Lemma 1

Let E be a subextension of N. Then N is a Galois extension of E and $\mathrm{Gal}(N/E)$ is a closed subgroup of $\mathrm{Gal}(N/K)$ with the induced topology.

Let $x \in N$; the minimal polynomial f of x over E divides in $E[X]$ the minimal polynomial $g$ of x over K (V, p. 17, Cor. 2). Since N is Galois over K, the polynomial g is a product in $N[X]$ of distinct factors of degree 1; hence the same is true of f and so N is Galois over E.

Let $\Gamma$ be the Galois group of N over K and A that of N over E. By definition A is the subgroup of $\Gamma$ consisting of all $\sigma$ such that $\sigma(x) = x$ for all $x \in E$. Now for each $x \in E$ the mapping $\sigma \mapsto \sigma(x)$ of $\Gamma$ into the discrete space N is continuous, hence D is closed in $\Gamma$. Let $\sigma \in \Gamma$; for $x_1, \ldots, x_n$ in N let $U(x_1, \ldots, x_n)$ be the set of all $\tau \in \Gamma$ such that $\tau(x_i) = \sigma(x_i)$ for $1 \leq i \leq n$; put

Then the family of sets $U(x_1, \ldots, x_n)$ (resp. $V(x_1, \ldots, x_n)$) is a base of neighbourhoods of $\sigma$ in $\Gamma$ (resp. A). Hence the topology on A is that induced by $\Gamma$.

B) The relation $A = g(k(\Delta))$ (for $A \in g$) is a consequence of the following more precise lemma: