Derivation-invariance is centralizer-closed

Verbal statement
The centralizer of a derivation-invariant subring of a Lie ring is also derivation-invariant.

Related facts

 * Ideal property is centralizer-closed

Weaker facts

 * Center is derivation-invariant

Proof
Given: A Lie ring $$L$$, a derivation-invariant subring $$S$$ of $$L$$. $$C = C_L(S)$$ is the set of $$a \in L$$ such that $$[a,s] = 0$$ for all $$s \in S$$.

To prove: $$C$$ is also derivation-invariant, i.e., $$d(C) \subseteq C$$ for any $$d \in \operatorname{Der}(L)$$.

Proof: Suppose $$d \in \operatorname{Der}(L)$$. For any $$c \in C$$ and $$s \in S$$, we need to show that $$[dc,s] = 0$$.

For this, note that, by the Leibniz rule property of derivations:

$$d([c,s]) = [dc,s] + [c,ds]$$.

Since $$[c,s] = 0$$, the left side is zero. Further, since $$S$$ is derivation-invariant, $$ds \in S$$, so $$[c,ds] = 0$$. This gives $$[dc,s] = 0$$ as required.