Completion $K \otimes_R \text{lim} R /I^n$ is a field Let $K$ be a field with valuation ring $R \subset K$. Let $\mathfrak{m}$ be the maximal ideal of this valuation ring.
Form the inverse limit $\hat{R} = \text{lim}_{I \text{ a nonzero ideal of } R} R/I$, and then put $\hat{K} = K \otimes_R \hat{R}$. I am looking for minimal conditions under which $\hat{K}$ is a field. In fact, I suspect it is under the provided conditions, though I can't seem to prove it.
Example: Let $K = \mathbb{Q}$. Take a prime $p \in \mathbb{Z}$, and put $R = \mathbb{Z}[p^{-1}]$. Then $\hat{R} = \mathbb{Z}_p$ ($p$-adic integers) and $\hat{Q} = \mathbb{Q}_p$ ($p$-adic numbers).
 A: Let $v:K^\times\to G$ be the valuation associated with $R$.  Suppose $x\in\hat{R}$ is nonzero, and write $x_I$ for its image in $R/I$ for each nonzero ideal $I$, and let $y_I$ be a lift of $x_I$ to $R$.  Since $x\neq 0$, there is some $I$ such that $x_I\neq 0$.  Note then that $y_I$ divides $y_J$ for all $J\subseteq I$, since $y_J-y_I\in I$ but $y_I\not\in I$ and thus $v(y_J-y_I)>v(y_I)$ so $v(y_J)=v(y_I)$.  Now let $z_J=y_J/y_I$.  Note that for $J'\subseteq J\subseteq I$, $v(z_{J'}-z_J)=v(y_J-y_{J'})-v(y_I)$.  As $J$ gets small, $v(y_J-y_{J'})$ gets arbitrarily large in $G$, and thus $v(z_{J'}-z_J)$ gets arbitrarily large as well.  In other words, the sequence $(z_J)_{J\subseteq I}$ is Cauchy with respect to $v$ and thus defines an element $z$ of $\hat{R}$ (explicitly, the image of $z$ in $R/J$ is the image of $z_{J'}$ in $R/J$ for all sufficiently small $J'$).  We moreover see that $y_Iz=x$, since $y_Iz_J=y_J$ approaches $x$ as $J$ gets small.
Thus, we have shown that $x$ is divisible by some nonzero element of $R$, namely $y_I$.  When we tensor with $K$ over $R$, we invert $y_I$, and so $x$ gets inverted.  Since $x$ was an arbitrary nonzero element of $\hat{R}$, this shows that $K\otimes_R \hat{R}$ is a field.
