Let $R$ be a discrete valuation ring containing an algebraically closed field $K$ of characteristic zero and let $L$ be a Lie algebra over $R$ whose underlying $R$-module is finitely generated and free.


Let $\mathfrak{m}$ denote the maximal ideal of $R$ and $\hat{R}$ the $\mathfrak{m}$-adic completion of $R$. While thinking about something else I came up with the following question, which I cannot answer:



> 
> Is it true that if $L \otimes\_R \hat{R}$ is isomorphic (as Lie algebras) to $\mathfrak{g} \otimes\_K \hat{R}$ for some simple Lie algebra $\mathfrak{g}$ over $K$, then $L$ is already isomorphic to $\mathfrak{g} \otimes\_K R$ ?
> 
> 
> 


I have the feeling that this is either well-known or false (or both), but since I am no expert in Lie-theory, I thought I might just ask it here. Thanks in advance.



Lie Algebras over DVRs and basechange to the completion
One can also give explicit high-rank counterexamples using special orthogonal groups of quadratic lattices over $R$ with non-degenerate reduction, and we can also arrange that $R$ has $K$ as its residue field (likewise for Jason Starr's examples). 


To give wider context, let $K$ be any field of characteristic 0, $R$ any $K$-algebra, $\mathfrak{g}$ any semisimple Lie algebra over $K$ (we'll use $\mathfrak{so}\_n$ in the end for any odd $n \ge 3$), and consider in place of $\widehat{R}$ any faithfully flat $R$-algebra $R'$. 


Now consider a projective $R$-module $\Lambda$ of constant odd rank $n \ge 3$ and a quadratic form $q:\Lambda \rightarrow R$ over $R$ that is non-degenerate at every point of ${\rm{Spec}}(R)$. Let $H = {\rm{SO}}(q)$ be the associated special orthogonal group scheme; this is a smooth affine $R$-group whose geometric fibers over Spec($R$) are the classical special orthogonal groups of the corresponding fibers of $q$. In particular, since $n$ is odd it follows that such fibers are of adjoint type (and $\mu\_2 \times H = {\rm{O}}(q)$ is the orthogonal group scheme of $q$). 


Let $L = \mathfrak{so}(q)$ be the Lie algebra of $H$, and let $\mathfrak{g}$ be the Lie algebra for the special orthogonal group ${\rm{SO}}\_n := {\rm{SO}}(q\_n)$ of the corresponding "split" rank-$n$ quadratic form
$$q\_n := x\_0^2 + x\_1 x\_2 + \dots + x\_{2m-1}x\_{2m}$$
where $2m+1=n$. Let $R'$ be a faithfully flat $R$-algebra such that $q$ becomes homothetic to $q\_n$ over $R'$; such an $R'$ always exists (can even be chosen to be etale over $R$ since $R$ is a $\mathbf{Z}[1/2]$-algebra). Thus, $L$ and $\mathfrak{so}\_n$ become isomorphic over $R'$. But I claim that if $L$ and $\mathfrak{so}\_n$ are $R$-isomorphic then necessarily $q$ and $q\_n$ are homothetic over $R$ (i.e., $q$ is isometric to $u q\_n$ for some $u \in R^{\times}$). 


It is a general fact that for odd $n \ge 3$, ${\rm{SO}}\_n$ represents the automorphism functor
of its Lie algebra on the category of all $K$-algebras, as may be checked
on $\overline{K}$-points since ${\rm{char}}(K)=0$. (This is a special case
of a general fact about connected semisimple groups of adjoint type whose Dynkin diagram has no nontrivial automorphisms.) Thus, $L$ as an fpqc $R$-form of $\mathfrak{so}\_n$ corresponds to an element in the cohomology set
${\rm{H}}^1(R'/R, {\rm{SO}}\_n)$. 


But since $n$ is odd, ${\rm{SO}}\_n$ represents the functor of conformal isometries of $q\_n$ (i.e., pairs $(T,u)$ consisting of a unit $u$ and 
$T \in {\rm{GL}}\_n$ such that $q\_n \circ T = u q\_n$). Unraveling everything, 
this says that $L := {\mathfrak{so}}(q)$ is $R$-isomorphic to $\mathfrak{so}\_n$ if and only if $q$ is homothetic to $q\_n$.


Now we can give explicit counterexamples over discrete valuation rings $R$ containing an algebraically closed field $K$ of characteristic 0 (and even with $K$ as its residue field). Letting $\mathfrak{m}$ be the maximal ideal of such an $R$, 
consider a quadratic lattice $(R^n, q)$ of odd rank $n \ge 3$ with $\overline{q} := q \bmod \mathfrak{m}$ non-degenerate over $R/\mathfrak{m}$ such that $\overline{q}$ is isometric to $q\_n$. There is no infinitesimal obstruction to lifting an isomorphism between non-degenerate quadratic spaces $(M,Q)$ and $(M',Q')$ over a $\mathbf{Z}[1/2]$-algebra $A$ (indeed, the Isom-scheme ${\rm{Isom}}(Q',Q)$ is an ${\rm{O}}(Q)$-torsor and hence is smooth since $A$ is a $\mathbf{Z}[1/2]$-algebra), so $q$ becomes isometric to $q\_n$ over the *completion* $R' := \widehat{R}$. Thus, $L := \mathfrak{so}(q)$ is a counterexample
(using $\mathfrak{g} = \mathfrak{so}\_n$) provided that $q$ is not homothetic to 
$q\_n$ over $R$.


Now writing $n=2m+1$, observe that the ring
$$S := K[t\_0, \dots, t\_{n-1},v\_1, \dots, v\_{n-1}]/(v\_i^2 + t\_0 v\_i - t\_i)$$ 
satisfies $S/t\_0 S = K[v\_1, \dots, v\_{n-1}]$ (in which 
$t\_i = v\_i^2$), so $t\_0 S$ is prime in $S$ and 
$S$ is normal (e.g., by Serre's criterion). Thus, $R = S\_{(t\_0)}$ is a dvr
and 
$$q = x\_0^2 + t\_1 x\_1^2 + \dots + t\_{n-1} x\_{n-1}^2$$
over $R$ has reduction isometric to $q\_n$. To show that $q$ is not homothetic to $q\_n$ over $R$ it suffices to prove that $q$ is even anisotropic over $F = {\rm{Frac}}(R) = {\rm{Frac}}(S) = K(t\_0, v\_1, \dots, v\_{n-1})$. This anisotropicity is "obvious", and is rigorously proved by induction on $n$ (allowed to have any parity) or perhaps in other ways too.


That $R$ does not have residue field $K$. But consider the field extension
$K(\!(t\_0)\!)$ of $K(t\_0)$. This has countably infinite transcendence degree, so we can find elements $t\_1, \dots, t\_{n-1} \in K(\!(t\_0)\!)$ algebraically independent over $K(t\_0)$ that are all 1-units. Thus, $X^2 + t\_0 X - t\_i$ has a root
$v\_i \in K[\![t\_0]\!]$. Thus, the ring $S$ above can be found inside $K[\![t\_0]\!]$, so the restriction to ${\rm{Frac}}(S)$ of the $t\_0$-adic valuation on $K(\!(t\_0)\!)$ is a discrete valuation whose valuation ring $R$ contains $S$ (it is not $S\_{(t\_0)}$!) and has the same fraction field as $S$, and has residue field $K$. Clearly $q$ makes sense as a quadratic form over $R$, and as such it does the job (by the same anisotropicity argument over ${\rm{Frac}}(R) = {\rm{Frac}}(S)$!). 


Remark. Theorem 3.5 of <http://arxiv.org/pdf/math/0406508v5.pdf> provides a much more far-reaching result: over *any* ring whatsoever (no field of char. 0 in the picture), the groupoid of semisimple group schemes of adjoint type is equivalent (via the Lie functor) to the groupoid of vector-bundle Lie algebras whose Killing form is fiberwise non-degenerate!


