Theorem 1

Let K be a (not necessarily commutative) locally compact field which is not discrete.

(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.