Let $RG$ be a group ring, where $R$ is a commutative ring. Let $I=\ker\epsilon$ be the augmentation ideal.
It is known that any element $\alpha\in I$ can be written in the form $$\alpha=\sum_{g\in G}a_g(g-1),$$ where $a_g\in R$.
Suppose further that $G$ is a free group with generators $g_1,g_2,\dots,g_n$. With this extra information, how can we improve on the previous result?
For example is it true that $\alpha$ can be written in the form $$\alpha=\sum_{i=1}^n\left( a_g(g_i-1)+b_g(g_i^{-1}-1)\right)?$$
Thanks.
Viewed as an $R$-module, the ideal $I$ is free, and a basis is given by $\mathcal{B} = \{ g-1 \ | \ g\in G\}$ (proof below). Therefore, if $G$ is any group (including free groups), it is not possible to "improve" on this basis in any sense that I can think of.
Proof that $\mathcal{B} = \{ g-1 \ | \ g\in G\}$ is a basis of $I$ as an $R$-module.
To see that $\mathcal{B}$ generates $I$, let $x=\sum_g b_g g \in I$. Since $x\in I$, we have that $b_1 = -\sum_{g\neq 1} b_g$. Therefore $$ x = \sum_{g\neq 1} b_g(g-1).$$ Thus $\mathcal{B}$ generates $I$.
To see that the elements of $\mathcal{B}$ are linearly independent, assume that $$ \sum_{g\neq 1} a_g(g-1) =0. $$ Written in the basis $\{g \ | \ g\in G\}$ of $RG$, this element is $$(-\sum_{g\neq 1}a_g) 1 + \sum_{g\neq 1}a_g g.$$ Thus, $a_g=0$ for all $g\in G \setminus \{1\}$.
Therefore, $\mathcal{B}$ is a basis of the $R$-module $I$.