Let $k$ be a commutative ring with unit and $L$ be a Lie $k$-algebra. 

Let $U(L)$ be the universal enveloping $k$-algebra of $L$ (one can define it as a quotient of the tensor algebra, as it is explained [in this MO question][1], or one can say that $U(-)$ is left adjoint to the forgetful functor sending an associative $k$-algebra to the Lie $k$-algebra obtained by taking the same underlying $k$-module and with Lie bracket being the commutator). 

The associative $k$-algebra $U(L)$ is filtered as a $k$-algebra, and there is a canonical epimorphism $S(L)\to gr\big(U(L)\big)$. 

If this epimorphism is an isomorphism, then we say that *$L$ has the PBW property*. 

All the examples of Lie $k$-algebras not satisfying the PBW property are constructed in the following way: one first finds an example of a Lie algebra for which the map $L\to U(L)$ (the unit of the adjunction) is not injective, and then it is quite clear that the PBW property can't hold. 

My question is then: 

> Is there any example of a Lie
> $k$-algebra $L$ such that the map
> $L\to U(L)$ is injective which does
> not satisfy the PBW property ?

Or is it that the PBW property is just equivalent to $L\to U(L)$ being injective (it would be great, but I have no idea why this would be true - EDIT: one might want to use that $L\to U(L)$ is injective to reduce to the cas when $k\supset\mathbb{Q}$)?

  [1]: http://mathoverflow.net/questions/61954/pbw-theorem-over-a-q-algebra-without-freeness-or-flatness