Theorem 5

Let $N'$ be an extension of $K$ generated by two subextensions $K'$ and $N$. Suppose that $N$ is Galois over $K$, with Galois group $\Gamma$ and that $K' \cap N = K$. Then the extension $N'$ of $K'$ is Galois and the canonical homomorphism $\varphi$ of $K' \otimes_K N$ into $N'$ is an isomorphism. Let $\sigma \in \mathrm{Gal}(N/K)$ and let $\sigma'$ be the element of $\mathrm{Gal}(N'/K')$ which corresponds to it under the translation isomorphism; then we have $\sigma' \circ \varphi = \varphi \circ (\mathrm{Id}_{K'} \otimes u)$.

We have $N' = K'(N)$ and $N$ is algebraic and separable over $K$; hence (V, p. 42, Prop. 10), the extension $N'$ of $K'$ is algebraic and separable. By Cor. 4 of V, p. 54 the extension $N'$ of $K'$ is quasi-Galois. Therefore the extension $N'$ of $K'$ is Galois. By c) above the mapping $\sigma \mapsto \sigma|_N$ is an homomorphism $\lambda$ of $\mathrm{Gal}(N'/K')$ onto $\mathrm{Gal}(N/K)$.

We have $N = K'[N]$ because $N$ is algebraic over $K$ (V, p. 18, Cor. 1), hence $\varphi$ is surjective. If $a$ belongs to $\mathrm{Gal}(N/K)$, we have

Therefore the kernel of $\varphi$ is stable under the mappings $\mathrm{Id}_{K'} \otimes \sigma$, hence of the form $K' \otimes_K N_0$ with $N_0 \subset N$ (V, p. 63, Cor.). For $x$ in $N_0$ we have $x = \varphi(1 \otimes x) = 0$, hence $N_0 = 0$ and so $\varphi$ is injective.