I've got a Dedekind domain $R$ with quotient field $K$, a non-zero prime ideal $P$ of $R$. I form the completion $\widehat{K}$ of $K$ wrt the valuation $v_P$ associated to $P$. Let $\widehat{R}$ be the valuation ring of $\widehat{K}$.
If $V$ is a $K$-vector space then how would I show that there is an isomorphism $ \widehat{R}\otimes_R V\cong \widehat{K}\otimes_K V$ of $\widehat{R}$-modules?
I've tried the map $r\otimes_R v \mapsto r\otimes_K v$ but this doesn't seem to be surjective.
I would be grateful if someone could tell me the best way to approach this problem.
There is a more general fact at work here. Let $A\to B$ be a ring map, $M$ an $A$-module, and $N$ a $B$-module. Then there is a unique $B$-module isomorphism $M\otimes_AN\simeq(M\otimes_AB)\otimes_BN$ sending $m\otimes n$ to $(m\otimes 1)\otimes n$ (the $B$-module structure on the source is via the $B$-module structure of $N$, whereas in the target, there is a $B$-module structure on both tensor factors).
In your case, take $A=R$, $B=K$, $M=\widehat{R}$, and $N=V$. Then by what I've said, there is a $K$-linear isomorphism $\widehat{R}\otimes_RV\simeq (\widehat{R}\otimes_RK)\otimes_KV$. It sends $r\otimes v$ to $(r\otimes 1)\otimes v$, and you can check that in this case it is a $\widehat{R}$-module isomorphism as well as a $K$-linear isomorphism. Now you just need to verify that the natural $\widehat{R}$-algebra map $\widehat{R}\otimes_RK\to\widehat{K}$ is an isomorphism. It is injective because every tensor in the source has the form $r\otimes x$ for some $r\in\widehat{R}$ and $x\in K$, and if this is mapped to zero, i.e. if $rx=0$ in $\widehat{K}$, then either $r=0$ or $x=0$, so $r\otimes x=0$ in either case. For surjectivity, reason as follows. Choose an element $\pi\in R$ with $v_P(\pi)=1$ (a uniformizer in the local ring). Then every non-zero element of $\widehat{K}$ has the form $\pi^n u$ with $n\in\mathbf{Z}$ and $u\in\widehat{R}^\times$. This element is the image of $u\otimes \pi^n\in\widehat{R}\otimes_RK$.
Note that the composite isomorphism $\widehat{R}\otimes_RV\simeq\widehat{K}\otimes_KV$ sends $r\otimes v$ to $r\otimes v$, as you expected.