UNIT GROUPS OF CUBE RADICAL ZERO COMMUTATIVE COMPLETELY PRIMARY FINITE RINGS

A completely primary finite ring is a ring R with identity 1≠0 whose subset of all its zero-divisors forms the unique maximal ideal J. Let R be a commutative completely primary finite ring with the unique maximal ideal J such that J3=(0) and J2≠(0). Then R/J≅GF(pr) and the characteristic of R is pk, where 1≤k≤3, for some prime p and positive integer r. Let Ro=GR(pkr,pk) be a Galois subring of R and let the annihilator of J be J2 so that R=Ro⊕U⊕V, where U and V are finitely generated Ro-modules. Let nonnegative integers s and t be numbers of elements in the generating sets for U and V, respectively. When s=2, t=1, and the characteristic of R is p; and when t=s(s


Introduction
In [6], Fuchs asked for a characterization of abelian groups which could be groups of units of a ring.This question was noted to be too general for a complete answer [12], and a natural course is to restrict the classes of groups or rings to be considered.
Let R be a ring and let R * denote its multiplicative group of unit elements.All local rings R with R * cyclic were determined by Gilmer [8] and this case was also considered by Ayoub [1] (also proofs are given in [10,11]).Pearson and Schneider have found all R where R * is generated by two elements.Clark [4] has investigated R * where the ideals form a chain and has shown that if p ≥ 3, n ≥ 2, and r ≥ 2, then the units of the Galois ring GR(p nr , p n ) are a direct sum of a cyclic group of order p r − 1 and r cyclic groups of order p n − 1 (this was also done independently by Raghavendran [11]).In fact, Raghavendran described the structure of the multiplicative group of every Galois ring.Stewart in [12] considered a related problem to that asked by Fuchs [6] by proving that for a given finite group G (not necessarily abelian), there are, up to isomorphism, only finitely many directly indecomposable finite rings having group of units isomorphic to G.
Ganske and McDonald [7] provided a solution for R * when the local ring R has Jacobson radical J such that J 2 = (0) by showing that where n = dim K (J/J 2 ), |K | = p t , and (π) denotes the cyclic group of order π.
In [5], Dolzan found all nonisomorphic rings with a group of units isomorphic to a group G with n elements, where n is a power of a prime or any product of prime powers, not divisible by 4; and also found all groups with n elements which can be groups of units of a finite ring, a contribution to Stewart's problem [12].More recently, X.-D.Hou et al. gave an algorithmic method for computing the structure of the group of units of a finite commutative chain ring and further strengthening the known result by listing a set of linearly independent generators for the group of units.
The present paper focuses on the group of units R * of a commutative completely primary finite ring R with unique maximal ideal J such that R/J ∼ = GF(p r ), J 3 = (0), and J 2 = (0) so that the characteristic of R is p k , where 1 ≤ k ≤ 3; and further identifies sets of generators for R * .
In particular, let R o = GR(p kr , p k ) be a Galois subring of R and let the annihilator of J be J 2 so that R = R o ⊕ U ⊕ V , where U and V are finitely generated R o -modules.Let nonnegative integers s and t be numbers of elements in the generating sets for U and V , respectively.When s = 2, t = 1, and charR = p, and when t = s(s + 1)/2, for any fixed s, the structure of the group of units R * of the ring R and its generators have been determined; these depend on the structural matrices (a i j ) and on the parameters p, k, r, and s.

Preliminaries
We refer the reader to [2] for the general background of completely primary finite rings R with maximal ideals J such that J 3 = {0} and J 2 = {0}.Let R be a completely primary finite ring with maximal ideal J such that J 3 = (0) and J 2 = (0).Then R is of order p nr and the residue field R/J is a finite field GF(p r ), for some prime p and positive integers n, r.The characteristic of R is p k , where k is an integer such that 1 ≤ k ≤ 3. Let GR(p kr , p k ) be the Galois ring of characteristic p k and order p kr , that is, Then, it can be deduced from the main theorem in [4] that R has a coefficient subring R o of the form GR(p kr , p k ) which is clearly a maximal Galois subring of R.Moreover, there Chiteng'a John Chikunji 581 exist elements m 1 ,m 2 ,...,m h ∈ J and automorphisms σ 1 ,...,σ h ∈ Aut(R o ) such that (as R o -modules), m i r = r σi m i , for every r ∈ R o and any i = 1,...,h.Further, σ 1 ,...,σ h are uniquely determined by R and R o .The maximal ideal of R is It is worth noting that R contains an element b of multiplicative order p r − 1 and that The following results will be useful.
Proposition 2.1.Let R be a completely primary finite ring (not necessarily commutative).
Then the group of units R * of R contains a cyclic subgroup b of order p r − 1, and R * is a semidirect product of 1 + J and b .
Proof.Obviously, the group of units , and φ : R → R/J induces a surjective multiplicative group homomorphism ϕ : R * → (R/J) * .Since ker φ = J, we have kerϕ = 1 + J.In particular, 1 + J is a normal subgroup of R * .Let β = (R/J) * , and let b o ∈ ϕ −1 (β).Then, the multiplicative order of b o is a multiple of p r − 1 and a divisor of |R − J| = p nr − p (n−1)r = p (n−1)r (p r − 1); hence, of the form p s (p r − 1).But then b = b p s o has multiplicative order p r − 1 and ϕ(b p s o ) = β p s , which is still a generator of (R/J) * , since (p s , p r − 1) = 1.
Finally, since Proposition 2.2.Let R be a completely primary finite ring (not necessarily commutative).Then the group of units R * is solvable.
Proof.That R * is a solvable group follows from the fact that 1 + J is a normal p-subgroup of R * , and Proof.This follows from key properties of p-solvable groups contained in the variation of Sylow's theorem, due to Philip Hall, since the order of G is prime to its index in R * (see, e.g., [9, Theorem 8.2 page 25]).
Proposition 2.4.Let R be a completely primary finite ring (not necessarily commutative).If R * contains a normal subgroup of order p r − 1, then the set K o = b ∪ {0} is contained in the center of the ring R.
it follows that b and 1 + J commute elementwise.Hence, b lies in the center of R. Proposition 2.5.Let R be a completely primary finite ring.Then, (1 + J i )/(1 + J i+1 ) ∼ = J i /J i+1 (the left-hand side as a multiplicative group and the right-hand side as an additive group).
Proof.Consider the map (2.3) defined by Then it is easy to see that η is an isomorphism.
Remark 2.6 (see [3,Result 2.7]).Let R be a completely primary finite ring of characteristic p k and with Jacobson radical J. Let R o be a Galois subring of R. If m ∈ J and p t is the additive order of m, for some positive integer t, then |R o m| = p tr .
Proof.Apply the fact that Now let R be a commutative completely primary finite ring with maximal ideal J such that J 3 = (0) and J 2 = (0).In [2], the author gave constructions describing these rings for each characteristic and for details, we refer the reader to [2,Sections 4 and 6].
If R is a commutative completely primary finite ring with maximal ideal J such that J 3 = (0) and J 2 = (0), then from Constructions A and B [2], where the R o -modules U, V , and W are finitely generated.The structure of R is characterized by the invariants p, n, r, d, s, t, and λ; and the linearly independent matrices (a k i j ) defined in the multiplication.Let ann(J) denote the two-sided annihilator of J in R. Notice that since J 2 ⊆ ann(J), we can write R = R o ⊕ U ⊕ M, and hence, J = pR o ⊕ U ⊕ M, where M = V ⊕ W, and the multiplication in R may be written accordingly.It is therefore easy to see that the description of rings of this type reduces to the case where ann(J) coincides with J 2 .Therefore, when investigating the structure of the group of units of this type of rings for a given order, say p nr , where ann(J) does not coincide with J 2 , we will first write all the rings of this type of order ≤ p nr , where ann(J) coincides with J 2 .
In what follows, we assume that ann(J) ) and let nonnegative integers s and t be numbers of elements in the generating sets {u 1 ,...,u s } and {v 1 ,...,v t } for finitely generated R o -modules U and V , respectively, where t ≤ s(s + 1)/2.Assume that u 1 ,u 2 ,...,u s and v 1 ,...,v t are commuting indeterminates.Then By Proposition 2.1, and since R is commutative, a direct product.
Our goal is to determine the structure and identify a set of generators of the multiplicative abelian p-group 1 + J.

The group 1 + J
Now let R be a commutative completely primary finite ring with maximal ideal J such that J 3 = (0) and J 2 = (0).Let 1 + J be the abelian p-subgroup of the unit group R * .
Remark 3.1.Notice that 1 + J 2 is a normal subgroup of 1 + J. But, in general, 1 + J does not have a subgroup which is isomorphic to the quotient (1 + J)/(1 + J 2 ) as may be illustrated by the following example.
Example 3.2.Let R = Z p 3 , where p is an odd prime.Then J = pZ p 3 , ann(J) = J 2 , and Remark 3.3.In view of the above remark and example, we investigate the structure of 1 + J by considering various subgroups of 1 + J.
3.1.The case when s = 2, t = 1, and char R = p.Suppose s = 2, t = 1, and charR = p.Let R o = F q = GF(p r ), the Galois field of q = p r elements.Then the Jacobson radical The multiplication in R is given by where a i j ∈ F q .The elements a i j form a nonzero symmetric matrix a 11 a 12 a 21 a 22 (3.5) since J 2 = (0).Since R * is a direct product of the cyclic group b of order p r − 1 and the group 1 + J of order p 3r , it suffices to determine the structure of 1 + J.
In this case, and since s and t are fixed, the structure of 1 + J now depends on the prime p, the integer r, and the structural matrix a11 a12 a21 a22 .We investigate this by considering cases depending on the type of the structural matrix.
Case (i).Suppose that a11 a12 a21 a22 = a 0 0 0 , with a = 0. Then (3.7) To see this, we consider the two cases separetely.So, suppose that p = 2.We first note the following results: For positive integers k i , l i , with k i ≤ 4, l i ≤ 2, we assert that will imply k i = 4 for all i = 1,...,r; and l i = 2 for all i = 1,...,r.
If we set F i = {(1 + ε i u 1 ) k |k = 1,...,4} for all i = 1,...,r; and G i = {(1 + ε i u 2 ) l |l = 1,2} for all i = 1,...,r, we see that F i , G i are all cyclic subgroups of the group 1 + J and that these are of the precise orders indicated by their definition.The argument above will show that the product of 2r subgroups F i and G i is direct.So, their product will exhaust the group 1 + J.
When p is an odd prime, we have to consider the equation and as each element in 1 + J raised to the power p equals 1, we see that 1 + J will be an elementary abelian group.
Case (ii).Suppose that a11 a12 a21 a22 = 0 a a 0 , with a = 0. Then Chiteng'a John Chikunji 585 for every p = char R. In this case, we consider the equation and the integers k i , l i , m i will imply k i = l i = m i = p for all i = 1,...,r.
If we set F i = {(1 + ε i u 1 ) k |k = 1,..., p} for all i = 1,...,r; G i = {(1 + ε i u 2 ) l |l = 1,..., p} for all i = 1,...,r; and H i = {(1 + ε i v) m |m = 1,..., p} for all i = 1,...,r, we see that F i , G i , and H i are all cyclic subgroups of the group 1 + J and that these are all of order p.The product of the 3r subgroups F i , G i , and H i is direct.So, their product will exhaust the group 1 + J.
Case (iii).Suppose now that a11 a12 a21 a22 = a b b 0 , with a and b being nonzero.Then (3.13) The argument is similar to that in Case (i).
If char R = 2, then in 1 + J, we see that o(1 + ε i u 1 ) = 4 and for each ε i , by considering the element 1 + ε i u 1 + ε i u 2 + ε i v of order 2, one obtains the direct product (3.15) Case (v).Finally, suppose that a11 a12 a21 a22 = a b b c , with a, b, and c being nonzero.Then u 2 1 = av, u 2 2 = cv, and u 1 u 2 = u 2 u 1 = bv.In this case, it is easy to verify that (3.16) The number of cases involved in determining the structure of 1 + J for larger values of s and for t < s(s + 1)/2 compels us to investigate the problem by considering the extreme case when the invariant t = s(s + 1)/2, and to leave the other cases for subsequent work.

3.2.
The case when t = s(s + 1)/2, for s fixed.Suppose that t = s(s + 1)/2 for a fixed nonnegative integer s.Let u 1 ,u 2 ,...,u s be commuting indeterminates over the Galois ring R o = GR(p kr , p k ), where 1 ≤ k ≤ 3. Then it is easy to verify that where is a commutative completely primary finite ring with Jacobson radical In this case, the linearly independent matrices (a k i j ) defined in the multiplication of R are the t = s(s + 1)/2, s × s symmetric matrices with 1's in the (i, j)th and ( j, i)th positions, and zeros elsewhere.
It follows clearly that R o u i u j , (3.21) and it can easily be deduced that every element x of 1 + J has a unique expression of the form Proof.We only show the case for charR = p 2 , the other case follows easily from this.Now, each element of 1 + pR o is of the form 1 + pr, for every r ∈ R o , and for any two elements 1 + pr 1 and 1 + pr 2 , we have which is clearly an element of 1 + pR o .
Proposition 3.5.For each pair u i , u j with i = j and u i u j = u j u i , 1 + R o u i u j is a subgroup of 1 + J.
Proof.It is easy to see that 1 + R o u i u j is a subgroup of 1 + J because for any two elements 1 + r 1 u i u j and 1 + r 2 u i u j in 1 + R o u i u j , we have i .In view of Remark 2.6 and Propositions 3.4, 3.5, and 3.6, we may now state the following.
i , and 1 + R o u i u j be the subgroups of 1 + J defined above.Then for every characteristic of R.
Proposition 3.8.The group 1 + J is a direct product of the subgroup 1 + pR o , s subgroups , and s(s − 1)/2 subgroups 1 + R o u i u j , where i = j and u i u j = u j u i .Proof.This follows from the fact that 1 i , and 1 + R o u i u j are subgroups of 1 + J, intersection of any pair of these subgroups is trivial (for every i, j = 1,...,s), and by Proposition 3.7, (3.35)

3.2.1.
The structure of 1 + pR o .The structure of 1 + pR o is completely determined by Raghavendran in [11].For convenience of the reader, we state here the results useful for our purpose.For detailed proofs, refer to [11,Theorem 9].We take r elements ε 1 ,...,ε r in R o with ε 1 = 1 such that the set {ε 1 ,...,ε r } is a basis of the quotient ring R o / pR o regarded as a vector space over its prime subfield GF(p).Then we have the following.Proposition 3.9 [11,Theorem 9].If char R o = p 2 , then 1 + pR o is a direct product of r cyclic groups 1 + pε j , each of order p, for any prime p. Proposition 3.10 [11, Theorem 9].Let charR o = p 3 .If p = 2, then 1 + pR o is a direct product of 2 cyclic groups −1 + 4ε 1 and 1 + 4ε 1 , each of order 2, and (r − 1) cyclic groups 1 + 2ε j ( j = 2,...,r), each of order 4. If p = 2, then 1 + pR o is a direct product of r cyclic groups 1 + pε j ( j = 1,...,r), each of order p 2 .

The structure of
i .We now consider the structure of the subgroup 1 + R o u i + R o u 2 i of the p-group 1 + J.We first note that if charR o = p, then R o = GF(p r ) the field of p r elements, if charR o = p 2 , then R o is the Galois ring GR(p 2r , p 2 ) of order p 2r , and if charR o = p 3 , R o = GR(p 3r , p 3 ) the Galois ring of order p 3r .
We choose r elements ε 1 ,...,ε r in R o with ε 1 = 1 such that the set {ε 1 ,...,ε r } is a basis of the quotient ring R o / pR o regarded as a vector space over its prime subfield GF(p).Then we have the following.
Chiteng'a John Chikunji 589 i is a direct product of r cyclic groups 1 + ε j u i ( j = 1,...,r), each of order 4.
i is a direct product of 2r cyclic groups 1 + ε j u i and 1 + 2ε j u i ( j = 1,...,r), each of order p.
i is a direct product of r cyclic groups 1 + 2ε j u i , each of order 2, and r cyclic groups 1 + 3ε j u i ( j = 1,...,r), each of order 4.
i is a direct product of r cyclic groups 1 + pε j u i , each of order p, and r cyclic groups 1 + ε j u i ( j = 1,...,r), each of order p 2 .
i is a direct product of r cyclic groups 1 + ε j u 2 i , each of order p, and r cyclic groups 1 + ε j u i ( j = 1,...,r), each of order p 2 .
Proof.Similar to the proofs of Propositions 3.11 and 3.12.

3.2.3.
The structure of 1 + R o u i u j .Choose r elements ε 1 ,...,ε r in R o with ε 1 = 1 such that the elements ε 1 ,...,ε r form a basis of the quotient ring R o / pR o regarded as a vector space over its prime subfield GF(p).Then we have the following.
Proposition 3.14.The group 1 + R o u i u j is a direct product of r cyclic groups 1 + ε l u i u j (l = 1,...,r), each of order p, for any characteristic p k (1 ≤ k ≤ 3) of R.
Proof.We first note that if the characteristic of R is p k , where 1 ≤ k ≤ 3, then pu i u j = 0. Hence, |1 + R o u i u j | = p r .Also, for any x ∈ 1 + R o u i u j , x p = 1.Now, for r elements ε 1 ,...,ε r ∈ R o defined above, since for any ν = µ, the result follows.
We now state the main results of this section.
Theorem 3.15.Let charR = p.If p = 2, then 1 + J is a direct product of (s(s − 1)/2)r cyclic groups, each of order 2, and sr cyclic groups, each of order 4. If p = 2, then 1 + J is a direct product of ((s 2 + 3s)/2)r cyclic groups, each of order p.
Proof.This follows from Propositions 3.11 and 3.14 and by the fact that the order of 1 + J is p ((s 2 +3s)/2)r .
Theorem 3.16.Let char R = p 2 .Then 1 + J is a direct product of ((s 2 + s + 2)/2)r cyclic groups, each of order p, and sr cyclic groups, each of order p 2 , for any prime p.
Theorem 3.17.Let char R = p 3 .If p = 2, then 1 + J is a direct product of 2 + ((s 2 + s)/2)r cyclic groups, each of order 2, and r − 1 + sr cyclic groups, each of order 4. If p = 2, then 1 + J is a direct product of ((s 2 + s)/2)r cyclic groups, each of order p, and (s + 1)r cyclic groups, each of order p 2 .

The Main theorem
By Proposition 2.1, the group of units R * of R contains a cyclic subgroup b of order p r − 1, and R * is a direct product of 1 + J and b .Moreover, the structure of 1 + J has been determined in Section 3 (Theorems 3.15, 3.16, and 3.17).We thus have the following result.
Theorem 4.1.The group of units R * , of a commutative completely primary finite ring R with maximal ideal J such that J 3 = (0) and J 2 = (0), and with invariants p, k, r, s, and t, where t = s(s + 1)/2, is a direct product of cyclic groups as follows: (iii) if char R = p 3 , then where γ = (s 2 − s)/2.
Remark 4.2.The structure of the multiplicative groups of commutative completely primary finite rings R with maximal ideals J such that J 3 = (0) and J 2 = (0), for which t < s(s + 1)/2 for a fixed nonnegative integer s, will be considered in subsequent work.