Proposition 10

*Let $x \in N$, then for $\{x\}$ to form a basis of the $\mathbf{K}^{(\Gamma)}$-module $N$ it is necessary and sufficient that $\det(\sigma \tau(x))_{\sigma, \tau \in \Gamma} \neq 0$.

Since $\mathbf{K}^{(\Gamma)}$ and $N$ have the same dimension over $K$, to say that $\{x\}$ is a basis of $N$ over $\mathbf{K}^{(\Gamma)}$ means that the mapping $a \mapsto ax$ of $\mathbf{K}^{(\Gamma)}$ into $N$ is injective. This in turn means that the mapping $b \mapsto b(1 \otimes x)$ of $N \otimes_K \mathbf{K}^{(\Gamma)}$ into $N \otimes_K N$ is injective (II, p. 306, Prop. 14). Now there is an $N \otimes_K \mathbf{K}^{(\Gamma)}$-module isomorphism of $N \otimes_K N$ onto $N \otimes_K \mathbf{K}^{(\Gamma)}$ which maps $1 \otimes x$ to $\sum \sigma^{-1}(x) \otimes \sigma$. It follows that $\{x\}$ is a basis of $N$ over $\mathbf{K}^{(\Gamma)}$ if and only if, for every non-zero family of elements $(n_\tau)$, of $N$ we have $\left( \sum n_\tau \otimes \tau \right) \left( \sum \sigma^{-1}(x) \otimes \sigma \right) \neq 0$. But this last relation means that there exists $\sigma \in \Gamma$ such that $\sum n_\tau \sigma^{-1}(x) \neq 0$, whence the proposition.

\* A) Suppose that $K$ is infinite; the mapping $x \mapsto \det(\sigma \tau(x))$ of $N$ into $N$ is a polynomial mapping over $K$ (IV, p. 54). By extension of scalars from $K$ to $N$ we get a corresponding mapping for the vector $N$-space $N \otimes_K N$ and we have just seen that the latter is not identically zero (because $N \otimes_K N$ is free of rank 1 over $N \otimes_K \mathbf{K}^{(\Gamma)}$). So there exists $x \in N$ such that $\det(\sigma \tau(x)) \neq 0$ (IV, p. 18, Th. 2); more generally, from the same reference we have: