Congruence condition on number of abelian subrings of prime index in nilpotent ring

Statement
Suppose $$L$$ is a nilpotent ring of order $$p^k$$, $$p$$ a prime number. Suppose further that $$L$$ has an abelian subring of index $$p$$. Then, the number of abelian subrings of $$L$$ of index $$p$$ is congruent to 1 mod $$p$$.

Moreover, if $$L$$ is non-abelian (i.e., the multiplication is not identically zero) then the number of abelian subrings of index $$p$$ is either 1 or $$p + 1$$.

Related facts

 * Congruence condition on number of abelian subgroups of prime index: The analogous statement for groups.