Thompson's replacement theorem for abelian Lie subrings

Statement
Suppose $$L$$ is a nilpotent Lie ring of order a prime power, with underlying prime $$p$$.

Let $$\mathcal{A}(L)$$ denote the set of abelian subrings of maximum order in $$L$$.

Suppose $$A \in \mathcal{A}(L)$$ and $$B$$ is an abelian subring of $$L$$ such that $$A$$ idealizes $$B$$ but $$B$$ does not idealize $$A$$. Then, there exists an abelian subring $$A^*$$ of $$AB$$ such that:


 * 1) $$A^* \in \mathcal{A}(L)$$
 * 2) $$A \cap B$$ is a proper subring in $$A^* \cap B$$
 * 3) $$A^*$$ is contained in $$\langle A^B \rangle = \langle A^{AB} \rangle$$, i.e., it is contained in the ideal closure of $$A$$ in $$AB$$. In particular, it is contained in the ideal closure $$\langle A^P \rangle$$ of $$A$$ in $$P$$.
 * 4) $$A^*$$ idealizes $$A$$.

We have the following additional conclusions:
 * 1) $$A^*$$ idealizes $$B$$: Note that since $$A^*$$ is contained in $$AB$$, and $$AB \le N_P(B)$$ by assumption, we see that $$A^*$$ idealizes $$B$$ as well.
 * 2) $$A^* \cap B$$ is a proper subring of $$B$$: This is because $$A^*$$ cannot contain $$B$$, since $$A^*$$ idealizes $$A$$ but $$B$$ does not idealize $$A$$.

Thus, all the conditions assumed for $$A$$ also hold for $$A^*$$, except possibly the fact that $$B$$ does not idealize $$A$$. Hence, the term replacement.