valuation ring, completeness

Question

Perhaps a trivial question: is there an example of a field $K$ and a valuation $v$ on $K$ such that the following holds:

• $K$ is not complete (with respect to the valuation topology)
• The valuation ring $A$ (corresponding to $v$) is complete.

Thanks in advance!

PS The definitions (Bourbaki, general topology) are as follows:

valuation:
map v: $K\to \Gamma \cup \{\infty\}$ where $\Gamma$ is a totally ordered abelian group such that ... (the usual properties)

valuation ring $A$ corresponding to $v$:
$A = \{x\in K: v(x)\geq 0\}\subseteq K$

Answer 1

After asking the question I realised the following:

• The fraction field of a valuation ring $A$ (corresponding to a valuation $v\colon K\to \Gamma\cup \infty$) is equal to $K$.
• The valuation $v$ extends to a valuation $\hat v\colon \hat K\to \Gamma\cup \infty$, where $\hat K$ is the completion of $K$ (which is a field).
• The valuation ring corresponding to $\hat v$ is the completion $\hat A$ of $A$.

So in particular, if $A$ is complete, we have

$\hat K = \text{fraction field of } \hat A = \text{fraction field of } A = K$

So I guess the answer is no.