Projections of Finite Rings

Let R and R^φ be associative rings with isomorphic subring lattices, and φ be a lattice isomorphism (or else a projection) of the ring R onto the ring R^φ. We call R^φ the projective image of a ring R and call R itself the projective preimage of a ring R^φ. The main result of the first part of the paper is Theorem 5, which proves that the projective image R^φ of a one-generated finite p-ring R is also one-generated if R^φ at the same time is itself a p-ring. In the second part, we continue studying projections of matrix rings. The main result of this part is Theorems 6 and 7, which prove that if R = M_n(K) is the ring of all square matrices of order n over a finite ring K with identity, and φ is a projection of the ring R onto the ring R^φ, then R^φ = M_n(K′), where K′ is a ring with identity, lattice-isomorphic to the ring K.

Introduction

Associative rings are considered. The lattice of all subrings of an arbitrary ring R is denoted by L(R). Two rings R and R^′ are said to be lattice-isomorphic if their subring lattices L(R) and L(R′), respectively, are isomorpic. A lattice isomorphism (or else a projection) L(R) ≅ L(R′) is denoted by the letter φ and the ring R′ is denoted as R^φ. We call R^φ the projective image of a ring R and call R itself the projective preimage of a ring R^φ. If the lattice isomorphism L(R) ^≅ L(R^φ) implies an isomorphism R ≅ R^φ, then we say that the ring R is lattice definable. We say that some class of rings $\mathcal{K}$ is lattice definable if the projective images of rings in the class $\mathcal{K}$ also belong to $\mathcal{K}$. In studying lattice isomorphisms of rings, we consider the problems of lattice definability of classes of rings and of lattice definability of rings.

Let k be a natural number. A finite ring R is said to be k-generated if any of its minimal sets of generators consists of k elements. For k = 1, R will be referred to as a one-generated ring. In [1, 2], it was proved that the property of being one-generated for rings is preserved under projections of nilpotent rings and finite fields. Lattice definability of Galois rings was stated in [3] (for a definition of a Galois ring, see [3, Def. 1]). However, one-generatedness of rings is not always preserved under projections [4, Example 2]. It is worth noting that the problem of preserving the property of being k-generated under lattice isomorphisms is important as for proving the lattice definability of rings as well as for constructing examples [5, Lemma 3, Example 2]. Let $\mathcal{K}$ _p be a class of rings whose additive groups are p-groups. The main results of the first part of the paper are Theorem 5 and Corollary 1, which prove that the property of being k-generated for a finite ring is preserved under projections in the class $\mathcal{K}$ _p. The relevance of these statements is strengthened by the fact that projective images of rings in $\mathcal{K}$ _p not infrequently themselves belong to $\mathcal{K}$ _p, which is confirmed in the second part of the paper devoted to projections of matrix rings.

Lattice isomorphisms of matrix algebras were first considered by D. W. Barnes [6], who proved that an algebra lattice-isomorphic to an algebra M_n(Δ), where n ≥ 2 and Δ is a finite-dimensional division algebra, is also a matrix algebra M_n(D) considered over some division algebra D lattice-isomorphic to Δ. A. V. Yagzhev [7] generalized Barnes’s result by lifting the restriction on the dimension of the algebra Δ. That matrix rings treated over different types of Galois rings are lattice definable was proved in [8, 9]. Lattice isomorphisms of matrix rings, when considered over finite local rings, were dealt with in [10]. The main results of the second part of this paper are Theorems 6 and 7, which prove that if R = M_n(K) is the ring of all square matrices of order n over a finite ring K with identity, and φ is a projection of the ring R onto the ring R^φ, then R^φ = M_n(K^′), where K′ is a ring with identity, lattice-isomorphic to the ring K. The question whether the ring R is lattice definable remains open.

We specify the notation used in the paper. Let S and T be subgroups in the additive group R⁺ of a ring R. If R = {s + t | s ∈ S ∧ t ∈ T}, then we use the notation R = S + T. And use R = S + T whenever R = S + T and S ∩ T = {0}. The equality R = S ⊕ T will mean that R = S + T and S and T are two-sided ideals in R. In this case we say that a ring R is decomposable into a direct sum of rings S and T. Other designations are standard: S ∨ T is a subring generated by subrings S and T in a ring R; Rad R is the Jacobson radical of a ring R; L(R) is the subring lattice of a ring R; l(R) is the length of a ring R, i.e., the length of its subring lattice L(R), where by the length of a finite lattice L is meant the greatest of the lengths of the chains in L; char R is the characteristic of a ring R; ℕ and ℤ are the sets of natural numbers and integers, respectively; $\langle {a}_{1},{a}_{2},... , {a}_{n}\rangle$⊃ is a ring generated by elements a₁, a₂,... , a_n; (r) is a principal ideal generated by an element r in R; o(r) is the additive order of an element r; ind r is the nilpotency index of an element r; the letters k, l, m, n, p, q with or without indices stand for natural numbers, and p and q also stand for prime numbers. Lower-case Greek letters, except φ, denote integers. The letter φ is used to denote a lattice isomorphism of the ring R onto the ring R^φ. In the cases when the projective image of a ring R generated by an element r is a one-generated ring, we denote $\langle r\rangle$ ^φ as $\langle r{\prime}\rangle$, in particular $\langle 0\rangle$ ^φ = $\langle 0{\prime}\rangle$.

1. PRELIMINARY INFORMATION

By a p-ring R we mean a ring whose additive group is a p-group. By an algebraic ring we mean a ring in which every element is a root of some primitive polynomial in the ring xℤ[x]. In proving a series of statements, we use a description of p-rings whose subring lattices are decomposable into direct products of lattices. For the reader’s convenience, we give this description.

THEOREM 1 [11, Thm. 1]. The subring lattice of a p-ring R is decomposable into a direct product of lattices if and only if R is an algebraic ring isomorphic to one of the rings

$$\begin{array}{l}{Q}_{1}=N\oplus P, L({Q}_{1}) \cong L(N)\times L(P),\\ {Q}_{2}=\left(N\oplus S\right)\dot{+}\langle e\rangle , L\left({Q}_{2}\right)\cong L\left(N\right)\times L\left(S \dot{+}\langle e\rangle \right),\\ {Q}_{3}={Q}_{2}\oplus P, L({Q}_{3})\cong L(N)\times L((S \dot{+}\langle e\rangle )\oplus P);\end{array}$$

here N, P are nonzero rings, N is a nil ring, P is a ring without nilpotent elements other than zero, S = {0} or S is a ring without nilpotent elements other than zero, and e is the identity element of order p in the ring Q₂.

As a criterion for one-generatedness of a finite commutative ring we use the following:

THEOREM 2 [4, Prop. 2]. Let a finite commutative p-ring R be representable as R = S + (r), where S is a subring decomposable into a direct sum of Galois rings and r is a nilpotent element. The ring R is generated by a single element if and only if the subring S is generated by a single element.

To study projective images of finite one-generated p-rings, we need a description of rings whose subring lattices are finite chains. We give this description.

THEOREM 3 [12, Thm. 1.6]. The subring lattice of a ring R is a finite chain if and only if R is isomorphic to one of the rings

$$\begin{array}{l}{C}_{1}=\langle r\rangle , \text{where} o\left(r\right)= {p}^{n},{r}^{n}={p}^{k}r,k=\overline{1,n}, \\ {C}_{2}=\langle r\rangle , \text{where} o\left(r\right)=p, \text{ind} r=3,\\ \begin{array}{l}{C}_{3}=\langle e\rangle , \text{where} o\left(e\right)={p}^{n}, {e}^{2}=e,\\ {C}_{4}=GF\left({p}^{{q}^{n}}\right).\end{array}\end{array}$$

2. PROJECTIONS OF k-GENERATED FINITE RINGS

Recall that by $\mathcal{K}$ _p we denoted a class of p-rings. Let k ∈ ℕ. It is easy to see that the property of a ring to be k-generated is preserved under projections in the class $\mathcal{K}$ _p if it is preserved for k = 1.

Therefore, in this section we mainly focus on the projections of one-generated finite p-rings.

PROPOSITION 1. Let R be a one-generated nilpotent ring of order pⁿ, and let φ be a lattice isomorphism of the ring R onto the ring R^φ. Then R^φ is a one-generated ring. If L(R) is not a chain, then R^φ is a nilpotent ring of order pⁿ; if, in addition, char R = char R^φ = p, then R ≅ R^φ.

Proof. Let R = $\langle r\rangle$ The ring R is finite and contains a unique maximal subring M = pR + R², so R^φ is a one-generated ring. Let R^φ = $\langle r{\prime}\rangle$.

Suppose that the lattice L(R) is not a chain. According to [1, Thm. 1], R^φ is a nilpotent p-ring. By [1, Lemma 7], |R^φ| = |R|. If char R = char R^φ = p, then R and R^φ are lattice-isomorphic finite- dimensional nilpotent algebras over a finite prime field GF (p). For finite-dimensional nilpotent algebras, the dimension of an algebra coincides with the length of its subalgebra lattice [6, p. 107]. Then ind R^φ − 1 = dim R^φ = l(R^φ) = l(R) = dim R = ind R − 1, whence ind R^φ = ind R. It is clear that indr = ind R = ind R^φ = ind R′, and so $\langle r\rangle$ ≅ $\langle r{\prime}\rangle$. The proposition is proved.

We move on to examining projections of one-generated nonnilpotent rings. Projections of one- generated finite rings with identity were studied in detail in [4], where sufficient conditions were found under which the projective image of a finite one-generated ring with identity likewise is a one-generated ring.

LEMMA 1. Let R be a finite nonnilpotent commutative ring without identity. The ring R is one-generated if and only if R is representable as

$$R=\langle t\rangle \oplus \langle r\rangle ,$$

(1)

where $\langle t\rangle$ is a subring with identity, decomposable into a finite direct sum of rings T_i = S_i + (r_i), i = $\stackrel{-}{1, n}$; moreover, S_i ≅ GR(pⁿⁱ, m_i), r_i is a nilpotent element, and the identity element of the subring S_i is also one in T_i, and r is a nonzero nilpotent element.

Proof. Necessity. Let R be one-generated. Since R is not a nil ring, it contains at least one nonzero idempotent element and, hence, contains a maximal orthogonal system of idempotents e′,..., e_n. Using the Peirce decomposition of R with respect to idempotents e_i, i = $\stackrel{-}{1, n}$, we represent R as R = T ⊕ N , where T = e′R ⊕ ··· ⊕ e_nR is a subring with identity and N is a nilpotent subring. As shown in [4, Lemma 1], if the ring R is one-generated, then so are its direct summands T and N : T = $\langle t\rangle$ and N = $\langle r\rangle$, where t, r ∈ R. Consequently, equality (1) holds. The structure of finite one-generated rings with identity is described in [4, Thm. 1]. According to that description, the ring $\langle t\rangle$ is decomposable into a finite direct sum of rings T_i = S_i + (r_i), i = $\stackrel{-}{1, n}$,, in which case S_i ≅ GR(pⁿⁱ, m_i), r_i is a nilpotent element, the identity of the subring S_i is also one in T_i.

Sufficiency. Let R be representable in form (1). Denote by e an identity element of the subring $\langle t\rangle$ and represent e as a polynomial in t, setting e = f (t), where f (x) ∈ xℤ[x]. Let v = t + r. Then f (v) = e + f (r). Let r^k = 0. Then (f (v))^k = e, and so e ∈ $\langle v\rangle$, whence $\langle v\rangle$ = R. The lemma is proved.

As follows from Lemma 1, the simplest one-generated rings not containing an identity are rings decomposable into a direct sum of two subrings, of which one is generated by an idempotent and the other, by a nilpotent element. A description of projective images of such rings is given in

PROPOSITION 2. Let φ be a lattice homomorphism of the p-ring R = $\langle e\rangle$ ⊕ $\langle r\rangle$, where o(e) = pⁿ, e² = e, and r is a nonzero nilpotent element, onto the ring R^φ. Then R^φ is a one- generated ring and the following conditions are satisfied:

(1) if n = 1 and $\langle r\rangle$ ≅ C_i, i = 1, 2 (cf. Theorem 3), then R^{φ ≅} P₁ or R^φ ≅ P₂, where P₁ = $\langle s{\prime}\rangle$⊕ $\langle t{\prime}\rangle$, o(s′) = p₁, $\langle t{\prime}\rangle$ is a p₂-ring, p₁, p₂ are primes, $\langle t{\prime}\rangle$ ^≅ C_j, and also either

(a) (s′)²= 0′ or (s′)² = s′ and j = $\stackrel{-}{1, 4}$, p₂ ≠ p₁, or

(b) (s′)² = 0′ and j = 3, 4, p₂ = p₁, or

(c) (s′)² = s′ and j = 1, 2, p₂ = p₁;

P₂ = $\langle s{\prime}\rangle$ + $\langle t{\prime}\rangle$, o(s′) = p₁, $\langle t{\prime}\rangle$ ≅ C_j, j = 1, 2, s′ is the identity of P₂;

(2) if n = 1 and the lattice L($\langle r\rangle$) is not a chain, then R^φ = $\langle e\rangle$ ^φ + $\langle r\rangle$ ^φ, $\langle r\rangle$ ^φ = $\langle r{\prime}\rangle$ is a nilpotent p-subring, $\langle e\rangle$ ^φ = $\langle s{\prime}\rangle$, o(s′) = q (q is a prime), and also

(a) for q = p, it is true that (s′)² = s′, s′r′ = r′s′ = 0′ or s′ is the identity of the ring R^φ;

(b) for q ≠ p, it is true that s′r′ = r′s′ = 0′, (s′)² = 0′ or (s′)² = s′;

(3) if n > 1, then R^φ is a q-ring, $\langle e\rangle$ ^φ = $\langle e{\prime}\rangle$, (e′)² = e′, o(e′) = qⁿ, $\langle r\rangle$ ^φ = $\langle r{\prime}\rangle$, r′ is a nilpotent element, and one of the following holds:

(a) R^φ = ˂e′˃ + ˂r′˃, where e′ is an identity, q = p;

(b) R^φ = ˂e′˃ ⊕ ˂v′˃, where v′ is a nilpotent element, q = p;

(c) R^φ ≅ GF (q^q1) ⊕ GF (q), and also n = 2, p = 2, q₁ is a prime.

Proof. The ring R satisfies the hypotheses of Theorem 3 in [13] and, therefore, contains exactly two maximal subrings. This means that the ring R itself and its projective image R^φ are one- generated. There are three cases to consider.

Case 1. Let n = 1 and $\langle r\rangle$ ≅ C_i, i = 1, 2. According to Theorem 1, L(R) ≅ L($\langle e\rangle$) × L($\langle e\rangle$). Let $\langle e\rangle$ ^φ = $\langle s{\prime}\rangle$ and $\langle r\rangle$ ^φ = $\langle t{\prime}\rangle$. By Theorem 3, o(s′) = p₁ and (s′)² = 0′ or (s′)² = s′, $\langle t{\prime}\rangle$ is a p₂-ring, and $\langle t{\prime}\rangle$ ≅ C_j, j = $\stackrel{-}{1, 4}$, p₁, p₂ are some primes. If p₂ ≠ p₁, then R^φ ≅ P₁ by [11, Thm. 2], and condition (1a) holds. Let p₂ = p₁. If R^φ does not contain an identity, then R^φ ≅ P₁ by Theorem 1, and condition (1b) or (1c) holds, and if R^φ is a ring with identity, then R^φ ≅ P₂.

Case 2. Let n = 1 and the lattice L($\langle r\rangle$) not be a chain. In view of Proposition 1, $\langle r\rangle$ ^φ = $\langle r{\prime}\rangle$ is a nilpotent p-subring. According to Theorem 1, L(R) is decomposable into a direct product of lattices L($\langle e\rangle$) and L($\langle r\rangle$). The subring $\langle e\rangle$ is an atom in the lattice L(R), so $\langle e\rangle$ ^φ = $\langle s{\prime}\rangle$, where s′ ∈ R^φ and o(s′) = q for some prime q.

Suppose q = p. Since L(R^φ) ≅ L($\langle s{\prime}\rangle$) × L($\langle r{\prime}\rangle$), and the subring $\langle r{\prime}\rangle$ is nilpotent, the subring $\langle e\rangle$ ^φ does not contain nonzero nilpotent elements, and we may therefore assume that (s′)² = s′. Applying Theorem 1, we conclude that either s′r′ = r′s′ = 0′ or s′ is the identity in the ring R^φ. In the latter case, it is obvious that o(r′) = p. If q ≠ p, then s′r′ = r′s′ = 0′, and either (s′)² = 0′ or (s′)² = s′. Conditions (2a) and (2b) hold.

Case 3. Let n > 1. If the subring $\langle r\rangle$ contains just p elements, then conditions (3a)-(3c) hold in view of [3, Lemma 7]. Therefore, below we assume that |$\langle r\rangle$| > p. Then a nilpotent subring S = $\langle pe\rangle$ ⊕ $\langle r\rangle$ contains more than p² elements. Furthermore, its subring lattice is not a chain, and hence the projective image S^φ is a p-ring, as follows by [1, Thm. 1]. Recall that R^φ is a commutative ring. Applying again [1, Thm. 1] to the subring S, we conclude that S^φ is a p-nil ring. This fact and Proposition 1 imply that $\langle r\rangle$ ^φ = $\langle r{\prime}\rangle$ is a nilpotent subring. In addition, R^φ is a p-ring because all minimal subrings of R^φ are contained in the subring S^φ. Since L($\langle e\rangle$) is a finite chain, it follows by Theorem 3 that the ring $\langle e\rangle$ ^φ either is a finite field or is generated by a nilpotent or idempotent element. The ring $\langle e\rangle$ ^φ cannot be a finite field since $\langle e\rangle$ ^φ ∩S^φ =$\langle pe\rangle$ ^φ is a nonzero nil ring. Suppose that $\langle e\rangle$ ^φ = $\langle s{\prime}\rangle$, where s′ is a nilpotent element. Then the ring R^φ is generated by two nilpotent elements s′ and r′, and R^φ being commutative implies that it will be nilpotent. We have arrived at a contradiction with [13, Lemma 2], which says that a finite p-ring that contains exactly two maximal subrings will contain a nonzero idempotent element. Hence $\langle e\rangle$ ^φ = $\langle e{\prime}\rangle$, where (e′)² = e′ and o(e′) = pⁿ. Thus the ring R^φ is generated by elements e′, r′ and contains exactly two maximal subrings. If the element e′ is an identity in the ring R^φ, then condition (3a) holds.

Assume that e′ is not an identity in R^φ and consider the Peirce decomposition: R^φ = e′R^φ ⊕ (1 − e′)R^φ. According to [13, Lemma 4], the subrings e′R^φ and (1 − e′)^φR each contains one maximal subring. The ring e′R^φ contains a nonzero idempotent element e′ of nonprime order. By [13, Thm. 1], one of the following two cases holds: either e′R^φ = $\langle e{\prime}\rangle$ or e′R^φ ≅ GR(pⁿ, ${q}_{2}^{m}$), where q₂ is a prime. If the second case holds, then [3, Thm. 3] says that the subring R contains a subring isomorphic to the ring GR(pⁿ, ${q}_{2}^{m}$). The ring R does not have such subrings since any subring of R containing an idempotent element e has the form $\langle e\rangle$ ⊕ T, where T ⊆ $\langle r\rangle$, and cannot be isomorphic to GR(pⁿ, ${q}_{2}^{m}$). Hence e′R^φ = $\langle e{\prime}\rangle$. If the subring (1 − e′)^φR is nonnilpotent, then it contains a nonzero idempotent element ${e}_{1}{\prime}$, and hence the ring R^φ will have two orthogonal idempotent elements e′ and ${e}_{1}{\prime}$. By virtue of [3, Lemmas 6, 8], the projective image of a subring $\langle e{\prime}\rangle$⊕ $\langle {e}_{1}{\prime}\rangle$ should also contain two nonzero orthogonal idempotent elements. Clearly, the ring R lacks such idempotent elements. Consequently, the subring (1 − e′)R^φ is generated by a nilpotent element. Let (1 − e′)R^φ = $\langle v{\prime}\rangle$. Then R^φ = $\langle e{\prime}\rangle$ ⊕ $\langle v{\prime}\rangle$, and condition (3b) holds. Proposition 2 is proved.

LEMMA 2. Let a finite one-generated p-ring R = S + (r) be given, where S = GR(pⁿ, m), m, n ∈ ℕ, r is a nilpotent element. Suppose also that φ is a projection from the ring R to the ring R^φ. Then R^φ is a one-generated ring.

Proof. Let r = 0. If, in addition, m = 1, then R = GR(pⁿ, 1) is a ring generated by an idempotent and L(R) is a finite chain. That the statement of the lemma is true in this case follows from Theorem 3. If n = 1 and m > 1, then R = GR(p, m) is a field of order p^m. In this case the statement of the lemma is true in virtue of [2, Thm. 2.1]. For n > 1 and m > 1, R ≅ R^φ according to [3, Thm. 4], and hence R^φ is one-generated.

Let r ≠ 0. Further we assume that m > 1 since R = GR(pⁿ, 1) ⊕ $\langle R\rangle$ for m = 1, and according to Proposition 2, the ring R^φ is one-generated.

Let e be an identity element of a ring S. By [4, Lemma 14], e is the unique nonzero idempotent element of the ring R. Suppose R is a ring with identity. Then e is the identity element in R. If n > 1, then [4, Lemma 17] implies that the statement of the lemma is true. Let n = 1. Then S ≅ GF (p^m), and by [5, Lemma 3], R^φ ≅ R. Hence the ring R^φ is generated by a single element.

Let R be a ring without identity. Consider the Peirce decomposition of R with respect to an idempotent e: R = eR ⊕ (1 − e)R. The subring (1 − e)R is nilpotent since e is the unique nonzero idempotent element of the ring R. Let r = s′ + r′, where s′ ∈ S and r′ ∈ (1 − e)R. Since R = S ∨ $\langle r\rangle$ = S ∨ $\langle {r}_{1}\rangle$ = S ⊕ $\langle {r}_{1}\rangle$, there is no loss of generality in assuming that er = 0, and so R = S ⊕ $\langle r\rangle$. There are two cases to consider.

Case 1. Let n = 1. Then S = GF (p^m). By Theorem 1, the subring lattice L(R) is decomposable into a direct product of lattices: L(R) ≅ L(S) × L($\langle r\rangle$). Consequently, L(R^φ) ≅ L(S^φ) × L($\langle r\rangle$ ^φ). In view of [2, Thm. 2.1], combined with Proposition 1, S^φ and $\langle r\rangle$ ^φ are one-generated rings. Let S^φ = $\langle s{\prime}\rangle$ and $\langle r\rangle$ ^φ = $\langle r{\prime}\rangle$. The rings $\langle s{\prime}\rangle$ and $\langle r{\prime}\rangle$ have primary additive groups since their subring lattices are not decomposable into direct products of lattices. Let $\langle s{\prime}\rangle$ be a p₁-ring and $\langle r{\prime}\rangle$ a p₂- ring, where p₁ and p₂ are primes. If p₁ ≠ p₂ then $\langle {s}{\prime}+r{\prime}\rangle$= $\langle s{\prime}\rangle$⊕ $\langle r{\prime}\rangle$ = R^φ. Let p₁ = p₂. The rings S^φ and $\langle r\rangle$ ^φ are one-generated, and by Theorem 1, R^φ is a commutative ring, and the set of its nilpotent elements is an ideal. Hence R^φ satisfies the hypotheses of Theorem 2 and is therefore one-generated.

Case 2. Let n > 1. By [3, Thm. 4], S^φ ≅ S, and by [4, Cor. 2], R^φ is a p-ring. Let a number m have the following canonical decomposition: m = ${q}_{1}^{{m}_{1}}$ ... ${q}_{k}^{{m}_{k}}$ , where q₁,... , q_k are primes.

According to [14, Lemma XVI.7], the ring S contains subrings G_i = GR(pⁿ, ${q}_{i}^{{m}_{i}}$ ), i = $\stackrel{-}{1, k}$. By [14, Thm. XVI.8], S = G₁ ∨ G₂ ∨ ··· ∨ G_k. In view of [13, Thm. 3], the ring S_i = G_i ⊕ $\langle r\rangle$, i = $\stackrel{-}{1, k}$, contains exactly two maximal subrings; hence the subring S^φ = G^φ ∨ $\langle r\rangle$ ^φ, i = $\stackrel{-}{1, k}$, is commutative, and so therefore is the ring R^φ itself. In accordance with Theorem 2, R^φ is a one-generated ring.

The lemma is proved.

It is well known that direct sums of Galois rings play an important role in finite ring theory. Projective images of finite rings decomposable into direct sums of different types of Galois rings were dealt with in [4]. The structure of projective images of one-generated finite rings decomposable into direct sums of Galois rings are described in the following:

THEOREM 4. Let R be a finite one-generated p-ring decomposable into a direct sum of n Galois rings. Suppose also that φ is a projection of the ring R onto the p-ring R^φ. Then R^φ is a one-generated ring, and if n > 1 and R ≅ GF (2^q) ⊕ GF (2), then R^φ is also decomposable into a direct sum of n Galois rings.

Proof. By hypothesis, R is a ring with identity. We represent R as follows: R = R₁ ⊕ R₂ ⊕ R₃, where R₁ = {0} or R₁ = GR(p^k1 , α₁) ⊕ ··· ⊕ GR(p^k , α_l) and (∀i = $\stackrel{-}{1, l}$) (k_i > 1, α_i > 1); R₂ = {0} or R₂ = GF (p^m1) ⊕ ··· ⊕ GF (p^ms) and (∀j = $\stackrel{-}{1, s}$) (m_j > 1); R₃ = {0} or R₃ = $\langle {v}_{1}\rangle$ ⊕ ··· ⊕ $\langle {v}_{n}\rangle$, ${v}_{i}^{2}$ = v_i, i = $\stackrel{-}{1, n}$. By hypothesis, the ring R is one-generated, and so are the subrings R₁, R₂, R₃ by [4, Lemma 1]. We prove that projective images of these subrings are one-generated.

Consider the ring R′. According to [4, Thm. 9], ${R}_{1}^{\varphi }$ = (GR(p^k1, α₁))^φ ⊕··· ⊕ (GR(p^kl, α_l))^φ and (∀i = $\stackrel{-}{1, l}$) = (GR(p^ki, α_i))^φ ≅ = (GR(p^ki, α_i)). Consequently, ${R}_{1}^{\varphi }$ ≅ R₁ and so ${R}_{1}^{\varphi }$ is a one-generated ring.

Consider the ring R₂. If s = 1, then R₂ is a finite field, and by [2, Thm. 2.1], ${R}_{2}^{\varphi }$ is a one generated ring. Let s > 1 and

$${R}_{2}={\underbrace{GF\left({p}^{{m}_{1}}\right)\oplus \cdots \oplus GF\left({p}^{{m}_{1}}\right)}_{{n}_{1}}}\oplus \cdots \oplus {\underbrace{GF\left({p}^{{m}_{s}}\right)\oplus \cdots \oplus GF\left({p}^{{m}_{s}}\right)}_{{n}_{s}}.}$$

In view of [4, Thm. 6],

$$\begin{array}{c}{R}_{2}^{\varphi }={\underbrace{{\left(GF\left({p}^{{m}_{1}}\right)\right)}^{\varphi }\oplus \cdots \oplus {\left(GF\left({p}^{{m}_{1}}\right)\right)}^{\varphi }}_{{n}_{1}}}\oplus \cdots \\ \oplus {\underbrace{{\left(GF\left({p}^{{m}_{s}}\right)\right)}^{\varphi }\oplus \cdots \oplus {\left(GF\left({p}^{{m}_{s}}\right)\right)}^{\varphi }}_{{n}_{s}},}\end{array}$$

with (∀i = $\stackrel{-}{1, s}$) = (GR(p^mi))^φ ≅ = (GR(${p}^{{m}_{i}{\prime}}$)), and if n_i > 1, then ${m}_{i}{\prime}$ = m_i by virtue of [4, Lemma 7]. The subring R₂ is generated by a single element, and according to [4, Prop. 1], for all i = $\stackrel{-}{1, s}$ the following condition is satisfied:

$${n}_{i}\le {N}_{p}\left({m}_{i}\right),$$

(2)

where N_p(m_i) is the number of all normed irreducible polynomials of degree m_i over the field GF(p). Applying [4, Prop. 1] to ${R}_{2}^{\varphi }$, we conclude that ${R}_{2}^{\varphi }$ is one-generated.

Consider the ring R₃. If R₃ = $\langle {v}_{1}\rangle$, then L(R₃) is a chain, and so ${R}_{3}^{\varphi }$ is a one-generated ring. Let R₃ = $\langle {v}_{1}\rangle$ ⊕ $\langle {v}_{2}\rangle$. If o(v₁) = o(v₂) = p, then, in view of R₃ being one-generated, we conclude that p ≠ 2. By hypothesis, R^φ is a p-ring. If we apply [6, Lemma 6] we conclude that ${R}_{3}^{\varphi }$ ≅ R₃. In all other cases, to R₃ we can apply [5, Prop. 2], according to which ${R}_{3}^{\varphi }$ ≅ R₃. The above argument implies that the theorem that we are proving is true if some two summands in the sum R = R₁ ⊕ R₂ ⊕ R₃ equal zero.

Let R₃ = {0} and R₁, R₂ be nonzero subrings. Then

$$R={\underbrace{GF\left({p}^{{k}_{1}},{m}_{1}\right)\oplus \cdots \oplus GF\left({p}^{{k}_{l}},{m}_{l}\right)}_{{R}_{1}}}\oplus {\underbrace{GF\left({p}^{{m}_{l+1}}\right)\oplus \cdots \oplus GF\left({p}^{{m}_{l+s}}\right)}_{{R}_{2}}.}$$

(3)

If among the positive integers m₁, . . .,m_l+s there are repetitions, then we number them anew, assigning all equal ones the same number. Let γ be the amount of the resulting numbers. Denote by n_i the amount of positive integers in the sequence m₁, . . .,m_l+s equal to m_i, 1 ≤ i ≤ γ. Taking into account that GF(p^m) = GR(p, m), we rewrite equality (3) in the following form:

$$R=\left({R}_{11}\oplus \cdots \oplus {R}_{{1n}_{1}}\right)\oplus \cdots \oplus \left({R}_{\gamma 1}\oplus \cdots \oplus {R}_{{\gamma n}_{\gamma }}\right),$$

(4)

where Rij ≅ GR(p^ki,m_i), i = $\stackrel{-}{1,\upgamma },$ j = $\stackrel{-}{1,{n}_{i}},$ and m₁, . . .,m_γ are pairwise distinct positive integers.

The ring R is one-generated, and [4, Prop. 1] says that the positive integers n_i, i = $\stackrel{-}{1,\gamma },$ satisfy equality (2). By [4, Thm. 11],

$${R}^{\varphi }=\left({R}_{11}^{\varphi }\oplus \cdots \oplus {R}_{1{n}_{1}}^{\varphi }\right)\oplus \cdots \oplus \left({R}_{\upgamma 1}^{\varphi }\oplus \cdots \oplus {R}_{\upgamma {n}_{\upgamma }}^{\varphi }\right),$$

(5)

with ${R}_{ij}^{\varphi }\cong {R}_{ij}$ or ${R}_{ij}^{\varphi }\cong GF\left({p}^{{m^{\prime}}_{j}}\right)$ for all i = $\stackrel{-}{1,\upgamma }$ and all j = $\stackrel{-}{1,{n}_{i}}.$ Consider an arbitrary summand T_i = $\left({R}_{i1}\oplus \cdots \oplus {R}_{{in}_{i}}\right)$ in the right part of (4) and its projective image ${T}_{i}^{\varphi }=\left({R}_{11}^{\varphi }\oplus \cdots \oplus {R}_{i{n}_{i}}^{\varphi }\right).$ Minimal polynomials of elements generating subrings ${R}_{ij}\left(j=\stackrel{-}{1,{n}_{i}}\right)$ have the same degree equal to mi. If n_i > 1, then [4, Lemma 7, Thm. 9, Lemma 13] says that minimal polynomials of elements generating subrings ${R}_{ij}^{\varphi }\left(j=\stackrel{-}{1,{n}_{i}}\right)$ have the same degree mi. If n_i = 1, then the degrees m_i and m′_i of minimal polynomials of elements generating rings T_i and ${T}_{i}^{\varphi },$ respectively, may differ; however, mi ≠ = m_k for i ≠ = k, and so m′_i ≠ m′_k, i.k = $\stackrel{-}{1,\gamma }.$ Consequently, inequality (2) for the ring R^φ holds, and by [4, Prop. 1], R^φ is a one-generated ring.

Let R₂ = {0} and R₁,R₃ be nonzero subrings. Then

$$R={\underbrace{GR\left({p}^{{k}_{1}},{m}_{1}\right)\oplus \cdots \oplus GR\left({p}^{{k}_{l}},{m}_{l}\right)}_{{R}_{1}}}\oplus {\underbrace{\langle {v}_{1}\rangle \oplus \cdots \oplus \langle {v}_{n}\rangle }_{{R}_{3}}},$$

(6)

where v₁, . . . , v_n are nonzero idempotent elements. Let e_i be the identity element of the ring $GR\left({p}^{{k}_{i}},{m}_{i}\right), i=\stackrel{-}{1,l}.$ By [3, property 9], ˂e_i˃^φ = ˂e′_i˃ and e′_i is the identity element in the ring $GR\left({p}^{{k}_{i}},{m}_{i}\right)$ ^φ. Consider a subring W = ˂e_i˃ ⊕⋯⊕ ˂e_l˃ ⊕ ˂v₁˃ ⊕⋯⊕ ˂v_n˃. By hypothesis, o(e₁) = ${p}^{{k}_{1}}$ and k₁ > 1. To the subring W, therefore, we can apply [5, Prop. 2], which says that W^φ ≅ W. Let ${W}^{\varphi }=\langle {e}_{1}{\prime}\rangle \oplus \cdots \oplus \langle {e}_{l}{\prime}\rangle \oplus \langle {v}_{1}{\prime}\rangle \oplus \cdots \oplus \langle {v}_{n}{\prime}\rangle ,$ where ${\left({v{\prime}}_{j}\right)}^{2}={v{\prime}}_{j}, j=\stackrel{-}{1,n}.$ At the beginning of the proof, it was noted that ${R}_{1}^{\varphi }\cong {R}_{1},$ so

$${R}^{\varphi }={\underbrace{{\left(GR\left({p}^{{k}_{1}},{m}_{1}\right)\right)}^{\varphi }\oplus \cdots \oplus {\left(GR\left({p}^{{k}_{l}},{m}_{l}\right)\right)}^{\varphi }}_{{R}_{1}^{\varphi }}}\oplus {\underbrace{\langle {v}_{1}{\prime}\rangle \oplus \cdots \oplus \langle {v}_{n}{\prime}\rangle }_{{R}_{3}^{\varphi }}\cong R.}$$

Consequently, being one-generated for R implies being one-generated for R^φ.

Let R₁ = {0} and R₂, R₃ be nonzero subrings. Then

$$R={\underbrace{GF\left({p}^{{m}_{1}}\right)\oplus \cdots \oplus GF\left({p}^{{m}_{s}}\right)}_{{R}_{2}}\oplus }{\underbrace{\langle {v}_{1}\rangle \oplus \cdots \oplus \langle {v}_{n}\rangle}_{{R}_{3}}},$$

(7)

where v₁, . . . , v_n are nonzero idempotent elements. We rewrite (7) in the form

$$\begin{array}{l}R={\underbrace{GF\left({p}^{{m}_{1}}\right)\oplus \cdots \oplus GF\left({p}^{{m}_{1}}\right)}_{{n}_{1}}\oplus \cdots \oplus }{\underbrace{GF\left({p}^{{m}_{s}}\right)\oplus \cdots \oplus GF\left({p}^{{m}_{s}}\right)}_{{n}_{1}}}\\ \oplus \langle {v}_{1}\rangle \oplus \cdots \oplus \langle {v}_{n}\rangle ,\end{array}$$

(8)

where, as above, n_i is the number of direct summands in the right part of (8) which are defined by a minimal polynomial of degree m_i, i = $\stackrel{-}{1,s}.$ Consider several versions.

(a) Let s = n₁ = n = 1, R = GF(p^m1 ) ⊕ ˂ v₁˃, and o(v₁) = p. If l(GF(p^m1 )) = 2, then it follows by [13, Thm. 3] that the ring R contains only two maximal subrings, and so both the rings R and R^φ are one-generated. If l(GF(p^m1 )) > 2, then l(R) > 3, and by [2, Cors. 3.1, 3.2]. the projective image R^φ is decomposable into a direct sum of two finite fields F′₁ and F′₂ of characteristic p. In view of [5, Lemma 7], the fields F′₁ and F′₂ are not isomorphic, and hence the ring R^φ satisfies the hypotheses of [4, Prop. 1] and is therefore one-generated.

(b) Let s = n₁ = n = 1, R = GF(p^m1 ) ⊕ ˂ v₁˃, o(v₁) = p^k, and k > 1. According to [4, Lemma 8, Remark 1], the rings R and R^φ are one-generated, with R^φ decomposable into a direct sum of two Galois rings.

(c) Let s = n₁ = 1, n > 1, R = GF(p^m1 ) ⊕ ˂ v₁˃ ⊕ · · · ⊕ ˂ v_n˃, and o(v_i) = p $\left(i=\stackrel{-}{1,n}\right).$ In this case [2, Cor. 3.1] implies that the ring R^φ is decomposable into a direct sum of n + 1 fields in characteristic p. Among these fields, only (GF(p^m1 )) ^φ is not a prime field, and the other n fields are isomorphic to GF(p). If the ring R is one-generated, then it satisfies [4, Cor. 1], and so n < p. This means that the ring R^φ satisfies the hypotheses of [4, Prop. 1] and is therefore also one-generated.

(d) Let s = n₁ = 1, n > 1, R = GF(p^m1) ⊕ ˂ v₁˃ ⊕· · · ⊕ ˂ v_n˃, o(v_i) = p^ki , k_i > 1, i = $\stackrel{-}{1,t},$ and 1 < t ≤ n. In this case, by [4, Lemma 9; 3, Lemma 6], R^φ = GF $\left({p}^{{m{\prime}}_{1}}\right) \oplus \langle {v}_{1}{\prime}\rangle \oplus \cdots \oplus \langle {v}_{n}{\prime}\rangle ,$ where $\langle {v}_{i}{\prime}\rangle \cong \langle {v}_{i}\rangle , i=\stackrel{-}{1,n}.$ By [4, Prop. 1], the rings R and R^φ are one-generated for n < p.

(e) Let s > 1. Suppose also that β = n₁ + · · · + n_s. In view of [4, Thm. 6],

$${R}_{2}^{\varphi }={\underbrace{GF\left({p}^{{m{\prime}}_{1}}\right)\oplus \cdots \oplus GF\left({p}^{{m{\prime}}_{1}}\right)\oplus \cdots \oplus GF\left({p}^{{m{\prime}}_{s}}\right)\oplus \cdots \oplus GF\left({p}^{{m{\prime}}_{s}}\right)}_{\beta },}$$

where $GF\left({p}^{{m{\prime}}_{i}}\right)={\left(GF\left({p}^{{m{\prime}}_{1}}\right)\right)}^{\varphi }, i=\stackrel{-}{1,\beta };$ moreover, in view of [4, Lemma 7], m′_i = m_i, for n_i > 1, and m′_i ≠ m′_k for ni = 1 and i ≠ k. Let e_i be the identity of a field GF(p^mi ), i = $\stackrel{-}{1,\beta },$ and ˂ e₁˃ ^φ = ˂ e′_i˃. Then e′_i is an identity of the field $GF\left({p}^{{m{\prime}}_{i}}\right),$ and e′₁,⋯, e′_β is an orthogonal system of idempotents. Consider a subring U = ˂ e₁˃ ⊕ ⋯ ⊕ ˂ e_β˃ ⊕ ˂ v₁˃ ⊕ ⋯ ⊕ ˂ v_n˃. Since β + n > 2, and R^φ is a p-ring, it follows by [5, Prop. 2] that R^φ contains idempotents v′_i, i = $\stackrel{-}{1,n},$ satisfying the following conditions: o(v′_i) = o(vi) (i = $\stackrel{-}{1,n}$) and U^φ = ˂ e′_i˃ ⊕ ⋯ ⊕ ˂ e′_β˃ ⊕ ˂ v′₁˃ ⊕ ⋯ ⊕ ˂ v′_n˃ Consequently,

$$\begin{array}{l}{R}^{\varphi }={\underbrace{GF\left({p}^{{m{\prime}}_{1}}\right)\oplus \cdots \oplus GF\left({p}^{{m{\prime}}_{1}}\right)}_{{n}_{1}}}\oplus \cdots \oplus {\underbrace{GF\left({p}^{{m{\prime}}_{s}}\right)\oplus \cdots \oplus GF\left({p}^{{m{\prime}}_{s}}\right)}_{{n}_{s}}}\\ \oplus \langle {v}_{1}{\prime}\rangle \oplus \cdots \oplus \langle {v}_{1}{\prime}\rangle .\end{array}$$

Numbers n_i, $i=\stackrel{-}{1,s}$, satisfy n_i ≤ Np(m′_i) and n < p, and by Theorem 2, R^φ is a one-generated subring.

Let R = R₁ ⊕ R₂ ⊕ R₃ and R_i ≠ = {0}, $i=\stackrel{-}{\text{1,3}}$. Consider a subring

$$K={\underbrace{GR\left({p}^{{k}_{1}},{m}_{1}\right)\oplus \cdots \oplus GR\left({p}^{{k}_{l}},{m}_{l}\right)}_{{R}_{1}}}\oplus {\underbrace{GF\left({p}^{{m}_{l+1}}\right)\oplus \cdots \oplus GF\left({p}^{{m}_{l+s}}\right)}_{{R}_{2}}}.$$

According to [4, Thm. 11],

$$\begin{array}{l}{K}^{\varphi }={\underbrace{{\left(GR\left({p}^{{k}_{1}},{m}_{1}\right)\right)}^{\varphi }\oplus \cdots \oplus {\left(GR\left({p}^{{k}_{l}},{m}_{l}\right)\right)}^{\varphi }}_{{R}_{1}^{\varphi }}}\\ { \oplus \underbrace{{\left(GF\left({p}^{{m}_{l+1}}\right)\right)}^{\varphi }\oplus \cdots \oplus {\left(GF\left({p}^{{m}_{l+s}}\right)\right)}^{\varphi }}_{{R}_{2}^{\varphi }}}.\end{array}$$

Let e_i be an identity of the ring GR(p^ki,m_i), $i=\stackrel{-}{1,l},$ and e_l+j be one of the field GF(p^ml+j ), $i=\stackrel{-}{1,s}$. Consider a subring E = ˂e₁˃ ⊕· · ·⊕ ˂ e_l+s ˃ ⊕ ˂v_i ˃ ⊕· · ·⊕ ˂v_n˃. Since R^φ is a p-ring, E^φ ≅ E by virtue of [5, Prop. 2]. Consequently, there exist idempotent elements v′_i ∈ R^φ, $i=\stackrel{-}{1,n},$ for which R^φ = K^φ ⊕ ˂v′₁˃ ⊕· · ·⊕ ˂v′_n˃. Rewriting the ring K in form (4) and applying equality (5), we obtain

$${R}^{\varphi }=\left({R}_{11}^{\varphi }\oplus \cdots \oplus {R}_{{1n}_{1}}^{\varphi }\right)\oplus \cdots \oplus \left({R}_{\gamma 1}^{\varphi }\oplus \cdots \oplus {R}_{{\gamma n}_{\gamma }}^{\varphi }\right)\oplus \langle {v}_{1}{\prime}\rangle \oplus \cdots \oplus \langle {v}_{n}{\prime}\rangle .$$

(9)

Numbers n_i, $i=\stackrel{-}{1,\gamma },$ in (9) satisfy the condition n_i ≤ N_p(m′_i), and the inequality n < p holds likewise. By [4, Prop. 1], R^φ is a one-generated ring. The theorem is proved.

One of the basic results of the paper is the following:

THEOREM 5. Let φ be a projection of a finite one-generated p-ring R onto a p-ring R^φ.

Then R^φ is a one-generated ring.

The proof is divided into two parts.

Part 1. Let R be a ring with identity. By [15, Thm. II.5], R is decomposable into a direct sum of n local rings R_i, $i=\stackrel{-}{1,n}$. In view of [4, Lemma 1], every subring R_i, $i=\stackrel{-}{1,n},$ is one-generated and is therefore representable as R_i = S_i +(r_i), where S_i = GR(p^ki,m_i), r_i is a nilpotent element, and the identity e_i of the subring S_i is also one in R_i. Let S = S₁ ⊕· · ·⊕ S_n and r = r₁ +· · · +r_n.

Then R = S + (r).

That the theorem that we are proving is true for n = 1 follows from Lemma 2.

Let n > 1. By Theorem 4, the subring S^φ is one-generated, and if r = 0, then we are done.

Let r ≠ 0. Consider the particular case where n = 2, S₁ ≅ GF(2^q), and S₂ ≅ GF(2). Then R = (S₁ ⊕ S₂)+(r). We prove that the ring (S₁ ⊕ S₂)^φ does not contain nonzero nilpotent elements. Assume the contrary. The subring S₁ ⊕ S₂ satisfies [3, Lemma 7]. Consequently, (S₁ ⊕ S₂)^φ contains nonzero nilpotent elements iff ${S}_{1}^{\varphi }$ is generated by an idempotent element of order 2². If e₁r = 0, then S₁r = {0}, and by Theorem 1, L(S₁ ⊕ ˂r˃) ≅ L(S₁) × L(˂r˃). In view of Theorem 1, a p-ring whose subring lattice decomposes into a direct product of lattices does not contain idempotent elements of nonprime additive order. If e₁r ≠ 0, then S₁ + (e₁r) is a local ring. According to [5, Lemma 3], (S₁ +(e₁r))^φ ≅ S₁ +(e₁r). This argument implies that the subring (S₁ ⊕ S₂)^φ contains no nonzero nilpotent elements and, therefore, decomposes into a direct sum of two finite fields.

Coming back to the general case and applying Theorem 4 to a subring S, we conclude that S^φ is a one-generated ring decomposable into a direct sum of n Galois rings.

Let $\left(\forall i=\stackrel{-}{1,n}\right) {S}_{i}^{\varphi }={S}_{i}{\prime}.$ The ring S^φ is one-generated, so $\left(\forall i,j=\stackrel{-}{1,n}\right) {S}_{i}{\prime}{S}_{j}{\prime}={S}_{j}{\prime}{S}_{i}{\prime}.$

Proposition 1 implies that ˂r˃^φ = ˂r′˃, where r′ ∈ R^φ. Then R^φ = ${S}_{1}{\prime}$ ∨ · · · ∨ ${S}_{n}{\prime}$ ∨ ˂r′˃. If we apply Lemma 2 to a subring S_i +(r) for every $i=\stackrel{-}{1,n}$ we conclude that ${S}_{1}{\prime}$ ∨ ˂r′˃ is a one-generated subring. Consequently, the ring R^φ is commutative.

By hypothesis, R is a ring with identity. Obviously, an identity in R is the element e = e₁ + · · · + e_n. Since er = r ≠ 0, there exists an index i ∈ {1, . . . , n} with which e_ir = r_i ≠ 0. Without loss of generality, we may assume that i = 1. Let ˂r₁˃^φ = $\langle {r}_{1}{\prime}\rangle .$ We prove that ${r}_{1}{\prime}$ is a nilpotent element. Consider a subring H = ˂e₁, r₁˃ ⊕ ˂e₂˃. By [13, Thm. 3], the subrings H₁ = ˂e₁, r₁˃ and H₂ = ˂e₂˃ ⊕ ˂r₁˃ each contains two maximal subrings, while H₃ = ˂e₁˃ ⊕ ˂e₂˃, in view of [3, Lemma 4], contains three maximal subrings. Let ˂e_i˃^φ = $\langle {e}_{i}{\prime}\rangle$, i = 1, 2. If o(e₁) > p or o(e₂) > p, then ${H}_{3}^{\varphi }\cong {H}_{3}$ by virtue of [3, Lemmas 6, 8]; so ${e}_{i}{\prime}$ are nonzero idempotents, and ${r}_{1}{\prime}$ is a nonzero nilpotent element by virtue of [13, Thm. 3]. Suppose o(e₁) = o(e₂) = p. Then o(r₁) = p, and so pH = {0}. According to Theorem 1, L(H) ≅ L(H₃) × L(˂r₁˃). Consequently, L(H^φ) ≅ L(${H}_{3}^{\varphi }$) × L(˂r₁˃^φ˃). In view of [11, Cor, 6], one of the subrings ${H}_{3}^{\varphi }$ or ˂r₁˃^φ is nilpotent and the other, on the contrary, contains no nonzero nilpotent elements. If the subring ${H}_{3}^{\varphi }$ is nilpotent, then p = 2 and ˂r₁˃^φ is a chain of length at most two. By Theorem 3, either ˂r₁˃^φ = GF(2) or ˂r₁˃^φ = GF(2q). According to [4, Cor. 4], the ring H is not generated by one element in either case. Therefore, R ≠ H. Hence two cases are possible.

Case (1). Let o(e₁) = o(e₂) = p and l(S₁) +l(S₂) > 2. Suppose l(S₁) > 1. Then the subring S₁ is a field, and S₁ + ˂r₁˃ is a local subring. In view of [16, Thm. 3], the subring ˂r₁˃^φ is nilpotent, which contradicts the assumption. If l(S₂) > 2, then the subring S₂ is a field. If we apply [2, Cor. 3.1] to a subring S₁ ⊕ S₂ we conclude that S₂ is a field of length 2. By Theorem 1, L(˂e₁, r₁˃ ⊕ S₂) ≅ L(˂e₁˃ ⊕ S₂) × L(˂r₁˃). The subrings ˂e₁˃^φ and ˂e₂˃^φ being nilpotent implies that so is ${S}_{2}^{\varphi }$ . Consequently, the subring (˂e₁˃ ⊕ S₂)^φ, too, is nilpotent. However, (˂e₁˃ ⊕ S₂) satisfies the hypotheses of [3, Lemma 7] and cannot be lattice-isomorphic to a nilpotent ring, a contradiction.

Hence Case (1) is impossible.

Case (2). Let n > 2. Consider a subring D = H ⊕ ˂e₃˃, where e₃ is the identity of the subring S₃. Let E = ˂e₁˃ ⊕ ˂e₂˃ ⊕ ˂e₃˃. By Theorem 1, L(H ⊕ ˂e₃˃) ≅ L(E) × L(˂r₁˃). According to [2, Lemma 3.1], E^φ ≅ E, and hence the assumption that the subring ${H}_{3}^{\varphi }$ is nilpotent is untrue. Consequently, nilpotent is the subring ˂r₁˃^φ . Thus the element r′₁ is nilpotent, as also are all elements r′₁ $\left(i=\stackrel{-}{1,n}\right).$

Let ˂r˃^φ = ˂r′˃ and ˂r_i˃^φ = ˂r′_i ˃, $i=\stackrel{-}{1,n}$. Since r ∈ ˂r₁, . . . , r_n˃, we have r′ ∈ ˂r′₁, . . . , r′_n˃, The ring R^φ is commutative, so the ring ˂r′₁, . . . , r′_n˃ is nilpotent. This implies that r′ is a nilpotent element. It is clear that R^φ = S^φ + (r′), and since S^φ is a one-generated subring, R^φ is a onegenerated ring, as follows by Theorem 2.

Part 2. Let R be a ring without identity. If R is nilpotent, then the result follows from Prop. 1.

Let R be nonnilpotent. By Lemma 1, the ring R is representable as R = ˂t˃ ⊕ ˂r˃, where ˂t˃ is a nonzero subring with identity, decomposable into a finite direct sum of rings Ti = Si+(ri), $i=\stackrel{-}{1,n}$, in which case S_i ≅ GR(pⁿⁱ,m_i), r_i is a nilpotent element, an identity element of a subring S_i is one in T_i, and r is a nonzero nilpotent element. The first part of the proof implies that the projective image of a subring ˂t˃ is one-generated. Let ˂t˃^φ = ˂t′˃. In view of Proposition 1, ˂r˃^φ = ˂r′˃ for some element r′ ∈ R^φ. Consequently, R^φ = ˂t′˃ ∨ ˂r′˃.

We prove that R^φ is a commutative ring. Consider a subring W_i = T_i⊕ ˂r˃ = (S_i + (r_i)) ⊕ ˂r˃ = S_i + (w_i), where w_i = r_i + r, $i=\stackrel{-}{1,n}$. Obviously, w_i is a nilpotent element. By hypothesis, every subring S_i, $i=\stackrel{-}{1,n},$ is one-generated, and so is the ring W_i in virtue of Theorem 2. By Lemma 2, the ring ${W}_{i}^{\varphi }$ is also one-generated. Since (∀$i=\stackrel{-}{1,n},$) (${W}_{i}^{\varphi }$ = ${T}_{i}^{\varphi }$ ∨ ˂r′˃ and ${T}_{1}^{\varphi }$ ∨· · ·∨ ${T}_{n}^{\varphi }$ = ˂t′˃), we have r′t′ = t′r′, and hence R^φ is commutative.

Let w = r₁+· · ·+r_n+r. Then w is a nilpotent element, and R = S+(w), where S = S₁ ⊕· · ·⊕ S_n. By hypothesis, the ring R is one-generated, and so is the ring S in virtue of Theorem 2. Since w is a nilpotent element, ˂w˃^φ is a one-generated ring. Let ˂w˃^φ = ˂w′˃. By Theorem 4, the ring S^φ is one-generated. If w′ is a nilpotent element, then R^φ = S^φ + (w′), and R^φ is a one-generated ring by Theorem 2.

Let N = <r₁> ⊕· · ·⊕ <r_n> ⊕ <r>. It is clear that w ∈ N. There are three cases to consider.

Case (a). Suppose (∀i = $\stackrel{-}{1,n}$) (r_i = 0). Then N = <r> and R = S ⊕ <r>. By Lemma 2, R^φ is a one-generated ring for n = 1.

Let n = 2 and S ≅ GF(2^q) ⊕ GF(2). According to Theorem 1, L(S ⊕ <r>) ≅ L(S)×L(<r>). By [11, Cor. 6], one of the subrings S^φ or <r> ^φ is nilpotent and the other contains no nonzero nilpotent elements. In view of [3, Lemma 7], no nil ring can be lattice-isomorphic to a ring S. Therefore, the subring <r> ^φ is nilpotent, and hence r′ is a nilpotent element. If n ≥ 2 and S ≇ GF(2^q) ⊕ GF(2), then it follows by Theorem 4 that S^φ is decomposable into a direct sum of n Galois rings, and so S^φ contains an orthogonal system of n nonzero idempotents. If the ring N^φ is nonnilpotent, then R^φ will contain an orthogonal system of n + 1 nonzero idempotents. Since n + 1 ≥ 3, it follows by [5, Prop. 2] that such an orthogonal system of idempotents should also be in R. Obviously, in the ring R = S₁ ⊕ · · · ⊕ S_n ⊕ <r> any orthogonal system of nonzero idempotents consists of not more than n elements. Thus if n > 1 and (∀i = $\stackrel{-}{1,n}$) (r_i = 0), then the subring N^φ = <r>^φ = <r′> is nilpotent, and hence, as shown above, R^φ is a one-generated ring.

Case (b). Let (∃i, j ∈ {1, . . . , n}) (i ≠ = j, r_i ≠ = 0, r_j ≠ = 0). By [1, Thm. 1], N^φ is a nil ring, and hence, as above, R^φ is a one-generated ring.

Case (c). Let (∃i ∈ {1, . . . , n}) (r_i ≠ = 0) and (∀j ∈ {1, . . . , n} \ {i}) (r_j = 0). There is no loss of generality in assuming that i = 1. Then R = (S₁ ∔ <r₁>) ⊕ · · · ⊕ S_n ⊕ <r>. For n = 1, R = (S₁ ∔ <r₁>) ⊕ <r> = S₁ +(w), where w = r₁ +r. By Lemma 2, R^φ is one-generated. Let n > 1. If we treat the subring R₁ = S₁ ⊕ · · · ⊕ S_n ⊕ <r> and apply to it the argument from Case (a) we conclude that r′ is a nilpotent element. Then the ring N^φ contains a nonzero nilpotent element r′, and by [1, Thm. 1], N^φ is a nil ring. Since w′ ∈ N^φ, the ring R^φ = S^φ + (w′) is one-generated by Theorem 2. The theorem is proved.

Using Theorem 5 as the base of induction on a variable k, we derive

COROLLARY 1. Let φ be a projection of a k-generated finite p-ring R onto a p-ring R^φ.

Then R^φ is a finite k-generated ring.

3. PROJECTIONS OF MATRIX RINGS

The theorem below likewise pertains to the basic results of the paper. Its proof relies essentially on the results and methods in [8,9,10].

THEOREM 6. Let R = M_n(K), where K is a finite p-ring with identity, n ≥ 2. Let φ be a lattice isomorphism of the ring R onto the ring R^φ. The following statements are valid:

(1) R^φ = M_n(K′), where K′ is a finite p-ring with identity;

(2) <u> ^φ = <u′>, where u and u′ are the identity elements of the rings R and R^φ, respectively;

(3) charR^φ = charR;

(4) the projective image Y^φ of a nilpotent ring Y ⊂ R is a nilpotent ring;

(5) (RadR) ^φ = RadR^φ;

(6) |RadR^φ| = |RadR|;

(7) if K is a semiprime ring then R^φ ≅ R;

(8) |R^φ| = |R|;

(9) (e_iiRe_ii) ^φ = e′_iiR^φe′_ii, where e_ii are diagonal matrix units in R, and <e_ii> ^φ= <e′_ii>, i =$\stackrel{-}{1,n}$;

(10) the ring K′ is matrix isomorphic to the ring K.

Proof. Let R = M_n(K), where n > 1, and K be a finite p-ring with identity e. Suppose also that o(e) = p^k. Let φ be a projection of R onto R^φ. Consider a subring S = M_n(<e>). By [8, Thm. 1.2], S^φ ≅ S. This, in particular, implies that the subring S^φ and hence the ring R^φ itself will be p-rings. Consequently, φ is a projection of the p-ring R onto the p-ring R^φ. According to Theorem 5, the projective image of any one-generated subring in R will be a one-generated subring in R^φ, and conversely, the projective preimage of any one-generated subring in R^φ will be a one-generated subring in R. In what follows, these facts will be used without further comments.

Let e_ij (i, j = $\stackrel{-}{1,n}$) be a system of matrix units in a ring S and u = e₁₁+· · ·+ e_nn be an identity in the ring R. By [10, Lemma 6], (∀i = $\stackrel{-}{1,n}$) (∃e′_ii ∈ R^φ) (<e_ii> ^φ = < e′_ii > ≅ <e_ii>); e′₁₁, e′₂₂, . . . , e′_nn is an orthogonal system of idempotents; u′ = e′₁₁ + e′₂₂ + · · · + e′_nn is an identity of the ring S^φ, and

$${\langle u\rangle }^{\varphi }=\langle {u}{\prime}\rangle .$$

(10)

By [10, Thm. 3], the system of idempotents e′₁₁, e′₂₂, . . . , e′_nn can be complemented to a full system of matrix units e′_ij , i, j = 1, n in the ring S^φ.

We prove that u′ is the identity in the ring R^φ. Assume the contrary and consider a two-sided Peirce decomposition of R^φ with respect to an idempotent u′:

$${R}^{\varphi }={u}{\prime}{R}^{\varphi }{u}{\prime}\dotplus {u}{\prime}{R}^{\varphi }\left(1-{u}{\prime}\right)\dotplus \left(1-{u}{\prime}\right){R}^{\varphi }{u}{\prime}\dotplus \left(1-{u}{\prime}\right){R}^{\varphi }\left(1-{u}{\prime}\right).$$

(11)

Note that S^φ ⊆ u′R^φu′ since u′R^φu′ is the greatest subring of R^φ in which the element u′ is an identity. In addition, the subring u′R^φu′ contains a system of matrix units e′_ij , i, j = 1, n, and according to [17, Prop. 6], u′R^φu′ = M_n(K′), where K′ is a subring of u′R^φu′ consisting of all elements commuting with all e′_ij , i, j = 1, n. The projective preimage of the ring u′R^φu′ contains a subring S; hence it contains an identity element and a system of matrix units and is therefore also the complete matrix ring over some subring B of K. Let T = M_n(B) and T^φ = u′R^φu′. By [10, Lemma 5], the ring T contains as a subring the Galois ring T₁ = GR(p^k, n). If k = 1, then T₁ = GF(pⁿ), and since the unique minimal subring of the field T₁ is the subring <u>, with equality (10) in mind, we conclude that u′ ∈ T^φ₁ . Projective images of finite fields were taken up in [2]. According to [2, Thm. 2.1], T^φ1 is a field. If k > 1, then T^φ₁ ≅ T₁ in view of [3, Thm. 4].

Suppose that u′R^φ (1 − u′) ≠ = {0′} and choose a nonzero element s′ ∈ u′R(1 − u′) of additive prime order. It is clear that u′s′ = s′, s′u′ = 0′, and (s′)² = 0′. For any integer α, therefore, the element u′ + αs′ is a nonzero idempotent, and hence a subring U′ = <u′> ∔ <s′> contains exactly p + 1 maximal subrings

$$\langle {u}{\prime}\rangle ,\langle {u}{\prime}+{s}{\prime}\rangle ,\cdots ,\langle {u}{\prime}+\left(p-1\right){s}{\prime}\rangle ,p\langle {u}{\prime}\rangle \dotplus \langle {s}{\prime}\rangle .$$

The ring R contains an element s of additive prime order such that <s> ^φ = <s′>. A subring U = <u> ∔ <s> is the projective preimage of the ring U′ under the lattice isomorphism φ. The subring U is commutative and is not a nil ring. In addition, it is not a direct sum of prime fields if k > 1 and, obviously, L(U) is not a chain. By virtue of [13, Thm. 4], the ring U contains a subring in which there are exactly two maximal subrings. In U′ every commutative subring either is generated by an idempotent element or is a nil ring, and by [13, Thm. 4], also, it does not contain a subring having exactly two maximal subrings, a contradiction. Consequently, u′R^φ (1−u′) = {0′} for k > 1.

Let k = 1. Again we address the subring T₁ = GF(pⁿ) and its projective image T^φ₁ . Let T^φ₁ = <x′>. Then x′s′, s′ are linearly independent nilpotent elements. A subring W′ = <u′, x′s′, s′> has order p³, and the length of its subring lattice equals three. The subring lattice of the projective preimage W of the ring W′ also has length three. According to [2, Lemma 3.1], the ring W cannot be decomposed into a direct sum of three prime fields, and W, being generated by its subrings of order p, lacks subfields of length 2. Consequently, the ring W contains a nonzero nilpotent element y of nilpotency index 2 and additive order p. Clearly, a subring <u, y> contains exactly two proper subrings <u> and <y>. By [13, Thm. 4], the ring W′ does not contain a subring in which there would be only two proper subrings, a contradiction. Consequently, for k = 1, too, the equality u′R(1 − u′) = {0′} holds.

The equality (1 − u′) R^φu′ = {0′} is proved in a similar way.

Suppose that (1−u′)R^φ (1−u′) ≠ = {0′} and choose in (1−u′)R^φ(1−u′) a nonzero element w′.

Let w′ be a nilpotent element of characteristic p and nilpotency index 2 in the subring (1 − u′)R^φ (1 − u′), w ∈ R, and <w>^φ = <w′>. Consider a subring V′ = T^φ₁ ⊕ <w′> and its projective preimage V = T₁ ∔ <w>. If k = 1, then, as noted, T^φ₁ is a nonprime field, and by Theorem 1, the subring lattice L(V′) decomposes into a direct product of lattices. It is not hard to see that a ring V does not satisfy the hypotheses of Theorem 1, and so the subring lattice L(V ) does not decompose into a direct product of lattices. If k > 1, then T^φ₁ ≅ T₁ = GR(p^k, n), and Theorem 2 says that V′ is a one-generated subring. Applying [4, Lemma 16] to the inverse projection φ⁻¹, we conclude that V ≅ V′. We are led to a contradiction since the ring V contains the identity of R, while the ring V′ has no identity element. Consequently, the ring (1− u′)R^φ (1 −u′) lacks nonzero nilpotent elements and, therefore, decomposes into a direct sum of fields.

Let y′ be a nonzero idempotent in a subring (1−u′)R^φ (1−u′), y ∈ R, and <y>^φ = <y′>. Applying [9, Thm. 4(2)] to a subring Y ′ = S^φ ⊕ <y′> and the inverse projection φ⁻¹, we conclude that y is a nonzero idempotent of characteristic p. According to [10, Lemma 5], a subring S^φ contains a Galois subring G′₁ = GR(p^k, n) which contains the identity element u′ of the subring S^φ. If k = 1, then G′₁ ≅ GR(p, n), i.e., G′₁ is a field of length n > 1. In this case the projective preimage G₁ of G′₁ is also a field since it cannot contain nonzero nilpotent elements in virtue of [10, Thm. 3(b)]. Furthermore, the field G₁ contains the identity of R, and so the subring G₁ ∨ <y>, in view of [2, Lemma 1.2, Cor. 3.1], cannot be lattice-isomorphic to ring G′₁ ⊕ <y′>. If k > 1, then G₁ ≅ G′₁ by [3, Thm. 4], and by [4, Lemma 12], the subring G₁ ∨ <y> too cannot be lattice-isomorphic to G′₁ ⊕ <y′>. Thus our supposition is invalid. Hence u′ is an identity element in the ring R^φ, and so R^φ = M_n(K′). Statements (1) and (2) are proved.

The truth of statement (3) follows from statement (2).

(4) Let N be a nilpotent subring in the ring R. Consider a local subring D = <u> + N and its projective image D^φ = <u′> +N^φ. The subring D and the projection φ of the ring D onto the ring D^φ satisfies the hypotheses of Lemma 4 in [10], which says that N^φ is a nilpotent subring.

(5) Let M be a maximal nilpotent subring in R. According to statement (4), M^φ is a nilpotent subring in R^φ. Suppose that the subring M^φ is not a maximal nilpotent subring in the ring R^φ. Let M^φ ⊂ M′₁, where M′₁ is a nilpotent subring in R^φ. A subring H′ = <u′> + M′₁ is a local ring. The projective preimage of the ring H′ is a ring H = <u> + M₁, where M₁ is the projective preimage of the subring M′₁. Applying [10, Lemma 4] to a subring H′ and the inverse projection φ⁻¹, we conclude that M₁ is a nilpotent subring. Since M is a proper subring of M₁, we arrive at a contradiction. Hence M^φ is a maximal nilpotent subring in R^φ. According to [18], RadR is the intersection of all maximal nilpotent subrings in R. Consequently, (RadR) ^φ = RadR^φ. Statement (6) follows from [1, Lemma 7].

(7) Let K be a semiprime ring, and namely: K = K₁ ⊕· · ·⊕K_m, where K_i is a finite prime ring, i.e., either K_i = GF(p^ki ) or K_i = M_ni(GF(p^ki )), with n_i > 1, i = $\stackrel{-}{1,m}$. Then R = M_n(K₁) ⊕· · ·⊕ M_n(K_m). By statement (1), (M_n(K_i)) ^φ = M_n(K′_i), where K′_i is a ring with identity, i = $\stackrel{-}{1,m}$. Let e_i be an identity in a ring K_i, i = $\stackrel{-}{1,m}$. Then e = e₁+· · ·+e_m, and so S = M_n(<e>) = M_n(<e₁>) ⊕· · ·⊕ M_n(<e_m>). In view of [10, Thm. 3.4] and statement (1), S^φ = M_n1(<e′₁>) ⊕· · ·⊕ M_nm(<e′_m>), where e′_i is an identity element of a ring K′_i, i = $\stackrel{-}{1,m}$. This implies that R^φ = M_n(K′₁) ⊕· · ·⊕ M_n(K′_m). Let i ∈ {1, . . . ,m}. For K_i = GF(p^ki), the lattice definability of a ring M_n(K_i) follows from [9, Thm. 1]. If K_i = M_ni(GF(p^ki)), then M_n(K_i) = M_n(M_ni (GF(p^ki)) = M_nni(GF(p^ki)), and so the lattice definability of M_n(K_i) again follows from [9, Thm. 1]. Statement (7) is proved.

Statement (5) implies that the factor rings $\overline{R }$ = R/RadR and $\overline{{R }^{\varphi }}$ = R^φ/RadR^φ are latticeisomorphic. In view of statement (7), $\overline{R }$ ≅ $\overline{{R }^{\varphi }}$, and in virtue of statement (6), |RadR^φ| = |RadR|.

Hence |R| = $\left|\overline{R }\right|$ · |RadR| = $\overline{{R }^{\varphi }}$ · |RadR^φ| = |R^φ|. Statement (8) is proved.

(9) Let |K| = p^κ. We use induction on κ.

For κ = 1, K = <e> , o(e) = p, and R = M_n(<e>). That statement (9) is true follows from [10, Thm. 3(d)].

Suppose statement (9) has been proven for all numbers 1 ≤ κ < ν. Let κ = ν. If K = <e>, then the truth of statement (9) again follows from [10, Thm. 3(d)]. Let K′ = <e>. Suppose that the ring K contains two distinct maximal subrings W₁ and W₂ containing an identity e. Let T_i = M_n(W_i), i = 1, 2. By the induction hypothesis, the following equalities hold: (e_jjT_ie_jj)^φ = e′_jjT^φ_i e′_jj for i = 1, 2 and j = 1, n. It is obvious that

$$\begin{array}{l}\left({e}_{jj}{T}_{1}{e}_{jj}\right) \bigvee \left({e}_{jj}{T}_{2}{e}_{jj}\right)={e}_{jj}\left({T}_{1} \bigvee { T}_{2}\right){e}_{jj}\\ ={e}_{jj}{M}_{n}\left({W}_{1} \bigvee { W}_{2}\right){e}_{jj}\\ ={e}_{jj}{M}_{n}\left(K\right){e}_{jj}\\ ={e}_{jj}R{e}_{jj}.\end{array}$$

Therefore,

$$\begin{array}{l}{\left({e}_{jj}R{e}_{jj}\right)}^{\varphi }=\left({e}_{jj}{\prime}{T}_{1}^{\varphi }{e}_{jj}{\prime}\right) \bigvee \left({e}_{jj}{\prime}{T}_{2}^{\varphi }{e}_{jj}{\prime}\right)\\ ={e}_{jj}{\prime}\left({T}_{1}^{\varphi } \bigvee {T}_{2}^{\varphi }\right){e}_{jj}{\prime}\\ ={e}_{jj}{\prime}{M}_{n}\left({W}_{1}^{\varphi } \bigvee {W}_{2}^{\varphi }\right){e}_{jj}{\prime}\\ ={e}_{jj}{\prime}{M}_{n}\left(K{\prime}\right){e}_{jj}{\prime}\\ ={e}_{jj}{\prime}{R}^{\varphi }{e}_{jj}{\prime}.\end{array}$$

Let W be the unique maximal subring in K containing an identity e. Obviously, (∀v ∈ K \ W) (<e, v> ⊈ W), and so <e, v> = K. This implies that K is a commutative ring. It is well known that a finite commutative ring with identity either is local or is decomposable into a direct sum of local rings [15, Chap. 2, Thm. 5]. Let K be decomposable into a direct sum of local rings, i.e., K = P₁ ⊕ · · · ⊕ P_l. Put R_j = M_n(P_j) (j = $\stackrel{-}{1,l}$). Then e_iiRe_ii = e_iiR₁e_ii ⊕ · · · ⊕ e_iiR_le_ii. By [10, Thm. 3], (∀j = $\stackrel{-}{1,l}$)((e_iiR_je_ii)^φ = e′_iiR^φ_j e′_ii. By [10, Cor. 2], R^φ = R^φ₁ ⊕ · · · ⊕ R^φ_l . Hence (∀i = $\stackrel{-}{1,n}$)((e_iiRe_ii)^φ = (e_iiR₁e_ii)^φ ⊕· · · ⊕(e_iiR_le_ii)^φ = e′_iiR^φ₁ e′_ii ⊕· · ·⊕e′_iiR^φ_l e′_ii = e′_iiR^φe′_ii.

Thus statement (9) is true also for κ = ν, and hence for all numbers κ.

(10) In view of [17, Prop. 6], there are isomorphisms e₁₁Re₁₁ ≅ K and e′₁₁R^φe′₁₁ ≅ K′. By virtue of statement (9), the rings e₁₁Re₁₁ and e′₁₁R^φe′₁₁ are lattice-isomorphic, and so L(K) ≅ L(K′). The theorem is proved.

THEOREM 7. Let R = M_n(K), where K is a finite ring with identity, n ≥ 2. Suppose also that φ is a lattice isomorphism of the ring R onto the ring R^φ. Then R^φ = M_n(K′), where K′ is a finite ring with identity, lattice-isomorphic to the ring K.

Proof. Let K = K₁ ⊕ · · · ⊕ K_m be a decomposition of K into a direct sum of p_i-rings K_i taken over distinct prime numbers p_i, i = $\stackrel{-}{1,m}$ . Then R = R₁ ⊕ · · · ⊕ R_m, where R_i = M_n(K_i), i = $\stackrel{-}{1,m}$ , and L(R) ≅ L(R₁)×···×L(R_m). Let φ be a lattice isomorphism of R onto R^φ. Clearly, L(R^φ) ≅ L(R^φ₁)×· · ·×L(R^φ_m). By Theorem 6, R^φ_i = M_n(K′_i), where K′_i is a p_i-ring with identity, lattice isomorphic to a ring K_i, i = $\stackrel{-}{1,m}$ . Since prime numbers p_i, i = $\stackrel{-}{1,m}$ , are pairwise distinct, we have R^φ = R^φ₁ ⊕· · ·⊕ R^φ_m . Let K′ = K′₁ ⊕· · ·⊕ K′_m. Then R^φ = M_n(K′) and K′, in this case, is a ring with identity, lattice-isomorpic to the ring K. The theorem is proved.

THEOREM 8. Let R = M_n1(K₁) ⊕ · · · ⊕ M_nl(K_l), where K_i is a finite local p-ring, |K_i/RadK_i| = p^mi , i = $\stackrel{-}{1,l}$. Suppose that R is not isomorphic to the rings GF(p^q) ⊕ GF(p) (q is a prime) and GF(p) ⊕ GF(p). Let φ be a lattice isomorphism of the ring R onto the ring R^φ.

Then the following statements hold:

(1) R^φ = R^φ₁ ∔ · · · ∔ R^φ_l (group sum);

(2) (∀i = $\stackrel{-}{1,l}$) (R_i is a prime ring ⇒ R^φ_i is a prime ring);

(3) if n_im_i> 1 for all i = $\stackrel{-}{1,l}$ , then R^φ = R^φ₁ ⊕· · ·⊕R^φ_l , and also (RadR)^φ = RadR^φ.

Proof. Let R_i = M_ni(K_i), i = $\stackrel{-}{1,l}$ . According to [14, Thm. XIX.4], every local p-ring K_i, i = $\stackrel{-}{1,l}$ , contains as a subring the Galois ring S_i = GR(p^ki,m_i), i = $\stackrel{-}{1,l}$ . In R we consider a subring T = T₁ ⊕· · ·⊕ T_l, where T_i = M_ni(S_i). This subring satisfies the hypotheses of [9, Thm. 4], which implies that statements (1) and (2) are true.

Suppose that, for all i ∈ {1, . . . , l}, the conditions that n_im_i > 1 are satisfied. By [9, Thm. 4], (∀i, j = $\stackrel{-}{1,l}$)(i ≠ = j ⇒ (T_i ⊕ T_j)^φ = T^φ_i ⊕ T^φ_j). These equalities imply the equality

$${R}^{\varphi }={R}_{1}^{\varphi }\oplus \cdots \oplus {R}_{l}^{\varphi }.$$

(12)

It is clear that RadR = RadR₁ ⊕ · · · ⊕ RadR_l. Let i ∈ {1, . . . , l}. According to [17, Thm. 3], RadR_i = M_ni(RadK_i), i = 1, l. For n_i = 1, R_i = K_i, and since m_i > 1, in view of [16, Thm. 3] it is true that

$${\left(\text{Rad} {R}_{i}\right)}^{\varphi }=\text{Rad }{R}_{i}^{\varphi }.$$

(13)

If n_i > 1 then, by [10, Thm. 3], equality (13) holds as well. The truth of statement (3) follows from equalities (12) and (13). The theorem is proved.