PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 95 (109) (2014), 215–220 DOI: 10.2298/PIM1409215C

UNIT GROUPS OF FINITE RINGS WITH PRODUCTS OF ZERO DIVISORS IN THEIR COEFFICIENT SUBRINGS

Chiteng’a John Chikunji

Communicated by Siniša Crvenković

Abstract. Let R be a completely primary finite ring with identity 1 ≠ 0 in which the product of any two zero divisors lies in its coefficient subring. We determine the structure of the group of units G_{R} of these rings in the case when R is commutative and in some particular cases, obtain the structure and linearly independent generators of G_{R}.

1. Introduction

All rings considered in this paper are associative (but not necessarily commutative) with identity element 1 ≠ 0. Let R be a completely primary finite ring with unique maximal ideal J. It is easy to see (cf. [3]) that |R| = p^{nr} , |J | = p^{(n−1)r} , and the characteristic of R is p^{k}, for some prime p and positive integers n, k and r with 1 ⩽ k ⩽ n. If k = n, then R is of the form ℤ_{p^{n}} [x]/(f) and R = ℤ_{p^{n}} [b], where ℤ_{p^{n}} is the ring of integers modulo p^{n}, f(x) is a monic polynomial over ℤ_{p^{n}} and irreducible modulo p, and b is an element of R of multiplicative order p^{r} − 1. These rings are uniquely determined by the integers p, n, r; they are called Galois rings and we shall denote them by GR(p^{n}, p^{nr} ).

Let R be a commutative completely primary finite ring. It is well known that any two coefficient subrings of R are conjugate (cf. [2]). Also if R_{0} is a coefficient subring of R, then there exist u_{1}, …, u_{h} in J such that

R = R_{0} ⊕ R_{0}u_{1} ⊕ ⋯⊕ R_{0}u_{h} (as R_{0}-modules)

and u_{i}r = ru_{i}, for all r in R_{0} and for all i = 1, …, h. (This is a direct consequence of Theorems 2-2 and 2-4 in [4]).

Throughout this paper, for a given commutative completely primary finite ring R with maximal ideal J, let 𝔽 = R/J, and let 𝔽^{∗} and G_{R} denote the multiplicative group of units of 𝔽 and R, respectively.

2010 Mathematics Subject Classification: Primary 16P10, 16U60; Secondary 20K01, 20K25.

Key words and phrases: Completely primary finite rings, Galois rings.

Let R_{0} = GR(p^{n}, p^{nr} ), R be a commutative completely primary finite ring with J^{2} contained in R_{0} and let u_{1}, …, u_{h} be elements in J. Since R_{0} ∩ R_{0}u_{i} = 0 and the product of any two zero divisors is in R_{0}, we have that pu_{i} = 0 for all i = 1, …, h. But u_{i}u_{j} is an element of pR_{0}; thus u_{i}u_{j} is an element of p^{n−1} R_{0}, for all i = 1, …, h. Suppose that u_{i}u_{j}, u_{i}u_{k} are non-zero elements of pR_{0} with j ≠ k. Then u_{i}u_{j} R_{0} = u_{i}u_{k}R_{0} = p^{n−1} R_{0} and we get u_{i}u_{j} = u_{i}u_{k}α, where α is an element of ⟨b⟩. Thus, u_{j} − u_{k}α is an element of ann(u_{i}), the annihilator of u_{i}, and subsequently it is contained in pR_{0} ⊕ R_{0}u_{1} ⊕ ⋯⊕ R_{0}u_{h} (j ≠ k). This implies that u_{j} is an element of pR_{0} ⊕ R_{0}u_{1} ⊕ ⋯⊕ R_{0}u_{h}, which is a contradiction. Therefore, for all i = 1, …, h, either u_{i}u_{j} is zero for all j = 1, …, h or u_{i}u_{j} is non-zero for only one j = 1, …, h. We assume w is the number of u_{i} such that u_{i}u_{j} is zero for all j = 1, …, h and λ is the number of the other u_{i}. Let us reindex u_{1}, …, u_{h} in such way that for each i = 1, …, λ, there exists only one j = 1, …, h with u_{i}u_{j} = p^{n−1} α_{ij} , where α_{ij} is an element of ⟨b⟩, and let θ be the function from {1, …, λ} to {1, …, h} determined by θ(i) = j. Clearly, θ is injective.

Let s be the number of i in {1, …, λ} such that θ(i) = i and t be λ − s. We reindex u_{1}, …, u_{λ} such that θ(i) = i for all i = 1, …, s and suppose α_{iθ(i)} = β_{i} for all i = 1, …, s. Put v_{e} = u_{e} for all i = 1, …, s and v_{e} = u_{e}α_{e} for all i = s+1, …, h, where if e is in the image of θ, say e = θ(i), then α_{e} = 1. Thus, u_{i}u_{θ(i)} = p^{n−1} for all i = s+1, …, λ. Hence, either u_{i}^{2} = 0, u_{i}^{2} = p^{n−1} or u_{i}^{2} = αp^{n−1} , α ∈ ⟨b⟩−{0, 1}; and u_{i}u_{j} = 0 for all i ≠ j.

2. Construction A

Let R_{0} be a Galois ring of the form GR(p^{n}, p^{nr} ) and 𝔽 be R_{0}/pR_{0}. Let U be an 𝔽-space which when considered as an R_{0}-module has a generating set {u_{1}, …, u_{h}} such that pu_{i} = 0 for all i = 1, …, u_{h}. Also assume that s, t, w are non-negative integers such that h = s + t + w and suppose that θ is an injective function from {s + 1, …, s + t} to {s + 1, …, h}. On the additive group

R = R_{0} ⊕ R_{0}u_{1} ⊕ ⋯⊕ R_{0}u_{h},

define the multiplication as follows:

u_{i}u_{j} = 0, for i ≠ j (1 ⩽ i, j ⩽ h);
u_{i}^{2} = α_{i}p^{n−1} , for i = 1, …, s;
u_{i}^{2} = p^{n−1} , for i = s + 1, …, s + t;
u_{i}^{2} = 0, for i = s + t + 1, …, h;
u_{i}r^{∗} = r^{∗}u_{i}, for all i = 1, …, h;

where α_{i} are non-trivial elements of 𝔽^{∗} and r^{∗} is the image of r under the canonical homomorphism from R_{0} to 𝔽 ≅ R_{0}/pR_{0}.

It can easily be verified that R is an associative ring with identity 1 ≠ 0.

Theorem 2.1. Let R be a commutative completely primary finite ring. Then the product of any two zero divisors is an element of its coefficient subring R_{0} if and only if R is one of the rings given by Construction A.

The proof follows from the discussion before Construction A; the converse that J^{2} lies in R_{0} is easy to check.

These rings were studied by Alkhamees [1], who gave their complete general construction for both commutative and non-commutative cases.

We notice that charR = p^{n}; J = pR_{0} ⊕ R_{0}u_{1} ⊕ ⋯⊕ R_{0}u_{h}, J^{2} = pR_{0}, and J^{n} = 0. Also, notice that |R| = p^{(n+h)r} , |J | = p^{(n+h−1)r} and hence, R/J ≅ 𝔽_{p^{r}} .

3. The group of units of R

There are many important results on the group of units of certain finite rings. For example, it is well known that the multiplicative group of the finite field GF (p^{r}) is a cyclic group of order p^{r}−1, and the multiplicative group of the finite ring ℤ/p^{k}ℤ, the ring of integers modulo p^{k}, for p a prime number, and k a positive integer, is a cyclic group of order p^{k−1} (p − 1).

Let G_{R_{0}} denote the group of units of the Galois ring R_{0} = GR(p^{n}, p^{nr} ). Then G_{R_{0}} has the following structure [3]:

Theorem 3.1. G_{R_{0}} = ⟨b⟩ × (1 + pR_{0}), where ⟨b⟩ is the cyclic group of order p^{r} − 1 and 1 + pR_{0} is of order p^{(n−1)r} whose structure is described below.

(i) If (a) p is odd, or (b) p = 2 and n ⩽ 2, then 1 + pR_{0} is the direct product of r cyclic groups each of order p^{(n−1)} .

(ii) When p = 2 and n ⩾ 3, the group 1 + pR_{0} is the direct product of a cyclic group of order 2, a cyclic group of order 2^{(n−2)} and r−1 cyclic groups each of order 2^{(n−1)} .

We now determine the structure of the group of units of this paper. We first recall that

G_{R} = ⟨b⟩ × (1 + J), |G_{R}| = |R| − |J | = p^{(n+h)r} − p^{(n+h−1)r}

and in fact |1 + J | = p^{(n+h−1)r} .

To simplify the problem, we split our study into two cases, namely,

(1) the case when u_{j}^{2} = 0 for every j = 1, …, h; and

(2) the case when u_{j}^{2} = α_{j} p^{n−1} , where α_{j} ∈ 𝔽^{∗} for every j = 1, …, h.

We shall use the information from the two cases in order to obtain the general structure of G_{R} (see Theorem 4.1). We treat the cases separately.

Let ε_{1}, ε_{2}, …ε_{r} be elements of R_{0} with ε_{1} = 1 so that \overline{ε_{1}}, \overline{ε_{2}}, …, \overline{ε_{r}} ∈ R_{0}/pR_{0} ≅ GF (p^{r}) form a basis of GF (p^{r}) over its prime subfield GF (p).

3.1. The case when u_{i}^{2} = 0 for every i = 1, …, h. In this subsection, we determine the structure of the group of units G_{R} of the ring R in the case when u_{i}^{2} = 0 for every i = 1, …, h.

Proposition 3.1. Let R be a ring given by construction A and suppose that u_{j}^{2} = 0 for every j = 1, …, h. Then

G_{R} ≅ \{ \begin{matrix}ℤ_{2^{r} −1} × ℤ_{2} × ℤ_{2^{n−2}} × ℤ_{2^{n−1}}^{r−1} × \underbrace{ℤ_{2}^{r} × ⋯× ℤ_{2}^{r}}_{h} & if p = 2, \\ ℤ_{p^{r} −1} × ℤ_{p^{n−1}}^{r} × \underbrace{ℤ_{p}^{r} × ⋯× ℤ_{p}^{r}}_{h} & if p is odd. \end{matrix}

Proof. We know that

R = R_{0} ⊕ R_{0}u_{1} ⊕ ⋯⊕ R_{0}u_{h}, J = pR_{0} ⊕ 𝔽u_{1} ⊕ ⋯⊕ 𝔽u_{h},

where u_{i} ∈ J, 𝔽 ≅ R_{0}/pR_{0}, J^{n−1} ≠ (0), and J^{n} = (0). Moreover,

G_{R} ≅ (⟨b⟩) × (1 + J),

where ⟨b⟩ is a cyclic group of order p^{r} − 1, for every prime number p. We need to determine the structure and linearly independent generators of 1 + J in order to complete the proof.

Since pu_{j} = 0 for all j = 1, …, h, u_{i}u_{j} = 0 for all 1 ⩽ i, j ⩽ h, and u_{j}^{2} = 0 for every j = 1, …, h, one easily sees that (1 + R_{0}u_{i}) ∩ (1 + R_{0}u_{j}) = {1}. Moreover, (1 + pR_{0}) ∩ (1 + R_{0}u_{j}) = {1}, for all j = 1, …, h. Further, it is easy to verify that 1 + R_{0}u_{1} ⊕ ⋯⊕ R_{0}u_{h} is a subgroup of 1 + J and hence,

1 + J = (1 + pR_{0}) × (1 + R_{0}u_{1} ⊕ ⋯⊕ R_{0}u_{h}),

a direct product.

The structure of 1 + pR_{0} is well known, for example, see Theorem 3.1. We now determine the structure of 1 + R_{0}u_{1} ⊕ ⋯⊕ R_{0}u_{h}. For any prime p and for each i = 1, …, r, we see that (1 + ε_{j} u_{1})^{p} = 1, (1 + ε_{j} u_{2})^{p} = 1, …, (1 + ε_{j} u_{h})^{p} = 1, and g^{p} = 1 for all g ∈ 1 + R_{0}u_{1} ⊕ ⋯⊕ R_{0}u_{h}.

For integers l_{ij} ⩽ p, we asset that ∏ _{i=1}^{r} ∏ _{j=1}^{h} {(1 + ε_{i}u_{j})^{l_{ij}} = 1, will imply l_{ij} = p, for all i = 1, …, r and j = 1, …, h.

If we set E_{ij} = {(1 + ε_{i}u_{j})^{l_{ij}} : l_{ij} = 1, …, p} for all i = 1, …, r, then we see that E_{ij} are all subgroups of 1 + R_{0}u_{1} ⊕ ⋯⊕ R_{0}u_{h} and that these are all of order p as indicated in their definition.

The argument above will show that the product of the hr subgroups E_{ij} is direct. Thus, their product will exhaust 1 + R_{0}u_{1} ⊕ ⋯⊕ R_{0}u_{h}, and this completes the proof. □

3.2. The case when u_{j}^{2} = α_{j} p^{n−1} for every j = 1, …, h; where α_{j} ∈ 𝔽^{∗}.

We now consider the second case.

Proposition 3.2. Let R be a ring given by construction A and suppose that u_{j}^{2} = α_{j} p^{n−1} for every j = 1, …, h, where α_{j} ∈ 𝔽^{∗}. Then

1 + J ≅ \{ \begin{matrix}ℤ_{2} × ℤ_{2^{n−3}} × ℤ_{2^{n−2}}^{r−1} × ℤ_{4}^{r} × \underbrace{ℤ_{2}^{r} × ⋯× ℤ_{2}^{r}}_{h−1} & if p = 2, \\ ℤ_{p^{n−1}}^{r} × \underbrace{ℤ_{p}^{r} × ⋯× ℤ_{p}^{r}}_{h} & if p is odd. \end{matrix}

Proof. The argument of the proof is similar to the proof of Proposition 3.2. Since pu_{j} = 0 for all j = 1, …, h, u_{i}u_{j} = 0 for all 1 ⩽ i, j ⩽ h, and u_{j}^{2} = αp^{n−1} for every j = 1, …, h, and a fixed α, one easily verifies that if p = 2, (1 + ε_{i}u_{j})^{4} = 1, for every i = 1, …, r and j = 1, …, h; and if p is odd, (1 + ε_{i}u_{j})^{p} = 1, for every i = 1, …, r and j = 1, …, h. This difference, in turn, breaks into two cases to consider.

Suppose first that p is an odd prime number. For each i = 1, …, r and j = 1, …, h, we see that for elements 1 + pε_{i}, 1 + ε_{i}u_{j} in 1 + J, (1 + pε_{i})^{p^{n−1}} = 1 and (1 + ε_{i}u_{j})^{p} = 1.

For positive integers m_{i} ⩽ p^{n−1} and l_{ij} ⩽ p, we assert that the equation

∏_{i=1}^{r} {(1 + pε_{i})^{m_{i}} } · ∏_{i=1}^{r} ∏_{j=1}^{h} {(1 + ε_{i}u_{j})^{l_{ij}} } = 1,

will imply m_{i} = p^{n−1} , for all i = 1, …, r, and l_{ij} = p, for all i = 1, …, r and j = 1, …, h.

If we set

E_{i} = {(1 + pε_{i})^{m_{i}} : m_{i} = 1, 2, …, p^{n−1} },
F_{ij} = {(1 + ε_{i}u_{j})^{l_{ij}} : l_{ij} = 1, …, p},

we see that E_{i}, and F_{ij} , are all cyclic subgroups of 1 + J and that these are all of the precise orders indicated by their definition.

The argument above shows that the product of the (1 + h)r subgroups E_{i}, F_{ij} is direct. So, their product will exhaust 1 + J; and we see that the proof for the case when p is odd is complete.

We now assume that p = 2. We remark that there exists at least one element β in R_{0} such that the equation x^{2}+x+β̄ = ¯0 over R_{0}/pR_{0} has no solution in R_{0}/pR_{0}. We then note that for elements (−1+2^{n−1} ε_{1}), (1+4β)^{2} = (1+8β+16β^{2}), (1+ε_{i}u_{j}) and (1 + ε_{i}u_{j} + ε_{i}u_{j+1} ) in 1 + J, (−1 + 2^{n−1} ε_{1})^{2} = 1, (1 + 8β + 16β^{2})^{2^{n−3}} = 1, (1 + ε_{i}u_{j})^{4} = 1 for all i = 1, …, r; and j = 1, …, h; and for a u_{j}^{2} = α2^{n−1} with α fixed for every j = 1, …, h; (1 + ε_{i}u_{j} + ε_{i}u_{j+1} )^{2} = 1, for all i = 1, …, r and j = 1, …, h − 1.

For positive integers k ⩽ 2, l ⩽ 2^{n−3} , m_{i} ⩽ 4 and n_{ij} ⩽ 2, we assert that the equation

(−1 + 2^{n−1} ε_{1})^{k} · (1 + 8β + 16β^{2})^{l}

· ∏_{i=1}^{r} {(1 + ε_{i}u_{1})^{m_{i}} } · ∏_{i=1}^{r} ∏_{j=1}^{h−1} {(1 + ε_{i}u_{j} + ε_{i}u_{j+1} )^{n_{ij}} } = 1,

will imply k = 2, l = 2^{n−3} , m_{i} = 4 for all i = 1, …, r; and n_{ij} = 2 for all i = 1, …, r and j = 1, …, h − 1.

If we set

E_{1} = {(−1 + 2^{n−1} ε_{1})^{k} : k = 1, 2},
E_{2} = {(1 + 8β + 16β^{2})^{l} : l = 1, …, 2^{n−3} },

E_{i1} = {(1 + ε_{i}u_{1})^{m_{i}} : m_{i} = 1, …, 4},
F_{ij} = {: (1 + ε_{i}u_{j} + ε_{i}u_{j+1} )^{n_{ij}} : n_{ij} = 1, 2},

then we see that E_{1}, E_{2}, E_{i1}, F_{ij} are all cyclic subgroups of 1 + J and that these are all of the precise orders indicated by their definition.

The argument above shows that the product of the 2 + hr subgroups E_{1}, E_{2}, E_{i1}, F_{ij} is direct. So, their product will exhaust 1 + J, and we see that the proof for the case when p = 2 is complete.

This completes the proof of the theorem. □

4. Conclusion

We now state the structure of the group of units G_{R} of the ring R in general.

Theorem 4.1. Let R be a ring given by construction A and suppose that u_{j}^{2} = α_{j} p^{n−1} , for every j = 1, …, s, where α_{j} ∈ 𝔽^{∗} and for j = s + 1, …, h, u_{j}^{2} = 0. Then

1 + J ≅ \{ \begin{matrix}ℤ_{2} × ℤ_{2^{n−3}} × ℤ_{2^{n−2}}^{r−1} × ℤ_{4}^{r} × \underbrace{ℤ_{2}^{r} × ⋯× ℤ_{2}^{r}}_{s−1}× \underbrace{ℤ_{2}^{r} × ⋯× ℤ_{2}^{r}}_{h−s} & if p = 2, \\ ℤ_{p^{n−1}}^{r} × \underbrace{ℤ_{p}^{r} × ⋯× ℤ_{p}^{r}}_{h} & if p > 2, \end{matrix}

and hence,

G_{R} ≅ \{ \begin{matrix}ℤ_{2^{r} −1} × (1 + J) & if p = 2, \\ ℤ_{p^{r} −1} × (1 + J) & if p is odd. \end{matrix}

Proof. Follows from Propositions 3.2 and 3.3. □

References

1. Y. Alkhamees, Finite completely primary rings in which the product of any two zero divisors of a ring is in its coefficient subring, Internat. J. Math. Math. Sci. 17:3 (1994), 463–468.

2. W. E. Clark, A coefficient ring for finite non-commutative rings, Proc. Amer. Math. Soc. 33 (1972), 25–28.

3. R. Raghavendran, Finite associative rings, Compositio Math. 21 (1969), 195–229.

4. B. R. Wirt, Finite non-commutative rings, PhD Thesis, University of Oklahoma, 1972.

Department of Basic Sciences (Received 27 08 2012) Botswana College of Agriculture Gaborone Botswana

    jchikunj@bca.bw