Theorem 6

Let $N$ be a Galois extension of finite degree over $K$ and let $\Gamma$ be the Galois group of $N$ over $K$. Then there exists a normal basis of $N$ over $K$; in other words, the $\mathbf{K}^{(\Gamma)}$-module $N$ is free of rank 1.

We shall give two proofs of this statement. The first uses the following lemma which will be proved in Chapter VIII (§ 2, No. 5).

\* Lemma 3. — Let $A$ be a $K$-algebra, $M_1$ and $M_2$ two $A$-modules of finite rank over $K$ and suppose that there exists an extension $L$ of $K$ such that the modules $L \otimes_K M_1$ and $L \otimes_K M_2$ over the ring $L \otimes_K A$ are isomorphic. Then the $A$-modules $M_1$ and $M_2$ are isomorphic.

We shall apply Lemma 3 in the case where $A = \mathbf{K}^{(\Gamma)}$, $M_1 = N$, $M_2 = A$, and $L = N$. By the Cor. of V, p. 64 there exists a $K$-isomorphism $\varphi$ of $N \otimes_K N$ onto $N \otimes_K \mathbf{K}^{(\Gamma)}$ which maps $x \otimes y$ to $\sum x u^1(y) \otimes \sigma$. It is clear that $\varphi$ is an isomorphism of $N \otimes_K \mathbf{K}^{(\Gamma)}$-modules and so the theorem follows from Lemma 3.

For the second proof we shall use the following proposition: