Congruence condition on number of ideals of given prime power order in nilpotent ring

Statement
Suppose $$L$$ is a finite fact about::nilpotent ring and $$p^r$$ is a prime power dividing the order of $$L$$. Then, the number of ideals (specifically, two-sided ideals) of $$L$$ of order $$p^r$$ is congruent to $$1$$ modulo $$p$$.

Similar facts

 * Congruence condition on number of subrings of given prime power order in nilpotent ring
 * Congruence condition on number of ideals of given prime power order in a given ideal in a nilpotent ring
 * Congruence condition on number of ideals of given prime power order and bounded exponent in nilpotent ring
 * Congruence condition on number of subrings of given prime power order and bounded exponent in nilpotent ring

Facts used

 * 1) uses::Congruence condition relating number of ideals containing minimal ideals and number of ideals in the whole ring
 * 2) uses::Fourth isomorphism theorem for rings
 * 3) uses::Formula for number of minimal ideals of nilpotent ring of prime power order

Reduction to the case of a nilpotent ring of prime power order
Any finite ring is a direct sum of its Sylow subrings, which are rings of prime power order, and all subrings of order $$p^r$$ in the ring and contained in the direct summand corresponding to the prime $$p$$. Further, this direct summand is also a nilpotent ring. Finally, a subring is an ideal in the whole ring if and only if it is an ideal in the Sylow subring.

It thus suffices to prove the statement for a nilpotent ring of prime power order.

Proof for a nilpotent ring of prime power order
Given: A finite nilpotent ring $$L$$ of order $$p^k$$, $$p$$ a prime. $$r \le k$$.

To prove: The number of ideals of $$L$$ of order $$p^r$$ is congruent to 1 mod $$p$$.

Proof: In this proof, we induct on $$k$$, i.e., we assume the statement is true inside groups of order $$p^l, l \le k$$.

Base case for induction: The case $$k = 0$$ is obvious.

Inductive step: If $$r = 0$$, the number of ideals is 1, so the statement is true. So we consider $$r > 0$$.

For any subring $$S$$ of $$L$$, denote by $$\nu(L,S)$$ the number of ideals of $$L$$ of order $$p^r$$ containing $$S$$. Denote by $$\nu(L)$$ the total number of ideals of $$L$$ of order $$p^r$$.