Let $\varphi: A \to B$ be a flat morphism of rings, $I$ an ideal in $A$ and $J= \varphi(I)B$. If $\hat{A}$ (resp. $\hat{B}$) denote the completion of $A$ (resp. $B$) with respect to $I$ (resp. $J$), is the induced morphism $\hat{A} \to \hat{B}$ flat?
I have a very simple motivation for this question: I actually want to know if $K[[t]]$ is flat over $R[[t]]$ when $R$ is a complete discrete valuation ring and $K$ is its fraction field. In this case, we would take $A=R[t]$, $B=K[t]$ and $I=(t)$.
If $B$ is a noetherian domain, by Krull intersection theorem it is Hausdorff for the adic topology definied by any ideal, and therefore if $B$ is flat over $A$, for any ideal $T$ of $A$ we have $T\otimes_AB \subset B$, and so $B$ is $I$-adically ideal-Hausdorff. Then by the local flatness theorem $\hat{A} \to \hat{B}$ flat. This answers only the particular case of $A=R[t]$, $B=K[t]$.