension of k. We have seen above that $\mathscr{G}'$ is identified with an automorphism group of $A^T$ and that $A^z$ is the ring of invariants of $\mathscr{G}'$. If $x \in A^T$, the polynomial $Q(X) = \prod_{\sigma' \in \mathscr{G}'} (X - \sigma'(x))$ therefore has its coefficients in $A^z$; the polynomial over $A^z / (p' \cap A^z)$ whose coefficients are the images of those of Q under $\pi'$ is of degree $q$ and has a root $\pi'(x) \in A^T / (p' \cap A^T )$. As $A^z / (p' \cap A^z) = k$ by Proposition 4 (ii), we see that every element of $k^T$ is of degree $\leq q$ over k.

This being so, let k, be the field of invariants of the group of k-automorphisms of the quasi-Galois extension $k^T$ of $k$; then $[k^T : k_1] = q$ (Algebra, Chapter V, § 10, no. 9, Proposition 14). Let $u$ be a primitive element of $k^T$ over $k_1$; as it is of degree $q$ over k, and of degree $\leq q$ over k, it is of degree $q$ over k and its minimal polynomial over k, has coefficients in $k$; this shows that $u$ is separable over k. On the other hand, for all $v \in k_1$, there exists a power $p'$ of the characteristic exponent p such that $v^{p'} \in k$. We conclude that $k(u - v)$, which contains

contains $u^{p'}$ and consequently $k(u^{p'})$. But as $u$ is separable over $k$, $k(u) = k(u^{p'})$ (Algebra, Chapter V, § 8, no. 3, Proposition 4), whence $k(u) \subset k(u - v)$. As $u$ is of degree $q$ over k and $u - v$ of degree $\leq q$, it follows that $k(u) = k(u - v)$, whence $v \in k(u)$. This shows that $v$ is separable over k, hence $k_1 = k$ and $k^T$ is separable over $k$.