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If $F$ is a free group then it has cohomological dimension one, which implies that the augmentation ideal $IF=\operatorname{ker}(\epsilon:\mathbb{Z}G\to \mathbb{Z})$ of its group ring is a projective $\mathbb{Z}F$-module. Hence $IF$ is a direct summand of a free $\mathbb{Z}F$-module $M$.

Question: Is it possible to give an explicit construction of such a free module $M$? Perhaps it has a basis related to a free basis of $F$?

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    $\begingroup$ The answer is yes. Use the chain complex for the Cayley graph of F to see it $\endgroup$
    – Benjamin Steinberg
    Commented Mar 10, 2022 at 11:33

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Let $X$ be a free basis for $F$. The Cayley graph of $F$ is a tree $T$ on which $F$ acts freely. The augmented chain complex gives a resolution $$0\to \mathbb ZF^{(X)}\to \mathbb ZF\to \mathbb Z\to 0$$ (since $T$ is a tree) where we identify $\mathbb ZF^{(X)}$ as the free abelian group on the edges of $T$ and the image of the boundary map sends the edge $(g,x)$ to $gx-x$ and is hence the augmentation ideal (which in any event is clearly the kernel of the augmentation map). So the augmentation ideal is a free module on $X$ with generators $x-1$ with $x\in X$.

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  • $\begingroup$ Wonderful, thank you! $\endgroup$
    – Mark Grant
    Commented Mar 10, 2022 at 12:04
  • $\begingroup$ I think more generally a free group is a FIR so all left ideals are free. But the free generating set is more complicated to describe $\endgroup$
    – Benjamin Steinberg
    Commented Mar 10, 2022 at 13:00

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