Throughout this article every ring is an associative ring with identity. Let R be a ring. An element u of R is right regular if ur = 0 implies r = 0 for r ∈ R. A left regular element is defined similarly. An element is regular if it is both left and right regular (and hence not a zero divisor). We use C(R) and U(R) to denote the monoid of regular elements and the group of units in R, respectively. The Wedderburn radical (i.e., sum of all nilpotent ideals), the upper nilradical (i.e., the sum of all nil ideals), the lower nilradical (i.e., the intersection of all prime ideals), the Jacobson radical, and the set of all nilpotent elements in R are denoted by N₀(R), N^*(R), N_*(R), J(R), and N(R), respectively. It is well-known that N₀(R) ⊆ N_*(R) ⊆ N^*(R) ⊆ N(R) and N^*(R) ⊆ J(R). The polynomial ring with an indeterminate x over R is denoted by R[x] and C_f(x) denotes the set of all coefficients of f(x) for f(x) ∈ R[x]. Denote the n by n full (resp., upper triangular) matrix ring over R by M_n(R) (resp., T_n(R)). D_n(R) denotes the subring {(a_ij) ∈ T_n(R)|a₁₁ = ⋯ = a_nn} of T_n(R). Let I_n and E_ij be the identity matrix and the matrix, with (i, j)-entry 1 and elsewhere 0, in M_n(R), respectively. Let ℤ (resp., ℤ_n) denote the ring of integers (modulo n), and ℝ denote the field of real numbers.

In this section we study the properties of regular-IFP rings as well as the relations between regular-IFP rings and ring properties that play important roles in noncommutative ring theory. Due to Bell [4], a ring is called IFP if ab = 0 for a, b ∈ R implies aRb = 0. Following Kim et al. [15], a ring R is called unit-IFP if ab = 0 for a, b ∈ R implies aU(R)b = 0. IFP rings are clearly unit-IFP, but not conversely by [15, Example 1.1]. A ring R is usually called reduced if N(R) = 0. Commutative rings are clearly IFP; and reduced rings are easily shown to be IFP, but not conversely because there exist many non-reduced commutative ring. A ring is usually called Abelian if every idempotent is central. Unit-IFP rings are Abelian by [15, Lemma 1.2(2)].

Definition 2.1

A ring R is called regular-IFP if ab = 0 for a, b ∈ R implies aC(R)b = 0.

Regular-IFP rings are clearly unit-IFP (hence Abelian), but not conversely by the following example.

Example 2.2

There exists a unit-IFP ring that is not regular-IFP. Let K be a field and A = K⟨a, b⟩ be the free algebra generated by the noncommuting indeterminates a, b over K. Let I be the ideal of A generated by b² and set R = A/I. Identify a, b with their images in R for simplicity. Then R is unit-IFP by [15, Example 1.1]. But R is not a regular-IFP ring because b² = 0 and bab ≠ 0 where a ∈ C(R).

Remark 2.3

• (1) The following conditions are equivalent, which can be proved by applying the regular-IFPness iteratively:

  • (i) A ring R is regular-IFP;

  • (ii) a₁C(R)a₂C(R)a₃ ⋯ a_n−1C(R)a_n = 0 whenever a₁a₂ ⋯ a_n = 0 for a₁, a₂, …, a_n ∈ R.

• (2) D₃(R) is (regular-)IFP over a reduced ring R by [15, Proposition 2.1]. However M_n(R) and T_n(R), over any ring R for n ≥ 2, cannot be regular-IFP since they are not Abelian, noting that unit-IFP (or regular-IFP) rings are Abelian.

• (3) There exists an Abelian ring that is not regular-IFP. Set R = D_n(S) for n ≥ 4 over an Abelian ring S. Then R is Abelian by [10, Lemma 2]. Let A = E₁₂, B = E₃₄ ∈ R. Then AB = 0. Consider C = I_n + E₂₃ ∈ R. Then C ∈ C(R) clearly. But ACB = E₁₄ ≠ 0, so that R is not regular-IFP.

• (4) Let R be a regular-IFP ring such that R = C(R) ∪ N(R). Let ab = 0 for a, b ∈ R. Then aC(R)b = 0 since R is regular-IFP. Moreover aN(R)b = 0 by [15, Lemma 1.2(2)]. Thus R is IFP.

Based on Armendariz [3, Lemma 1], a ring R is called Armendariz if ab = 0 for all a ∈ C_f(x) and b ∈ C_g(x) whenever f(x)g(x) = 0 for f(x), g(x) ∈ R[x], by Rege and Chhawchharia [19]. Reduced rings are Armendariz by [3, Lemma 1]. The concepts of Armendariz rings and commutative rings are independent of each other by Example 2.2 and [19, Example 3.2], noting that the ring R in Example 2.2 is Armendariz by [2, Example 4.8].

By Goodearl [7], a ring R is called (von Neumann) regular if for every a ∈ R there exists b ∈ R such that a = aba, and a ring R is called strongly regular if a ∈ a²R for every a ∈ R. It is easily checked that J(R) = 0 for every regular ring R, and note that a ring is strongly regular if and only if it is Abelian regular, by [7, Theorems 3.2 and 3.5]. Recall that unit-IFP rings are Abelian, and Armendariz rings are also Abelian by [12, Corollary 8]. So for a regular ring R, we have that R is reduced if and only if R is Armendariz if and only if R is IFP if and only if R is regular-IFP if and only if unit-IFP if and only if R is Abelian, by [7, Theorem 3.2].

Following [11], a ring is called locally finite if every finite subset generates a finite multiplicative semigroup. Finite rings are clearly locally finite, but not conversely by the existence of algebraic closures of finite fields. It is shown that a ring is locally finite if and only if every finite subset generates a finite subring, in [11, Theorem 2.2(1)].

Proposition 2.4

• (1) Let R be a locally finite ring.

  • (i) If R is an Armendariz ring, then it is regular-IFP.

  • (ii) If R is a regular-IFP ring, then R/J(R) is strongly regular with J(R) = N(R).

• (2) Let R be a right or left Artinian ring. If R is regular-IFP, then R/J(R) is a strongly regular ring with J(R) = N(R).

Proposition 2.5

• (1) Let R_λ (λ ∈ Λ) be Abelian rings. Then R_λ is regular-IFP for each λ ∈ Λ if and only if Π_λ∈ΛR_λ is regular-IFP if and only if the subring of Π_λ∈ΛR_λ generated by ⊕_λ∈ΛR_λ and 1_Πλ∈ΛRλis regular-IFP.

• (2) Let R be an Abelian ring and e² = e ∈ R. Then R is regular-IFP if and only if both eR and (1 − e)R are regular-IFP.

Corollary 2.6

A ring R is semiperfect regular-IFP if and only if R is a finite direct product of local regular-IFP rings.

Example 2.7

There exists a non-regular-IFP ring R whose factor rings are regular-IFP. Consider R = T₂(D) over a division ring D. Then every non-trivial factor ring is one of R/J(R) ≅ D ⊕ D, R/I ≅ D and R/K ≅ D, where $J(R)=\left(\begin{array}{cc}0& D\\ 0& 0\end{array}\right),I=\left(\begin{array}{cc}D& D\\ 0& 0\end{array}\right),K=\left(\begin{array}{cc}0& D\\ 0& D\end{array}\right)$. These factor rings are reduced and so (regular-)IFP. But R cannot be regular-IFP because R is non-Abelian.

In this section we examine the regular-IFP property of ring extensions that play roles in noncommutative ring theory.

Regarding Remark 2.3(3), we have the following.

Proposition 3.1

For a ring R the following conditions are equivalent:

• (1) R is a reduced ring;

• (2) D₃(R) is an IFP ring;

• (3) D₃(R) is a regular-IFP ting;

• (4) D₃(R) is a unit-IFP ring;

• (5) AN(D₃(R))B = 0 whenever AB = 0 for A, B ∈ D₃(R).

Example 3.2

We refer to the construction and argument in [16, Example 2.1]. Let

be the free algebra generated by noncommuting indeterminates a₀, a₁, a₂, b₀, b₁, b₂, c over ℤ₂. Next, let I be the ideal of A generated by

where the constant terms of r, r₁, r₂, r₃, r₄ ∈ A are zero. Now set R = A/I. We identity

with their images in R for simplicity. Then R is reversible by [16, Example 2.1] but not reduced clearly.

Now, consider

Then AB = 0. But

because a₀cb₁ + a₁cb₀ ∉ I, noting $C=\left(\begin{array}{ccc}c& 0& 0\\ 0& c& 0\\ 0& 0& c\end{array}\right)\in N(D_3(R))$. Thus D₃(R) does not satisfy the condition (5) of Proposition 3.1.

Remark 3.3

• (1) Note that D₂(R) over a reduced ring R is IFP by [16, Proposition 1.6] and so it is regular-IFP. Moreover, there exists a non-reduced non-commutative reversible ring R over which D₂(R) is regular-IFP by [15, Example 2.2]. However, the ring S is always regular-IFP when D₂(S) is regular-IFP. For, suppose that D₂(S) is regular-IFP and let ab = 0 for a, b ∈ S. For $A=\left(\begin{array}{cc}a& 0\\ 0& a\end{array}\right),B=\left(\begin{array}{cc}b& 0\\ 0& b\end{array}\right)\in D_2(S)$, we have AB = 0 and so AC(D₂(S))B = 0 by assumption. Set $C=\left(\begin{array}{cc}c& 0\\ 0& c\end{array}\right)$ for any c ∈ C(S). Then C ∈ C(D₂(S)) and ACB = 0, entailing acb = 0. Thus S is regular-IFP.

• (2) Related to (1) above, there exists a reversible ring R such that D₂(R) is not regular-IFP. Let ℍ be the Hamilton quaternions over ℝ and R = D₂(ℍ). Then R is reversible [16, Proposition 1.6]. We refer to the argument in [16, Example 1.7]. Consider

  $$A=\left(\begin{array}{cc}\left(\begin{array}{cc}0& i\\ 0& 0\end{array}\right)& \left(\begin{array}{cc}j& 0\\ 0& j\end{array}\right)\\ \left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)& \left(\begin{array}{cc}0& i\\ 0& 0\end{array}\right)\end{array}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\text{and\hspace{0.17em}}B=\left(\begin{array}{cc}\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)& \left(\begin{array}{cc}k& 0\\ 0& k\end{array}\right)\\ \left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)& \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\end{array}\right)$$

  in D₂(R). Then AB = 0.

  Note that

  $$C=\left(\begin{array}{cc}\left(\begin{array}{cc}j& 0\\ 0& j\end{array}\right)& \left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\\ \left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)& \left(\begin{array}{cc}j& 0\\ 0& j\end{array}\right)\end{array}\right)\in C(D_2(R))$$

  by [13, Lemma 2.1] because $\left(\begin{array}{cc}j& 0\\ 0& j\end{array}\right)\in C(D_2(ℍ))$. But ACB ≠ 0, hence D₂(R) is not regular-IFP.

• (3) For a ring R and n ≥ 2, let V_n(R) be the ring of all matrices (a_ij) in D_n(R) such that a_st = a_(s+1)(t+1) for s = 1, …, n − 2 and t = 2, …, n − 1. Note that $V_n(R)\cong \frac{R[x]}{x^{n}R[x]}$. If R is a reduced ring, then V_n(R) is (regular)-IFP by [17, Lemma 2.3 and Proposition 3.3], but the converse does not hold in general as can be seen by the commutative ring V_n(R) over a non-reduced commutative ring (e.g., ℤ_n^l for n, l ≥ 2) R for n ≥ 2.

Proposition 3.4

• Let M be a multiplicatively closed subset of a ring R consisting of central regular elements. Then R is regular-IFP if and only if M⁻¹R is regular-IFP.

• Let R be a ring. Then R[x] is regular-IFP if and only if R[x, x⁻¹] is regular-IFP.

Proposition 3.5

Let R be a ring.

• {c + c₁x + … + c_tx^t ∈ R[x] | c ∈ C(R) for t ≥ 1} ⊆ C(R[x]) and {d₀ + d₁x + … + d_s−1x^s−1 + dx^s ∈ R[x] | d ∈ C(R) for s ≥ 1} ⊆ C(R[x]).

• If R[x] is regular-IFP, then so is R.

• Let R be a regular-IFP ring such that C(R[x]) = {c + xN(R)[x] | c ∈ C(R)}. If R is Armendariz then R[x] is regular-IFP.

Example 3.6

We use the ring and argument in [12, Example 2]. Let A = ℤ₂⟨a₀, a₁, a₂, b₀, b₁, b₂, c⟩ be the free algebra with noncommuting indeterminates a₀, a₁, a₂, b₀, b₁, b₂, c over ℤ₂. Let B be the set of all polynomials with zero constant terms in A. Next, consider the ideal I of A generated by

where r ∈ A and r₁, r₂, r₃, r₄ ∈ B. Set R = A/I, and identify a₀, a₁, a₂, b₀, b₁, b₂, c with their images in R for simplicity. Then R is (regular-)IFP but not Armendariz, by [12, Example 2] and [19, Proposition 4.6].

Notice that (a₀ + a₁x + a₂x²)(b₀ + b₁x + b₂x²) = 0. But

because a₀cb₁ + a₁cb₀ ∉ I, noting 1 + c ∈ C(R[x]) as in [15, Example 2.7]. Thus R[x] is not regular-IFP.

Considering Proposition 3.5, it is natural to ask whether C_f(x) ∩ C(R) ≠ ∅︀ when f(x) ∈ C(R[x]), where R is a given ring. But the answer is negative by the following.

Example 3.7

Let A be any ring and R = A × A. Consider a polynomial f(x) = (1, 0) + (0, 1)x in R[x]. Suppose f(x)g(x) = 0 for $g(x)={\sum }_{i=0}^m{a}_i{x}^i\in R[x]$ with a_i = (b_i, c_i). Then from f(x)g(x) = 0, we obtain f₁(x)g₁(x) = 0 and f₂(x)g₂(x) = 0, where

This implies ${\sum }_{i=0}^m{b}_i{x}^i=0$ and ${\sum }_{i=0}^m{c}_i{x}^{i+1}=0$, so that b_i = 0 and c_i = 0 for all i. Thus a_i = 0 for all i, and hence f(x) ∈ C(R[x]). But (1, 0), (0, 1) ∉ C(R).

We consider next some equivalent conditions to the regular-IFP property in relation to the sum of coefficients of polynomials which satisfy some property of inserting regular polynomials. For f(x) ∈ R[x], let f(1) denote the sum of all coefficients of f(x).

Proposition 3.8

For a ring R the following conditions are equivalent:

• (1) R is regular-IFP;

• (2) If f₁(x)f₂(x) ⋯ f_n(x) = 0 for f₁(x), f₂(x), …, f_n(x) ∈ R[x], then the sum of all coefficients of every polynomial in

  $$f_1(x)C(R)[x]f_2(x)C(R)[x]\cdots f_{n-1}(x)C(R)[x]f_n(x)$$

  is zero;

• (3) If f(x)g(x) = 0 for f(x), g(x) ∈ R[x], then the sum of all coefficients of every polynomial in f(x)C(R)[x]g(x) is zero;

• (4) If f(x)g(x) = 0 for linear polynomials f(x), g(x) in R[x], then the sum of all coefficients of every polynomial in f(x)C(R)[x]g(x) is zero;

• (5) f(x)g(x) = 0 implies f(x)C(R)[x]g(x) = 0 for linear polynomials f(x), g(x) in R[x].

Based on Proposition 3.1(5), a ring R is called nilpotent-IFP [8] if aN(R)b = 0 whenever ab = 0 for a, b ∈ R. Every unit-IFP ring is nilpotent-IFP by the proof of Proposition 3.1 and this direction is irreversible by [15, Example 2.5]. For a unit-IFP (or regular-IFP) ring R, we have N₀(R) = N_*(R) = N^*(R) by [15, Theorem 1.3(1)]. But there exists a unit-IFP (hence nilpotent-IFP) ring R such that N₀(R) ⊊ N(R) and N(R) ⊈ J(R), by [15, Example 1.1]. The following partially extends the result of [15, Theorem 1.3].

Theorem 4.1

For a nilpotent-IFP ring R, we have the following.

• (1) N₀(R) = N_*(R) = N^*(R).

• (2) J(R[x]) = N₀(R[x]) = N_*(R[x]) = N^*(R[x]) = N₀(R)[x] = N_*(R)[x] = N^*(R)[x] = N₀(R)[x] ⊆ J(R)[x]. Moreover, J(R[x]) = J(R)[x] when J(R) is nil.

Proposition 4.2

Every one-sided DR ring is regular-IFP.

Example 4.3

Let R be the Hamilton quaternions over ℤ. Then R is a domain and so regular-IFP. But R is not right DR by [9, Example 2.5(1)]. Moreover R is also not left DR by a similar argument to [9, Example 2.5(1)].

Consider the group ring KQ₈, where K is a field and Q₈ denotes the quaternion group.

Corollary 4.4

For a field K of characteristic 0 and R = KQ₈, the following conditions are equivalent:

• (1) R is DR;

• (2) R is regular-IFP;

• (3) R is unit-IFP;

• (4) R is nilpotent-IFP;

• (5) R is Abelian;

• (6) The equation 1 + x² + y² = 0 has no solutions in K;

• (7) R is isomorphic to a finite direct product of division rings;

• (8) R is reduced.