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](https://mathoverflow.net/questions/413347/):



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> **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')?
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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](https://en.wikipedia.org/wiki/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



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> I.N. Herstein, *A Commutativity Theorem*, J. Algebra 38 (1976), No. 1, 112-118.
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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](https://en.wikipedia.org/wiki/Nil_ideal). 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](https://mathoverflow.net/questions/413347/) 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 $X$.



Rings s.t. each element has a power lying in the center (and their completely prime ideals)
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](https://doi.org/10.1007/978-1-4419-8616-0)” 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.


