Nilperiod rings and additive periods of power maps
Henry Zweiman
Abstract
A ring is nilperiod if every nilpotent element is an additive period of some power map. We prove that every nilperiod ring is an NI-ring, resolving a conjecture of Burnette without finiteness, polynomial-identity, characteristic, or identity assumptions. The proof shows that each nilpotent element generates a nil ideal, using a square-zero substitution and descent through nil ideals. We also prove that finite unital nilperiod rings are exactly the finite direct products of finite local rings. Finally, we determine the additive periods of all power maps on Corbas rings and obtain a corrected count for the cases with nontrivial field automorphism.
1. Introduction
All rings are associative; an identity is assumed only when explicitly stated. For a ring , let denote its set of nilpotent elements. For write and
This is an additive subgroup of , and substituting shows that every one of its elements is nilpotent. Following Burnette [1], the ring is nilperiod if
The exponent may depend on the nilpotent element. A ring is an NI-ring if its nilpotent elements form a two-sided ideal.
Burnette conjectured that every nilperiod ring is an NI-ring [1, Conjecture 2.6], and proved the assertion for PI algebras [1, Theorem 2.9]. We prove the conjecture in its full stated generality.
Theorem 1.
In a nilperiod ring, every nilpotent element generates a nil two-sided ideal. Consequently every nilperiod ring is an NI-ring.
The proof is elementary. If is an additive period of a power map, the substitution forces to be nilpotent for every . Additive closure of nilpotents then makes the ideal generated by nil. For an arbitrary nilpotent , induction first treats the ideal generated by , after which the square-zero argument applies in the quotient.
The period condition also forces every idempotent to be central. This gives an exact finite unital classification.
Theorem 2.
A nonzero finite unital ring is nilperiod if and only if it is a finite direct product of finite local rings.
Here, as usual, a unital ring is local when its nonunits form a proper ideal. The zero ring is nilperiod and can be included in Theorem 2 as the empty direct product. In the final section we apply this viewpoint to the twisted rings considered in [1, Example 2.2] and compute their periodic power maps exactly.
2. Nil ideals generated by nilpotents
An ideal is called nil if each of its elements is nilpotent; no common nilpotency exponent is required. To handle rings without identity, write for the standard unitization , with multiplication
The ring embeds as a two-sided ideal in . If already has an identity, one may instead take . The ideal generated by is the set of finite sums of terms with . We do not assume that is nilperiod.
The following observation is already used in the proof of [1, Theorem 2.9].
Lemma 3.
The nilpotents of a nilperiod ring form an additive subgroup.
Proof.
Negatives of nilpotents are nilpotent. Let be nilpotent. Choose with for every , and choose with . Then , so .
Lemma 4.
If is nilperiod and is a nil ideal, then is nilperiod.
Proof.
Suppose is nilpotent. Then for some . As is nil, for some . Hence is nilpotent in . A period identity for descends to the required identity for in .
Lemma 5.
If is nilperiod and , then is a nil ideal.
Proof.
There is nothing to prove if . Otherwise choose with for all . Let and set . Since , the noncommutative expansion has only two possibly nonzero words:
Indeed, any occurrence of before the final position is followed by or , and the resulting word vanishes. The period identity therefore gives . Right multiplication by shows that
For , it follows that
Thus every two-sided multiple is nilpotent. By Lemma 3, every finite sum of such multiples is nilpotent, which proves the assertion.
We prove by strong induction on that, in every nilperiod ring, each element satisfying generates a nil ideal. The cases and follow from and Lemma 5, respectively. Let , and assume the assertion for all smaller positive integers. Then
The induction hypothesis makes a nil ideal. By Lemma 4, is nilperiod, and the image of has square zero. Lemma 5 applied in shows that is nil. Since , this ideal is .
For , there is consequently an with . Since is nil, for some . Hence is nilpotent, completing the induction. Finally, Lemma 3 gives additive closure of , and multiplication of a nilpotent by any ring element stays in the nil ideal . Thus is a two-sided ideal.
Corollary 6.
If is nilperiod, then is its largest nil ideal and is reduced.
Proof.
Every nil ideal is contained in . If is nilpotent, then for some , so is nilpotent and its image is zero.
3. Idempotents and finite rings
Proposition 7.
Every idempotent in a nilperiod ring is central.
Proof.
Let and . The element satisfies , , and . Thus is idempotent. A period identity for , evaluated at , gives
Hence . Similarly, is square-zero and is idempotent, so the same argument gives . Therefore .
Lemma 8.
Every element of a finite ring has an idempotent positive power.
Proof.
For in a finite ring there exist positive integers such that . Choose divisible by . Eventual periodicity of the powers then gives .
Suppose first that is finite, unital, and nilperiod. By Proposition 7, its idempotents are central. Successively split the identity into orthogonal nonzero central idempotents until each component contains no idempotents except zero and its identity. The process terminates since is finite. It gives
Each direct factor is nilperiod: a nilpotent in it lifts to a nilpotent with zero coordinates in the other factors, and the corresponding period identity restricts to that factor. Its nilpotents therefore form an ideal by Theorem 1.
For , Lemma 8 gives or . Thus is either nilpotent or a unit. The nonunits of are exactly its nilpotents, which form a proper ideal, so is local.
Conversely, let be finite and local, with nonunit ideal . Its only idempotents are zero and one: if , one of is a unit, and gives the conclusion. For , an idempotent power remains in , so it must be zero. Thus . Choose at least the maximum of the nilpotency indices of the elements of , and divisible by the exponent of the finite group . Then
Translation by any element of preserves membership in . Consequently every element of is a period of . For a finite direct product of finite local rings, choose one satisfying these requirements in all factors.
Example 9.
An NI-ring need not be nilperiod, even when finite. The upper triangular matrices over have nilpotents , a square-zero ideal. But for and , the distinct matrices and are both idempotent. Hence for every .
4. Power maps on Corbas rings
Let , where is prime and , and choose . Let be the corresponding automorphism of , and put . We allow , so the identity automorphism has . Consider the Corbas ring
This is an associative unital ring, with identity and square-zero ideal . If , the element has inverse
Thus is local and .
The following calculation corrects the assertion in [1, Example 2.2 and Table 1] that nonidentity twists have no periodic power maps. It also determines exactly which exponents occur.
There are exactly distinct periodic power maps on .
Proof.
Direct induction gives
Every additive period must lie in , by evaluation at zero. For , the element is a period exactly when for all . At this forces , and hence . The condition at then holds. If and , then , which also vanishes. For the geometric sum gives
Its vanishing is equivalent to . The homomorphism on the cyclic group has kernel and therefore image of order . Consequently vanishes everywhere if and only if and . Since , this is . The argument also determines the full period group as stated.
For the count, the unit group of has exponent . The displayed power formula shows that every unit has th power one. Elements realize order , while with have order , so the exponent is exactly . All nonunits have square zero. Therefore, for , the maps and agree exactly when : sufficiency follows from the unit and nonunit cases, and necessity from the definition of the unit-group exponent. There are thus distinct maps with exponent at least two, and among their exponent classes exactly
are periodic. The identity map is distinct from these maps on nonzero elements of and has no nonzero additive period.
Example 11.
For and the nontrivial automorphism , one has , , , and . The power map equals on all units and zero on , so its period group is . It is the unique distinct periodic power map on this noncommutative sixteen-element ring.
References
[1] C. Burnette, On power maps over periodic rings, Graduate J. Math. 10, no. 1 (2025), 5–16. Published article. Earlier preprint under the title On power maps over weakly periodic rings: arXiv:2207.14283.