Lemma 1

Let $x, y$ be two permutable elements of $A$ such that $x - y \in m^j \ (j \geq 1)$; then $x^{p^n} - y^{p^n} \in m^{j+n}$ for every integer $n \geq 0$.

By induction on $n$, it is reduced to proving the lemma for $n = 1$. Then

the second factor is a sum of $p$ terms congruent to one another mod. $m$ and, as $A/m$ is of characteristic $p$, $p \cdot 1 \in m$ in $A$ and hence

whence $x^p - y^p \in m^{j+1}$.

We know that the multiplicative group $(A/m)^*$ is a cyclic group with $q - 1$ elements (Algebra, Chapter V, § 11, no. 1, Theorem 1); let $x$ be a representative in $A$ of a generator of this group; then $x^q - x \in m$, whence, by Lemma 1, $x^{q^{n+1}} - x^{q^n} \in m^{1+fn}$, since $x^q$ and $x$ are permutable. This proves that $(x^{q^n})_{n \geq 0}$ is a Cauchy sequence in $A$; as $A$ is compact and hence complete, this sequence has a limits in $A$ which obviously satisfies $s \equiv x \pmod{m}$ and $s^q = s$. As $s \neq 0$, $s^{q-1} = 1$, more precisely $s$ is a primitive $(q-1)$*-th root of unity* in $A$. Clearly the set $S$ consisting of $0$ and the powers $s^j \ (0 \leq j \leq q-2)$ is a system of representatives of the classes of $A$ mod. $m$ and is invariant under multiplication in $A$.

Now let $a$ be a uniformizer for $v$ and consider the inner automorphism $y \mapsto a^{-1}ya$ of $K$; it maps $A$ to itself, $m$ to itself and therefore, taking quotients, it defines an automorphism of the field $A/m$; it is known (Algebra, Chapter V, § 11, no. 4, Proposition 5) that such an automorphism is of the form $z \mapsto z^{p^r}$, where $0 \leq r \leq f-1$. Then $a^{-1}s^j a \equiv s^{jp^r} \pmod{m}$ for $0 \leq j \leq q-2$; as $a \in m$ and $s \notin m$, this implies that $s^{-j}as^{jp^r} \equiv a \pmod{m^2}$.

Let us write

Then $u \equiv (q - l) a \equiv -a (\text{mod. } m^2)$ since $p . 1 \in m$; we conclude that $u$ is also a uniformizer for $v$; moreover (3)

whence we deduce that $u^{-1} S u = S$.

Finally, for all $x \in A$ there exists a unique sequence $(s_i)$ ($i \in \mathbf{N}$) such that $s_i \in S$ for all $i$ and $x \equiv \sum s_i u^i$ (mod. $m^{n+1}$) for all $n \geq 0$: it is immediate by induction on $n$, every element $t$ of $m^{n+1}$ satisfying a relation of the form $t \equiv t' u^{n+1}$ (mod. $m^{n+2}$), where $t'$ is a uniquely determined element of $S$. Then $x = \sum_{i=0}^{\infty} s_i u^i$ and the family $(s_i)$ satisfying this relation and such that $s_i \in S$ for all $i$ is determined uniquely.