Units, zero-divisors and idempotents in rings graded by torsion-free groups

Problem 1 (Higman/Kaplansky)

Let 𝐾 be a field, let 𝐺 be a torsion-free group and denote by K ⁢ [ G ] the corresponding group ring.

1. Is every unit in K ⁢ [ G ] necessarily trivial, i.e. a scalar multiple of an element of 𝐺?

2. Is K ⁢ [ G ] necessarily a domain?

3. Is every idempotent in K ⁢ [ G ] necessarily trivial, i.e. either 0 or 1?

Let 𝐺 be a torsion-free group and 𝑅 a unital 𝐺-graded ring equipped with a non-degenerate (see Definition 2.1) 𝐺-grading such that R e is a domain.

1. Under the assumption that char ⁡ ( R e ) = 0 , is every unit in 𝑅 necessarily homogeneous with respect to the given 𝐺-grading?

2. Is 𝑅 necessarily a domain?

3. Is every idempotent in 𝑅 necessarily trivial?

(a) 𝑅 is said to have a non-degenerate𝐺-grading (cf. [11, 39]) if, for each g ∈ G and each nonzero r g ∈ R g , we have r g ⁢ R g − 1 ≠ { 0 } , R g − 1 ⁢ r g ≠ { 0 } .

(b) 𝑅 is said to have a fully component regular𝐺-grading if r g ⁢ s h ≠ 0 for any g , h ∈ G , r g ∈ R g ∖ { 0 } and s h ∈ R h ∖ { 0 } .

(a) Every strong 𝐺-grading on a unital ring 𝑅 is non-degenerate. Indeed, take g ∈ G and r g ∈ R g , and suppose that r g ⁢ R g − 1 = { 0 } . Then

Thus, r g = 0 . Similarly, R g − 1 ⁢ r g = { 0 } implies r g = 0 .

(b) The term fully component regular has been chosen to capture the essence of those gradings. There is no immediate connection to the component regular gradings considered by Passman in e.g. [43, p. 16], which are in fact special types of non-degenerate gradings.

In the following two examples, the grading group is G := ( Z , + ) .

(a) Consider the polynomial ring R := R ⁢ [ t ] in one indeterminate. We may define a ℤ-grading on 𝑅 by putting R n := R ⁢ t n for n ≥ 0 , and R n := { 0 } for n < 0 . Clearly, this grading is fully component regular, but it is degenerate.

(b) Consider the ring of 2 × 2 -matrices with real entries, R := M 2 ⁢ ( R ) . We may define a ℤ-grading on 𝑅 by putting

whenever | n | > 1 . This grading is neither fully component regular, nor strong, but one easily sees that it is non-degenerate.

Let 𝐻 be a subgroup of 𝐺. If a ∈ R and b ∈ R H , then we have π H ⁢ ( a ⁢ b ) = π H ⁢ ( a ) ⁢ b and π H ⁢ ( b ⁢ a ) = b ⁢ π H ⁢ ( a ) .

Proof

Take a ∈ R and b ∈ R H . Put a ′ := a − π H ⁢ ( a ) . Clearly, a = a ′ + π H ⁢ ( a ) and Supp ⁡ ( a ′ ) ⊆ G ∖ H . If g ∈ G ∖ H and h ∈ H , then g ⁢ h ∉ H . Thus, we have Supp ⁡ ( a ′ ⁢ b ) ⊆ G ∖ H . Hence,

Analogously, one may show that π H ⁢ ( b ⁢ a ) = b ⁢ π H ⁢ ( a ) . ∎

Suppose that the 𝐺-grading on 𝑅 is non-degenerate. Let 𝐻 be a subgroup of 𝐺 and let r ∈ R H . The following assertions hold:

1. 𝑟 is a left (right) zero-divisor in R H if and only if 𝑟 is a left (right) zero-divisor in 𝑅;

2. 𝑟 is left (right) invertible in R H if and only if 𝑟 is left (right) invertible in 𝑅.

Proof

The “only if” statements are trivial. We only need to show the “if” statements. The proofs of the right-handed claims are treated analogously and are therefore omitted.

(i) Suppose that r ⁢ s = 0 for some nonzero s ∈ R . Then, by Lemma 2.4,

Without loss of generality we may assume that Supp ⁡ ( s ) ∩ H ≠ ∅ , and thus we have 0 ≠ π H ⁢ ( s ) ∈ R H . For otherwise, we may take some g ∈ Supp ⁡ ( s ) and some nonzero x g − 1 ∈ R g − 1 such that e ∈ Supp ⁡ ( s ⁢ x g − 1 ) . Note that r ⁢ ( s ⁢ x g − 1 ) = 0 and Supp ⁡ ( s ⁢ x g − 1 ) ∩ H ≠ ∅ .

(ii) Suppose that s ⁢ r = 1 R for some s ∈ R . Then, by Lemma 2.4,

in R H . ∎

If the 𝐺-grading on 𝑅 is non-degenerate and R e is a domain, then the 𝐺-grading is fully component regular.

Proof

Take g , h ∈ G , r g ∈ R g and s h ∈ R h . Suppose that r g ⁢ s h = 0 . Then

Using the fact that R e is a domain, we get that R g − 1 ⁢ r g = { 0 } or s h ⁢ R h − 1 = { 0 } . By the non-degeneracy of the grading, we conclude that r g = 0 or s h = 0 . ∎

There are large classes of group graded rings whose gradings are non-degenerate but not necessarily strong. For example, crystalline graded rings [35], (nearly) epsilon-strongly graded rings [37, 38], and in particular Leavitt path algebras and crossed products by unital twisted partial actions. However, it should be noted that an epsilon-strongly 𝐺-graded ring 𝑅, and in particular a partial crossed product, for which R e is a domain, is necessarily strongly graded by a subgroup of 𝐺.

(a) The support of the 𝐺-grading on 𝑅 is defined as the set

(b) If Supp ⁡ ( R ) = G , then the 𝐺-grading on 𝑅 is said to be fully supported.

If the 𝐺-grading on 𝑅 is non-degenerate and R e is a domain, then Supp ⁡ ( R ) is a subgroup of 𝐺.

Proof

Take g , h ∈ Supp ⁡ ( R ) . Using Proposition 2.6, we conclude that

Thus, g ⁢ h ∈ Supp ⁡ ( R ) . Moreover, by the non-degeneracy of the grading, it is clear that R g − 1 ≠ { 0 } . Thus, g − 1 ∈ Supp ⁡ ( R ) . This shows that Supp ⁡ ( R ) is a subgroup of 𝐺. ∎

If the 𝐺-grading on 𝑅 is non-degenerate and R e is a domain, then 𝑅 has a natural grading by the subgroup H := Supp ⁡ ( R ) of 𝐺. This 𝐻-grading is non-degenerate and R e G = R e H is a domain. Moreover, the 𝐻-grading is fully supported and fully component regular.

Let 𝐺 be a group.

1. 𝐺 is said to be a unique product group if, given any two non-empty finite subsets 𝐴 and 𝐵 of 𝐺, there exists at least one element g ∈ G which has a unique representation of the form g = a ⁢ b with a ∈ A and b ∈ B .

2. 𝐺 is said to be a two unique products group if, given any two non-empty finite subsets 𝐴 and 𝐵 of 𝐺 with | A | + | B | > 2 , there exist at least two distinct elements 𝑔 and ℎ of 𝐺 which have unique representations of the form g = a ⁢ b , h = c ⁢ d with a , c ∈ A and b , d ∈ B .

Lemma 3.2 (Strojnowski [52])

A group 𝐺 is a unique product group if and only if it is a two unique products group.

Every unique product group is necessarily torsion-free.

Let 𝐺 be a unique product group and 𝑅 a unital 𝐺-graded ring. Furthermore, suppose that the 𝐺-grading on 𝑅 is non-degenerate and that R e is a domain. The following assertions hold:

1. every unit in 𝑅 is homogeneous;

2. 𝑅 is a domain;

3. every idempotent in 𝑅 is trivial.

Proof

(i) Take x , y ∈ R which satisfy x ⁢ y = 1 R . Put

By assumption, | A | and | B | are positive. We want to show that | A | = | B | = 1 . Seeking a contradiction, suppose that | A | > 1 . Then | A | + | B | > 2 . Using the fact that 𝐺 is a unique product group and hence, by Lemma 3.2, a two unique products group, there are two distinct elements g , h ∈ A ⁢ B such that g = a ⁢ b and h = c ⁢ d with a , c ∈ A and b , d ∈ B . We must have x a ⁢ y b = 0 or x c ⁢ y d = 0 since | Supp ⁡ ( x ⁢ y ) | = | Supp ⁡ ( 1 R ) | = | { e } | = 1 . But, by Proposition 2.6, 𝑅 is fully component regular, and hence neither of the two equalities can hold. This is a contradiction. We conclude that | A | = 1 , i.e. 𝑥 is homogeneous. From the equality x ⁢ y = 1 R and the full component regularity of the grading, we get that 𝑦 is also homogeneous.

(ii) Take two nonzero elements x , y ∈ R . Seeking a contradiction, suppose that x ⁢ y = 0 . Using the fact that 𝐺 is a unique product group, there is some a ∈ Supp ⁡ ( x ) and some b ∈ Supp ⁡ ( y ) such that x a ⁢ y b = 0 . By Proposition 2.6, this is a contradiction.

(iii) This follows from (ii) since u 2 = u ⇔ u ⁢ ( u − 1 R ) = 0 . ∎

Note that Theorem 3.4 (ii) also holds if 𝑅 is non-unital. We want to point out that Malman has already proved Theorem 3.4 (ii) in [29, Lemma 3.13].

The proof of Theorem 3.4 (i) yields a seemingly stronger conclusion than the one we aim to prove. But in fact, note that, for a 𝐺-graded ring 𝑅 with a fully component regular grading, the following three assertions are equivalent:

1. every left invertible element in 𝑅 is homogeneous;

2. every right invertible element in 𝑅 is homogeneous;

3. every unit in 𝑅 is homogeneous.

Typical examples of unique product groups are the diffuse groups (see [6, 24]) and in particular the right (or left) orderable groups, including e.g. all free groups, all torsion-free nilpotent groups, and hence all torsion-free abelian groups.

For many years, it was not known whether every torsion-free group necessarily had the unique product property. However, in 1987, Rips and Segev [46] presented an example of a torsion-free group without the unique product property. Since then, a growing number of examples of torsion-free non-unique product groups have surfaced (see e.g. [1, 9, 17, 44, 50]). Notably, Passman [42, p. 606] showed that the so-called (Passman) fours group

is torsion-free and non-right-orderable. Promislow [44] later showed that 𝑃 is not a unique product group. And more than 30 years later, Gardam [16] showed that the group ring K ⁢ [ P ] has non-trivial units, where 𝐾 is the field of two elements. It is interesting to note that 𝑃 is a solvable group, and hence K ⁢ [ P ] is in fact a domain (see [25, Theorem 1.4]).

Let 𝐺 be a torsion-free group and 𝑅 a unital 𝐺-graded ring whose 𝐺-grading is non-degenerate. If R e is a domain, then every unit in R Z ⁢ ( G ) is homogeneous, R Z ⁢ ( G ) is a domain and every idempotent in R Z ⁢ ( G ) is trivial.

Let 𝐺 be a torsion-free group and 𝑅 a unital commutative 𝐺-graded ring whose 𝐺-grading is non-degenerate. If R e is a domain, then every unit in 𝑅 is homogeneous, 𝑅 is an integral domain and every idempotent in 𝑅 is trivial.

Proof

By Corollary 2.10, there is a torsion-free group 𝐻 such that 𝑅 may be equipped with an 𝐻-grading which is fully supported and fully component regular. Take g , h ∈ H . There are nonzero homogeneous elements r g ∈ R g and r h ∈ R h such that r g ⁢ r h = r h ⁢ r g ≠ 0 . Thus, R g ⁢ h ∩ R h ⁢ g ≠ ∅ , which yields g ⁢ h = h ⁢ g . This shows that 𝐻 is a torsion-free abelian group. The result now follows from Theorem 3.4. ∎

(a) Let 𝔽 be a field and consider the first Weyl algebra

It is an easy exercise to show that 𝑅 is a domain, but this elementary fact is also an immediate consequence of Theorem 3.4. Indeed, note that, by assigning suitable degrees to the generators, deg ⁡ ( x ) := 1 and deg ⁡ ( y ) := − 1 , 𝑅 becomes graded by the unique product group ( Z , + ) . Moreover, R 0 = F ⁢ [ x ⁢ y ] is a domain and the ℤ-grading is non-degenerate. Thus, the first Weyl algebra 𝑅 is a domain.

(b) More generally, let 𝐷 be a ring, let σ := ( σ 1 , … , σ n ) be a set of commuting automorphisms of 𝐷, and let a := ( a 1 , … , a n ) be an 𝑛-tuple with nonzero entries from Z ⁢ ( D ) satisfying σ i ⁢ ( a j ) = a j for i ≠ j . Given this data, it is possible to define the corresponding generalized Weyl algebra R := D ⁢ ( σ , a ) (see [3] or [35]). One may show that 𝑅 is Z n -graded with R e = D . If 𝐷 is a domain, then the Z n -grading is non-degenerate and Theorem 3.4 yields that 𝑅 is a domain. Thus, we have recovered [3, Proposition 1.3 (2)].

(c) Any crystalline graded ring R := A ⋄ σ α G (see [35]) is equipped with a non-degenerate 𝐺-grading with R e = A . If 𝐴 is a domain and 𝐺 is a unique product group, then 𝑅 is a domain by Theorem 3.4.

Lemma 4.1 (Neumann [36])

Every torsion-free FC-group is abelian.

Lemma 4.2 (Passman [41])

Let 𝐿 be a group and let H 1 , H 2 , … , H n be a finite number of subgroups of 𝐿. Suppose that there exists a finite collection of elements s i , j ∈ L for i ∈ { 1 , … , n } and j ∈ { 1 , … , m } such that L = ⋃ i , j H i ⁢ s i , j . Then for some k ∈ { 1 , … , n } , we have [ L : H k ] < ∞ .

Let 𝐺 be a group and consider the subgroup H := Δ ⁢ ( G ) . Suppose that 𝐹 is a non-empty finite subset of 𝐻 and that A , B are non-empty finite subsets of G ∖ H . Let f ∈ F and h ∈ H be arbitrary. There exists some g ∈ C G ⁢ ( F ) such that f ⁢ h ∉ g − 1 ⁢ A ⁢ g ⁢ B .

Proof

Put L := C G ⁢ ( F ) and suppose that A = { a 1 , … , a n } , B = { b 1 , … , b m } . Let f ∈ F and h ∈ H be arbitrary. Seeking a contradiction, suppose that

For each i ∈ { 1 , … , n } , we define the subgroup H i := L ∩ C G ⁢ ( a i ) .

For ( i , j ) ∈ { 1 , … , n } × { 1 , … , m } , if a i is conjugate to f ⁢ h ⁢ b j − 1 by an element of 𝐿, then choose s i , j ∈ L such that s i , j − 1 ⁢ a i ⁢ s i , j = f ⁢ h ⁢ b j − 1 . Otherwise, choose s i , j = e . Note that L ⊇ ⋃ i , j H i ⁢ s i , j .

Now, let g ∈ L be arbitrary. By our assumption, there exist 𝑖 and 𝑗 such that f ⁢ h = g − 1 ⁢ a i ⁢ g ⁢ b j . That is, f ⁢ h ⁢ b j − 1 = g − 1 ⁢ a i ⁢ g = s i , j − 1 ⁢ a i ⁢ s i , j . From the last equality, we get that g ⁢ s i , j − 1 ∈ L ∩ C G ⁢ ( a i ) = H i . Thus, g ∈ H i ⁢ s i , j . Since 𝑔 was arbitrarily chosen, it is clear that L ⊆ ⋃ i , j H i ⁢ s i , j . This shows that L = ⋃ i , j H i ⁢ s i , j .

By Lemma 4.2, there is some k ∈ { 1 , … , n } such that [ L : H k ] < ∞ . Consider the chains of subgroups G ⊇ L ⊇ H k and G ⊇ C G ⁢ ( a k ) ⊇ H k . Recall that [ G : C G ( r ) ] < ∞ for each r ∈ F . Thus, [ G : L ] < ∞ , and hence [ G : H k ] < ∞ . This shows that [ G : C G ( a k ) ] < ∞ . But this is a contradiction since a k ∈ G ∖ H . ∎

Let 𝐺 be a torsion-free group and 𝑅 a 𝐺-graded ring. If the 𝐺-grading on 𝑅 is non-degenerate and R e is a domain, then 𝑅 is a prime ring.

Proof

By Corollary 2.10, G ′ := Supp ⁡ ( R ) is a torsion-free subgroup of 𝐺. Moreover, 𝑅 can be equipped with a non-degenerate and fully supported G ′ -grading. Thus, we will without loss of generality assume that the 𝐺-grading on 𝑅 is fully supported.

Put H := Δ ⁢ ( G ) . Note that 𝐻 is a torsion-free FC-group. Thus, by Lemma 4.1, 𝐻 is torsion-free abelian. Theorem 3.4 now yields that R H is a domain.

Seeking a contradiction, suppose that there are nonzero ideals 𝐼 and 𝐽 of 𝑅 such that I ⋅ J = { 0 } . By Proposition 2.6, the 𝐺-grading on 𝑅 is fully component regular. Using this and Lemma 2.4, it is clear that π H ⁢ ( I ) and π H ⁢ ( J ) are nonzero ideals of R H .

Choose some x ∈ I and y ∈ J such that π H ⁢ ( x ) ≠ 0 and π H ⁢ ( y ) ≠ 0 . Put

Note that F := Supp ⁡ ( x ′ ) ⊆ H and A := Supp ⁡ ( x ′′ ) ⊆ G ∖ H . Similarly, put

and note that Supp ⁡ ( y ′ ) ⊆ H and B := Supp ⁡ ( y ′′ ) ⊆ G ∖ H .

Choose some f ∈ F and h ∈ Supp ⁡ ( y ′ ) . Put L := C G ⁢ ( F ) and let g ∈ L be arbitrary. Choose some nonzero elements r g ∈ R g and r g − 1 ∈ R g − 1 . Then we have r g − 1 ⁢ x ⁢ r g ⊆ I , and thus

Now, by combining the facts that the 𝐺-grading on 𝑅 is fully supported and fully component regular and that R H is a domain, it is not difficult to see that r g − 1 ⁢ x ′ ⁢ r g ⁢ y ′ ≠ 0 . In fact, f ⁢ h ∈ Supp ⁡ ( r g − 1 ⁢ x ′ ⁢ r g ⁢ y ′ ) = Supp ⁡ ( x ′ ⁢ y ′ ) ⊆ H . Using the fact that 𝐻 is a subgroup of 𝐺 which is closed under conjugation, we note that Supp ⁡ ( r g − 1 ⁢ x ′′ ⁢ r g ⁢ y ′ ) ∩ H = ∅ and Supp ⁡ ( r g − 1 ⁢ x ′ ⁢ r g ⁢ y ′′ ) ∩ H = ∅ , and hence we must have f ⁢ h ∈ Supp ⁡ ( r g − 1 ⁢ x ′′ ⁢ r g ⁢ y ′′ ) = g − 1 ⁢ A ⁢ g ⁢ B . But g ∈ L may be chosen arbitrarily, and thus Lemma 4.3 yields a contradiction. This shows that 𝑅 is prime. ∎

Let 𝐺 be a torsion-free group and 𝑅 a unital 𝐺-graded ring. If the 𝐺-grading on 𝑅 is non-degenerate and R e is a domain, then Z ⁢ ( R ) is an integral domain. In particular, every central idempotent in 𝑅 is trivial.

Let 𝐺 be a torsion-free group and 𝑅 a 𝐺-graded ring whose 𝐺-grading is non-degenerate. Then 𝑅 is a domain if and only if 𝑅 is reduced and R e is a domain.

Proof

The “only if” statement is trivial. We proceed by showing the “if” statement. To this end, suppose that R e is a domain and that 𝑅 is not a domain. We need to show that 𝑅 is not reduced. By Theorem 4.4, we conclude that 𝑅 is a prime ring. Choose some nonzero elements x , y ∈ R which satisfy x ⁢ y = 0 . By primeness of 𝑅, we have y ⁢ R ⁢ x ≠ { 0 } . Note that ( y ⁢ R ⁢ x ) 2 ⊆ y ⁢ R ⁢ x ⁢ y ⁢ R ⁢ x = { 0 } . Thus, there is some nonzero z ∈ y ⁢ R ⁢ x which satisfies z 2 = 0 . This shows that 𝑅 is not reduced. ∎

Let 𝐺 be a torsion-free group and 𝑅 a unital 𝐺-graded ring. Furthermore, suppose that the 𝐺-grading on 𝑅 is non-degenerate and that R e is a domain. Consider the following assertions:

1. every unit in 𝑅 is homogeneous;

2. 𝑅 is reduced;

3. 𝑅 is a domain;

4. every idempotent in 𝑅 is trivial.

Proof

(i) ⇒ (ii) Suppose that every unit in 𝑅 is homogeneous. Let x ∈ R be an element which satisfies x 2 = 0 . Note that

This shows that 1 R − x is a unit in 𝑅, and hence, by assumption, 1 R − x ∈ R g for some g ∈ G . Put r g := 1 R − x and H := ⟨ g ⟩ , the subgroup of 𝐺 generated by 𝑔. Consider the subring R H whose 𝐻-grading is non-degenerate. Using the fact that 1 R ∈ R e , we note that x = 1 R − r g ∈ R H . We claim that R H is a domain. If we assume that the claim holds, then x 2 = 0 implies x = 0 and we are done. Now we show the claim.

1. ( g = e ): By assumption, R H = R e is a domain.

2. ( g ≠ e ): 𝐻 is an infinite cyclic group which can be ordered. The desired conclusion follows from Theorem 3.4 (ii).

(ii) ⇔ (iii) This follows from Proposition 5.1.

(iii) ⇒ (iv) This is trivial. ∎

If u = u 2 ∈ R is an idempotent, then

Thus, if 2 is invertible in R e , then one can directly, without invoking a primeness argument, show that (i) ⇒ (iv) in Theorem 5.2 by proceeding as in the proof of (i) ⇒ (ii).

If 𝐺 is a unique product group and 𝐴 is a unital commutative ring, then the group ring A ⁢ [ G ] has only trivial units if and only if 𝐴 is reduced and indecomposable (see [51, Proposition 2.1]).

Let 𝐺 be a unique product group and let A := C ∞ ⁢ ( R ) be the algebra of all smooth functions R → R with pointwise addition and multiplication. Note that 𝐴 is not a domain. However, 𝐴 is reduced and indecomposable. Thus, A ⁢ [ G ] has only trivial units.

(a) While torsion-freeness of 𝐺 is clearly a necessary condition for a group ring K ⁢ [ G ] to be a domain, this is not the case for strongly 𝐺-graded rings in general. Indeed, consider for instance the real quaternion algebra ℍ which is a division ring, and which is strongly graded by the finite group Z / 2 ⁢ Z × Z / 2 ⁢ Z .

(b) Example 5.4 shows that Theorem 5.2 fails to hold if the assumption on R e is dropped.

(c) Non-degeneracy of the grading is not a necessary condition for a group graded ring to be a domain. To see this, consider e.g. Example 2.3 (a).

Let 𝐺 be a group and 𝑅 a unital 𝐺-graded ring whose 𝐺-grading is non-degenerate. Furthermore, suppose that R e is a domain and that the 𝐺-grading is fully supported. If 𝑥 is a central element in 𝑅, then the subgroup of 𝐺 generated by Supp ⁡ ( x ) is an FC-group.

Proof

Let x = ∑ g ∈ G x g be a central element in 𝑅. Take s ∈ G . Choose some nonzero r s ∈ R s and, using Proposition 2.6, note that

This shows that Supp ⁡ ( x ) is closed under conjugation by elements of 𝐺. Thus, by finiteness of Supp ⁡ ( x ) , we get that Supp ⁡ ( x ) ⊆ Δ ⁢ ( G ) . Let 𝐻 be the subgroup of 𝐺 generated by Supp ⁡ ( x ) . Using the fact that 𝐻 is finitely generated, we conclude that 𝐻 is an FC-group. ∎

Let 𝐺 be a torsion-free group and 𝑅 a unital 𝐺-graded ring. If the 𝐺-grading on 𝑅 is non-degenerate and R e is a domain, then the following assertions hold:

1. every central unit in 𝑅 is homogeneous;

2. 𝑅 has no non-trivial central zero-divisor;

3. 𝑅 is indecomposable, i.e. every central idempotent in 𝑅 is trivial.

Proof

By Corollary 2.10, G ′ := Supp ⁡ ( R ) is a torsion-free subgroup of 𝐺. Moreover, 𝑅 can be equipped with a non-degenerate and fully supported G ′ -grading. Thus, we will without loss of generality assume that the 𝐺-grading on 𝑅 is fully supported.

(i) Let x ∈ Z ⁢ ( R ) be a unit in 𝑅. Denote by 𝐻 the subgroup of 𝐺 generated by Supp ⁡ ( x ) . Note that, by Lemma 2.5 (ii), 𝑥 is a unit in R H . By Proposition 6.1, 𝐻 is a torsion-free FC-group. Thus, using Lemma 4.1 and Theorem 3.4, we conclude that 𝑥 is homogeneous.

(ii) Let x ∈ Z ⁢ ( R ) be nonzero. Suppose that x ⁢ y = 0 for some y ∈ R . Denote by 𝐻 the subgroup of 𝐺 generated by Supp ⁡ ( x ) . By Proposition 6.1, 𝐻 is a torsion-free FC-group. Using Lemma 4.1 and Theorem 3.4, we conclude that R H is a domain. Thus, 𝑥 is not a zero-divisor in R H , and by Lemma 2.5 (i), we conclude that y = 0 .

(iii) This follows from (ii) or from Corollary 4.5. ∎

(a) Malman has essentially proved Theorem 6.2 (ii) in [29, Proposition 3.14]. Thanks to Neumann’s lemma (Lemma 4.1), our proof is shorter.

(b) Note that we can immediately recover Corollary 3.10 from Theorem 6.2.

If 𝑅 is a reduced ring, then every idempotent in 𝑅 is central in 𝑅.

Proof

Let u ∈ R be an idempotent. Take any r ∈ R . Note that ( u ⁢ r − u ⁢ r ⁢ u ) 2 = 0 and ( r ⁢ u − u ⁢ r ⁢ u ) 2 = 0 . Using the fact that 𝑅 is reduced, we conclude that

Hence, u ⁢ r = r ⁢ u . This shows that u ∈ Z ⁢ ( R ) . ∎

Let 𝐺 be a torsion-free group and 𝑅 a unital 𝐺-graded ring. Furthermore, suppose that the 𝐺-grading on 𝑅 is non-degenerate and that R e is a domain. If 𝑅 is reduced, then every idempotent in 𝑅 is trivial.

Let 𝐺 be a group and 𝑅 a 𝐺-graded ring. If 𝑁 is a normal subgroup of 𝐺, then 𝑅 may be viewed as a G / N -graded ring. Indeed, by writing

it is easy to see that this yields a G / N -grading.

Let 𝐺 be a group and 𝑅 a 𝐺-graded ring whose 𝐺-grading is non-degenerate. If 𝑁 is a normal subgroup of 𝐺, then the canonical G / N -grading on 𝑅 is non-degenerate.

Let 𝐺 be a group and 𝑅 a unital 𝐺-graded ring whose 𝐺-grading is non-degenerate. If 𝑁 is a normal subgroup of 𝐺 and R e is a domain, then the following assertions hold:

1. If R N := ⨁ n ∈ N R n is a domain and G / N is a unique product group, then 𝑅 is a domain.

2. Suppose that 𝑁 is torsion-free, that G / N is a unique product group and that every unit in 𝑅 which is contained in R g ⁢ N for some g ∈ G must be homogeneous with respect to the 𝐺-grading. Then every unit in 𝑅 is homogeneous with respect to the 𝐺-grading.

Proof

(i) We will view

as a G / N -graded ring. By Lemma 7.2, the G / N -grading is non-degenerate, and by assumption, R N is a domain. The desired conclusion now follows immediately from Theorem 3.4.

(ii) We begin by noting that, by assumption, every unit in R N must be homogeneous. Thus, by Theorem 5.2, R N is a domain. Take x , y ∈ R which satisfy x ⁢ y = 1 R . Let Supp ⁡ ( x ) and Supp ⁡ ( y ) denote the support of 𝑥 and 𝑦, respectively, with respect to the 𝐺-grading. Let ϕ : G → G / N denote the quotient homomorphism. Define 𝑎 and 𝑏 to be the cardinalities of ϕ ⁢ ( Supp ⁡ ( x ) ) and ϕ ⁢ ( Supp ⁡ ( y ) ) , respectively. If a + b > 2 , then by the unique product property of G / N (and Lemma 3.2), we will reach a contradiction in the same way as in the proof of Theorem 3.4, by instead considering the G / N -grading. Thus, a = b = 1 . This means that there is some g ∈ G such that x ∈ R g ⁢ N and y ∈ R g − 1 ⁢ N . Now, by assumption, both 𝑥 and 𝑦 must be homogeneous with respect to the 𝐺-grading. ∎

By taking N = { e } , note that, from Proposition 7.3 (i), we recover Theorem 3.4 (ii), and from Proposition 7.3 (ii), we recover Theorem 3.4 (i).

Let 𝐺 be a group and 𝑅 a 𝐺-crossed product. Suppose that 𝑁 is a torsion-free normal subgroup of 𝐺, that G / N is a unique product group and that R e is a domain. The following two assertions are equivalent:

1. every unit in R = ⨁ g ∈ G R g is homogeneous with respect to the 𝐺-grading;

2. every unit in R N := ⨁ n ∈ N R n is homogeneous with respect to the 𝑁-grading.

Proof

(i) ⇒ (ii) This is trivial.

(ii) ⇒ (i) The 𝐺-grading on 𝑅 is non-degenerate. Thus, the first part of the proof may be carried out in the same way as the proof of Proposition 7.3 (ii). Indeed, for elements x , y ∈ R which satisfy x ⁢ y = y ⁢ x = 1 R , we get that x ∈ R g ⁢ N and y ∈ R g − 1 ⁢ N for some g ∈ G . Using the fact that 𝑅 is a 𝐺-crossed product, we may choose homogeneous units x ′ and y ′ of degree g − 1 and 𝑔, respectively, such that x ′ ⁢ y ′ = y ′ ⁢ x ′ = 1 R . Note that

where x ′ ⁢ x ∈ R N and y ⁢ y ′ ∈ R N . By assumption, x ′ ⁢ x and y ⁢ y ′ are homogeneous with respect to the 𝑁-grading on R N . Using the fact that x ′ and y ′ are homogeneous, we conclude that 𝑥 and 𝑦 must be homogeneous with respect to the 𝐺-grading. ∎

Let 𝐺 be a torsion-free group and 𝑅 a 𝐺-crossed product for which R e is a domain. If 𝑁 is an abelian normal subgroup of 𝐺 such that G / N is a unique product group, then every unit in 𝑅 is homogeneous with respect to the 𝐺-grading. Moreover, 𝑅 is a domain and every idempotent in 𝑅 is trivial.

Let 𝐺 be a group and suppose that 𝐺 has a finite subnormal series

with quotients G i + 1 / G i all of which are torsion-free abelian. If 𝑅 is a 𝐺-crossed product with R e a domain, then every unit in 𝑅 is homogeneous with respect to the 𝐺-grading, and 𝑅 is a domain. In particular, every idempotent in 𝑅 is trivial.

Proof

Using that G 0 is a torsion-free normal subgroup of G 1 , that R G 0 = R e is a domain and that G 1 / G 0 is a unique product group, we get by Proposition 7.5 that every unit in R G 1 is homogeneous. More generally, if G i is torsion-free and every unit in R G i is homogeneous, then, by Theorem 5.2, R G i is a domain. Using the fact that G i is normal in G i + 1 and that G i + 1 / G i is a unique product group, Proposition 7.5 yields that every unit in R G i + 1 is homogeneous. Furthermore, since both G i + 1 / G i and G i are torsion-free, we notice that G i + 1 is also torsion-free. By induction over 𝑖, we conclude that every unit in 𝑅 is homogeneous with respect to the 𝐺-grading. Clearly, 𝐺 is torsion-free. Thus, Theorem 5.2 yields that 𝑅 is a domain and that every idempotent in 𝑅 is trivial. ∎

It is also possible to obtain Theorem 7.7 directly from Theorem 3.4. Indeed, one can show that the group 𝐺 in Theorem 7.7 is right-ordered (see [42, Lemma 13.1.6]), and hence a unique product group.

Let 𝐺 be a torsion-free group and 𝑅 a unital 𝐺-graded ring whose 𝐺-grading is non-degenerate. If R e is a domain with char ⁡ ( R e ) = 0 , then the following assertions hold:

1. every unit in 𝑅 is homogeneous;

2. 𝑅 is a domain;

3. every idempotent in 𝑅 is trivial.

Using Lemma 2.5, it is not difficult to see that, in order to resolve Conjecture 8.1, it is enough to consider the case where 𝐺 is finitely generated.

In their work, Dykema, Heister and Juschenko [12], and independently Schweitzer [49], identified certain classes of finitely presented torsion-free groups. Amongst other results, they showed that, in order to prove the zero-divisor conjecture for group rings over the field of two elements, it is sufficient to prove the conjecture for groups coming from the aforementioned classes of finitely presented groups. Further investigation is needed to determine whether a similar reduction is possible for group rings in general, and even more generally for the group graded rings appearing in Problem 2.