Proposition 13

— Let $N$ be a quasi-Galois extension of $K$. We denote by $N_r$, the field of invariants of the group of all $K$-automorphisms of $N$ and by $N_s$, the relative separable algebraic closure of $K$ in $N$ ($V$, p. 44). Then:

a) $N_r$ is the relative $p$-radical closure of $K$ in $N$ ($V$, p. 25).

b) $N$, is a Galois extension of $K$ and every $K$-automorphism of $N_s$ extends in a unique fashion to an $N_r$-automorphism of $N$.

c) The fields $N_r$ and $N$, are linearly disjoint over $K$ and we have $N = K[N, \cup N_s]$; in other words, the canonical homomorphism of $N_r \otimes_K N$, into $N$ is an isomorphism.

Let $\Omega$ be an algebraic closure of $K$ containing $N$ as subextension ($V$, p. 23, Th. 2). Every $K$-automorphism of $\Omega$ induces an automorphism of $N$ because $N$ is quasi-Galois. Therefore every element of $N_r$ is invariant under the group of $K$-automorphisms of $\Omega$, hence $p$-radical over $K$ ($V$, p. 53, Cor. 3). Conversely every element of $N$ which is $p$-radical over $K$ is clearly invariant under every $K$-automorphism of $N$ and hence belongs to $N_r$. This proves a).

Every $K$-automorphism of $\Omega$ maps $N$ into $N$, hence $N_r$ into $N_r$, so $N_r$ is a quasi-Galois extension of $K$ ($V$, p. 54, Prop. 3). It follows that $N_s$ is a Galois extension of $K$. Every element of $N_r \cap N$, is separable algebraic and $p$-radical over $K$, hence belongs to $K$ ($V$, p. Cor. 3); we thus have $K_r \cap K_s = K$. Now $N$ is $p$-radical over $N_r$ ($V$, p. 44, Prop. 13) and separable algebraic over $N_r$ ($V$, p. 56, Th. 1) hence both $p$-radical and separable over $K(N_r \cup N_s)$. Hence we have $N = K(N_r \cup N_s)$ ($V$, p. 39, Cor. 3) and the assertions b), c) follow from Th. 5 ($V$, p. 71).