Congruence condition on number of subrings of given prime power order in nilpotent ring
Contents
Statement
Suppose is a finite nilpotent ring (not necessarily a Lie ring or an associative ring or a commutative ring) and is a prime power dividing the order of . Then, the number of subrings of of order is congruent to 1 mod .
Related facts
Similar facts
- Congruence condition on number of ideals of given prime power order in nilpotent Lie ring
- Congruence condition on number of subrings of given prime power order in nilpotent Lie ring
- Congruence condition on number of subgroups of given prime power order
Opposite facts
Facts used
- Congruence condition relating number of subrings in maximal subrings and number of subrings in the whole ring
- Formula for number of maximal subrings in nilpotent ring of prime power order
Proof
Reduction to the case of prime power order
Any finite ring is a direct sum of rings of prime power order (Sylow subrings) and any ring of prime power order is contained in the corresponding Sylow subring. Thus, it suffices to prove the result for the case of nilpotent rings of prime power order.
Proof for the prime power order case
This proof uses the principle of mathematical induction in a nontrivial way (i.e., it would be hard to write the proof clearly without explicitly using induction).
Given: A nilpotent ring of order , a prime number.
To prove: For any , the number of subrings in of order is congruent to modulo .
Proof: In this proof, we induct on , i.e., we assume the statement is true inside nilpotent rings of order .
Base case for induction: The case is obvious.
Inductive step: If , the number of subrings is 1, so the statement is true. So we consider .
For a subring of , denote by the number of subrings of of order .
Step no. | Assertion/construction | Facts used | Given data/assumptions used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | . Here, means that is a maximal subring of . | Fact (1) | [SHOW MORE] | ||
2 | For every , . | inductive hypothesis | [SHOW MORE] | ||
3 | The number of maximal subrings of is congruent to 1 mod . | Fact (2) | Fact-direct. | ||
4 | Steps (1)-(3) | [SHOW MORE] |