This follows from § 6, no. 1, Proposition 1, § 5, no. 1, Proposition 2 and Integration, Chapter VII, § 1, no. 10, Propositions 12 and 13.

The centre $Z$ of a locally compact field $K$ is closed in $K$ and therefore locally Compact.

In general it is known that in a (left or right) vector space of finite dimension $n$ over $K$, the homothety with ratio $x \in K$ has modulus equal to $(\mathrm{mod}_K(x))^n$; it suffices to apply this to $K'$.

We shall use the following lemma:

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)

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.

(i) If K is of characteristic 0 and mod, is not an ultrametric absolute value then K is isomorphic to one of the fields $\mathbf{R}, \mathbf{C}$ or $\mathbf{H}$.

(ii) If K is of characteristic 0 and mod, is an ultrametric absolute value, K is an algebra of finite rank over a p-adic field $\mathbf{Q}_p$.

(iii) If K is of characteristic $p \neq 0$, it is isomorphic to a field with centre a field of formal power series $\mathbf{F}_q((T))$ (where $q$ is a power of $p$) and of finite rank over its centre.

(i) It follows from Ostrowski's Theorem (§ 6, no. 4, Theorem 2) that K is a topological field isomorphic to an everywhere dense subfield of $\mathbf{R}, \mathbf{C}$ or $\mathbf{H}$ and, as K is complete, it is isomorphic to $\mathbf{R}, \mathbf{C}$ or $\mathbf{H}$.

(ii) Let A be the ring of the absolute value mod, and m its maximal ideal. We know that A/m is a finite field (§ 5, no. 1, Proposition 2) and hence the absolute value induced by mod, on $\mathbf{Q}$ has a finite residue field, which is only possible if it is equivalent to a $p$-adic absolute value ($§ 6$, no. 3, Proposition 4); the closure of $\mathbf{Q}$ in $K$ is therefore isomorphic to $\mathbf{Q}_p$ and is contained in the centre of $K$ since the latter is closed in $K$; we conclude using Proposition 2 of no. 1.

(iii) The second assertion follows from the first and the Corollary to Proposition 2 of no. 2. To show the first assertion, note that mod, is necessarily an ultrametric absolute value ($§ 6$, no. 2, Corollary to Proposition 3); in the notation of the proof of Proposition 3 of no. 2, the centre $Z$ of $K$ consists of the elements which commute with both $s$ and $u$; but by virtue of (3),

so that $u^q \in Z$ and we conclude that $Z$ is not discrete. As $Z$ is locally compact, we may confine our attention to the case where $K$ is commutative. The sub-$\mathbf{F}_p$-algebra $\mathbf{F}_p[s]$ in $K$ is then a finite field since $s^{q-1} = 1$ and obviously $y^q = y$ for every element of this field, which is therefore identical with $S$ and isomorphic to $\mathbf{F}_q$ since $S \subset \mathbf{F}_p[s]$ has $q$ elements. Since the sum of two elements of $S$ is in $S$, the mapping which maps each formal power series $\sum_{i=0}^\infty s_i T^i \in \mathbf{F}_q[[T]]$ to the element $\sum_{i=0}^\infty s_i u^i$ is a bijective homomorphism of the ring $\mathbf{F}_q[[T]]$ onto the ring $A$, whence immediately the conclusion.

If the topology on a field $K$ is defined by an ultrametric absolute value, $K$ is totally disconnected with this topology.