Pretty straitforward: If a field has a metric in which it is complete can it have a metric in which it is not complete? By metric I mean field norm of course
2 Answers
How about the algebraic closure of the $p$adics $\mathbb{Q}_p^{\mathrm{alg}}$. This is not complete under the $p$adic metric, but it is isomorphic as a field to the complex numbers $\mathbb{C}$ which is complete under the standard metric (as both fields are algebraically closed of characteristic zero with the same transcendence degree over $\mathbb{Q}$ (assuming AC of course)).
But if you want the second field to have nonarchimedean norm, take the $p$adic complex field $\mathbb{C}_p$ (the completion of $\mathbb{Q}_p^{\mathrm{alg}}$).

12$\begingroup$ How about Q_p? That's complete with its usual norm, but you can embed it into C (using the iso Q_pbar=C you use above) and, because the image contains Q and the square root of a negative integer, it will be dense in C but not all of C, so not complete with the usual metric induced from C. $\endgroup$ Commented Aug 16, 2010 at 12:28

$\begingroup$ Thanks alot! Of course I was asking about Qp to begin with... $\endgroup$– Adam GalCommented Aug 16, 2010 at 16:57
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://alpha.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.
 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).
 (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://alpha.math.uga.edu/~pete/8410Chapter3.pdf
 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.