Centralizer-free ideal implies automorphism-faithful

Statement
Suppose $$L$$ is a Lie ring and $$I$$ is a centralizer-free ideal of $$L$$, i.e., $$I$$ is an ideal of $$L$$ and its centralizer in $$L$$ is zero. Then, $$I$$ is an automorphism-faithful Lie subring (and hence an automorphism-faithful ideal) of $$L$$: Any non-identity automorphism of $$L$$ that restricts to an automorphism of $$I$$ restricts to a non-identity automorphism of $$I$$.

Similar facts for Lie rings

 * Centralizer-free ideal implies derivation-faithful

Analogues in groups

 * Normal and centralizer-free implies automorphism-faithful
 * Normal and self-centralizing implies coprime automorphism-faithful

Proof
Given: A Lie ring $$L$$, a centralizer-free ideal $$I$$ of $$L$$. An automorphism $$\sigma$$ of $$L$$ such that the restriction of $$\sigma$$ to $$I$$ is the identity map.

To prove: $$\sigma(l) = l$$ for all $$l \in L$$.

Proof: By the definition of automorphism, we have, for every $$a \in A$$:

$$\! \sigma([l,a]) = [\sigma(l),\sigma(a)]$$.

Since $$A$$ is an ideal and $$a \in A$$, $$[l,a] \in A$$. Thus, $$\sigma([l,a]) = [l,a]$$. Also, $$\sigma(a) = a$$. We thus have:

$$\! [l,a] = [\sigma(l),a]$$.

By the biadditivity of the Lie bracket, this gives:

$$\! [\sigma(l) - l,a] = 0$$.

In other words, $$\sigma(l) - l \in C_L(A)$$. By assumption, $$C_L(A) = 0$$, so $$\sigma(l) = l$$, completing the proof.