$\newcommand{\im}{\operatorname{im}}$Given a set $X$ and non-zero unital commutative ring $R$, let: \begin{align} A &= \mbox{free unital, associative algebra on $X$ with coefficients in $R$},\\ L &= \mbox{free Lie algebra on $X$ with coefficients in $R$}. \end{align} Regarding $A$ as a Lie algebra via the ring commutator, the universality of $L$ applied to $X \hookrightarrow A$, yields a natural morphism of Lie algebras: \begin{align} \rho : L \to A. \end{align} I don't know whether $\rho$ is always injective but I suspect not. My question is thus:

Can someone provide an example of a ring $R$ (and set $X$) for which $\rho$ fails to be injective?

Identify $X$ with its image in $A$, let $\im \rho$ be the image of $\rho$ and note that:

1. $X \subseteq \im \rho$.
2. It follows from the uniqueness of the universal property of $L$, that the image of $\rho$ is the smallest Lie subalgebra of $A$ containing by $X$.
3. $\rho$ is injective when $R$ is a field of characteristic zero. To see this is to note that $X \hookrightarrow \im \rho$, satisfies the universal property of the free Lie algebra. Indeed if we have $f : X \to K$ for some Lie algebra $K$ and $\mu : K \to U_K$ is the map from $K$ to its universal enveloping algebra, then the universal property of $A$ applied to $\mu \circ f$ gives a map $\hat f : A \to U_K$ and using 2 above, it is easy to see that $\hat f(\im \rho) \subseteq \im \mu$. Since $R$ is a field of characteristic zero, $\mu$ is injective, so we can define $\mu^{-1} \circ \hat f|_{\im\rho} : \im\rho \to K$ and it is easily checked that this construction satisfies the required universality.

$\newcommand{\im}{\operatorname{im}}$Given a set $X$ and non-zero unital commutative ring $R$, let: \begin{align} A &= \mbox{free unital, associative algebra on $X$ with coefficients in $R$},\\ L &= \mbox{free Lie algebra on $X$ with coefficients in $R$}. \end{align} Regarding $A$ as a Lie algebra via the ring commutator, the universality of $L$ applied to $X \hookrightarrow A$, yields a natural morphism of Lie algebras: \begin{align} \rho : L \to A. \end{align} I don't know whether $\rho$ is always injective but I suspect not. My question is thus:

Can someone provide an example of a ring $R$ (and set $X$) for which $\rho$ fails to be injective?

Identify $X$ with its image in $A$, let $\im \rho$ be the image of $\rho$ and note that:

1. $X \subseteq \im \rho$.
2. It follows from the uniqueness of the universal property of $L$, that the image of $\rho$ is the smallest Lie subalgebra of $A$ containing $X$.
3. $\rho$ is injective when $R$ is a field of characteristic zero. To see this is to note that $X \hookrightarrow \im \rho$, satisfies the universal property of the free Lie algebra. Indeed if we have $f : X \to K$ for some Lie algebra $K$ and $\mu : K \to U_K$ is the map from $K$ to its universal enveloping algebra, then the universal property of $A$ applied to $\mu \circ f$ gives a map $\hat f : A \to U_K$ and using 2 above, it is easy to see that $\hat f(\im \rho) \subseteq \im \mu$. Since $R$ is a field of characteristic zero, $\mu$ is injective, so we can define $\mu^{-1} \circ \hat f|_{\im\rho} : \im\rho \to K$ and it is easily checked that this construction satisfies the required universality.

Given a set $X$ and non-zero unital commutative ring $R$, let: \begin{align} A &= \mbox{free unital, associative algebra on $X$ with coefficients in $R$},\\ L &= \mbox{free Lie algebra on $X$ with coefficients in $R$}. \end{align} Regarding $A$ as a Lie algebra via the ring commutator, the universality of $L$ applied to $X \hookrightarrow A$, yields a natural morphism of Lie algebras: \begin{align} \rho : L \to A. \end{align} I don't know whether $\rho$ is always injective but I suspect not. My question is thus:

Can someone provide an example of a ring $R$ (and set $X$) for which $\rho$ fails to be injective?

Identify $X$ with its image in $A$, let $im \rho$ be the image of $\rho$ and note that:

1. $X \subseteq im \rho$.
2. It follows from the uniqueness of the universal property of $L$, that the image of $\rho$ is the smallest Lie subalgebra of $A$ containing by $X$.
3. $\rho$ is injective when $R$ is a field of characteristic zero. To see this is to note that $X \hookrightarrow im \rho$, satisfies the universal property of the free Lie algebra. Indeed if we have $f : X \to K$ for some Lie algebra $K$ and $\mu : K \to U_K$ is the map from $K$ to its universal enveloping algebra, then the universal property of $A$ applied to $\mu \circ f$ gives a map $\hat f : A \to U_K$ and using 2 above, it is easy to see that $\hat f(im \rho) \subseteq im \mu$. Since $R$ is a field of characteristic zero, $\mu$ is injective, so we can define $\mu^{-1} \circ \hat f|_{im\rho} : im\rho \to K$ and it is easily checked that this construction satisfies the required universality.

$\newcommand{\im}{\operatorname{im}}$Given a set $X$ and non-zero unital commutative ring $R$, let: \begin{align} A &= \mbox{free unital, associative algebra on $X$ with coefficients in $R$},\\ L &= \mbox{free Lie algebra on $X$ with coefficients in $R$}. \end{align} Regarding $A$ as a Lie algebra via the ring commutator, the universality of $L$ applied to $X \hookrightarrow A$, yields a natural morphism of Lie algebras: \begin{align} \rho : L \to A. \end{align} I don't know whether $\rho$ is always injective but I suspect not. My question is thus:

Can someone provide an example of a ring $R$ (and set $X$) for which $\rho$ fails to be injective?

Identify $X$ with its image in $A$, let $\im \rho$ be the image of $\rho$ and note that:

1. $X \subseteq \im \rho$.
2. It follows from the uniqueness of the universal property of $L$, that the image of $\rho$ is the smallest Lie subalgebra of $A$ containing by $X$.
3. $\rho$ is injective when $R$ is a field of characteristic zero. To see this is to note that $X \hookrightarrow \im \rho$, satisfies the universal property of the free Lie algebra. Indeed if we have $f : X \to K$ for some Lie algebra $K$ and $\mu : K \to U_K$ is the map from $K$ to its universal enveloping algebra, then the universal property of $A$ applied to $\mu \circ f$ gives a map $\hat f : A \to U_K$ and using 2 above, it is easy to see that $\hat f(\im \rho) \subseteq \im \mu$. Since $R$ is a field of characteristic zero, $\mu$ is injective, so we can define $\mu^{-1} \circ \hat f|_{\im\rho} : \im\rho \to K$ and it is easily checked that this construction satisfies the required universality.