Characteristic Lie subring not implies ideal

Statement
A characteristic subring of a Lie ring need not be an ideal of the Lie ring.

Similar facts

 * Characteristic not implies derivation-invariant
 * Derivation-invariant not implies characteristic

Analogues in other algebraic structures

 * Characteristic implies normal (in groups)
 * Characteristic not implies normal in loops

Example of the simple Witt algebra
Suppose $$p$$ is a prime number greater than 3. Let $$\mathbb{F}_p$$ be the prime field for the prime $$p$$. Denote by $$L$$ the simple Witt algebra for $$\mathbb{F}_p$$ corresponding to the prime $$p$$; explicitly, this means that:


 * The additive group has basis $$e_{-1},e_0,\dots,e_{p-2}$$. Explicitly, it is $$\bigoplus_{i=-1}^{p-2} Fe_i$$.
 * The Lie bracket is defined as follows on the basis:

$$[e_i,e_j] := \left \lbrace\begin{array}{rl} (j - i)e_{i+j}, & -1 \le i + j \le p - 2 \\0, & \mbox{otherwise}\\\end{array}\right.$$

Consider the "sandwich" Lie subring $$S$$ of $$L$$ given by:

$$S = \bigoplus_{2i > p} \mathbb{F}_pe_i$$


 * $$S$$ is clearly a subring of $$L$$.
 * $$S$$ is characteristic in $$L$$, because $$x \in S \iff [x,[x,y]] = 0 \ \forall \ y \in L$$ (note that the set with this description is not always a subring, but in this case it is).
 * $$S$$ is not an ideal in $$L$$: for instance, $$[S,e_{-1}]$$ is not in $$S$$. In fact, $$L$$ is simple, so it has no proper nonzero ideal.