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, 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 I am aware of 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 $k\supset\mathbb{Q}$)?

QFT and Strings, but flipping through I can't find it. The idea is to prove that the canonical symmetrization map (which requires dividing by all $n!$s) is a filtered isomorphism $S(L) \to U(L)$ covering the canonical map $S(L)\to gr(U(L))$. – Theo Johnson-Freyd Nov 27 '12 at 23:18