The Zero Divisor Conjecture and Self-Injectivity for Monoid Rings

The Zero Divisor Conjecture and Self-Injectivity for Monoid Rings Joe Sullivan May 2, 2011 1 Background 1.1 Monoid Rings Definition 1.1. Let G be a set and let : G G G be a binary operation on G. Then the ordered pair (G, ) is a monoid provided: 1. a b G for all a, b G, 2. a (b c) = (a b) c for all a, b, c G, and 3. there exists 1 G such that a 1 = 1 a = a for all a G. We shall refer to the monoid (G, ) as G when no confusion may arise. Note that if we require G to be closed under inverses (i.e. if for each a G there exists b G such that a b = b a = 1) then G is necessarily a group under. Definition 1.2. Let G be a monoid and let k be a field. Denote is the set of all maps α : G k with finite support as kg. Define addition + : kg kg kg on kg by (α + β)(x) = α(x) + β(x) (1) for all α, β kg and x G. Define multiplication : kg kg kg on kg by (α β)(z) = α(x)β(y) (2) xy=z for all α, β kg and x, y, z G. Then the triple (kg, +, ) is called the monoid ring of G over k. By finite support we mean the set supp α = {g G α(g) 0} is finite. It is straightforward to check that (kg, +, ) is indeed a ring with the operations defined above. Again when no confusion may arise, we shall refer to the monoid ring (kg, +, ) as kg. Also, if α, β kg, we shall let αβ denote α β. As a point of notation, suppose α kg. We shall henceforth write α = x G a x x (3) 1

where α(x) = a x k for all x G. Thus if α, β kg, sums and products will be written α + β = x G(a x + b y )x (4) αβ = (a x b y )xy (5) x,y G where, as before, α(x) = a x and β(x) = b x for all x G. As with products in kg, we have written the product x y G as xy G. These sums are always finite since the elements in kg have finite support. 1.2 The Zero Divisor Problem Definition 1.3. Let G be a monoid. Then G is torsion-free if x G and x n = 1 implies x = 1 for every positive integer n. The Classical Zero Divisor conjecture [3] is stated in terms of groups. Conjecture 1.4 (Zero Divisor Conjecture). Suppose G is a torsion-free group and k is a field. If 0 α kg and 0 β kg then αβ 0. In other words, if G is a torsion-free group, then the group ring kg is a domain (i.e. kg has no zero divisors). In this project, we discuss a generalization of the problem (and other related questions) to monoids: if G is a torsion-free monoid, is the monoid ring kg necessarily a domain? 2 Zero Divisors In regards to the previous question, if G is a torsion-free monoid, then the monoid ring kg is not necessarily a domain, as is shown by the following example. Example 2.1. Let k = Z/2Z, G = {1, x}, and suppose : G G G is defined by the following multiplication table: 1 x 1 1 x x x x Then (G, ) is a torsion-free (commutative) monoid. Let kg denote the monoid ring over G, and let α, β kg with α = x β = 1 + x 2

Thus α 0 and β 0, but αβ = x(1 + x) = x + x 2 = x + x = 0 and so kg is not a domain. This particular example may be generalized to yield the following result. Theorem 2.1. Let G be a monoid and k be a field. If kg is a domain, then G is cancellative. Proof. We shall use a contrapositive argument: suppose that G is not cancellative. Without loss of generality, we may assume G is not left-cancellative. This means there exists x G such that xy = xz for some y, z G, and y z. Let α = x kg and β = y z kg. We have that α 0 and β 0, but and so kg is not a domain. αβ = x(y z) = xy xz = 0 As we can see from this result, if G is any non-cancellative monoid (torsionfree or not), then kg necessarily has zero divisors. It is known that for a commutative monoid G, G is cancellative if and only if G can be embedded in a group. Thus in the case where G is commutative and not cancellative, the Zero Divisor Problem returns to the realm of group-ring theory where it remains unsolved. 3 Nilpotents and Units Given a (torsion-free) monoid G, we investigate whether or not kg has nontrivial nilpotent elements and/or nontrivial units for some field k. Example 3.1. Let G be the monoid from Example 2.1. Then for any field k, the only nilpotent element of kg is zero. To see why, let a, b k and let n Z +. 3

Then n ( ) n (a + bx) n = a n j (bx) j j j=0 n ( ) n = a n + a n j b j x j j j=1 n = a n + j=1 ( ) n j a n k b j x since x j = x for all j Z +. Suppose a + bx kg is nilpotent. Then for some n Z +, (a + bx) n = 0, i.e. n ( ) a n + n a n k b j x = 0 j j=1 and so a n = 0 hence a = 0 (since k is a field). But then b n = 0 so b = 0 as well. Thus the only nilpotent elements of kg are trivial for any field k. Now suppose that k is a field with char(k) 2. Then kg has nontrivial units, since char(k) 2 implies (1 2x) 1 in kg, so (1 2x)(1 2x) = 1 4x + 4x 2 = 1 4x + 4x thus 1 2x is a nontrivial unit of kg. However, if char(k) = 2 (namely, k = Z/2Z), then let a, b, c, d k. We have = 1 (a + bx)(c + dx) = ac + (ad + bc + bd)x and if this quantity were 1, we would have that ac = 1 and ad + bc + bd = 0. But if ac = 1, then a = c = 1 since char(k) = 2. But this means b + d + bd = 0 which implies b = c = 0 since again char(k) = 2. So the only units of kg take the form 1 + 0x, i.e. kg has no nontrivial units. Example 3.2. Let G = {1, x, y} and suppose : G G G is defined by the following multiplication table: 1 x y 1 1 x y x x y y y y y y Then for any field k, kg does have nontrivial nilpotent elements, one of which is x y, since (x y) 2 = x 2 xy yx y 2 = y y y + y = 0 4

We have found monoids for which there are either no nontrivial nilpotents over any field, or there are always nontrivial nilpotents over any field. We ve also found a monoid which has nontrivial units over some fields but no nontrivial units over others. 4 Semisimplicity Theorem 4.1 (Maschke s Theorem). Let G be a finite group and let k be a field. Then kg is semisimple if and only if char(k) G. We show that Maschke s Theorem fails in the case where G is a finite monoid in both directions. First, note that when G is a monoid and k is a field, kg is commutative if and only if G is commutative. Also, when G is finite, kg is finitely generated. Now when kg is commutative and finitely generated, J(kG) = { α kg α n = 0 for some n Z + } (6) where J(kG) is the Jacobson radical of kg. This leads to a useful characterization of semisimplicity: Theorem 4.2. Suppose G is a finite, commutative monoid and k is a field. Then kg is semisimple if and only if J(kG) = 0. Example 4.1. Let kg be the monoid ring from Example 2.1 above. We have that G is finite and commutative; in particular, G = 2 and char(k) = 2 so char(k) G. But kg is semisimple, since J(kG) = 0. This is because the only nilpotent element of kg is 0 as was shown in Example 3.1. So we have a finite monoid G and field k such that char(k) G and kg is semisimple. Example 4.2. Let G be the monoid from Example 3.2 above. Then for any field k, kg is not semisimple (in particular, CG is not semisimple). Since G is commutative and finite, kg is again commutative and finitely generated. Thus J(kG) = { α kg α n = 0 for some n Z + } Recall that 0 x y kg is nilpotent from Example 3.2 above. Thus x y J(kG) and so J(kG) 0 for any field k; e.g. if k = Z/2Z then char(k) = 2 and G = 3, so char(k) G but kg is not semisimple. Also, CG is not semisimple even though char(c) = 0 G. 5 Self-injectivity Theorem 5.1. If k is a field and G is a finite group then kg is self-injective; that is, kg is an injective kg-module. 5

This is not the case with finite monoids: we provide an example of a field k and finite monoid G such that kg is not self-injective. We shall henceforth let k = Z/2Z and G = {1, x, y} with the following multiplication : G G G: 1 x y 1 1 x y x x x x y y y y Theorem 5.2. kg is not an injective left kg-module. Proof. Denote the left ideals of kg as A = {0, x, y, x + y} = kgx = kgy B = {0, 1 + x, 1 + y, x + y} = (1 + x)kg = (1 + y)kg C = {0, x + y} = kg(x + y) D = {0, 1 + x} = kg(1 + x) E = {0, 1 + y} = kg(1 + y) and define θ : kg kg ϕ : kg kg ψ : kg kg by θ(α) = αx ϕ(α) = α(1 + x) ψ(α) = α(x + y) for all α kg. Then θ, ϕ, and ψ are left kg-module homomorphisms with im θ = A, im ϕ = D, im ψ = C, ker θ = D ker ϕ = A ker ψ = A Thus a projective resolution of kg/c is θ kg ϕ kg θ kg ψ kg kg/c 0 which yields the corresponding sequence 0 Hom kg (kg/c, kg) Hom kg (kg, kg) ψ Hom kg (kg, kg) θ Hom kg (kg, kg) ϕ Hom kg (kg, kg) θ 6

where θ (α) = xα ϕ (α) = (1 + x)α ψ (α) = (x + y)α and hence im θ = F, im ϕ = B, im ψ = C, ker θ = B ker ϕ = F ker ψ = B where F = {0, x}. We have that Ext 1 kg(kg/c, kg) = ker θ im ψ = B/C 0 and since Ext 1 kg(kg/m, kg) 0 for some left kg-submodule M of kg, we have that kg is not an injective left kg-module. Since this particular monoid ring kg is not self-injective, the obvious question to ask is what is the injective hull of kg. Definition 5.3. Let E be an R-module and M E be an R-submodule of E. We say that E is an essential extension of M if for every R-submodule N of E, we have that N = 0 whenever M N = 0. The injective hull of M, denoted E(M), is an essential extension of M that is injective. Theorem 5.4. The injective hull of kg has dimension 6 over k; that is, dim k E(kG) = 6. In particular, E(kG) = Hom k (kg, C) Hom k (kg, C). (7) Proof. Note that x kg is idempotent and that x + (1 + x) = 1, so kg = kgx kg(1 + x) = A D. (8) Thus E(kG) = E(A) E(D) (9) Also note that C, D, and E are all submodules of B, B = 4, and B = C D = C E = D E, so C = D = E. Thus E(C) = E(D) = E(E), so E(kG) = E(A) E(C) (10) 7

We show that A is not injective by calculating Ext 1 kg(a/c, kg) to be nonzero. A projective resolution of A/C is θ kg ϕ kg θ kg ψ A A/C 0 where θ, ϕ, and ψ are as in the previous proof. This yields the corresponding sequence 0 Hom kg (A/C, kg) Hom kg (A, kg) ψ Hom kg (kg, kg) θ Hom kg (kg, kg) ϕ Hom kg (kg, kg) θ where θ, ϕ, and ψ are also as in the previous proof. Thus Ext 1 kg(a/c, kg) = ker θ im ψ = B/C 0 We have that A is an essential extension of C, since the only nontrivial left submodule of A is C. This means that E(A) = E(C), in particular, and hence E(kG) = E(C) E(C) (11) dim k E(kG) = 2 dim k E(C) and dim k A = 2 and A is not injective, so dim k E(A) 3, thus dim k E(C) 3. Recall that if Q is an injective k-module (which is always the case when k is a field), then Hom k (kg, Q) is an injective kg-module, and that Q = Hom kg (kg, Q) Hom k (kg, Q) (12) This implies that C is contained in the injective kg-module Hom k (kg, C). Since dim k C = 1, we have that dim k Hom k (kg, C) = 3. But Hom k (kg, C) is injective, so dim k E(C) 3. Thus dim k E(C) = 3, so dim k E(kG) = 6. In particular, E(C) = Hom k (kg, C) (13) and so E(kG) = Hom k (kg, C) Hom k (kg, B). 6 Injective and Projective kg-modules 6.1 Injective kg-modules We wish to classify all injective kg-modules. Note that every injective kgmodule is the direct sum of indecomposable injective kg-modules. Which submodules of kg are injective? 8

We ve shown that A is not an injective kg-module and E(A) = E(C) = E(D) = E(E), so C, D, and E are not injective kg-modules. A similar calculation can be used to show that B is not an injective kg-module as well: a projective resolution of B/C is θ kg ϕ kg θ kg ψ B B/C 0 (14) where θ, ϕ, and ψ are as before. This yields the corresponding sequence 0 Hom kg (B/C, kg) Hom kg (B, kg) ψ Hom kg (kg, kg) θ Hom kg (kg, kg) ϕ Hom kg (kg, kg) θ where θ, ϕ, and ψ are also as before. Thus Ext 1 kg(b/c, kg) = ker θ im ψ = B/C 0 (15) and so B is not injective. Also since kg = A C = A D = A E, we have that A = kg/c = kg/d = kg/e and so kg/c, kg/d, and kg/e are not injective kg-modules as well. We will show that kg/b is injective, however. We make use of the following theorem [2]. Theorem 6.1 (Baer s Criterion). Let R be a ring with a 1. Then the R- module Q is injective if and only if for every left ideal I of R, any R-module homomorphism f : I Q can be extended to an R-module homomorphism F : R Q. Theorem 6.2. kg/b is an injective left kg-module. Proof. Recall from the proof of Theorem 5.2 that the left ideals (i.e. the left kg-submodules) are A, B, C, D, and E. We must show that for any M {A, B, C, D, E} and any f Hom kg (M, kg/b), there exists F Hom kg (kg, kg/b) making the following diagram commute: M ι kg f kg/b F where ι : M kg denotes the natural inclusion. We do not need to consider the cases when M = 0 or M = kg since F = 0 and F = f suffice as lifts respectively: 0 kg kg id kg 0 f f kg/b kg/b 9

Suppose M = A. Then there are precisely two elements in Hom kg (A, kg/b), the zero map and f : A kg/b described by f(0) = B f(x) = 1 + B f(y) = 1 + B f(x + y) = B This is because for any θ Hom kg (A, kg/b), θ(0) must equal θ(x + y): if θ(x + y) θ(0) = B, then θ(x + y) = 1 + B and hence 1 + B = x + B = x(1 + B) = xθ(x + y) = θ(x(x + y)) = θ(0) = B which is impossible. This implies θ(x) = θ(y), since B = θ(0) = θ(x + y) = θ(x) + θ(y) and θ(x) + θ(y) B if θ(x) θ(y). So we are left with exactly two options for f Hom kg (A, kg/b). If f is the zero map, it can be trivially extended to the zero map F : kg kg/b, F (α) = 0 for all α kg. If f Hom kg (A, kg/b) is not the zero map, then f can be extended to F = π : kg kg/b, the natural projection: F (1) = 1 + B F (1 + x) = B F (1 + y) = B F (1 + x + y) = 1 + B and F (α) = f(α) for α {0, x, y, x + y}. Suppose M = B. Then there is precisely one element in Hom kg (B, kg/b), the zero map. This is because if θ Hom kg (B, kg/b), then θ(0) = θ(x + y) as before, also implying θ(1 + x) = θ(1 + y) as before. Notice, however, that if θ(1 + x) = 1 + B, then B = θ(0) = θ(x(1 + x)) = xθ(1 + x) = x(1 + B) = x + B = 1 + B which is impossible, so θ(1 + x) = B = θ(1 + y). Thus Hom kg (B, kg/b) = 0 and 0 Hom kg (B, kg/b) can be trivially extended to 0 Hom kg (kg, kg/b). Suppose M = C. Again, there is precisely one element in Hom kg (C, kg/b), the zero map. Suppose 0 θ Hom kg (C, kg/b). Then since θ(0) = B, we must have that θ(x + y) = 1 + B. But this would mean B = θ(0) = θ(x(x + y)) = xθ(x + y) = x(1 + B) = x + B = 1 + B 10

which is impossible. So Hom kg (C, kg/b) = 0 and again 0 can be extended to 0 Hom kg (kg, kg/b). Suppose M = D. Again, there is precisely one element in Hom kg (D, kg/b), the zero map. Suppose 0 θ Hom kg (D, kg/b). Then since θ(0) = B, we must have that θ(1 + x) = 1 + B. But this would mean B = θ(0) = θ(x(1 + x)) = xθ(1 + x) = x(1 + B) = x + B = 1 + B which is impossible. So Hom kg (D, kg/d) = 0, and 0 again 0 can be extended to 0 Hom kg (kg, kg/b). Suppose M = E. Again, there is precisely one element in Hom kg (E, kg/b), the zero map. Suppose 0 θ Hom kg (E, kg/b). Then since θ(0) = B, we must have that θ(1 + y) = 1 + B. But this would mean B = θ(0) = θ(x(1 + y)) = xθ(1 + y) = x(1 + B) = x + B = 1 + B which is impossible. So Hom kg (E, kg/d) = 0, and 0 again 0 can be extended to 0 Hom kg (kg, kg/b). We have exhausted all cases: for any left ideal M of kg, every f Hom kg (M, kg/b) can be lifted to some F Hom kg (kg, kg/b) making the following diagram commute: M ι kg f kg/b F Thus by Baer s Criterion, kg/b is an injective left kg-module. Note that dim k kg/b = 1; in particular, kg/b is an indecomposable (injective) kg-module. Since kg/b is injective, the injective hull of kg/b is itself: E(kG/B) = kg/b. Also the only indecomposable left submodule of kg is C = D = E, and E(C) = Hom k (kg, C). Thus every injective kg-module can be written as a direct sum of isomorphic copies of Hom k (kg, C) and kg/b, the indecomposable injective kg-modules. 6.2 Projective kg-modules A similar classification for the projective left kg-modules can be obtained by invoking the following theorem on Artinian rings (which kg is). Theorem 6.3. Every indecomposable projective module over an Artinian ring R is isomorphic to Re for some primitive idempotent e R. 11

The primitive idempotents of kg are x, y, 1 + x, and 1 + y. Now kgx = kgy = A kg(1 + x) = D kg(1 + y) = E and D = E, so the only indecomposable projective kg-modules are isomorphic to A and D. Thus every projective kg-module can be written as a direct sum of isomorphic copies of A and D. 6.3 Uniserial kg-modules The indecomposable projective kg-modules are A and C = D = E as was just shown. The indecomposable injective kg-modules are Hom k (kg, C) and kg/b, also shown above. We wish to determine whether or not these modules are uniserial (i.e. whether or not they have a unique composition series). Obviously the one-dimensional modules (C and kg/b) are uniserial: their composition series are 0 C 0 kg/b. We also have that A is uniserial since its only submodule is C: 0 C A. Claim 6.4. Hom k (kg, C) is not a uniserial kg-module. Proof. Suppose Hom k (kg, C) were uniserial. Then kg would be a Nakayama algebra (i.e. each indecomposable projective and injective kg-module would be uniserial). This would imply that Hom k (kg, C) is a factor of some indecomposable projective kg-module [1], which in our case is either A or C. This is impossible, however, since dim k A = 2 and dim k C = 1, but dim k Hom k (kg, C) = 3. 6.4 Left Semi-hereditarity Definition 6.5. A ring R is hereditary if every submodule of a projective R-module is projective. A noncommutative ring R is left semi-hereditary if every finitely generated projective left R-module has projective dimension less than or equal to 1. The following criterion will be useful to determine if kg is left semi-hereditary. 12

Proposition 6.6. If every irreducible left R-module has projective dimension less than or equal to 1, then R is left semi-hereditary. Theorem 6.7. kg is a left semi-hereditary ring. Proof. We have precisely 2 irreducible left kg-modules, C and kg/b. The projective dimension of C is certainly less than or equal to 1 since C is projective (because it is a submodule of the projective module kg). kg/b is not projective, however we do have that kg/b = A/C and that the following sequence is exact: 0 C A A/C 0 and C and A are both projective, so A/C = kg/b has projective dimension less than or equal to 1 as well. Thus kg is left semi-hereditary. References [1] Maurice Auslander, Idun Reiten, and Sverre O. Smalø. Representation Theory of Artin Algebras. Cambridge University Press, 1995. [2] David S. Dummit and Richard M. Foote. Abstract Algebra. John Wiley and Sons, Inc., third edition, 2004. [3] Donald S. Passman. The Algebraic Structure of Group Rings. John Wiley and Sons, Inc., 1977. 13