When is every non central-unit a sum of two nilpotents?
aa r X i v : . [ m a t h . R A ] J a n WHEN IS EVERY NON CENTRAL-UNIT A SUM OF TWONILPOTENTS?
SIMION BREAZ AND YIQIANG ZHOU
Abstract.
Motivated by the notions of 2-good rings and fine rings, we definea 2-nilgood ring as a ring whose every non central-unit is the sum of two nilpo-tents. This is a study of 2-nilgood rings. The following results are obtained:(a) a ring is 2-nilgood iff it is either a 2-nilgood simple ring or a commutativelocal ring with nil Jacobson radical; (b) a 2-nilgood simple ring is a field ifit is right Goldie; (c) a 2-nilgood ring containing a uniform right ideal is acommutative local ring with nil Jacobson radical; (d) a ring whose every noncentral-unit is a sum of a square-zero element and a nilpotent is a commutativelocal ring with nil Jacobson radical. We conclude with the conjecture that a2-nilgood simple ring is a field. Introduction
Throughout, rings are associative with identity. We start by recalling threespecial types of elements in a ring. An element in a ring is 2-good if it is a sumof two units (see [19]), is fine if it a sum of a unit and a nilpotent (see [6]), andis 2-nilgood if it is a sum of two nilpotents (see [5]). By the terminology ofV´amos [19], a ring is 2-good if each of its elements is 2-good. The study of 2-good rings (also called rings with the 2-sum property in the literature), initiatedby Wolfson [21] and Zelinsky [22], has attracted considerable interest (see, forexample, [2],[14], [18], [19] and the references there). A ring is called a fine ringif each nonzero element is fine. The zero element is excluded because the zero Mathematics Subject Classification.
Key words and phrases.
Nilpotent, unit, 2-nilgood-ring, matrix ring, simple ring. This terminology should not be confused with the notion of a 2-nil-good element in [M.S.Abdolyousefi, N. Ashrafi and H. Chen, On 2-nil-good rings,
J. Algebra Appl. (6) (2018),1850110, 13 pp.] element in a ring being fine implies that the ring is a trivial one. Fine rings wereintroduced and extensively investigated by C˘alug˘areanu and Lam [6].We will say that an element in a ring is 2 -nilgood if it is a sum of two nilpotentelements. One easily sees that if a central unit in a ring is 2-nilgood, then thering must be trivial. Therefore, we define a ring to be a 2 -nilgood ring if everyelement that is not a central unit is 2-nilgood. For instance, any field is a 2-nilgood ring, and the converse holds for any reduced ring (i.e., a ring withoutnonzero nilpotents). The problem which asks to identify the additive subgroupof a ring R which is generated by all nilpotent elements of R has a long history. Itis also connected to the question which asks to study the subgroup generated bythe commutator elements of the ring. We refer to [7] and [11] for more details andreferences. In particular, Harris proved that if the additive group of a divisionring D is generated by commutators then every quadratic matrix over D is a sumon nilpotent elements, [11]. The techniques from this paper were used in [4] toprove that every endomorphism of an infinitely generated free module is a sumof four square-zero endomorphisms. For other results about sums of square-zeroendomorphisms we refer to [17] and [20]. Also, in [15] the author provides anexample of a family of division rings D such that all 3 × D aresums of the form X + Y + Z with X = Y = Z = 0. These examples are usedto prove that any unital algebra can be embedded as a unital subring of a simpleunital algebra which is a sum of three nilpotent subalgebras of degree 3. Thisimprove Bokut’s Theorem proved in [3].Here a study of 2-nilgood rings is carried out as motivated by the work on2-good rings and fine rings. In Section 2, a reduction theorem is obtained: a ringis 2-nilgood iff it is either a 2-nilgood simple ring or a commutative local ringwith nil Jacobson radical. In Section 3, it is proved that a 2-nilgood simple ringis a field if it is right Goldie and that a 2-nilgood ring containing a uniform rightideal is a commutative local ring with nil Jacobson radical. In Section 4, we showthat a ring whose every non central-unit is a sum of a square-zero element and -NILGOOD RINGS 3 a nilpotent is a commutative local ring with nil Jacobson radical and concludewith the conjecture that every 2-nilgood simple ring is a field.For a ring R , we denote by C ( R ), J ( R ) and nil( R ) the center, the Jacobsonradical and the set of nilpotents of R , respectively. We write M n ( R ) for the ringof n × n matrices over R whose identity is denoted by I n .2. The reduction
Following [5], an element in a ring is called 2-nilgood if it is a sum of 2 nilpo-tents. Since a central unit in a non-trivial ring can not be 2-nilgood, the nextdefinition is motivated.
Definition 2.1.
A ring is called a 2-nilgood ring if every non central-unit is 2-nilgood.Here a non central-unit means an element that is not a central unit. Examplesof 2-nilgood rings include: fields and commutative local rings with nil Jacobsonradical. A reduced ring is 2-nilgood iff it is a field.
Lemma 2.2.
Let R be a -nilgood ring. The following hold: (1) R is indecomposable. (2) J ( R ) is the unique largest proper ideal of R . (3) J ( R ) is nil and J ( R ) ⊆ C ( R ) . (4) C ( R ) is a local ring with nil Jacobson radical. (5) For any ideal I of R , central units of R/I can be lifted to central units of R . (6) R/I is -nilgood for every ideal I of R .Proof. (1) If e = 1 is a central idempotent, then e = b + b , where b , b ∈ nil( R ).Since e is centra, b and b commute. So b + b ∈ nil( R ), and hence e = 0.(2) Let I be an proper ideal of R . If a ∈ I is not contained in J ( R ), then1 + ar is not a unit for some r ∈ R , so 1 + ar = b + b where b , b are nilpotent. BREAZ AND ZHOU
Then, in
R/I , ¯1 = ¯ b + ¯ b , a sum of two nilpotents. This is a contradiction. So I ⊆ J ( R ). Hence (2) holds.(3) Let j ∈ J ( R ). Assume that j / ∈ C ( R ). Then 1 + j = b + b , where b , b ∈ nil( R ). Then, in R/J ( R ), ¯1 = ¯ b + ¯ b , a sum of two nilpotents. This is acontradiction. So J ( R ) ⊆ C ( R ). Thus, each j ∈ J ( R ) is a sum of two commutingnilpotents, so j is nilpotent.(4) Every nonunit a in C ( R ) is a nonunit in R , so is a sum of two commutingnilpotents. Hence a is nilpotent. So (4) holds.(5) Let ¯ u := u + I be a central unit of R/I . We claim that u ∈ R is a centralunit. Otherwise, u is a sum of two nilpotents in R . Hence ¯ u is a sum of twonilpotents in R/I , a contradiction.(6) If ¯ a := a + I is a non central-unit of R/I , then a is a non central-unit of R . So a is a sum of two nilpotents in R . Hence ¯ a is a sum of two nilpotents in R/I . (cid:3) Corollary 2.3. If R is a -nilgood ring and nil( R ) is an ideal, then R is acommutative local ring with J ( R ) nil.Proof. The assumption implies that all elements in R \ nil( R ) are central units.Moreover, nil( R ) = J ( R ) by Lemma 2.2(3). (cid:3) The next result reduces the study of 2-nilgood rings to that of 2-nilgood simplerings.
Theorem 2.4. ( Reduction Theorem ) A ring R is -nilgood iff R is either a -nilgood simple ring or a commutative local ring with nil Jacobson radical.Proof. The sufficiency is clear. For the necessity, suppose that R is a 2-nilgoodring that is not simple. Then, by Lemma 2.2(2), J ( R ) = 0. Let 0 = x ∈ J ( R ). If a, b ∈ R then, noting J ( R ) ⊆ C ( R ) by Lemma 2.2(3), we have( ab ) x = a ( bx ) = ( bx ) a = b ( xa ) = b ( ax ) = ( ba ) x, -NILGOOD RINGS 5 hence ( ab − ba ) x = 0. It follows that ab − ba belongs to the ideal Ann ( x ) = { r ∈ R | rx = 0 } . Since Ann ( x ) = R , Ann ( x ) ⊆ J ( R ) by Lemma 2.2(2). Itfollows that R/J ( R ) is commutative. Since R/J ( R ) is simple by Lemma 2.2, itis a field. Thus, by Lemma 2.2(5), for any a ∈ R \ J ( R ), a = u + j where u is acentral unit and j ∈ J ( R ); so a is central. Hence, R is commutative, and so R isa commutative local ring with J ( R ) nil by Lemma 2.2(4) or Corollary 2.3. (cid:3) Remark 2.5.
In [6] an element of a ring R is called fine if it is sum of a unit anda nilpotent, and it is proved in [6, Theorem 3.2] that if all non-zero elements of R are fine then R is simple. Almost all elements in a 2-nilgood ring R are fine:: if x − ∈ R is not a central unit of R then then exist two nilpotents n and n suchthat x = (1 + n ) + n , hence x is fine. However, each central nilpotent elementis not fine. 3. Simple rings
In light of Theorem 2.4, the study of 2-nilgood rings is reduced to 2-nilgoodsimple rings. An important step is to determine simple artinian rings that are2-nilgood. For an n × n nilpotent matrix B over a field F , A is similar to a strictupper triangular matrix, so tr( A ) = 0 because the trace is a similarity-invariant. (cid:0) The characteristic polynomial of A being x n also implies that tr( A ) = 0. (cid:1) Thus,a 2-nilgood matrix in M n ( F ) ( n ≥
2) must have trace 0, and hence M n ( F ) is not2-nilgood. This argument does not work for matrix rings over a division ring,because the trace is no longer a similarity-invariant in this situation. Example 3.1.
Let A = (cid:18) i j − j i (cid:19) ∈ M ( H ) , where H is the ring of real quater-nions. Then A = 0 and (cid:18) k (cid:19) A (cid:18) − k (cid:19) = (cid:18) j (cid:19) . But tr( A ) = 2 i = . We did not see a reference for the following well-known result. A simple proofis given for the reader’s convenience.
Lemma 3.2.
Let D be a division ring. If A ∈ M n ( D ) is nilpotent, then A n = 0 . BREAZ AND ZHOU
Proof.
We identify A with a linear transformation of the right vector space V := D n over D . Then the chain Im ( A ) ⊃ Im ( A ) ⊃ Im ( A ) ⊃ · · · should be strictly decreasing (otherwise, A is not nilpotent). Since dim( V D ) = n ,it follows that A n = 0. (cid:3) Lemma 3.3.
Let R be a ring and n ≥ . If A = a · · · · · · a n . . . . . . .... . . . . . ... a nn ∈ M n ( R ) ,then A k = ∗ ∗ P ki =1 a ii · · · · · · · · · . . . . . . .... . . . . . ... P ni = n − k +1 a ii ∗ , where the lower left block has size ( n − k + 1) × ( n − k + 1) . So the ( k, -entryof A k is a + . . . + a kk for k = 1 , . . . , n ; in particular, the ( n, -entry of A n is tr( A ) .Proof. The claim can be proved by induction on k . (cid:3) With the help of Lemma 3.2 and Lemma 3.3, we can show the following keylemma.
Lemma 3.4.
Let D be a division ring and n ≥ . If A = (cid:18) a α (cid:19) ∈ M n ( D ) where = a ∈ D , then A is not a sum of two nilpotents.Proof. We prove the claim by induction on n . The claim is true if n = 1. Let n > k with 1 ≤ k < n . Let A = (cid:18) a α (cid:19) ∈ M n ( D ) where 0 = a ∈ D , and assume to the contrary that A = X + Y is a sum of two nilpotents in M n ( D ). We next show that there is acontradiction. -NILGOOD RINGS 7 Write X = (cid:18) x γ β X (cid:19) in block form where X is an ( n − × ( n −
1) matrix.Note that β = 0; otherwise, a = x + ( a − x ) is a sum of two nilpotents in D ,a contradiction. So, there exists an invertible matrix V in M n − ( D ) such that V β = . Then U = (cid:18) V (cid:19) is invertible in M n ( D ) and U AU − = U XU − + U Y U − , where U AU − = (cid:18) a αV − (cid:19) and U XU − = (cid:18) x γ V − V β V X V − (cid:19) . Replacing A by U AU − and X by U XU − , we can assume that the first column of X is x , and write X = (cid:18) x x x (cid:19) γ β X , where X is an ( n − × ( n − β = 0; otherwise A ∈ M ( D ) is a sum of two nilpotents, acontradiction. So, there exists an invertible matrix V in M n − ( D ) such that V β = . Then U = (cid:18) I V (cid:19) is invertible in M n ( D ) and U AU − = U XU − + U Y U − , where U AU − = (cid:18) a ∗ (cid:19) and U XU − = (cid:18) x x x (cid:19) γ V − V β V A V − . Replacing A by U AU − and X by U XU − , we can assume that the block consisting ofthe first two columns of X is V β = x x x . Continuing this process, wecan find an ( n − × ( n −
1) invertible matrix V such that, with U = (cid:18) V (cid:19) , BREAZ AND ZHOU
U AU − = (cid:18) a (cid:0) a · · · a n (cid:1) (cid:19) (for some a , a , . . . , a n in D ) and U XU − = x · · · · · · x n x nn . Thus, U AU − = U XU − + U Y U − , where U Y U − = − − a + x x − a · · · x n − a n x · · · x n . . . . . . ...1 x nn By Lemma 3.2, (
U XU − ) n = ( U Y U − ) n = 0 . So, by Lemma 3.3, P ni =1 x ii = tr( U XU − ) = 0 and − a = − a + P ni =1 x ii =tr( − U Y U − ) = 0. This is a contradiction. (cid:3) Since a 2-nilgood division ring is a field, Theorem 3.5 below follows from Lemma3.4.
Theorem 3.5.
A simple artinian ring is -nilgood iff it is a field. Remark 3.6.
It was proved in [15] that there exists a division ring D such thatthe ring M ( D ) of all 3 × D are sums of three nilpotent matrices. Corollary 3.7.
Let V D be a non-trivial vector space over a division ring D . Then End( V D ) is -nilgood iff dim( V ) = 1 and D is a field.Proof. The sufficiency is clear. For the necessity, the assumption ensures thatEnd( V D ) is a simple ring by Theorem 2.4, as J (End( V D )) = 0, and so dim( V ) < ∞ . It follows from Theorem 3.5 that dim( V ) = 1 and D is a field. (cid:3) A ring is semipotent if every nonzero right (or left) ideal not contained inthe Jacobson radical contains a nonzero idempotent. A ring R is said to be of -NILGOOD RINGS 9 bounded index (of nilpotence) if there is a positive integer n such that a n = 0 forall nilpotent elements a of R . Corollary 3.8.
The following statements hold: (1) If R is a commutative ring and n ≥ , then M n ( R ) is not -nilgood. (2) Let R be a semipotent ring of bounded index or a semilocal ring. Then R is -nilgood iff R is a commutative local ring with J ( R ) nil.Proof. (1) Let I be a maximal ideal of R . Then R/I is a field, so M n ( R/I ) is not2-nilgood by Theorem 3.5. Hence, M n ( R ) is not 2-nilgood by Lemma 2.2.(2) We only need to show the necessity, so let R be a 2-nilgood semipotent ringof bounded index or a 2-nilgood semilocal ring. We may assume that R is simpleby Theorem 2.4. If R is semilocal, then R is simple artinian. If R is semipotentof bounded index, then R is also simple artinian by [1, Corollary 6]. So R is afield by Theorem 3.5. (cid:3) The Weyl algebra is an example of a simple ring that is not right Artinian, butit is right Noetherian. We next determine simple right Noetherian rings that are2-nilgood. Lemma 3.4 can be used to prove the next result. A ring is called aright Goldie ring if it satisfies the ascending chain condition on right annihilatorsand contains no infinite direct sums of right ideals. For i, j ∈ { , , . . . , n } , wedenote by E ij the n × n matrix whose ( i, j )-entry is 1 and all other entries are 0. Theorem 3.9.
A prime right Goldie ring is -nilgood iff it is a field.Proof. The sufficiency is clear.For the necessity, let R be a prime right Goldie ring that is 2-nilgood. ByGoldie’s Theorem, R is a right order of M n ( D ) where n ≥ D is a divisionring. To show that R is a field, it suffices to show that n = 1. Indeed, if n = 1,then R is a subring of a division ring. Since nil( R ) = 0, the only non central-unitof R is 0, so R is a field. Assume to the contrary that n ≥
2. First note that if r := · · · · · · u i · · · u in · · · · · · ∈ R ∩ M n ( D ) with u ii = 0, then, in M n ( D ), r is similar to r ′ = (cid:18) u ii ∗ (cid:19) , which isnot 2-nilgood in M n ( D ) by Lemma 3.4; so r is not 2-nilgood in M n ( D ) and hence r is not 2-nilgood in R . As r is a non central-unit of R , R is not 2-nilgood. So,in our discussion below, we assume that R does not contain matrices like r .For 1 ≤ k ≤ n , write E kk = r k s − k where r k , s k ∈ R and where r k = ( r ( k ) ij ) and s k = ( s ( k ) ij ) are matrices over D . Since s k is invertible in M n ( D ), row k of s k isnonzero. We now have r = E s = s (1)11 · · · s (1)1 n · · · · · · ,r = E s = · · · s (2)21 · · · s (2)2 n · · · · · · , · · · , and r n = E nn s n = · · · · · · s ( n ) n · · · s ( n ) nn . By our assumption above, s (1)11 = s (2)22 = · · · = s ( n ) nn = 0. -NILGOOD RINGS 11 Note that 0 = r ∈ R ∩ M n ( D ). Let i be the minimal number such that thereis a matrix a i := (cid:18) ,i − (cid:0) a ∗ (cid:1) (cid:19) ∈ R for some 0 = a ∈ D and, of course, 1 < i ≤ n . For any matrix r in R ∩ M n ( D ) (including r i ) whose only nonzero row is row i , which is denoted by (cid:0) x · · · x n (cid:1) , we have a i r = ax · · · ax n · · · · · · ∈ R , so the minimality of i ensures that x = · · · = x i − = x i = 0 (note that x i is the ( i , i )-entry of r , somust be 0). Now0 = r i = n,i i − ,n − i (cid:16) s ( i ) i i +1 · · · s ( i ) i n (cid:17) n − i ,n − i ∈ R ∩ M n ( D ) . Let i be the minimal number such that there is a matrix a i i := i ,i − (cid:18) i − , i − ,n − i a ∗ (cid:19) n − i ,i − n − i ,n − i +1 ∈ R for some 0 = a ∈ D , and we see i < i . For any matrix r in R ∩ M n ( D ) (including r i ) whose only nonzero row is row i , which is denoted by (cid:0) x · · · x n (cid:1) , we have a i i r = · · · · · · ax · · · ax n · · · · · · ∈ R , so the minimality of i ensures that x = · · · = x i − = x i = 0 (note that x i is the ( i , i )-entry of r , so must be 0). Let i bethe minimal number such that there is a matrix a i i := i ,i − (cid:18) i − , i − ,n − i a ∗ (cid:19) n − i ,i − n − i ,n − i +1 ∈ R for some 0 = a ∈ D , and we see i < i . An induction argument shows that thereexists a series of integers 1 < i < i < · · · < i k < i k +1 = n where i k +1 = n is theminimal integer such that there is a matrix a i k n := i k ,n − (cid:18) i k − , a (cid:19) n − i k ,n − n − i k , ∈ R for some 0 = a ∈ D . We have a i k n r n = · · · · · · as ( i n ) i n · · · as ( i n ) i n n ∈ R, and the minimality of i k +1 shows that s ( i n ) i n = · · · = s ( i n ) i n i n = 0. That is, row n of s n is a zero row. This is a contradiction. (cid:3) Corollary 3.10.
A right noetherian simple ring is -nilgood iff it is a field.Proof. It follows from Theorem 3.9. (cid:3)
A nonzero right ideal I of a ring R is called a uniform right ideal if the inter-section of any two nonzero right ideals contained in I is nonzero. Corollary 3.11.
Suppose that R contains a uniform right ideal. If R is -nilgood,then R a commutative local ring with nil Jacobson radical.Proof. We may assume that R is simple by Theorem 2.4. It follows from [10]that R contains no infinite direct sums of right ideals (i.e., R R is of finite uniformdimension). Moreover, from the last remark of [12] it follows that the maximalright ring of quotients of R is simple artinain. Thus, R is right Goldie by [13,Proposition 13.41], and hence R is a field by Theorem 3.9. (cid:3) Some special cases and a conjecture
Definition 4.1.
Let p, q be integers with p ≥ q ≥
2. A ring R is called a2-nilgood ring of ( p, q )-type if, for each non central-unit a in R , a = b + c where -NILGOOD RINGS 13 b p = c q = 0, and a 2-nilgood ring of ( p, ∞ )-type if, for each non central-unit a in R , a = b + c where b p = 0 and c is nilpotent. Lemma 4.2.
Let R be a -nilgood ring of (2 , ∞ ) -type. Then R has only thetrivial idempotents.Proof. By the hypothesis, every 1 = e = e ∈ R is a sum of a square-zero elementand a nilpotent. So, e = 0 by [8, Proposition 2]. (cid:3) Theorem 4.3.
A simple ring R is -nilgood ring of (2 , ∞ ) -type iff R is a field.Proof. We only need to show the necessity. Since R is a simple ring, 0 is the onlycentral nilpotent. Assume to the contrary that R is not a field. Then, since R is 2-nilgood, nil( R ) = 0. Hence, there exists a square-zero element x that is notcentral. Since R is of (2 , ∞ )-type, there exist y, z ∈ R such that1 = x + y + z, where y = 0 and z n = 0 for some n ≥
1. Then (1 − x − y ) n = 0. It follows that(4.1) 1 = n ( x + y ) + n ( xy + yx ) + n ( xyx + yxy ) + n ( xyxy + yxyx ) + · · · , where n = n and each n i is an integer. Multiplying both sides of (4.1) from theleft by x gives x = n xy + n xyx + n xyxy + n xyxyx + · · · (4.2)Multiplying both sides of (4.2) from the right by y gives xy = n xyxy + n xyxyxy + · · · = a ( xy ) = ( xy ) a, where a = n + n xy + · · · . It follows that a ( xy ) is an idempotent. Since x = 0, a ( xy ) = 1. So, by Lemma 4.2, a ( xy ) = 0. Hence xy = 0. It follows from (4.2)that x = 0. So 1 = y + z , a contradiction. (cid:3) Corollary 4.4.
A ring R is -nilgood of (2 , ∞ ) -type iff R is a commutative localring with J ( R ) nil. Proof.
It follows from Theorem 2.4 and Theorem 4.3. (cid:3)
Lemma 4.5. [16, Proposition 9]
The equality x + y + z where x = y = z = 0 does not hold for any K -algebra A of characteristic with unit and x, y, z ∈ A . Proposition 4.6.
Let R be a simple ring of characteristic . If R is a -nilgoodring of (3 , q ) -type with ≤ q ≤ , then R is a field.Proof. Since R is a simple ring, 0 is the only central nilpotent. Assume to thecontrary that R is not a field. Then, since R is 2-nilgood, nil( R ) = 0. Hence,there exists a square-zero element x that is not central. Since R is of (3 , q )-type,there exist y, z ∈ R such that 1 − x = y + z and y = z = 0. This a contradictionto Lemma 4.9. (cid:3) Every 2-nilgood ring we know so far is a commutative local ring with nil Ja-cobson radical. We make the following conjecture.
Conjecture 4.7.
A ring is -nilgood iff it is a commutative local ring with nilJacobson radical. That is, a simple ring is -nilgood iff it is a field. Acknowledgment
Simion Breaz was supported by the UEFISCDI grant PN-III-P4-ID-PCE-2020-0454.Yiqiang Zhou was supported by a Discovery Grant (RGPIN-2016-04706) fromNSERC of Canada.
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Email address : [email protected] Department of Mathematics and Statistics, Memorial University of New-foundland, St.John’s, NL A1C 5S7, Canada
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