ac.commutative algebra - Is completeness of a field an algebraic property?

I claim that, for a field $K$, the following are equivalent:
(i) $K$ can be given a nontrivial norm -- i.e., there exists $x in K$ with $|x| neq 0,1$.
(ii) $K$ admits a nontrivial rank one valuation $v$.
(iii) $K$ admits infinitely many inequivalent rank one valuations $v$ such that $(K,v)$ is not complete.
(iv) $K$ is not an algebraic extension of a finite field.

Some of these facts are proved in

http://math.uga.edu/~pete/8410Chapter2v2.pdf

(see e.g. Theorem 1).

Let me prove here that (iv) $implies$ (iii), which answers the OP's question in a rather definitive way.

1) Suppose first that $K$ has characteristic $0$. Then $K$ contains $mathbb{Q}$, which admits the $p$-adic valuations $v_p$. By Theorem 1 of loc. cit., each $v_p$ extends to a valuation on $K$.

Now suppose that $K$ has characteristic $p$ and contains an element $t$ which is not algebraic over $mathbb{F}_p$. Thus $K$ contains the rational function field $mathbb{F}_p(t)$, which carries infinitely many inequivalent nontrivial valuations $v_P$ corresponding to the irreducible polynomials $P in mathbb{F}_p[t]$ (and one more corresponding to the point at infinity on the projective line).

2) (F.K. Schmidt) If a field $K$ is complete with respect to two inequivalent rank one valuations, it is algebraically closed and uncountable. See e.g. Theorem 24 of

http://math.uga.edu/~pete/8410Chapter3.pdf

3) So we are reduced to the case in which $K$ is algebraically closed and uncountable. Then $K$ is isomorphic to the algebraic closure of $K(t)$. If we give $K$ the trivial valuation and $K(t)$ the Gauss norm $v$, then the algebraic closure of $K(t)$ has infinite degree over $K(t)$ so any extension of $v$ to the algebraic closure is not complete. The image of the Gauss norm $v$ under the group $PGL_2(K)$ of linear fractional transformations gives us infinitely more pairwise inequivalent valuations.

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