I asked this question on MSE but no answer so far, so I'm also asking it here.

Let $L$ be a Lie ring (a Lie algebra over $\Z$) with generators $x_1,\dots,x_n$ and relations of the form $$[x_i,\sum x_j]=0$$ where the sum is over some subset of $\{1,\dots,n\}$ (in my particular case I have exactly one of those relation for each $x_i$, if that helps). In particular, in those relations every generator which appears does so with coefficient 1. Note also that those are quadratic, in particular homogeneous, so $L$ is graded and the $x_i$'s form a free basis of its degree 1 part.

Is there an efficient way to check whether $L$ is free as a $\Z$-module ?

On the one hand it seems to me that this should be a difficult question in general. On the other hand, I have a quotient of a free Lie algebra by an homogeneous ideal generated by monic Lie polynomials, and the little I understand of this paper introducing Gröbner-Shirshov basis seems to imply that this always produces a free $\Z$-module, but that sounds to good to be true.

Said differently, the left hand side of the relations above generate a pure submodule/a direct summand of the degree two part of the free lie algebra on the $x_i$'s, and I'm wondering whether the ideal they generate is again a direct summand.

Edit To be slightly more precise about the kind of result I'd like: so far I understand if an ideal $I$ in a free (Lie) ring $F$ has a Gröbner basis of monic polynomials then this essentially provides a canonical basis of the quotient $F/I$ and an algorithm to compute the multiplication in this basis. In particular it does imply that the quotient is free as a module, but it's much much stronger (indeed this is trivial when working over a field, so Grôbner basis really are about the "algorithm" part).

Now while constructing Gröbner basis is hard, checking whether a given set of polynomials is a Gröbner basis can be done in a finite amount of time. But again this is asking much more than what I need.

So I'm looking for something similar in spirit, i.e. a finite amount of things to check on some generating set of my ideal, some property much weaker than being a Gröbner basis (hence more likely to be satisfied in my case) which would guarantee that the quotient is a free $\Z$-module.

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    $\begingroup$ Note that since $L$ is canonically graded $L=\bigoplus_{i\ge 1} L_i$ and all $L_i$ are finitely generated abelian groups, $L$ is free $\Leftrightarrow$ $L$ is torsion-free as $\mathbf{Z}$-module $\Leftrightarrow$ each $L_i$ is torsion-free as $\mathbf{Z}$-module. (You certainly know this but this was to clarify). $\endgroup$ – YCor Mar 7 '17 at 14:24
  • $\begingroup$ Indeed, thanks for pointing this out. This was kinda implicit in my question, what I really do care about is torsion-freeness but as you say in that situation this is equivalent to just being free. $\endgroup$ – Adrien Mar 7 '17 at 15:05

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