Suppose that $A$ is a Noetherian regular (added later) local domain. Moreover $\widehat A$ is $\mathfrak m$-adic completion $\widehat A$ w.r.t the maximal ideal and $K$ is the fraction field of $A$.
Now, could you explain why the following two claims are true?
1) If $A$ has Krull dimension $1$ then $\widehat A\otimes_A K$ is a field (in particular it is $\widehat K$).
2) If $A$ has Krull dimension $\ge 2$ then $\widehat A\otimes_A K$ is not a field.
Edit: The question arises from the following geometric viewpoint:
$S$ is an algebraic smooth (added later) surface on a perfect field, $A=\mathcal O_{S,x}$ and $K$ is the function field. In $1)$ is $x$ a point of codimension $1$, and in $2$) $x$ is a closed point. I was told that in the first case the result is a field, but this is not true in the second case.
Let $\widehat{A}=B$. If $A$ is smooth, $B$ is a regular domain and in particular $B\otimes_AK$ is a field if and only if for any non-zero prime ideal $P$ of $B$, $P\cap A\neq 0$. If $\dim A=1$, $B$ has only one non-zero prime ideal, which is maximal and the intersection with $A$ is the maximal ideal, hence non-zero.
For the second case, let me write down one example in characteristic zero, the general case is not very different. Let $A=k[x,y]_{(x,y)}$, so that $B=k[[x,y]]$. Consider the prime ideal generated by $y-xe^x\in B$. You can easily check that its intersection with $A$ is zero.