# Rings s.t. each element has a power lying in the center (and their completely prime ideals)

Let $$R$$ be a ring (throughout, all rings are associative and unital). We say $$R$$ satisfies condition (C) if, for every $$a \in R$$, there exists an integer $$n \ge 1$$ (depending on $$a$$) such that $$a^n$$ lies in the center $$\mathcal Z(R)$$ of $$R$$; and condition (C') if there exists an integer $$n \ge 1$$ such that $$a^n \in \mathcal Z(R)$$ for every $$a \in R$$ (so, the difference with condition (C) is that now $$n$$ does not depend on $$a$$). The following questions are motivated by those in a related thread here:

Questions. (i) Is there any special/standard name for a ring satisfying either (C) or (C'), or any keyword I can use to read about them in the literature? (ii) Is it true that, if $$R$$ satisfies condition (C), then every non-unit of $$R$$ is contained in a completely prime$${}^{(1)}$$ (two-sided) ideal? (iii) If not, what about the case where $$R$$ satisfies condition (C')?

Of course, (C') implies (C). Note also that (C') is equivalent to the existence of an integer $$n \ge 1$$ with the property that, for each $$a \in R$$, there is an integer $$k$$ between $$1$$ and $$n$$ (with $$k$$ depending on $$a$$) such that $$a^k \in \mathcal Z(R)$$.

Edit 1. It's perhaps worth remarking that a ring satisfying (C) is Dedekind-finite$${}^{(2)}$$, which is a necessary condition for every non-unit of $$R$$ to lie in a proper (and, in particular, in a completely prime) ideal. In fact, assume $$ab = 1_R$$ for some $$a, b \in R$$; we need to check that $$bu = 1_R$$ for some $$u \in R$$. To begin, we have that $$(ba)^k = b(ab)^{k-1} a = ba$$ for every $$k \in \mathbf N^+$$. On the other hand, assuming $$R$$ satisfies (C) implies that $$(ba)^n \in \mathcal Z(R)$$ for a certain $$n \in \mathbf N^+$$. It follows that $$ybax = y(ba)^n x = (ba)^n yx = bayx, \qquad \text{for all }x, y \in R,$$ which ultimately shows (by taking $$x = b$$ and $$y = a$$) that $$ba^2b = (ab)^2 = 1_R$$. []

Edit 2. In a comment to the OP, Benjamin Steinberg asked for non-commutative examples of rings satisfying, say, condition (C'). Luckily enough, the first idea that comes to mind seems to work just fine.

Let $$\mathscr F(X)$$ be the free monoid on a non-empty set $$X$$, and let $$F\langle X \rangle$$ be the monoid ring of $$\mathscr F(X)$$ over the two-element field $$F$$ (i.e., $$F\langle X \rangle$$ is the free $$F$$-algebra on the basis $$X$$) and $$\mathfrak S(X)$$ be the group of permutations of $$X$$. We assume $$X \subseteq \mathscr F(X)$$ and use $$\varepsilon$$ for the identity of $$\mathscr F(X)$$ (namely, the empty $$X$$-word). Moreover, given $$\mathfrak u \in \mathscr F(X)$$, we denote by $$\delta_{\mathfrak u}$$ the "Kronecker delta" function $$H \to F$$ that maps $$\mathfrak u$$ to $$1_F$$ and every $$X$$-word $$\mathfrak v \ne \mathfrak u$$ to $$0_F$$.

The quotient ring $$R := F\langle X \rangle/\mathfrak i$$ of $$F\langle X \rangle$$ by the (two-sided) ideal $$\mathfrak i$$ generated by the set $$\bigcup_{\sigma \in \mathfrak S(X)} \{\delta_x \delta_y \delta_z - \delta_{\sigma(x)} \delta_{\sigma(y)} \delta_{\sigma(z)}: x, y, z \in X\}$$ is commutative if and only if $$|X| = 1$$; and we are going to show that $$R$$ satisfies condition (C') with $$n = 4$$. A couple of remarks are in order before proceeding:

• If $$k$$ is an integer $$\ge 3$$ and $$\mathfrak z_1, \ldots, \mathfrak z_k$$ are non-empty $$X$$-words, then it is straighforward from the definition of the ideal $$\mathfrak i$$ and the factoriality$${}^{(3)}$$ of $$\mathscr F(X)$$ that $$\delta_{\mathfrak z_1} \cdots \delta_{\mathfrak z_k} \equiv \delta_{\mathfrak z_{\sigma(1)}} \cdots \delta_{\mathfrak z_{\sigma(k)}} \bmod \mathfrak i$$ for every permutation $$\sigma$$ of the discrete interval $$[\![1, k ]\!]$$.
• By the previous remark, $$\delta_\mathfrak{z} \bmod \mathfrak i \in \mathcal Z(R)$$ for every $$X$$-word $$\mathfrak z$$ whose length is $$\ge 3$$.

Now, pick $$f \in F\langle X \rangle$$; we need to check that $$f^4 \bmod \mathfrak i \in \mathcal Z(R)$$. For, note that, since $$F$$ has characteristic $$2$$ and $$\delta_\varepsilon$$ lies in the center of $$F \langle X \rangle$$, we have $$(f \pm \delta_\varepsilon)^2 = f^2 + \delta_\varepsilon$$ and hence $$(f-\delta_\varepsilon)^4 = f^4 + \delta_\varepsilon$$. It follows that $$f^4 \bmod \mathfrak i \in \mathcal Z(R)$$ if and only if $$(f-\delta_\varepsilon)^4 \bmod \mathfrak i \in \mathcal Z(R)$$, and we may therefore assume without loss of generality (as we do) that $$f(\varepsilon) = 0_R$$. In consequence, the support $$\text{s}(f) := \mathscr F(X) \setminus f^{-1}(0_R)$$ of $$f$$ is, by the very definition of $$F\langle X \rangle$$, a finite subset of $$\mathscr F(X) \setminus \{\varepsilon\}$$ with $$f = \sum_{\mathfrak z \in \text{s}(f)} \delta_\mathfrak{z}$$. Then $$f^4 = \sum_{(\mathfrak z_1, \mathfrak z_2, \mathfrak z_3, \mathfrak z_4) \in \text{s}(f)^{\times 4}} \delta_{\mathfrak z_1} \delta_{\mathfrak z_2} \delta_{\mathfrak z_3} \delta_{\mathfrak z_4};$$ and since $$\varepsilon \notin \text{s}(f)$$, we see that every $$X$$-word in the support of $$f^4$$ has length $$\ge 4$$. We thus conclude from the second remark above that $$f^4 \bmod \mathfrak i \in \mathcal Z(R)$$. []

Edit 3. A domain satisfying condition (C) is necessarily commutative. As noted by Benjamin Steinberg in a comment to the OP, this follows from the main theorem of

I.N. Herstein, A Commutativity Theorem, J. Algebra 38 (1976), No. 1, 112-118.

The result states that, if $$R$$ is a ring with the property that, for all $$a, b \in R$$, there exist positive integers $$m$$ and $$n$$ (depending on $$a$$ and $$b$$) such that $$a^m b^n = b^n a^m$$, then the commutator ideal of $$R$$ (that is, the two-sided ideal generated by the elements of the form $$xy-yx$$ with $$x, y \in R$$) is nil. But a domain has no non-zero nil ideals; and on the other hand, it is obvious that every ring satisfying condition (C) also satisfies the hypothesis of Herstein's theorem. Therefore, a domain satisfies condition (C) if and only if it is commutative.

Notes.

(1) An ideal $$\mathfrak p$$ of $$R$$ is completely prime if it is proper (in the sense that $$\mathfrak p \subsetneq R$$) and $$ab \in \mathfrak p$$ for some $$a, b \in R$$ implies $$a \in \mathfrak p$$ or $$b \in \mathfrak p$$; and is prime if it is proper and $$aRb \subseteq \mathfrak p$$ for some $$a, b \in R$$ implies $$a \in \mathfrak p$$ or $$b \in \mathfrak p$$ (cf. the article on prime ideals on Wiki.en).

(2) $$R$$ is Dedekind-finite if every left- or right-invertible element is a unit (equivalently, if $$ab = 1_R$$ for some $$a, b \in R$$, then $$ba = 1_R$$).

(3) I mean the fact that every non-empty $$X$$-word factors uniquely, in $$\mathscr F(X)$$, as a product of elements of the basis $$X$$.

• Do you have noncommutative examples of such rings? I believe it follows from one of Herstein's commutativity theorems that your condition plus no nilpotent ideals implies commutative. Commented Jan 11, 2022 at 11:05
• @BenjaminSteinberg Unless I made a mistake, there are entire families of non-commutative domains that satisfy (C'). Let me try and add the details directly in the OP. Commented Jan 11, 2022 at 11:07
• The results of Herstein mentioned at the bottom of the second page of reader.elsevier.com/reader/sd/pii/… seem to imply a noncommutative domain can't have that property Commented Jan 11, 2022 at 13:06
• @BenjaminSteinberg Thanks for the reference (it's quite useful). In the example I've just finished adding to the OP, I had somehow overlooked that $\delta_x$ is a zero divisor for every $x \in X$. Commented Jan 11, 2022 at 16:32

## 1 Answer

Assume (C), so that for each $$a\in R$$, there exists some integer $$n\geq 1$$ (possibly depending on $$a$$) such that $$a^n\in Z(R)$$. The equivalence of conditions (1) and (3) in Theorem 12.11 from Lam's “A First Course in Noncommutative Rings” tells us that $$R/J(R)$$ is commutative; here, $$J(R)$$ is the Jacobson radical. [This result is attributed to Herstein and Kaplansky.] Thus, any maximal one-sided ideal of $$R$$ (which, of course contains $$J(R)$$) is two-sided, and is completely prime. This answers your second (and hence third) question.

Regarding your first question, I'm not familiar with any terminology, probably because if $$J(R)=0$$, then the condition is equivalent to being commutative. I would recommend a literature search for those papers that cite Herstein's and Kaplansky's works, to see if the condition is studied more fully.

• This also suggests that R/J(R) is commutative is the more natural condition to impose. Commented Jan 11, 2022 at 21:04
• It's hard to resist the (misleading) suggestion "$R$ is radically commutative". Commented Jan 12, 2022 at 14:22