Lemma 4

Let $\Gamma$ be a cyclic group of order $n$, $\sigma$ a generator of $\Gamma$ and M a commutative group on which $\Gamma$ operates according to the rule $\gamma . (m + m') = \gamma . m + \gamma . m'$ for all $\gamma \in \Gamma$ and $m, m' \in M$. Let Z be the set of all mappings of $\Gamma$ into M satisfying the relation

Put $u(f) = f(a)$ for $f \in Z$ and $t(m) = \sum_{\tau \in \Gamma} \tau . m$ for $m \in M$. Then the sequence

is exact.

Let $f \in \mathbf{Z}$; taking $\tau = \tau' = 1$ in (10) we find that $f(1) = 0$. Further, by induction on $m \geqslant 0$ we deduce the relation

We have $\sigma^n = 1$, whence $f(\sigma^n) = 0$; the preceding relation with $m = n$ is equivalent to the equality $t(u(f)) = 0$, whence $\operatorname{Im}\ u \subset \operatorname{Ker}\ t$. Moreover, from (11) it follows that $\operatorname{Ker}\ u = 0$.

Let $m \in M$ be such that $t(m) = 0$, that is, $m + \sigma \cdot m + \cdots + \sigma^{n-1} \cdot m = 0$. Let us define the mapping $f$ of $\Gamma$ into $M$ by

for $0 \leqslant j \leqslant n - 1$. It may be left to the reader to establish the relation (10). We clearly have $m = f(a)$, whence $\operatorname{Im}\ u \supset \operatorname{Ker}\ t$.

With the lemma now proved, let us take $y \in E^*$ and $x = y/\sigma(y)$; we have $N_{E/K}(\sigma(y)) = N_{E/K}(y)$, whence $N_{E/K}(x) = 1$. Conversely, let $x$ in $E^*$ be such that $N_{E/K}(x) = 1$; by Lemma 4, applied to $M = E^*$, there exists a family of elements $(c_\tau)_{\tau \in \Gamma}$ of $E^*$ satisfying the relation $c_{\tau \tau'} = c_\tau \cdot \tau(c_{\tau'})$ for $\tau, \tau'$ in $\Gamma$ and $c_\sigma = x$. By Cor. 1 of Prop. 9 (V, p. 65) there exists $y \in E^*$ with $c_\tau = y/\tau(y)$ for all $\tau \in \Gamma$, whence in particular $x = c_\sigma = y/\sigma(y)$. If $y_1 \in E^*$ satisfies $x = y_1/\sigma(y_1)$, then we have

hence $y_1 y^{-1}$ belongs to $K^*$ because $a$ generates the Galois group of $E$ over $K$. This proves a).

The assertion b) follows in similar fashion from Cor. 2 of Prop. 9 (V, p. 65).