Additive periods of power maps lie in the prime radical

Henry Zweiman

Abstract

We prove that every additive period of every power map on an associative ring belongs to its prime radical. No identity, characteristic, finiteness, or polynomial-identity assumption is required. The proof combines a substitution using the highest nonzero power of a nilpotent element with the classical bounded-index one-sided nil-ideal lemma. It follows that every nilperiod ring is 2-primal and that its nilpotents form a locally nilpotent ideal. This strengthens the NI conclusion conjectured by Burnette. We also characterize finite unital nilperiod rings as finite products of local rings, and determine the periodic power maps on Corbas rings, correcting the count for nonidentity twists.

1. Introduction

All rings are associative and may lack an identity unless otherwise stated. For a ring , let be its set of nilpotent elements, and let be its prime radical, also called the lower nilradical. Thus is the intersection of the prime ideals, or equivalently the smallest ideal for which is semiprime. The prime radical is locally nilpotent: every finite subset generates a nilpotent subring. In particular, is a nil ideal; see, for example, [4, Section 1].

For an integer , set and define its additive period group by

Evaluation at zero gives . Our main result locates these periods more precisely.

Theorem 1.

For every associative ring and every positive integer ,

Consequently, every power map on a semiprime ring has trivial additive period group.

Following Burnette [2], a ring is nilperiod if every nilpotent element belongs to for some , which may depend on the element. A ring is NI if its nilpotents form a two-sided ideal, and it is 2-primal if .

Corollary 2.

Every nilperiod ring is 2-primal. Its nilpotent elements form a locally nilpotent two-sided ideal, which is both its prime radical and its largest nil ideal. In particular, every nilperiod ring is an NI-ring.

Proof.

Theorem 1 gives , while the reverse containment holds in every ring. The other conclusions follow from the standard properties of the prime radical.

Burnette’s Conjecture 2.6 asks whether every nilperiod ring is NI, and his Theorem 2.9 establishes this for PI algebras. Corollary 2 proves the conjecture and strengthens its conclusion. Theorem 1 also treats individual periods in rings that need not be nilperiod. In particular, it extends the absence of nonzero periods in matrix rings over division rings used in [2, Lemma 2.8] to all semiprime rings.

The main proof uses the classical fact that a semiprime ring has no nonzero one-sided nil ideal of bounded index. The new reduction is short: given a period of nilpotency index , evaluate its identity at and multiply by . This produces the bounded-index right ideal to which the classical result applies.

The remaining results concern finite rings and power-map counts. As usual, a unital ring is local if its nonunits form a proper ideal.

Theorem 3.

A nonzero finite unital ring is nilperiod if and only if it is a finite direct product of finite local rings.

The zero ring may be included as the empty product. The structural decomposition behind this result is standard; we explicitly credit it in Section 3. Section 4 gives an exact computation for the twisted rings in [2, Example 2.2].

2. Periods in semiprime rings

We use the following classical bounded-index lemma. Its left-ideal form is stated in [1, p. 93, item (V)]; the right-ideal form follows by passing to the opposite ring. The bounded nilradical discussion in [4, Section 1] also records the result in the nonunital setting.

Lemma 4 (Bounded-index one-sided nil ideals).

Let be an associative ring, not necessarily unital. If has a nonzero right ideal and a fixed positive integer such that for every , then has a nonzero nilpotent two-sided ideal. In particular, a semiprime ring has no nonzero nil right ideal of bounded index.

This lemma requires neither a chain condition on ideals nor a restriction on the characteristic. The uniform bound concerns the elements of , not all nilpotents of .

Proof of Theorem 1.

First let be semiprime and let . Since , a nonzero has a minimal nilpotency index satisfying . Suppose that this is the case, and put .

Let be the standard unitization, with multiplication

If already has an identity, we may take . For any , the element lies in , so the period identity applies to . Also . In the word expansion of , a word vanishes whenever an occurs before an , since such a word contains adjacent factors . The remaining words have the form , and terms with vanish. Hence

Right-multiply (1) by , interpreting in if . The term with becomes

while all terms with vanish because . Therefore

The nonzero principal right ideal of is thus nil of bounded index . This contradicts Lemma 4, so . Notice that we have only evaluated the period identity at elements of ; no period property of is assumed.

Now take an arbitrary ring and . Its period identity descends to the quotient . Since this quotient is semiprime, the first part of the proof shows that the image of is zero. Thus .

Remark 5.

Local nilpotence in Corollary 2 does not supply a uniform nilpotency exponent for the entire ideal . Nor does the proof assert that each nilpotent element generates a nilpotent two-sided ideal. It does imply that every such generated ideal is nil, since it lies in .

3. Idempotents and finite rings

Proposition 6.

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 7.

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 .

The forward structural implication is an instance of the standard fact that an Abelian semiperfect ring is a finite direct product of local rings [3, Proposition 2.6]. Finite rings are semiperfect. We include an elementary proof in this finite setting.

Proof of Theorem 3.

Suppose first that is finite, unital, and nilperiod. By Proposition 6, 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 Corollary 2.

For , Lemma 7 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 8.

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 [2, Example 2.2 and Table 1] that nonidentity twists have no periodic power maps. It also determines exactly which exponents occur.

Theorem 9.

For the ring (2), set

Then

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 10.

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

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[2] C. Burnette, On power maps over periodic rings, Graduate J. Math. 10 (2025), no. 1, 5–16. Published article. Earlier preprint: On power maps over weakly periodic rings, arXiv:2207.14283.

[3] J. Han, Y. Lee, and S. Park, Structure of Abelian rings, Front. Math. China 12 (2017), no. 1, 117–134. doi:10.1007/s11464-016-0586-z.

[4] P. P. Nielsen, Bootstrapping the bounded nilradical, J. Pure Appl. Algebra 217 (2013), no. 9, 1711–1715. doi:10.1016/j.jpaa.2012.12.002.