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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2022; 62(2): 213-227

Published online June 30, 2022

### The Relation Between Units and Nilpotents

Jeoung Soo Cheon, Tai Keun Kwak*, Yang Lee, Young Joo Seo

Department of Mathematics, Pusan National University, Busan 46241, Korea
e-mail : jeoungsoo@pusan.ac.kr

Department of Data Science, Daejin University, Pocheon 11159, Korea
e-mail : tkkwak@daejin.ac.kr

Department of Mathematics, Yanbian University, Yanji 133002, China Institute for Applied Mathematics and Optics, Hanbat National University, Daejeon 34158, Korea
e-mail : ylee@pusan.ac.kr

Department of Data Science, Daejin University, Pocheon 11159, Korea
e-mail : jooggang@daejin.ac.kr

Received: November 17, 2021; Revised: March 8, 2022; Accepted: March 8, 2022

We discuss the relation between units and nilpotents of a ring, concentrating on the transitivity of units on nilpotents under regular group actions. We first prove that for a ring R, if U(R) is right transitive on N(R), then Köthe's conjecture holds for R, where U(R) and N(R) are the group of all units and the set of all nilpotents in R, respectively. A ring is called right UN-transitive if it satisfies this transitivity, as a generalization, a ring is called unilpotent-IFP if aU(R)⊆ N(R) for all a∈ N(R). We study the structures of right UN-transitive and unilpotent-IFP rings in relation to radicals, NI rings, unit-IFP rings, matrix rings and polynomial rings.

Keywords: transitivity of units, right UN-transitive ring, unilpotent-IFP ring, unit, nilpotent, nilradical, NI ring

### 1. Preliminaries

All rings considered in this article are associative with identity unless otherwise stated. Let R be a ring. The group of all units and the set of all idempotents in R are written by U(R) and I(R), respectively. A nilpotent element of R is said to be a nilpotent for simplicity. 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), and the set of all nilpotents in R are denoted by N0(R), N*(R), N*(R), and N(R), respectively. Write N(R)=N(R)\{0}. It is well-known that N0(R)N*(R)N*(R)N(R). The polynomial ring with an indeterminate x over a ring R is denoted by R[x]. We denote by n the ring of integers modulo n. Denote the n by n (n≥2) full (resp., upper triangular) matrix ring over R by Matn(R) (resp., Tn(R)). Use eij for the matrix with (i, j)-entry 1 and elsewhere 0.

A ring R is usually called reduced if N(R)=0. Due to Bell [2], a ring R is called IFP if ab=0 for a, b∈ R implies aRb=0. Both commutative rings and reduced ring are IFP clearly. There are many non-reduced commutative rings (e.g., nl for n,l2), and many noncommutative reduced rings (e.g., direct products of noncommutative domains). A ring is usually called abelian if every idempotent is central. IFP rings are abelian by a simple computation. A ring R is called NI [14] if N*(R)=N(R). It is clear that R is NI if and only if N(R) forms an ideal if and only if R/N*(R)is reduced. It is easily checked that IFP rings are NI but not conversely.

Following [11], a ring R is said to be unit-IFP if ab=0 for a, b∈ R implies aU(R)b=0. Unit-IFP rings are abelian by [11, Lemma 1.2]. IFP rings are clearly unit-IFP. The rings below show that the classes of u

Example 1.1. (1) Let K=2 and A=Ka,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. Note that the ring coproduct R=R1*KR2 with R1=K[a] and R2=K[b]K[b]b2K[b], where K[a] (resp., K[b]) is the polynomial ring with an indeterminate a (resp., b) over K. Then R is unit-IFP but not NI by [11, Example 1.1] and [1, Example 4.8].

(2) Consider Tn(R) over an NI ring R for n≥ 2. Then Tn(R) is NI by [8, Proposition 4.1] but not unit-IFP by [11, Lemma 1.2] since Tn(R) is not abelian.

### 2. Transitivity and Unilpotent-IFP Rings

In this section we introduce two kinds of ring properties through which we study the relation between units and nilpotents. The first is related to the transitivity of units on nilpotents under regular group actions and the second is related to the property of inserting units into nilpotent products of elements.

Recall first the following definitions. Let R be a ring and suppose that there exist two (left and right) regular actions of U(R) on N(R). The orbit of a∈ N(R) is or(a)={auuU(R)}=aU(R) (resp., ol(a)={uauU(R)}=U(R)a) under the right (resp., left) regular action of U(R) on N(R). We write o(a) when ol(a)=or(a).

For a ring R, U(R) shall be called right(resp., left) transitive on $N(R)$ provided that N(R)=0 or else, if there exists a ∈ N(R)' such that or(a)=N(R) (resp., ol(a)=N(R)) under the right (resp., left) regular action. If U(R) is both right and left transitive on N(R), then U(R) is said to be transitive on N(R). Observe that U(R) in a reduced ring R is transitive on N(R) by definition. However there exists a unit-IFP ring that does not satisfy the transitivity as in the part (2) of the following remark.

Remark 2.1. (1) Let R be a non-reduced ring. We claim that if U(R) is right (resp., left) transitive on N(R) then or(q)=N(R) (resp., ol(q)=N(R)) for any qN(R). Let $U(R)$ be right transitive on $N(R)$. Then or(a)=aU(R)=N(R) for some aN(R). For any qN(R), there exists u∈ U(R) such that au=q. Then a=qu1, hence it implies that or(a)=qu1U(R)=or(q). Thus or(q)=N(R) for any qN(R). The argument for the left case is done by symmetry.

(2) Consider the unit-IFP ring R in Example 1.1(1). Then U(R)={k0+k1b+bfbk0K\{0},k1K and fR} by [13, Theorem 1.2], and N(R)={kb+bfbkK and fR} with N(R)2=0 by [13, Theorem 1.3]. So or(kb+bfb)={k0kb+b(k0f)bk0K\{0}}=ol(kb+bfb)N(R) for any 0kb+bfbN(R). In fact, if k=0 then bor(kb+bfb)={b(k0f)b}; if bfb=0 then babor(kb+bfb)={k0kb}; and if k≠ 0, bfb≠ 0 then b,babor(kb+bfb)={k0kb+b(k0f)bk0k0 and b(k0f)b0)}. Thus U(R) is neither right nor left transitive on N(R).

We shall call a ring R right (resp., left) UN-transitive if U(R) is right (resp., left) transitive on N(R), and R is called UN-transitive if U(R) is transitive on N(R). We first have the following by Remark 2.1(1).

Lemma 2.2. A non-reduced ring R is right (resp., left) UN-transitive if and only if or(a)=N(R) (resp., ol(a)=N(R)) for any a ∈ N(R)'.

Given a ring R and k ≥ 1, write Nilk(R)={aRak=0}. Note N(R)=i=1Nili(R). Note that Köthe's conjecture (i.e., the sum of two nil left ideals is nil) holds for a given ring R when N(R) is additively closed.

Proposition 2.3. Let R be a right or left UN-transitive ring. Then we have the following assertions.

• (1) N(R)2=0.

• (2) N(R)=Nil2(R) and N(R) is a subring of R.

• (3) Köthe's conjecture holds for R.

Proof. (1) If N(R)=0, then we are done. Assume N(R)0. Let qN(R), say qn=0 with n2. Then N(R)=or(q) by Lemma 2.2, whence qn1=qu for some uU(R). Multiplying by q on the left, we get 0=q2u and so q2=0. This concludes N(R)=Nil2(R).

Further, we claim N(R)2=0. Let p,qN(R). Then p2=0=q2 as above. Since or(p)=N(R)=or(q) by Lemma 2.2, pu1=q and qu2=p for some u1,u2U(R). So qpu1=q2=0 and pqu2=p2=0, whence pq=0=qp.

(2) N(R)=Nil2(R) by (1). Let p,qN(R). Then (p+q)2=p2+pq+qp+q2=0 by (1), so that p+qN(R). Consequently, N(R) is a subring of R.

(3) This is evident from (2).

The proof for the left UN-transitive case is similar.

Each converse of Proposition 2.3 needs not hold by Remark 2.1(2). The rings below shall provide the motivation for the main argument in this article.

Example 2.4. (1) Let D be a division ring and R=T2(D). Then R is NI with N(R)=0D00 and U(R)={(aij)Ra11,a22U(D)}, noting U(D)=D\{0}. So, for any M=0a00N(R) (i.e., a0), we have

ol(M)=0U(D)a00=U(D)e12 and or(M)=0aU(D)00=U(D)e12.

Thus ol(M)=or(M)=N(R) and so R is UN-transitive.

(2) We follow the construction in [6, Example 1.2(2)] which applies [15, Definition 1.3]. Let A be a commutative ring with an endomorphism σ and M be an A-module. For AM, the addition and multiplication are given by (r1,m1)+(r2,m2)=(r1+r2,m1+m2) and (r1,m1)(r2,m2)=(r1r2,r1m2+m1σ(r2)). Then this construction also forms a ring.

For a field K, let K(x) be the quotient field of the polynomial ring K[x] and σ be the non-surjective monomorphism of K(x) defined by σf(x)g(x)=f(x2)g(x2). Let R=K(x)K(x) with the preceding multiplication. Note that U(R)=U(K(x))K(x)=K(x)K(x), where K(x)=K(x)\{0}. Then R is isomorphic to the subring

h(x)k(x)0σ(h(x))h(x),k(x)K(x)ofT2(K(x)),
via(h(x),k(x))h(x)k(x)0σ(h(x)),

by the argument in [6, Example 1.2(2)]. Since T2(K(x)) is clearly NI, R is NI by [8, Proposition 2.4]. Note that N(R)={0}K(x)=N*(R) and R/N*(R)K(x).

For any a=(0,f(x))N(R), we have

ol(a)=(K(x)K(x))(0,f(x))={0}K(x)f(x)={0}K(x)(=N(R))

and or(a)=(0,f(x))(K(x)K(x))={0}f(x)σ(K(x)). Here if f(x)σ(K(x)) then f(x)σ(K(x))=σ(K(x)); and if f(x)σ(K(x)) then 1f(x)σ(K(x)). Thus f(x)σ(K(x))K(x), and or(a)ol(a) follows. These also imply that R is left UN-transitive but not right UN-transitive.

(3) Let R=K(x)K(x) be the ring and σ be the non-surjective monomorphism of K[x], as in (2). Give R the multiplication (r1,m1)(r2,m2)=(r1r2,σ(r1)m2+m1r2) which is defined by [15, Definition 1.3]. Then this construction also forms a ring, and R is isomorphic to the subring

h(x)0k(x)σ(h(x))h(x),k(x)K(x)ofT2(K(x)),
via(h(x),k(x))h(x)0k(x)σ(h(x)),

by the argument in [6, Example 1.2(1)], where Tn(R) denotes the n by n lower triangular matrix ring over R. Then, by a similar argument to (2), R is an NI ring with U(R)=K(x)K(x) and N(R)={0}K(x)=N*(R).

Let a=(0,f(x)) be arbitrary in N(R)'. Then we can show that ol(a)or(a) by the symmetric computation to (2), and moreover R is right UN-transitive but not left UN-transitive on N(R).

Next we consider a generalized condition of one-sided UN-transitivity by considering "⊆", in place of "=".

Proposition 2.5. (1) For a ring R, the following conditions are equivalent:

• (i) ab ∈ N(R) for a, b ∈ R implies aU(R)b ⊆ N(R);

• (ii) a ∈ N(R) implies ras ∈ N(R) for all r, s ∈ U(R);

• (iii) ol(a) ⊆ N(R) for all a ∈ N(R);

• (iv) or(a) ⊆ N(R) for all a ∈ N(R);

• (v) If a1⋯ an ∈ N(R) for a1, …, an∈ R and n ≥ 2, then for all u1, …, un+1 ∈ U(R), u1a1u2a2⋯ unanun+1∈ N(R).

(2) Let R be a ring which satisfies any of the preceding conditions. Then u+aU(R) for all uU(R) and aN(R).

Proof. (1) (i) (ii): Assume that (i) holds. Let aN(R). Then we have the following implications: a=1aN(R)1raN(R) for all rU(R)ra=(ra)1N(R)(ra)s1N(R) for all sU(R)rasN(R).

(v) (i), (ii) (iii) and (iii) (iv) are obvious. The proof of (i) (v) is done by using the condition (i), iteratively.

(iv) (i): Assume that (iv) holds. Let abN(R) for a,bR. Then baN(R) and so or(ba)=baU(R)N(R) by assumption. Thus aU(R)bN(R), showing that the condition (i) is satisfied.

(2) Note that 1+nU(R) for all nN(R). Let uU(R) and aN(R). Then u+a=u(1+u1a) and we have u1(u+a)=1+u1a. Since u1aN(R) by (1), 1+u1aU(R). Thus u+a=u(1+u1a)U(R).

Based on the facts above, we consider the following as a generalization of not only one-sided UN-transitive rings, but unit-IFP rings and NI rings.

A ring shall be called unilpotent-IFP if it satisfies any of the conditions in Proposition 2.5(1). In this case, we will usually use the condition (i).

Every NI ring R is unilpotent-IFP since R/N*(R) is reduced, and every unit-IFP ring is also unilpotent-IFP by [11, Lemma 1.2], but each converse needs not hold by Example 1.1. Moreover it is clear that the class of unilpotent-IFP rings contains right (left) UN-transitive rings, but not conversely by Example 1.1 or Example 2.4(2, 3).

The non-unilpotent-IFP rings below provide useful manner to argue about the unilpotent-IFPness of given rings.

Example 2.6. (1) Consider Matn(A) over any ring A for n2, and take α=e21,γ=e22 and β=e21+e12+e33++enn in Matn(A). Then αγ=0 but αβγ=e22I(Matn(A)), noting βU(Matn(A)). Hence αU(Matn(A))γN(Matn(A)).

(2) Consider the rings constructed in [10, Example 2(2)]. Let K=2, the field of integers modulo 2, A=K[a], and B=K[b], where K[a] and K[b] be the polynomial rings with indeterminates a and b over K, respectively. Let J be the ideal of B generated by b2. Set C=AA and D=B/J. Identify b with its image for simplicity. Let R=C*KD which stands for the ring coproduct of C and D over K. Then 1+b is a unit and (1,0)b(0,1) is a nilpotent of R, in fact, [b(1,0)b(0,1)]2=0. But

[(1+b)(1,0)b(0,1)]2n=(1,0)b(0,1)[b(1,0)b(0,1)]2n1+[b(1,0)b(0,1)]2n,

which is not a nilpotent, for all n. Hence [b(1,0)b(0,1)]U(R)[b(1,0)b(0,1)]N(R).

Based on the structures of rings in Example 2.4, a ring R shall be called left unilpotent-duo (resp., right unilpotent-duo) if or(a)ol(a), i.e., aU(R)U(R)a (resp., ol(a)or(a), i.e., U(R)aaU(R))) for all aN(R). A ring is said to be unilpotent-duo if it is both left and right unilpotent-duo. Reduced rings are clearly unilpotent-duo. This unilpotent-duo property is not left-right symmetric as we see in Example 2.4(2, 3). In fact the ring R in Example 2.4(2) is left unilpotent-duo but not right unilpotent-duo; while, the ring R in Example 2.4(3) is right unilpotent-duo but not left unilpotent-duo.

Theorem 2.7. (1)Every right or left unilpotent-duo ring is unilpotent-IFP.

(2) Let R be a ring with an involution *. Then

• (i) R is left unilpotent-duo if and only if it is right unilpotent-duo.

• (ii) R is left UN-transitive if and only if it is right UN-transitive. Especially, if R is non-reduced right UN-transitive, there exists aN(R) such that

ol(a)=ol(a*)=or(a)=or(a*)=N(R).

(3) Let R be a non-reduced ring. If R is UN-transitive with o(a)=N(R) for some aN(R), then R is unilpotent-duo.

Proof. (1) Let R be a right unilpotent-duo ring and suppose aN(R) for aR. Then an=0 for some n1. Let uU(R). Since R is right unilpotent-duo, ua=au1 for some u1U(R) and u1ua=au2 for some u2U(R). Continuing in this manner, there exists unU(R) such that un1ua=aun where un1U(R) for n2. Then

(ua)n=ua(ua)n1=a(u1ua)(ua)n2=a2(u2ua)(ua)n3=      =an1(un1ua)=anun=0,

and hence uaN(R). Thus R is unilpotent-IFP. The proof for the left unilpotent-duo ring is similar.

2-(i) Let R be left unilpotent-duo and suppose that aN(R) and uU(R). Then a*N(R) and u*U(R). Since R is left unilpotent-duo, a*u*=va* for some vU(R). This yields

ua=((ua)*)*=(a*u*)*=(va*)*=av*aU(R),

implying that R is right unilpotent-duo. The converse is shown similarly.

2-(ii) First, if N(R)= 0 then we are done. We assume N(R)≠ 0 and let R be left UN-transitive. Then ol(a)=U(R)a=N(R) for some aN(R). Let bN(R) with b=ua for some u∈ U(R). Then a*u*=b*N(R). Since N(R)=U(R)a, b*=a*u*=va for some vU(R), and so

b=(b*)*=(va)*=a*v*a*U(R).

This implies N(R)=a*U(R)=or(a*), noting a*U(R)N(R) by Proposition 2.5. Therefore R is right UN-transitive. The converse is similar to above.

Next, suppose that R is non-reduced UN-transitive. Then, by the preceding argument, there exists aN(R) such that

U(R)a=ol(a)=N(R)=or(a*)=a*U(R).

From this, we obtain a*=ua and a=a*v for some u,vU(R); hence a=u1a* and a*=av1. Thus we also have

N(R)=ol(a)=U(R)a=U(R)u1a*=U(R)a*=ol(a*)

and

N(R)=or(a*)=a*U(R)=av1U(R)=aU(R)=or(a).

Therefore ol(a*)=ol(a)=or(a*)=or(a)=N(R).

(3) Suppose that R is UN-transitive with o(a)=N(R) for some aN(R). Consider U(R)b with bN(R) and let c=ubU(R)b. Since N(R)=o(a)=U(R)a=aU(R), b=ga=ag1 for some g,g1U(R). So we have

c=ub=uga=ag2=ag1g11g2=bg11g2bU(R),

where (ug)a=ag2 for some g2U(R) because N(R)=o(a). Thus R is right unilpotent-duo. The proof of left unilpotent-duo case is similar.

The converse of Theorem 2.7(1) does not hold in general by Example 3.8(1) to follow.

Let K be a commutative ring and G be any group. Recall the standard involution * on the group ring KG in [3], i.e., (aigi)*=aigi1 for all aiK and giG. Thus we obtain the following as a corollary of Theorem 2.7.

Corollary 2.8. Let K be a commutative ring and G be any group. Then we have the following results.

• (1) The group ring KG is left unilpotent-duo if and only if it is right unilpotent-duo.

• (2) The group ring KG is left UN-transitive if and only if it is right UN-transitive.

One may ask whether if R is a commutative ring then R is UN-transitive. However the answer is negative by the following.

Example 2.9. Consider the infinite direct product R= i=12i, and the subring S of R generated by the direct sum i=12i and the identity of R.

Let (ai)N(R) such that or((ai))=(ai)U(R)=N(R). Then (ai)k=0 for some k ≥ 2, and (b(i)j)(ai)U(R) for all i1, where b(i)i=2 and b(i)j=0 for ij. From (ai)U(R)=N(R), we obtain that [(ai)(ui)]k=(ai)k(ui)k=0 for all (ui)U(R); hence (ci)k=0 for all (ci)N(R). However (b(i)j)k0 for all ik+1, and so (b(i)j)N(R) for all ik+1. This induces a contradiction because (b(i)j)i=0. Thus there cannot exist (ai)N(R) such that or((ai))=N(R). That is, R is not UN-transitive.

Let (ai)N(S) such that or((ai))=(ai)U(S)=N(S). Then there exists h2 such that ai=0 for all ih. So, letting (dj)S such that dh=2 and dj=0 for all jh, (dj)(ai)U(S) and (dj)N(S) follows. This induces a contradiction because (dj)h=0. Thus there cannot exist (ai)N(S) such that or((ai))=N(S). That is, S is not UN-transitive.

Recall that unilpotent-IFP rings need not be NI by Example 1.1(1). We see conditions under which unilpotent-IFP rings may be NI. Note that for a ring R, if R is one-sided UN-transitive, then R is unilpotent-IFP.

Theorem 2.10. Let R be a non-reduced ring and suppose that there exists aN(R) such that aRN(R). If R is right or left UN-transitive, then R is an NI ring such that N0(R)=N*(R)=N*(R)=N(R)=aR=RaR and N*(R)2=0.

Proof. By hypothesis, aRN(R) with aN(R). Suppose that R is right UN-transitive. Then or(a)=N(R) by Lemma 2.2, whence we have

N(R)=aU(R)aRN(R),

from which we infer that aR=N(R)=aU(R){0}.

Now consider RaR. Since aR is nil, we see that rasN(R) for all r,sR. By Proposition 2.3, N(R) is a trivial subring of R (i.e., N(R)2=0) and so RaR is also nil, entailing RaR=N(R)=N*(R). Furthermore, aR=bR for any bN(R) since or(a)=N(R)=or(b) by Lemma 2.2. Consequently we now have

RaR=N(R)=aR=bR=RbR=N*(R) for any bR,

concluding that R is NI. Moreover since N(R)2=0 by Proposition 2.3(1), we see that N0(R)=N*(R)=N*(R)=N(R) and N*(R)2=0.

The proof of the left transitive case can be done by symmetry.

Regarding Theorem 2.10, it is evident that if R is an NI ring then R is a unilpotent-IFP ring such that aRN(R) for all aN(R). But Example 2.9 illuminates that there exists an NI ring which is neither right nor left UN-transitive.

Polynomial rings over NI rings need not be NI by Smoktunowicz [16]. But if given a ring R satisfies the condition of Theorem 2.10, then R[x] is NI as we see in the following.

Corollary 2.11. Let R be a non-reduced ring and suppose that there exists aN(R) such that aRN(R). If R is left or right UN-transitive, then R[x] is an NI ring such that N(R[x])=N*(R[x])=R[x]aR[x].

Proof. By hypothesis and Theorem 2.10, we first obtain that N*(R)2=0 and N0(R)=N*(R)=N*(R)=N(R)=aR=RaR. Then R[x] is an NI ring such that N(R[x])=N*(R[x])=N*(R)[x], by help of the proof of [8, Proposition 4.4]. This yields

N(R[x])=N*(R[x])=N*(R)[x]=(aR)[x]=(RaR)[x]=R[x]aR[x].

Remark 2.12. Let R be a ring and let aN(R). Then ak=0 for some k2. Assume N*(R)=aR=RaR. Then we have

N*(R)k=(aR)k=(aR)(aR¯)(aR)k2=(aaR¯)(aR)k2=  =(ak1R)(aR¯)=ak1aR¯=0.

Next argue about the actual form of elements in N*(R)k. Let biN*(R) for i=1,2,,k. Then bi=aci for some ciR. So we have

### 3. Structure of Unilpotent-IFP Rings

In this section we study the structure of unilpotent-IFP rings as well as the relations between unilpotent-IFP rings and related rings, studying the structures of some kinds of unilpotent-IFP rings which are considered ordinarily in (noncommutative) ring theory.

We first note that the class of unilpotent-IFP rings is not closed under homomorphic images, since every ring is a homomorphic image of a free ring (which is reduced and therefore unilpotent-IFP). But we obtain elementary properties for unilpotent-IFP rings as follows. The direct product of rings Ri (iΛ) is denoted by iΛRi.

Proposition 3.1. (1) If S is a subring of a unilpotent-IFP ring R with the identity of R, then S is unilpotent-IFP.

(2) Let I be a nil ideal of a ring R. Then R/I is unilpotent-IFP if and only if so is R.

(3) Let {RiiΛ} be a family of rings and R= iΛRi, where Λ is finite. Then Ri is a unilpotent-IFP ring for all iΛ if and only if R is unilpotent-IFP.

Proof. (1) Note that U(S)U(R) and N(S)=SN(R). Suppose that R is unilpotent-IFP and let aN(S) and uU(S).

Then auN(R)S=N(S), and so S is unilpotent-IFP.

(2) Let I be a nil ideal of R. Note that

N(R/I)=a+IaN(R) and U(R/I)=u+IuU(R).

For aN(R) and uU(R), au+IN(R/I) if and only if auN(R). Thus the proof can be shown easily.

(3) Since Λ is finite, N(R)= iΛN(Ri). Note U(R)= iΛU(Ri). For a=(ai)iN(R) and u=(ui)iU(R), au=(aiui)iN(R) if and only if aiuiN(Ri) for all iΛ since Λ is finite. Thus the proof can be shown easily.

We next study some sorts of unilpotent-IFP rings which are able to provide plentiful information to related studies. For a ring R and n2, let Dn(R) be the ring of all matrices in Tn(R) whose diagonal entries are all equal and Vn(R) be the ring of all matrices (aij) in Dn(R) such that ast=a(s+1)(t+1) for s=1,,n2 and t=2,,n1. Note that Vn(R) is isomorphic to R[x]/xnR[x].

Let R be any ring and n2. Then Matn(R) is not unilpotent-IFP by Example 2.6(1), and Tn(R) cannot be unit-IFP by help of [11, Lemma 1.2(2)], but we have the following.

Proposition 3.2. Let R be a ring and n ≥ 2. The following conditions are equivalent:

• (1) R is unilpotent-IFP;

• (2) Tn(R) is unilpotent-IFP;

• (3) Dn(R) is unilpotent-IFP;

• (4) Vn(R) is unilpotent-IFP.

Proof. It suffices to show (1) (2) by Proposition 3.1(1). Let R be unilpotent-IFP. Consider the nil ideal I={(aij)Tn(R)aii=0 for all i} of Tn(R). Then Tn(R)/I is isomorphic to an n-copies of R; hence Tn(R)/I is unilpotent-IFP by Proposition 3.1(3). Thus Tn(R) is unilpotent-IFP by Proposition 3.1(2).

By the same idea as in the proof of Proposition 3.2, we consider a similar proposition which also provides examples of unilpotent-IFP rings, being concerned with modules.

Proposition 3.3. Let R, S be rings and RMS be an (R, S)-bimodule. Then T=RM0S is unilpotent-IFP if and only if R and S are unilpotent-IFP.

The following is an application of Proposition 2.5(1). Let R be an algebra over a commutative ring S. Following Dorroh [4], the Dorroh extension of R by S is the Abelian group R×S with multiplication given by (s1,r1)(s2,r2)=(s1s2,s1r2+s2r1+r1r2) for riR and siS.

Proposition 3.4. Let R be an algebra with identity over a commutative reduced ring S. Then R is unilpotent-IFP if and only if so is the Dorroh extension D of R by S.

Proof. It suffices to show the necessity by help of Proposition 3.1(1). Clearly s is identified with s1 in R for all s ∈ S. Note that R={r+s(r,s)D} and N(D)=(0,N(R)) since S is a commutative reduced ring.

Suppose that R is unilpotent-IFP. Let (u,b)U(D), then $u\in U(S)$. Say (u,b)1=(u1,c). Since (u,b)(u1,c)=(1,0)=(u1,c)(u,b), uc+u1b+bc=0, and u1b+uc+cb=0. Thus (u+b)(u1+c)=1=(u1+c)(u+b) in $R$, and so u+bU(R). Now consider (0,a)N(D). Then a(u+b)N(R) since R is unilpotent-IFP, and it implies that (0,a)(u,b)=(0,a(u+b))N(D). Therefore D is unilpotent-IFP by Proposition 2.5(1).

In what follows we consider some conditions under which the set of all nilpotent elements in unilpotent-IFP rings forms a subring, which it is compared with Proposition 2.3.

Proposition 3.5. Let R be a unilpotent-IFP ring with N(R)=Nil2(R). Then we have the following results.

• (1) N(R) is a subring of R, and ab=-ba for all a,bN(R).

• (2) N(R) is a commutative subring of R, when R is of characteristic 2.

Proof. (1) Let a,bN(R)=Nil2(R). Since R is unilpotent-IFP, (ab)2=a2b+abab=a(1+b)abN(R) from a(ab)=0N(R) and 1+bU(R). Thus abN(R) and baN(R) follows. Since (ab)2=0 and (ba)2=0, (a+b)4=(ab+ba)2=abab+baba=(ab)2+(ba)2=0 and so a+bN(R). Hence N(R) forms a subring of R. Moreover, (a+b)2=0 implies that ab+ba=0.

(2) It is an immediate consequence of (1).

The following elaborates Proposition 3.5.

Example 3.6. (1) We recall the unit-IFP (and so unilpotent-IFP) ring R in Example 1.1. Then N(R)=Kb+bRb=Nil2(R) as mentioned earlier; hence N(R) is a commutative subring of R.

(2) We recall the ring R as in Example 2.6(2). Then it is obvious U(C)={(1,1)} by [10, Example 2(2)]. So we have that:

N(R)={(c,0)r(0,d),(0,e)s(f,0),btb,br,s,tR and c,d,e,fA}

and U(R)={1+kwkK and wN(R)}, entailing N(R)=Nil2(R). The characteristic of R is 2, but N(R) is not closed under multiplication as can be seen by (a,0)b(0,a)bN(R). So R is not unilpotent-IFP by Proposition 3.5. In fact, (a,0)(0,a)b=0 but (a,0)b(0,a)b=(a,0)(1+b)(0,a)bN(R) (in spite of 1+bU(R)).

Recall that Köthe's conjecture holds for a given ring R when N(R) is additively closed. So Köthe's conjecture holds for a unilpotent-IFP ring R with N(R)=Nil2(R) by Proposition 3.5 as well as for a left or right UN-transitive ring by Proposition 2.3.

Theorem 3.7. (1) A ring R is unilpotent-IFP and satisfies Köthe's conjecture if and only if R/N*(R) is a unit-IFP ring.

(2) A ring R is NI if and only if N(R) is a subring of R such that ab∈ N(R) for a, b∈ R implies a(R\N(R))bN(R).

Proof. (1) Suppose that R is unilpotent-IFP and satisfies Köthe's conjecture. Then S = R/N*(R) clearly satisfies Köthe's conjecture, and S is unilpotent-IFP by Proposition 3.1(2). Assume that ab = 0 for a, b ∈ S. Then bxaN(S) for all x∈ S. Note U(S)=u+N*(R)uU(R). Since S unilpotent-IFP, bxauN(S) for all u ∈ U(S) by Proposition 2.5(1). This yields aubxN(S), and thus aub generates a nil right ideal in S. But N*(S) = 0 and S satisfies Köthe's conjecture, so S contains no nonzero nil one-sided ideals. Therefore aub = 0, proving that S is unit-IFP.

Conversely, suppose that R/N*(R) is unit-IFP. Then it is unilpotent-IFP and satisfies Köthe's conjecture by definition and [11, Theorem 1.3(1)], respectively. So R obviously satisfies Köthe's conjecture and is unilpotent-IFP by Proposition 3.1(2).

(2) The necessity is obvious. For the converse, let abN(R) for a, b∈ R. Since N(R) is a subring of R and ba∈ N(R), we get aN(R)bN(R). Consequently we have aRbN(R) by the condition that ab∈ N(R) for a, b∈ R implies a(R\N(R))bN(R). Then R is NI by [12, Corollary 1.4].

Regarding Theorem 3.7(1), nilpotents always form a subring in a unilpotent-IFP ring satisfying Köthe's conjecture. Indeed, if R is unilpotent-IFP satisfying Köthe's conjecture, then R/N*(R) is unit-IFP, so nilpotents there form a subring, which in turn implies that nilpotents of R form a subring of R.

Consider the necessity of Theorem 3.7(2). If a ring R satisfies the condition that abN(R) for a, b∈ R implies a(R\N(R))bN(R). Then R is unilpotent-IFP. For, if the ring R above is not unilpotent-IFP, then there exist a, b∈ R such that abN(R) and aU(R)b\nsubseteq N(R). Then aubN(R) for some u∈ U(R), contrary to auba(R\N(R))bN(R). So one may ask whether if R is a unilpotent-IFP ring such that N(R) is a subring of R then R is NI. But the answer is negative by the unilpotent-IFP ring R in Example 1.1(1) that is not NI. Note that N(R) is a subring of R by [1, Theorem 4.7 and Corollary 3.3], and that bba=0, babaN(R) and babab(R\N(R))ba.

The following elaborates upon the relations among the concepts above.

Example 3.8. (1) There exists a unilpotent-IFP ring that is neither right nor left unilpotent-duo. Let R1 be the ring R in Example 2.4(2) that is left unilpotent-duo but not right unilpotent-duo; and let R2 be the ring R in Example 2.4(3) that is right unilpotent-duo but not left unilpotent-duo. Set R=R1×R2. Then R is unilpotent-IFP by Theorem 2.7(1) and Proposition 3.1(3). Note U(R)=U(R1)×U(R2) and N(R)=N(R1)×N(R2). So R is neither right nor left unilpotent-duo.

(2) There exists an IFP ring that is neither right nor left unilpotent-duo. We use the ring in [7, Example 2]. Let A=2a0,a1,a2,b0,b1,b2,c be the free algebra with noncommuting indeterminates a0,a1,a2,b0,b1,b2,c over 2; and let B={fA the constant term of f is zero}. Let I be the ideal of A generated by

a0b0,a1b2+a2b1,a0b1+a1b0,a0b2+a1b1+a2b0,a2b2,a0rb0,a2rb2,(a0+a1+a2)r(b0+b1+b2), and r1r2r3r4,

where r,r1,r2,r3,r4B; and set R=A/I. Then R is IFP by the argument in [7, Example 2]. Identify a0,a1,a2,b0,b1,b2,c with their images in R for simplicity. Note U(R)={1+ggB}=1+B since B4=0. Consider or(c) and ol(c). Then or(c)ol(c) and or(c)ol(c) because c(1+a0)ol(c) and (1+a0)cor(c) for (1+a0)U(R). For assuming c(1+a0)=(1+g1)c and (1+a0)c=c(1+g2) for some giB, we get 0ca0=g1c and 0a0c=cg2 (i.e., ca0g1c,a0ccg2I), contrary to the construction of I. Therefore R is neither right nor left unilpotent-duo.

The following diagram shows all implications among the concepts above.

In what follows we consider a condition under which the ring properties mentioned above coincide. Following [5], a ring R is said to be von Neumann regular if for each a ∈ R there exists b ∈ R such that a=aba.

Proposition 3.9. For a von Neumann regular ring R, the following conditions are equivalent:

(1) R is reduced; (2) R is IFP; (3) R is unit-IFP; (4) R is abelian; (5) R is right(left) UN-transitive; (6) R is unilpotent-IFP; (7) R is right or left unilpotent-duo.

Proof. The implications (1) (2), (1) (7), (2) (3) and (3) (6) are obvious. (3) (4), (1) (4) and (7) (6) are shown by [11, Lemma 1.2(2)], [5, Theorem 3.2] and Theorem 2.7(1), respectively.

The implications (1) (5) and (5) (6) are obvious.

(6) (4): Let R be unilpotent-IFP. Assume on the contrary that there exist e2=e,rR such that er(1e)0. Say a=er(1e). Then a2=0 and so 1-a∈ U(R). Since R is von Neumann regular, a=aba for some b∈ R. Then bab(1ab)=0. Since R is unilpotent-IFP, bab(1a)(1ab)N(R) because 1aU(R). But

bab(1a)(1ab)=(babba)(1ab)=ba+ba2b=baN(R),

contrary to bab(1a)(1ab)N(R). Thus R is abelian.

Recall that a ring R is said to be directly finite (or Dedekind finite) if ab=1 implies ba=1 for a, b ∈ R. It is well-known that abelian rings are directly finite. NI rings are directly finite by [8, Proposition 2.7(1)]. We also obtain this result as a corollary of the following.

Proposition 3.10. Every unilpotent-IFP ring is directly finite.

Proof. Let R be a unilpotent-IFP ring and assume on the contrary that R is not directly finite. Then ab=1 and ba1 for some a, b∈ R. In what follows we refer to the argument for one-sided inverses in [9, page 1]. Consider x=1-ba and y=bb2a=b(1ba). Then xy=0. Since z=(1ba)aN(R), 1+zU(R). Then x(1+z)y=xzyN(R) because R is unilpotent-IFP. But

xzy=(1ba)((1ba)a)(b(1ba))=1baN(R),

contrary to xzyN(R). Thus R is directly finite.

The converse of Proposition 3.10 needs not hold as can be seen by Matn(D), over a division ring D for n ≥ 2, which is Artinian (hence directly finite) but not unilpotent-IFP by Example 2.6(1).

### Acknowledgments.

The authors thank the referee for very careful reading of the manuscript and many valuable suggestions that improved the paper by much.

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