The strong nil-cleanness of semigroup rings

Lemma 2.1

[8, Theorem 3.4.1] Let M 1 = ℳ 0 ( G ; I , Λ ; P ) and M 2 = ℳ 0 ( G ; I , Λ ; P ) be two completely 0-simple semigroups. Then M 1 ≅ M 2 if and only if there exist a Λ × Λ diagonal matrix X over G 0 and an I × I diagonal matrix Y over G 0 such that P = X Q Y .

Lemma 2.2

[11, Lemma 2.15(ii)] Let U be a ring and a ∈ U . Then a ∈ J ( U ) if and only if Ua is a left quasi-regular left ideal of U.

Lemma 3.1

[1, Proposition 2.9] A left (resp., right) ideal of a strongly nil-clean ring is also strongly nil-clean.

Lemma 3.2

[2] A ring U is strongly nil-clean iff U / J ( U ) is boolean and J ( U ) is nil.

Lemma 3.3

[12] Let M = ℳ 0 ( G ; I , Λ ; P ) be a completely 0-simple semigroup. Then

Lemma 3.4

Let U be a ring and f be any idempotent in U. If U is strongly nil-clean, then the ring fUf is also strongly nil-clean.

Proof

Since Uf is a left ideal of U and fUf is a right ideal of Uf, we apply Lemma 3.1 to obtain first that Uf is a strongly nil-clean ring, and then that fUf is a strongly nil-clean ring. The lemma is proved.□

Lemma 3.5

Let M be a completely 0-simple semigroup with exactly two ℛ -classes and exactly two ℒ -classes. Then its contracted semigroup ring R 0 [ M ] is not strongly nil-clean.

Proof

By hypothesis, the completely 0-simple semigroup M is of the form ℳ 0 ( G ; 2 , 2 ; P ) , where G is a group, 2 = { 1 , 2 } and P is a 2 × 2 -matrix over G 0 . Since M is a regular semigroup, by Lemma 2.1, the sandwich matrix P must be one of the matrices in the set Δ , N 1 , N 1 ′ , N 2 , N 2 ′ , N 3 , up to isomorphic, where

and where e is the identity of the group G and g ∈ G . We claim that the contracted semigroup ring R 0 [ M ] is not strongly nil-clean, no matter which of the above normalized matrix P is chosen. There are five cases to be considered. For convenience, let ℱ denote the set consisting of all the fields of characteristic 2.

Case 1. P = Δ . Then R 0 [ M ] ≅ M ( R [ G ] ; 2 , 2 ; P ) ≅ M 2 ( R [ G ] ) , and thus R 0 [ M ] is not strongly nil-clean, see [3].

Case 2. P = N 1 , P = N 1 ′ , P = N 2 or P = N 2 ′ . It is clear that P is an invertible matrix, and thus the map R 0 [ M ] → M 2 ( R [ G ] ) defined by A ↦ A P is an isomorphism of rings. Then, by Case 1, R 0 [ M ] is not strongly nil-clean.

Case 3. P = N 3 with g ≠ e , and R ∉ ℱ . Noting that each element in the Munn ring R 0 [ M ] = M ( R [ G ] ; 2 , 2 ; P ) is of the form a b c d , where a , b , c , d ∈ R [ G ] . By Lemma 3.3, it is easy to verify that

By contrary, suppose that R 0 [ M ] is strongly nil-clean. This, together with Lemma 3.2, yields that R 0 [ M ] / J ( R 0 [ M ] ) is boolean. Let c be an arbitrary element of R [ G ] , define N = 0 0 0 c ∈ R 0 [ M ] . Then N ∘ N + J ( R 0 [ M ] ) = ( N + J ( R 0 [ M ] ) ) 2 = N + J ( R 0 [ M ] ) , which is equivalent to say that N ∘ N − N ∈ J ( R 0 [ M ] ) . Note that N ∘ N = N N 3 N = 0 0 0 c g c , hence N ∘ N − N = 0 0 0 c g c − c . It follows that ( c g c − c ) g ∈ J ( R [ G ] ) by (3.1). Next, we show that J ( R [ G ] ) is nil. For this, note that the group ring R [ G ] ≅ e 0 0 0 R 0 [ M ] e 0 0 0 is a corner subring of R 0 [ M ] . Since R 0 [ M ] is strongly nil-clean by assumption, R [ G ] is strongly nil-clean by Lemma 3.4. Then it follows from Lemma 3.2 that J ( R [ G ] ) is nil. Since g is an invertible element in G, we deduce that ( c g − e ) c is a nilpotent element. Since c is arbitrary, we can take c = − g − 1 , it follows that − 2 g − 1 ∈ R [ G ] is nilpotent. This, together with the fact R ∉ ℱ , implies that the element g − 1 ∈ G is nilpotent, that is, ( g − 1 ) m = 0 for some positive integer m, which is a contradiction. Consequently, R 0 [ M ] is not strongly nil-clean, as required.

Case 4. P = N 3 and g = e , and R ∉ ℱ . By Case 3, if J ( R [ G ] ) is not nil, then R 0 [ M ] is not strong nil-clean. Suppose now that J ( R [ G ] ) is nil. We claim that R 0 [ M ] is not strongly nil-clean. By Lemma 3.2, it is sufficient to prove that R 0 [ M ] / J ( R 0 [ M ] ) is not boolean. Noting that, by Lemma 3.3, we have

Let c , d be two arbitrary elements of R [ G ] . Define N = 0 0 c d ∈ R 0 [ M ] . Then

thus

By contrary, assume that N ∘ N − N ∈ J ( R 0 [ M ] ) . It follows that ( c + d ) ( c + d − e ) ∈ J ( R [ G ] ) . If we take c + d = − e , then 2 e = ( c + d ) ( c + d − e ) ∈ J ( R [ G ] ) . Since J ( R [ G ] ) is nil, there exists a positive integer n such that 2 n e = 0 , which is a contradiction, because R ∉ ℱ . Therefore, R 0 [ M ] is not strongly nil-clean.

Case 5. P = N 3 and R ∈ ℱ . By contrary, suppose that R 0 [ M ] is a strongly nil-clean ring. By similar arguments to the above, we infer that J ( R [ G ] ) is nil, and R 0 [ M ] / J ( R 0 [ M ] ) is boolean. It is easy to check that N 2 ∘ N 2 − N 2 = 0 e + g e e + g . Since char R = 2 , we have that e + g + ( e + g ) + e = e , which does not belong to J ( R [ G ] ) . This, together with (3.2), implies that N 2 + J ( R 0 [ M ] ) is not an idempotent in R 0 [ M ] / J ( R 0 [ M ] ) . This is a contradiction, because R 0 [ M ] / J ( R 0 [ M ] ) is boolean. Therefore, R 0 [ M ] is not strongly nil-clean.

Consequently, in either case, the contracted semigroup ring R 0 [ M ] is not strongly nil-clean, as required.□

Proposition 3.6

Let M = ℳ 0 ( G ; I , Λ ; P ) be a completely 0-simple semigroup. If R 0 [ M ] is strongly nil-clean, then either | I | = 1 or | Λ | = 1 .

Proof

Suppose that R 0 [ M ] is strongly nil-clean. Noting that | I | ≥ 1 and | Λ | ≥ 1 . Then in order to show either | I | = 1 or | Λ | = 1 , assume on the contrary that | I | ≥ 2 and | Λ | ≥ 2 . It would then follow that the Λ × I -matrix P may be written in the block form P 1 P 2 P 3 P 4 , where P 1 is a 2 × 2 -matrix over G 0 . Define a subset A = ( a i λ ) ∈ R 0 [ M ]   |   a i λ = 0 for all i ≠ 1 , 2 of R 0 [ M ] . It is clear that A is closed under multiplication. Moreover, let ( a i λ ) ∈ A and ( b j μ ) ∈ R 0 [ M ] , then the entry in the ( i , μ ) -position of the product ( a i λ ) ∘ ( b j μ ) is ∑ λ ∈ Λ , j ∈ I a i λ p λ j b j μ . Note that ∑ λ ∈ Λ , j ∈ I a i λ p λ j b j μ = 0 for all i ∉ { 1 , 2 } because a i λ = 0 for all such i. It follows that ( a i λ ) ∘ ( b j μ ) ∈ A , whence A is a right ideal of R 0 [ M ] . Since R 0 [ M ] is strongly nil-clean, A is strongly nil-clean by Lemma 3.1. Moreover, A is isomorphic to the Munn ring M R [ G ] ; 2 , Λ ; P 1 P 2 . Then a similar argument shows that the Munn ring U = M ( R [ G ] ; 2 , 2 ; P 1 ) is isomorphic to a left ideal of A, and hence U is strongly nil-clean by Lemma 3.1 again. This is a contradiction, because U ≅ R 0 [ ℳ 0 ( G ; 2 , 2 ; P 1 ) ] is not strongly nil-clean by Lemma 3.5. Therefore, we must have either | I | = 1 or | Λ | = 1 .□

Lemma 3.7

Let M = ℳ 0 ( G ; I , Λ ; P ) be a completely 0-simple semigroup. If | I | = 1 or | Λ | = 1 , then R 0 [ M ] is strongly nil-clean if and only if R [ G ] is strongly nil-clean.

Proof

It is sufficient to consider the case of | Λ | = 1 , since the case of | I | = 1 is dual. Assume that | Λ | = 1 . Before moving on, we explore the structure of M and determine J ( R 0 [ M ] ) . For this, since M is a completely 0-simple semigroup, S is regular, whence every ℛ -class of M contains at least one idempotent. Since M has only one ℒ -class, it follows that each ℛ -class of M is a maximal group in M, and hence the 1 × I -matrix P must be of the form [ e , e , … , e , … ] whose entries are all equal to e, where e is the identity of the group G. Denote N ∗ by the set { 1 , 2 , … } consisting of all positive integers. Let x = x 1 , x 2 , … , x n , 0 , … T ∈ R 0 [ M ] with x 1 , x 2 , … , x n ∈ R [ G ] and whose other entries are all equal to 0, where n ∈ N ∗ . Then it is easy to check that P x P = ∑ i = 1 n x i e , e , … , e , … . Thus, by Lemma 3.3,

Now we suppose that R [ G ] is strongly nil-clean. By Lemma 3.2, R [ G ] / J ( R [ G ] ) is boolean and J ( R [ G ] ) is nil. We want to show that R 0 [ M ] is strongly nil-clean. It suffices to prove that J ( R 0 [ M ] ) is nil and R 0 [ M ] / J ( R 0 [ M ] ) is boolean, by Lemma 3.2 again. For this, let x = x 1 , x 2 , … , x n , 0 , … T ∈ J ( R 0 [ M ] ) , where x 1 , x 2 , … , x n ∈ R [ G ] and n ∈ N ∗ . By (3.3), ∑ i = 1 n x i ∈ J ( R [ G ] ) , and because J ( R [ G ] ) is nil, it follows that ∑ i = 1 n x i is a nilpotent element.

Note that, for any element y = y 1 , y 2 , … , y t , 0 , … T ∈ R 0 [ M ] , where y 1 , y 2 , … , y t ∈ R [ G ] and t ∈ N ∗ , we have y ∘ y = y P y = y ∑ i = 1 t y i , and thus for any integer k ≥ 2 , y k = y k − 2 ∘ y ∘ y = y k − 2 ∘ y ∑ i = 1 t y i = … = y ⋅ ∑ i = 1 t y i k − 1 .

Therefore, x is a nilpotent element when x ∈ J ( R 0 [ M ] ) , which yields that J ( R 0 [ M ] ) is nil.

On the other hand, for any z = z 1 , z 2 , … , z m , 0 , … T ∈ R 0 [ M ] \ J ( R 0 [ M ] ) , where z 1 , … , z m ∈ R [ G ] and m ∈ N ∗ , we have ∑ j = 1 m z j ∉ J ( R [ G ] ) , whence ∑ j = 1 m z j ∈ R [ G ] \ J ( R [ G ] ) . Note that

where e is the identity of G. Note also that ∑ i = 1 m z i ∑ j = 1 m z j − e = ∑ j = 1 m z j 2 − ∑ j = 1 m z j ∈ J ( R [ G ] ) , because ∑ j = 1 m z j ∈ R [ G ] \ J ( R [ G ] ) and R [ G ] / J ( R [ G ] ) is boolean. Then, by (3.3), z 2 − z ∈ J ( R 0 [ M ] ) . It follows that R 0 [ M ] / J ( R 0 [ M ] ) is boolean.

Conversely, suppose that R 0 [ M ] is strongly nil-clean. We claim that R [ G ] is strongly nil-clean. Indeed, let f denote the element [ e , 0 , … , 0 , … ] T in R 0 [ M ] , where e is the identity of G. Then f is an idempotent, and moreover f ∘ R 0 [ M ] ∘ f ≅ R [ G ] . This, together with Lemma 3.4, implies that R [ G ] is strongly nil-clean.□

Theorem 3.8

Let M = ℳ 0 ( G ; I , Λ ; P ) be a completely 0-simple semigroup. Then R 0 [ M ] is strongly nil-clean if and only if R [ G ] is strongly nil-clean, and either | I | = 1 or | Λ | = 1 .

Proof

Note that if R 0 [ M ] is strongly nil-clean, then | I | = 1 or | Λ | = 1 by Lemma 3.6. Then the result follows immediately from Lemma 3.7.□

Corollary 3.9

Let M = ℳ ( G ; I , Λ ; P ) be a completely simple semigroup. Then R [ M ] is strongly nil-clean if and only if R [ G ] is strongly nil-clean, and either | I | = 1 or | Λ | = 1 .

Proof

Note that the set M 0 ≔ ℳ ( G ; I , Λ ; P ) ∪ { 0 } forms a completely 0-simple semigroup with R 0 [ M 0 ] ≅ R [ M ] . The result then follows from Theorem 3.8.□

Corollary 3.10

Let S be a locally inverse semigroup with idempotent set locally finite and finitely many D -classes. Then R 0 [ S ] is strongly nil-clean if and only if the following two conditions hold.

1. Each D -class D of S has either a unique ℛ -class or a unique ℒ -class;

2. For each maximal subgroup G of S, R [ G ] is strongly nil-clean.

Proof

By [13, Theorem 5.1], R 0 [ S ] is a finite direct product of the contracted completely 0-simple semigroup rings R 0 D α ¯ 0 , where α runs over the set ( S / D ) ∗ consisting all non-zero elements in S / D . Then by [1, Proposition 3.13], R 0 [ S ] is strongly nil-clean if and only if R 0 D α ¯ 0 is strongly nil-clean for each α ∈ ( S / D ) ∗ . Furthermore, for each α ∈ ( S / D ) ∗ , the maximal subgroup of D α ¯ 0 is isomorphic to the maximal subgroup in D α , and the number of ℒ -classes (resp., ℛ -classes) in D α ¯ 0 is equal to the number of ℒ -classes (resp., ℛ -classes) in D α , see [13]. The result then follows from Theorem 3.8.□

Lemma 4.1

Every idempotent of R 0 [ S ] / R 0 [ T ] can be lifted to an idempotent of R 0 [ S ] .

Proof

Suppose that x 1 + x 2 + R 0 [ T ] ∈ R 0 [ S ] / R 0 [ T ] is an idempotent, where x 1 ∈ R S β and x 2 ∈ R 0 [ T ] . Then x 1 2 − x 1 + x 1 x 2 + x 2 x 1 + x 2 2 − x 2 ∈ R 0 [ T ] . Since β is maximal in Y and R S β is a subring of R 0 [ S ] , we infer that s u p p x 1 2 − x 1 ∩ supp x 1 x 2 + x 2 x 1 + x 2 2 − x 2 = ∅ , which yields that x 1 = x 1 2 ∈ R S β is an idempotent. Since x 1 + x 2 + R 0 [ T ] = x 1 + R 0 [ T ] , the result follows.□

Lemma 4.2

Let x = x 1 + x 2 ∈ R 0 [ S ] , where x 1 ∈ R S β and x 2 ∈ R 0 [ T ] . If x ∈ J ( R 0 [ S ] ) , then x 1 ∈ J R S β .

Proof

Suppose that 0 ≠ x = x 1 + x 2 ∈ J ( R 0 [ S ] ) . We claim that x 1 ∈ J R S β . For this, let z be an arbitrary element in R S β , then by hypothesis, there exists an element y ∈ R 0 [ S ] such that z x + y + y ( z x ) = 0 . Write y = y 1 + y 2 , where y 1 ∈ R S β and y 2 ∈ R 0 [ T ] . It follows from the maximality of β that z x 1 + y 1 + y 1 z x 1 = 0 , whence R S β x 1 is a left quasi-regular left ideal of R S β . By Lemma 2.2, x 1 ∈ J R S β , as required.□

Lemma 4.3

[14] Let α be a non-zero element in Y. If A is an ideal of R S α , then A ∗ = { a ∗   |   a ∈ A } is an ideal of R 0 [ S ] . Moreover, the map A → A ∗ : a ↦ a ∗ is a ring isomorphism from A to A ∗ .

Lemma 4.4

J ( R [ S β ] ) + R 0 [ T ] ⊆ J ( R 0 [ S ] ) + R 0 [ T ] .

Proof

If J ( R 0 [ S ] ) = 0 , then J ( R [ S α ] ) = 0 for each α ∈ Y , because J ( R 0 [ S ] ) = 0 if and only if J ( R [ S α ] ) = 0 for each α ∈ Y by [14, Theorem 2]. Hence, in this case, J ( R [ S β ] ) + R 0 [ T ] ⊆ J ( R 0 [ S ] ) + R 0 [ T ] holds true. If J ( R 0 [ S ] ) ≠ 0 , then we claim that J ( R [ S β ] ) + R 0 [ T ] ⊆ J ( R 0 [ S ] ) + R 0 [ T ] . Indeed, for any element 0 ≠ x ∈ J ( R [ S β ] ) , the element x ∗ ∈ R 0 [ S ] defined by (4.1) is non-zero, and furthermore belongs to J ( R [ S β ] ) ∗ . Since J ( R [ S β ] ) ∗ is an ideal of R 0 [ S ] and is isomorphic to J ( R [ S β ] ) by Lemma 4.3, we infer that J ( R [ S β ] ) ∗ is a left quasi-regular ideal of R 0 [ S ] . Because J ( R 0 [ S ] ) contains any left quasi-regular left ideal of R 0 [ S ] , J ( R [ S β ] ) ∗ ⊆ J ( R 0 [ S ] ) . It follows that x ∗ ∈ J ( R 0 [ S ] ) . Therefore, by the definition of x ∗ again, x + R 0 [ T ] = x ∗ + R 0 [ T ] ∈ J ( R 0 [ S ] ) + R 0 [ T ] . Consequently, we have J ( R [ S β ] ) + R 0 [ T ] ⊆ J ( R 0 [ S ] ) + R 0 [ T ] .□