$\newcommand{\QQ}{\Bbb Q}$ Let $G$ be a free group on the symbols $x_1, \dots, x_n$, with $\QQ[G]$ its rational group algebra.

Write $\varepsilon: \QQ[G] \to \QQ$ for the augmentation, and for $\alpha \in \QQ[G]$ define $d(\alpha) = \alpha - \varepsilon(\alpha)$. Then there is an exact sequence of $\QQ[G]$-modules $$ 0 \to \bigoplus_{i=1}^n \QQ[G] \cdot dx_i \to \QQ[G] \xrightarrow{\varepsilon} \QQ \to 0. $$ One can see that this is exact in the middle term by using the relation $d(\alpha \beta) = d(\alpha) \varepsilon(\beta) + \alpha d(\beta)$ to inductively write any element in $\ker(\varepsilon) = \mathrm{Im}(d)$ as a linear combination of the $dx_i$. Exactness at the left-hand term is special to free groups, and proving it makes use of the monomial basis.

Let $S$ be the set of elements in the group algebra such that $\varepsilon(s) \neq 0$. Then we can construct the localization $S^{-1} \QQ[G]$ by inverting the elements of $S$ (a Cohn localization defined by a universal property, since this ring is very far from having a calculus of fractions). The map $\varepsilon$ sends the elements in $S$ to units. The definition of $d$ therefore extends to $S$, and one can show that the sequence $$ \bigoplus_{i=1}^n S^{-1} \QQ[G] \cdot dx_i \to S^{-1} \QQ[G] \xrightarrow{\varepsilon} \QQ \to 0 $$ is still exact using the same techniques (this did not depend on $S$). When $n=1$, the rings involved are commutative and so the sequence is still exact. However, localizations of noncommutative rings are not flat, and so the left-hand map is not guaranteed to be injective (I suspect that it isn't).

Question: Is the left-hand map above injective? If not, does the kernel have a concrete description?

One approach might be to make use of the Magnus transformation: an injective ring homomorphism from $\QQ[G]$ to the noncommutative power series ring $\QQ\langle\!\langle t_1, \dots, t_n \rangle\!\rangle$ sending $x_i$ to $1 + t_i$. The Magnus transformation sends the elements of $S$ to units, and so factors through a map $$ \phi: S^{-1} \QQ[G] \to \QQ\langle\!\langle t_1, \dots, t_n \rangle\!\rangle. $$ The corresponding sequence $$ 0 \to \bigoplus_{i=1}^n \QQ\langle\!\langle t_1, \dots, t_n \rangle\!\rangle \cdot t_i \to \QQ\langle\!\langle t_1, \dots, t_n \rangle\!\rangle \to \QQ \to 0 $$ for the power series ring is exact, and so we would be done if we knew that the extension $\phi$ was injective.

Question: Is the extension $\phi$ of the Magnus transformation still injective?

Unfortunately, this is not guaranteed in the noncommutative case either, because the ideals in the localization are not controlled by the ideals in the original ring. For example, the two-sided ideal $J$ generated by $y x^{-1} y$ in the ring $x^{-1} \QQ\langle x,y\rangle$ intersects $\QQ\langle x,y\rangle$ trivially, and so we get an injection $$\QQ\langle x,y\rangle \to x^{-1} \QQ\langle x,y\rangle / J$$ whose extension to $x^{-1} \QQ\langle x,y\rangle$ is not injective.

For context, this question came up in studying the properties of Bousfield localization of spaces with respect to rational homology. If a space $X$ is $H_*(-;\QQ)$-local, then one can show that the action of the fundamental group $G$ on the higher homotopy groups $\pi_n(X)$ factors through the localization $S^{-1} \QQ[G]$. One can ask whether this property is equivalent to Bousfield's definition of the higher homotopy groups being $\QQ$-local with respect to this action, and so I'm looking at homological properties of this localization.


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