Let $k$ be a commutative ring with $1$. Let $L$ be a $k$-Lie algebra, which is not necessarily free as a $k$-module. Let $S\left(L\right)$ denote the symmetric algebra of $L$ (over $k$), constructed as a quotient of the tensor algebra $T\left(L\right)$ of $L$. Let $U\left(L\right)$ denote the universal enveloping algebra of $L$ (over $k$), constructed as the quotient of the tensor algebra $T\left(L\right)$ modulo the two-sided ideal generated by all elements of the form $x\otimes y-y\otimes x-\left[x,y\right]$ with $x\in L$ and $y\in L$.

The canonical filtration of the tensor algebra $T\left(L\right)$ descends to a filtration of the universal enveloping algebra $U\left(L\right)$. The associated graded algebra of $U\left(L\right)$ - let us call it $GU\left(L\right)$ - is commutative (as is easily seen) and generated by the elements $\overline{\sigma\left(v\right)}\in U_1\left(L\right) / U_0\left(L\right)$ for $v\in L$ (where $\sigma$ denotes the map $L\to T\left(L\right)\to U\left(L\right)$). Thus, there exists a surjective $k$-algebra homomorphism $S\left(L\right)\to GU\left(L\right)$ which maps $v$ to $\overline{\sigma\left(v\right)}$ for every $v\in L$ (according to the universal property of the symmetric algebra).

One version of the Poincaré-Birkhoff-Witt theorem (abbreviated PBW theorem) says that under certain conditions, this homomorphism is actually an isomorphism. The Wikipedia page says that it is so if any of the following four cases holds (here I am quoting Wikipedia):

(1) $L$ is a flat $k$-module,

(2) $L$ is torsion-free as an abelian group,

(3) $L$ is a direct sum of cyclic modules (or all its localizations at prime ideals of $k$ have this property), or

(4) $k$ is a Dedekind domain.

A reference is given to a paper which I have no access to:

P.J. Higgins, Baer Invariants and the Birkhoff-Witt theorem, J. of Alg. 11, 469-482, (1969)

Most internet sources which prove PBW only prove it under the condition that $L$ is a free $k$-module. (Out of these proofs I consider Garrett's version most readable.) I am interested in a proof in case (2). I know that it is enough to consider the case when $k$ is a $\mathbb Q$-algebra.

The following two sources might give such a proof, if only I could understand them:

Source 1:

T. Ton-That, T.-D. Tran, Poincaré's proof of the so-called Birkhoff-Witt theorem Rev. Histoire Math., 5 (1999), pp. 249-284. As this is only formulated for $k$ a field, this needs some modifications, but that's not my main worry. I fail to understand this paragraph on pages 277-278:

"The first four chains are of the form

$U_1 = XH_1,\ U'_1 = H'_1Z,\ U_2 = YH_2,\ U'_2 = H'_2T$,

where each chain $H_1$, $H'_1$, $H_2$, $H'_2$ is a closed chain of degree $p - 1$; therefore by induction, each is the head of an identically zero regular sum. It follows that $U_1$, $U'_1$, $U_2$, $U'_2$ are identically zero, and therefore each of them can be considered as the head of an identically zero regular sum of degree $p$."

I don't understand the "It follows that $U_1$, $U'_1$, $U_2$, $U'_2$ are identically zero" part. This seems to be equivalent to $H_1 = H'_1 = H_2 = H'_2 = 0$, which I don't believe (the head of an identically zero regular sum isn't necessarily zero), but the authors are only using the weaker assertion that each of $U_1, U'_1, U_2, U'_2$ is the head of an identically zero regular sum of degree $p$ - which, however, is still far from being obvious to me.

Source 2:

P.M. Cohn, A remark on the Birkhoff-Witt theorem, J. London Math. Soc. 38, 197-203, (1963). This gives a rather strange argument, which doesn't really rhyme for me. Probably I don't understand it though. If anybody could write it up in modern terms I would be very thankful. (If you want to know where exactly I am stuck, it's "$1w=w\in M_n$" on page 202, but I fear that there are also some things I have not really grasped before that point.)

Is there any accessible (I'm not at a university campus right now, and I need this rather soon) source for a proof of PBW in case (2)?