3 added 248 characters in body

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). Here is a sketch of

Let me prove here that (iv) $\implies$ (iii):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 $K$. (If

Now suppose that $K$ has characteristic $p$ and contains an element $t$ which is not algebraic over $\mathbb{F}_p$, then \mathbb{F}_p$. Thus$K$contains the rational function field$\mathbb{F}_p(t)$\mathbb{F}_p(t)$, which also carries infinitely many inequivalent nontrivial valuations $v_P$ corresponding to the irreducible polynomials $P \in \mathbb{F}_p[t]$.)mathbb{F}_p[t]$(and one more corresponding to the point at infinity on the projective line). 2) (F.K. Schmidt) If any a field$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)$of linear fractional transformations gives us infinitely more pairwise inequivalent valuations. 2 added 1 characters in body I claim that, for a field$K$, the following are equivalent: (i)$K$can be given a nontrvial nontrivial norm -- i.e., there exists$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$is not an algebraic extension of a finite field. Some of these facts are proved in http://math.uga.edu/~pete/8410Chapter2v2.pdf (see e.g. Theorem 1). Here is a sketch of (iv)$\implies$(iii): 1) Suppose$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$K$. (If$K$has characteristic$p$and is not algebraic over$\mathbb{F}_p$, then$K$contains$\mathbb{F}_p(t)$which also carries infinitely many valuations$v_P$corresponding to the irreducible polynomials$P \in \mathbb{F}_p[t]$.) 2) (F.K. Schmidt) If any 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. 1 I claim that, for a field$K$, the following are equivalent: (i)$K$can be given a nontrvial 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). Here is a sketch of (iv)$\implies$(iii): 1) Suppose$K$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$K$. (If$K$has characteristic$p$and is not algebraic over$\mathbb{F}_p$, then$K$contains$\mathbb{F}_p(t)$which also carries infinitely many valuations$v_P$corresponding to the irreducible polynomials$P \in \mathbb{F}_p[t]$.) 2) (F.K. Schmidt) If any 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.