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Let $R$ be a graded commutative unital ring and $M$ a graded $R$-module (all gradings are over $\mathbb{Z}$). I'm looking for a reference for the following statement:

If $M$ is $R$-free, then the free graded Lie algebra over $R$ generated by $M$ is also $R$-free.

For the purposes of this question, a graded Lie algebra over $R$ is a graded $R$-module $L$ with an $R$-linear map $[-, - ] : L\otimes_RL\to L$ satisfying the following identities for all homogeneous elements $a,b,c\in L$:

(1) $[a,b]+(-1)^{|a||b|}[b,a]=0$

(2) $(-1)^{|a||c|} [a,[b,c]] + (-1)^{|b||a|}[b,[c,a]] + (-1)^{|c| |b|} [c,[a,b]] = 0 $

(3) $[a,a]=0$ if $a$ is of even degree

(4) $[a,[a,a]]=0$.

I should also say that I'm aware of Reutenauer's book, and, as far as I can tell, it does not deal with the graded case.

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  • $\begingroup$ Is that really the statement? Does the supposed $R$-basis of $M$ not need to be connected to the grading in any way for this to be true? $\endgroup$
    – Johannes Hahn
    Commented Apr 26, 2018 at 17:42
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    $\begingroup$ I don't think I understand what you mean by "connected to the grading." Can you elaborate? To be sure that we're talking about the same thing, I'm calling a graded $R$-module free if it is a direct sum of shifts of $R$. $\endgroup$
    – Ben Knudsen
    Commented Apr 26, 2018 at 17:47
  • $\begingroup$ I meant that there exists a basis consisting of homogeneous elements or some similar condition. Your definition for freeness is perfectly reasonable in a graded setting and I really should have thought of it myself! $\endgroup$
    – Johannes Hahn
    Commented Apr 26, 2018 at 17:52
  • $\begingroup$ Graded in what? an arbitrary abelian group? in $\mathbf{Z}$? in $\mathbf{N}_{>0}$? in $\mathbf{N}_{\ge 0}$? $\endgroup$
    – YCor
    Commented Apr 26, 2018 at 21:36
  • $\begingroup$ Question edited to specify grading. $\endgroup$
    – Ben Knudsen
    Commented Apr 26, 2018 at 22:39

1 Answer 1

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I will answer under the assumption that $2$ is invertible in $R$, the general case of which follows after extension of scalars from the case $R=\mathbb{Z}[1/2]$, which is a particular case of Proposition 8.5.1 in Neisendorfer’s “Algebraic Methods in Unstable Homotopy Theory.” The assumption that $2$ is invertible is necessary to ensure that Definition 8.1.1 coincides with the definition stated above (see the following remark on p. 262). I do not know if this assumption is necessary.

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