Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 41 (2000), No. 2, 489-511.

Polynomial and Transformation Composition Rings

S. Veldsman

Department of Mathematics, University of Port Elizabeth, South Africa and presently on leave at the Department of Mathematics, Sultan Qaboos University, Al-Khodh, Sultanate of Oman.

Abstract. The base of an arbitrary composition ring is defined. This is used, amongst others, to identify polynomial type composition rings and to describe the maximal ideals of certain types of composition rings. MSC 2000: 16S34, 16Y30

1. Introduction

Polynomials are one of the earliest entities studied in mathematics. As an algebraic entity, they have been studied for more than a century, and then mainly within the framework of a ring with respect to the usual operations of addition and multiplication of polynomials. But with any polynomial ring is associated a polynomial near-ring with respect to the operation of composition. These polynomial near-rings have also been studied, albeit much less extensively and only during the last few decades. But not much attention has been given to their natural algebraic home, that of a ring together with the additional operation of composition. Sporadically this has been done under the more general guise of a composition ring (or composition near-ring). But this approach is mostly too general to capture their significant properties for an abstract investigation. For example, if R is any ring, then the set M(R) of all mappings from R to R, together with pointwise addition, pointwise multiplication and composition is a composition ring. Moreover, such a composition ring is the prototype of all composition rings, since any composition ring can be embedded in an M(R) for a suitable R.

The study of the composition rings R[x], R_{0}[x] and R_{N}[[x]] has inevitably led to the relationship between the properties of the composition ring and those of the underlying base

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2000 Heldermann Verlag

ring R. In the case where constants are allowed (e.g. R[x]), the base ring is easily identified as all those elements c for which c∘0 = c. The same procedure captures R in the composition ring M(R). But for R_{0}[x] and R_{N}[[x]] things are more problematic. For these two cases, a different approach is required and one is led to consider D(c) ∘ 0 where D(c) denotes the formal derivation of c (which takes one outside the considered composition ring). But for the composition ring M_{0}(R) (zero preserving functions from R to R) there may be no realistic derivation in sight to enable us to capture R.

It is these two problems that will be addressed here. We will define the base of an arbitrary composition ring. It will be seen that the base for each of the composition rings R[x], R_{0}[x], R_{N}[[x]], M_{0}(R) and M(R) is a ring isomorphic to R. As can be expected, the relationship between the composition ring and its base will not be the same for all composition rings, but at least the homological structure will be the same: The base B of a composition ring C is a subring of C (with respect to the composition) for which there is (at least) a group homomorphism β : (C, +) → (B, +).

To realize the existence of the mapping β and to obtain more structure on the composition rings, we will define and investigate mainly two types of composition rings. These are the composition p-rings (p standing for polynomial, power series or related structures) and composition t-rings (t standing for transformation). This will enable us, amongst others, to determine the maximal ideals of a composition p-ring in terms of those of the underlying base ring (irrespective of whether constants are allowed or not).

We should point out that Kautschitsch and Mlitz [2] have given a general method to describe the maximal ideals of R_{0}[x], R[x] and R_{N}[[x]]. This was done within the framework of composition subrings of R[[x]]. The results presented here will be more general and really only serve as a guideline on how to study polynomials and power series under this more general setting.

In the sequel, we use C to denote the composition ring (C, +, ·, ∘), i.e. C_{1} := (C, +, ·) is a ring, C_{2} := (C, +, ∘) is a near-ring and ab ∘ c = a ∘ c · b ∘ c for all a, b, c ∈ C. Unless indicated otherwise by brackets, juxtaposition (which represents multiplication) has a higher priority than composition (in the order of executing the operations) and composition has a higher priority than multiplication when the latter is denoted by “·”. For example, ab∘c = a∘c·b∘c means (ab) ∘ c = (a ∘ c) · (b ∘ c).

The foundation of C, denoted by F, is the constant part of the near-ring C_{2}, i.e. F = F ound(C) = {c ∈ C | c ∘ 0 = c}. F is a subcomposition ring of C, but usually when we refer to the foundation, we mean the ring F = (F, +, ·). C_{0} denotes the 0-symmetric part of of the near-ring C_{2}, i.e. C_{0} = {c ∈ C | c ∘ 0 = 0}. It can be verified that C_{0} is an ideal of the ring C_{1} and a left ideal of the near-ring C_{2}.

The near-ring C_{2} may have an identity which we will denote by x ∈ C_{2}. Often it is required that x be a commuting composition identity, i.e. c ∘ x = c = x ∘ c and xc = cx for all c ∈ C. The ring C_{1} may have an identity i ( i.e. ci = c = ic for all c ∈ C ), denoted by i ∈ C_{1}. In case i ∈ F, it will be called a constant multiplicative identity. The ring (C_{0}, +, ·) may also have an identity, denoted by e ∈ C_{0} (which may differ from i) and is called a 0-symmetric multiplicative identity.

For c ∈ C and k ∈ ℕ, c^{k} denotes c^{k} = ccc ⋯ c (k factors) and c^{(k)} = c ∘ c ∘ c ∘ ⋯ ∘ c (k times). For a subset A ⊆ C, A^{k} = {a_{1}a_{2} ⋯ a_{k} | a_{i} ∈ A} and A^{(k)} = {a_{1}∘a_{2}∘⋯∘a_{k} | a_{i} ∈ A}.

When the ring R is used in the context of R[x], R_{0}[x] or R_{N}[[x]], it will be assumed to be commutative and to have an identity 1 ∈ R. As usual, R[x] is the composition ring of all polynomials over R, R_{0}[x] is the 0-symmetric part of R[x] and R_{N}[[x]] is the composition ring of all formal power series over R ( i.e. infinite sums of the form ∑_{n=0}^{∞} a_{n}x^{n}) with constant term a_{0} from N where N is a fixed nil ideal of R.

2. The base of a composition ring

Let C be a composition ring. The base of C, denoted by B, is defined by B = Base(C) = {v ∈ C | v is left distributive over (C, +) and v ∘ ab = (v ∘ a)b for all a, b ∈ C}. Let \overline{F B} = \{∑ n_{i}v_{i} | n_{i} ∈ F and v_{i} ∈ B\} and \overline{C_{0}B} = \{∑ u_{i}v_{i} | u_{i} ∈ C_{0} and v_{i} ∈ B\} be the subgroups of (C, +) generated by F B and C_{0}B respectively.

2.1. Proposition. (1) (B, +, ∘) is a subring of the near-ring (C_{0}, +, ∘)

(2) F B ⊆ B and F B ∘ B + B ∘ F B ⊆ F B

(3) \overline{F B} ⊲ (B, +, ∘)

(4) If nc = cn for all n ∈ F and c ∈ C, then BF ⊆ B.

(5) If C has a constant multiplicative identity i, then F B = B.

(6) Any composition left identity of C is contained in B.

(7) (0 : C)_{C_{2}} := {d ∈ C | d ∘ c = 0 for all c ∈ C} ⊆ B.

Proof. (1) B ⊆ C_{0} since every element of B is left distributive over (C, +). Let v_{1}, v_{2} ∈ B and a, b ∈ C. Then:

(v_{1} − v_{2}) ∘ (a + b) = v_{1} ∘ a + v_{1} ∘ b − v_{2} ∘ a − v_{2} ∘ b
= (v_{1} − v_{2}) ∘ a + (v_{1} − v_{2}) ∘ b
(v_{1} − v_{2}) ∘ ab = v_{1} ∘ ab − v_{2} ∘ ab
= (v_{1} ∘ a)b − (v_{2} ∘ a)b = ((v_{1} − v_{2}) ∘ a)b
(v_{1} ∘ v_{2}) ∘ (a + b) = v_{1} ∘ (v_{2} ∘ a + v_{2} ∘ b) = (v_{1} ∘ v_{2}) ∘ a + (v_{1} ∘ v_{2}) ∘ b and
(v_{1} ∘ v_{2}) ∘ ab = v_{1} ∘ (v_{2} ∘ ab) = v_{1} ∘ ((v_{2} ∘ a)b) = (v_{1} ∘ v_{2} ∘ a)b.

Thus B is a subnear-ring of (C_{0}, +, ∘). Clearly B is a ring.

(2) Let v, u ∈ B, n ∈ F. Then nv ∘(a+b) = n·(v ∘(a+b)) = n·v ∘a+n·v ∘b = nv ∘a+nv ∘b and nv ∘ ab = n · v ∘ ab = n · (v ∘ a)b = (nv ∘ a)b. Thus F B ⊆ B. Also, nv ∘ u = n · v ∘ u ∈ F B and u ∘ nv = (u ∘ n)v ∈ F B.

(3) Follows from (2) above.

(4) By the assumption, BF = F B. (5), (6) and (7) are clear. □

2.2. Proposition. Let x ∈ C_{2} be a commuting composition identity. Then:

(1) d ∘ c · c = c · d ∘ c for all c, d ∈ C. In particular, for all n ∈ F and c ∈ C, nc = cn.

(2) x ∈ B

(3) (v ∘ c)x = vc for all v ∈ B, c ∈ C.

(4) F x = \overline{F B} = F B = BF ⊲ (B, +, ∘). If B is a simple ring, then F x = B or F x = 0.

(5) For all k ≥ 1, v ∘ x^{k+1} = vx^{k} for all v ∈ B and B^{k+1} = Bx^{k}.

(6) If J ⊲ B, then JF + F J ⊆ J.

(7) v ∘ nx = nx ∘ v for all n ∈ F, v ∈ B.

(8) C_{0}x = {ux | u ∈ C_{0}} is a subcomposition ring of C.

Proof. (1) d ∘ c · c = d ∘ c · x ∘ c = dx ∘ c = xd ∘ c = c · d ∘ c. If d = n ∈ F, then nc = cn.

(2) follows from Proposition 2.1.

(3) (v ∘ c)x = v ∘ cx = v ∘ xc = (v ∘ x)c = vc.

(4) F x ⊆ F B and for n ∈ F and v ∈ B, n = (v ∘ n)x ∈ F x (by (3) above). Thus F x ⊆ F B = BF ⊆ F x. Then F x = F B = \overline{F B} ⊲ B follows from Proposition 2.1.

(5) For v ∈ B, v ∘ x^{2} = (v ∘ x)x = vx. If v ∘ x^{k+1} = vx^{k}, then v ∘ x^{k+2} = (v ∘ x^{k+1})x = vx^{k}x = vx^{k+1}. Clearly Bx ⊆ B^{2} and if v, w ∈ B, then vw = (v ∘ w)x ∈ Bx. Thus B^{2} = Bx. If B^{k} = Bx^{k−1}, then B^{k+1} = BB^{k} = B(Bx^{k−1}) = B^{2}x^{k−1} = (Bx)x^{k−1} = Bx^{k}.

(6) For j ∈ J and n ∈ F, jn = (j ∘ n)x = j ∘ nx ∈ J ∘ F x ⊆ J ∘ B ⊆ J and nj = jn ∈ J.

(7) v ∘ nx = (v ∘ n)x = vn = nv = nx ∘ v.

(8) For ux, vx ∈ C_{0}x, ux − vx = (u − v)x ∈ C_{0}x; ux · vx = (uxv)x ∈ C_{0}x and ux ∘ vx = u ∘ vx · vx = (u ∘ vx · v)x ∈ C_{0}x. □

2.3. Proposition. Let x ∈ C_{2} be a commuting composition identity and let i ∈ C_{1} be a multiplicative identity. Then:

(1) B = {v ∈ C | v ∘ ab = (v ∘ a)b for all a, b ∈ C}.

(2) (0 : x)_{C_{1}} ∩ F = 0 (i.e. nx = 0, n ∈ F, implies n = 0).

(3) If (0 : x)_{C_{1}} = 0 (i.e., cx = 0, c ∈ C, implies c = 0), then B = {v ∈ C | (v ∘ a)x = va for all a ∈ C}

(4) If i is a constant multiplicative identity, then (B, +, ∘) = (F x, +, ∘) ≅ (F, +, ·).

Proof. (1) Suppose v ∈ C is such that v ∘ ab = (v ∘ a)b for all a, b ∈ C. We show v is left distributive. For, a, b ∈ C, v ∘ (a + b) = v ∘ (i(a + b)) = (v ∘ i)(a + b) = (v ∘ i)a + (v ∘ i)b = v ∘ ia + v ∘ ib = v ∘ a + v ∘ b; hence v ∈ B.

(2) If nx = 0, n ∈ F, then n = ni = nx ∘ i = 0.

(3) Suppose v ∘ ab = (v ∘ a)b for all a, b ∈ C. For any c ∈ C, (v ∘ c)x = v ∘ cx = v ∘ xc = (v ∘ x)c = vc. Conversely, suppose (v ∘ c)x = vc for all c ∈ C. For any a, b ∈ C, (v ∘ ab)x = v(ab) = (va)b = ((v ∘ a)x)b = (v ∘ a)bx. Thus v ∘ ab − (v ∘ a)b ∈ (0 : x)_{C_{1}} = 0; hence v ∘ ab = (v ∘ a)b. Then (3) follows from (1).

(4) By Proposition 2.2(4) we know F x ⊆ B. Conversely, for v ∈ B, v = vi = (v ∘ i)x ∈ F x since by assumption i ∈ F. Thus B = F x. Define ψ : (F, +, ·) ⟶ (F x, +, ∘) by ψ(n) := nx. Then it is straightforward to verify that ψ is a surjective (ring) homomorphism. If nx = mx, then n = m follows from (2) above. Thus ψ is an isomorphism. □

We note that (1) above holds without the assumption of a composition identity.

2.4. Proposition. Let x ∈ C_{2} be a commuting composition identity. Suppose there is an s ∈ C_{0} such that x = sx. Then B ⊆ C_{0}x.

Proof. Let v ∈ B. Then v = v ∘ x = v ∘ sx = (v ∘ s)x ∈ C_{0}x. □

2.5. Proposition. Let C and D be composition rings with θ : C ⟶ D a surjective composition ring homomorphism and let S a subcomposition ring of C. Then

(1) θ(Base(C)) ⊆ Base(D),

(2) Base(C) ∩ S ⊆ Base(S).

Proof. Straightforward. □

2.6. Examples. Below we will give several examples of composition rings and describe the base of each. In most cases we will give additional properties that will be referred to in the sequel.

(1) For each of R[x], R_{0}[x] and R_{N}[[x]], x is a commuting composition identity. In R[x], i = 1 ∈ R is a constant multiplicative identity and thus Base(R[x]) = F x = (Rx, +, ∘) ≅ (R, +, ·) (Proposition 2.3(4)). Also, for each of R_{0}[x] and R_{N}[[x]], the base is (Rx, +, ∘) ≅ (R, +, ·). We will verify this, for example, for R_{N}[[x]]. Let r ∈ R and a, b ∈ R_{N}[[x]]. Clearly rx is left distributive over + and rx ∘ ab = r(ab) = (ra)b = (rx ∘ a)b. Conversely, suppose v = v_{1}x + v_{2}x^{2} + v_{3}x^{3} + ⋯ ∈ Base(R_{N}[[x]]), v_{i} ∈ R. Then v ∘ x^{2} = (v ∘ x)x = vx, i.e.

v_{1}x^{2} + v_{2}x^{4} + v_{3}x^{6} + ⋯ = v_{1}x^{2} + v_{2}x^{3} + v_{3}x^{4} + · · · .

The left hand side has only even powers of x; hence the coefficients of all the odd powers of x on the right hand side are 0, i.e. v_{2k} = 0 for all k ≥ 1. Thus the term containing x^{4k} on the left hand side is 0, which means that the corresponding term on the right hand side is zero which yields v_{4k−1} = 0 (for all k ≥ 1). Since v_{4k−1} = 0, the term containing x^{2(4k−1)} on the left hand side is 0, which makes v_{2(4k−1)−1} = v_{8k−3} = 0. Continuing in this way, we get v_{m} = 0 where m = 2^{n}k − (2^{n−1} − 1) for all n, k ≥ 1. But any m ≥ 2 is of the form 2^{n}k − (2^{n−1} − 1) for some n and k. Indeed, if m is even, take n = 1. Then m = 2k for a suitable k. If m is odd and greater than 1, say m = 2m′ + 1 (m′ ≥ 1), choose n ∈ ℕ maximal such that 2^{n−2} | m′ (if m′ is odd, then n = 2). Hence m′ = 2^{n−2}p for some odd p, say p = 2k − 1, k ∈ ℕ. Then m′ = 2^{n−2}(2k − 1) from which m = 2m′ + 1 = 2^{n}k − (2^{n−1} − 1) follows. Thus 0 = v_{2} = v_{3} = v_{4} ⋯ and so v = v_{1}x, v_{1} ∈ R.

(2) For any ring R, we will use the following notation for certain maps in the composition ring M(R) (or M_{0}(R) if applicable). Let a ∈ R. Then \overline{a}, â and ã denotes the following self maps of R:

\overline{a}(t) = at for all t ∈ R
â(t) = a for all t ∈ R and
ã(t) = \{ \begin{matrix}a & if & t ≠ 0 \\ 0 & if & t = 0. \end{matrix}

Let R be a commutative ring with identity 1 ∈ R. Both M_{0}(R) and M(R) have commutative composition identity x = 1_{R}, M(R) has constant multiplicative identity ˆ1 and M_{0}(R) has a 0-symmetric multiplicative identity ˜1. By Proposition 2.3(4), B = Base(M(R)) = F x = Rx = \{\overline{a} | a ∈ R\} = \{c | c is an R - endomorphism of (R, +)\}, for the latter we regard (R, +) as a right R -module. But also Base(M_{0}(R)) = \{\overline{a} | a ∈ R\} = B. Indeed, for any a ∈ R it is easy to see that \overline{a} ∈ Base(M_{0}(R)). Conversely, let v ∈ Base(M_{0}(R)). Then a := v(1) ∈ R and from (v ∘ ˜1)x = v ∘ ˜1x we get v = \overline{a}. For both cases (M(R) and M_{0}(R)), we have (B, +, ∘) ≅ (R, +, ·) and since x = ˜1x, we have B ⊆ M_{0}(R)x (cf. Proposition 2.4).

More generally, let R and S be rings with α : S → R a fixed function. Let C = M(R, S, α) be the sandwich composition ring determined by R, S and α. This means C consists of all the functions from R to S, the addition and multiplication are componentwise and the composition, here denoted by ∗, is defined by: f ∗ g = f ∘ α ∘ g. Then it can be shown that B = Base(C) = {f ∈ C | f ∘ α ∈ End_{S}(S, +)}.

(3) For rings R without identity, the base of M_{0}(R), or M(R), need not be isomorphic to R. Let R = 2ℤ and let C = M_{0}(R). In this case, Base(M_{0}(R)) = \{\overline{a} | a ∈ ℤ\} ≅ ℤ ≇ R: Clearly any \overline{a}, a ∈ ℤ, is in B = Base(C). Let v ∈ B and suppose v(2) = 2a, a ∈ ℤ. From (v ∘ ˜2)x = (v ∘ ˜2)1_{R} = v ∘ (˜21_{R}) = v ∘ (1_{R}˜2) = (v ∘ 1_{R})˜2, we get v(2)t = v(t)2 for all t ∈ R. Then 2\overline{a}(t) = 2at = v(2)t = v(t)2; hence \overline{a} = v. The isomorphism (B, +, ∘) ≅ (ℤ, +, ·) is clear. Note that here we have B ∩ C_{0}x = \{\overline{a} | a ∈ 2ℤ\}. Indeed, if \overline{a} = ux ∈ B ∩ C_{0}x (a ∈ ℤ, u ∈ C_{0}), then 2a = u(2) · 2. Thus a = u(2) ∈ 2ℤ. The converse inclusion is clear.

(4) Let R be a commutative ring with S ≠ R a subring of R with identity 1_{S} ∈ S. Let C = {f ∈ M(R) | f(R) ⊆ S}. Then C is a subcomposition ring of M(R). Note that C does not have a composition identity. However, it has many left identities: Let s_{0} ∈ S and let

x(t) = \{ \begin{matrix}t & if & t ∈ S \\ s_{0} & if & t ∈ R\S. \end{matrix}

Then x is a left identity. Here we have F ound(C) = S and B = Base(C) = {a_{g} | a ∈ S

and g ∈ C\} where a_{g} : R ⟶ R is defined by a_{g}(t) = \{ \begin{matrix}at & if & t ∈ S \\ g(t) & if & t ∈ R\S. \end{matrix}

Indeed, every a_{g} is in C and also in B. Let v ∈ B and let a := v(1_{s}). We show v = a_{v} ∈ B. By definition of C, a ∈ S. Define b, c : R ⟶ R by b(t) = 1_{s} for all t ∈ R and

c(t) = \{ \begin{matrix}t & if & t ∈ S \\ 1_{s} & if & t ∈ R\S. \end{matrix}

Then b, c ∈ C and from v∘bc = (v∘b)c, we get v(t) = (v∘bc)(t) = ((v∘b)c)(t) = at = a_{v}(t) for any t ∈ S. Thus v = a_{v}.

(5) Let C = \overline{R[x]} where \overline{R[x]} denotes all the polynomial functions. Then C is a subcomposition ring of M(R) and B = Base(C) = {rx | r ∈ R}. Note that here we may have B ∩ C_{0}x ≠ 0. For example, if R = ℤ_{4}, then 0 ≠ 2x = 2x^{2} ∈ B ∩ C_{0}x.

(6) Let C = (R, +, ·, ∘) where (R, +, ·) is a ring and (R, +, ∘) is any near-ring.

(a) If (R, +, ·) is a zero ring (i.e. R^{2} = 0) then Base(C) = {c ∈ C | c is left distributive }. If (R, +, ∘) is constant, then Base(C) = 0.

(b) If (R, +, ∘) is a zero ring, then Base(C) = (C, +, ∘).

(7) Let R be a commutative ring with identity 1 ∈ R. Let (R[x, y], +, ·) be the ring of all polynomials in the two commuting indeterminates x and y. As is well-known, we can define a near-ring multiplication ∘ on (R[x, y], +) by: f(x, y) ∘ g(x, y) := f(g(x, y), g(x, y)). Then C = (R[x, y], +, ·, ∘) is a composition ring with constant multiplicative identity 1. But C has no composition identity. Both x and y are composition left identities; in fact so are c+x−c∘x for all c ∈ C. These all are in B = Base(C) and also 0 ≠ (0 : C)_{C_{2}} ⊆ B. It can be shown that B = {v ∈ C | v = ∑_{i=1}^{k} ∑_{j=0}^{i} c_{ij}x^{i−j}y^{j} for some k ≥ 1 and c_{ij} ∈ R for all i, j where ∑_{j=0}^{i} c_{ij} = 0 for all i = 2, 3, ..., k}.

(8) With (R[x, y], +, ·) as in (7) above, define ∘ by f(x, y) ∘ g(x, y) = f(g(x, x), g(y, y)). Then C = (R[x, y], +, ·, ∘) is a composition ring with right identities x and y and F ound(C) = R. In this case, it can be shown that B = Base(C) = 0. Indeed, this follows by considering (v ∘ 1)x = v ∘ x = v and (v ∘ 1)y = v ∘ y = v for v ∈ B.

(9) Let (R[x, y], +, ·) be as in (7) above. On the direct sum of two copies of this ring with itself, define ∘ by: (f(x, y), g(x, y)) ∘ (h(x, y), k(x, y)) := (f(h(x, y), k(x, y)), g(h(x, y), k(x, y))).

Then C := (R[x, y] × R[x, y], +, ·, ∘) is a composition ring with constant multiplicative identity (1, 1) and commuting composition identity (x, y). Here F ound(C) = R × R and Base(C) = (R × R)(x, y) = Rx × Ry.

2.7. Remark. Let C be a composition ring with F = F ound(C) and B = Base(C). Let S = F + B. Then (S, +, ∘) is a subnear-ring of (C, +, ∘) and it can easily be verified that S is an abstract affine near-ring.

3. Composition p-rings

A composition ring C is called a composition p-ring if it has a commuting composition identity x and (C_{0}, +) = (B, +) ⊕ (C_{0}x, +).

As is well-known, (C, +) = (C_{0}, +) ⊕ (F, +); hence (C, +) = (F, +) ⊕ (B, +) ⊕ (C_{0}x, +). Thus every c ∈ C has a unique representation as c = n + v + ux with n ∈ F, v ∈ B and u ∈ C_{0}. This give rise to a surjective group homomorphism β : (C, +) ⟶ (B, +) defined by β(c) = v with kerβ = F + C_{0}x. For c = n_{c} + v_{c} + u_{c}x and d = n_{d} + v_{d} + u_{d}x ∈ C, cd = n_{c}n_{d} +(v_{c}n_{d} +n_{c}v_{d})+w where w = n_{c}u_{d}x+v_{c}v_{d} +v_{c}u_{d}x+u_{c}xn_{d} +u_{c}xv_{d} +u_{c}xu_{d}x ∈ C_{0}x since C_{0} ⊲C_{1}, B^{2} = Bx ⊆ C_{0}x and x is a commuting composition identity. From F B = BF ⊆ F x we get β(cd) = v_{c}n_{d} + n_{c}v_{d} ∈ F x. Note that if c, d ∈ C_{0}, then β(cd) = 0. Furthermore,

c ∘ d = n_{c} + v_{c} ∘ n_{d} + v_{c} ∘ v_{d} + v_{c} ∘ u_{d}x + u_{c}x ∘ d

where n_{c} + v_{c} ∘ n_{d} ∈ F, v_{c} ∘ v_{d} ∈ B and v_{c} ∘ u_{d}x = (v_{c} ∘ u_{d})x ∈ C_{0}x. Let a = u_{c} ∘ d. Then u_{c}x ∘ d = u_{c} ∘ d · d = ad. Hence

β(c ∘ d) = β(c) ∘ β(d) + β(ad), i.e.
β(c ∘ d) − β(c) ∘ β(d) = β(ad) ∈ F x from the above.

If d ∈ C_{0}, then a ∈ C_{0} and β(c ∘ d) = β(c) ∘ β(d). We record these properties of β in the next number. But before we do so, note that (B, +, ∘) can be made into a composition ring (B, +, ∗, ∘) where v ∗ w = 0 for all v, w ∈ B.

3.1. Proposition. Let C be a composition p-ring. Then β : (C, +) ⟶ (B, +), as defined above, is a surjective group homomorphism with kerβ = F + C_{0}x which satisfies:

(1) β(v) = v for all v ∈ B.

(2) β(cd) = β(c)n_{d} + n_{c}β(d) ∈ F x for all c, d ∈ C.

(3) β(c ∘ d) = β(c) ∘ β(d) + β(ad) for all c, d ∈ C where a = u_{c} ∘ d. If β_{0} denotes the restriction of β to C_{0}, then β_{0} : (C, +, ·, ∘) ⟶ (B, +, ∗, ∘) is a surjective composition ring homomorphism with kerβ_{0} = C_{0}x. □

From this result we know that C_{0}x is an ideal of (C_{0}, +, ·, ∘). But one can say more:

3.2. Proposition. Let C be a composition p-ring. Then for all k ∈ ℕ

(1) C_{0}^{k+1} = C_{0}x^{k},

(2) C_{0}x^{k} ⊲ (C_{0}, +, ·, ∘).

Proof. (1) (by induction on k). For k = 1, C_{0}x ⊆ C_{0}^{2}. If w_{1}, w_{2} ∈ C_{0}, say w_{i} = v_{i} +u_{i}x ∈ B + C_{0}x, i = 1, 2, then w_{1}w_{2} = v_{1}v_{2} + v_{1}u_{2}x + u_{1}xv_{2} + u_{1}xu_{2}x ∈ C_{0}x since v_{1}v_{2} ∈ B^{2} = Bx ⊆ C_{0}x (cf. Proposition 2.2.(5)). Thus C_{0}^{2} = C_{0}x. If C_{0}^{k} = C_{0}x^{k−1}, then C_{0}^{k+1} = C_{0} · C_{0}^{k} = C_{0}C_{0}x^{k−1} = C_{0}xx^{k−1} = C_{0}x^{k}.

(2) Let v, w, u ∈ C_{0}. Then vx^{k} − ux^{k} = (v − u)x^{k} ∈ C_{0}x^{k}, (vx^{k})u = (vu)x^{k} ∈ C_{0}x^{k}, v(ux^{k}) = (vu)x^{k} ∈ C_{0}x^{k} and vx^{k} ∘ u = v ∘ u · u^{k} ∈ C_{0}C_{0}^{k} = C_{0}^{k+1} = C_{0}x^{k}. Finally we have to show that v ∘ (w + ux^{k}) − v ∘ w ∈ C_{0}x^{k}. Suppose v = v_{1} + u_{1}x ∈ B + C_{0}x. Then u_{1} ∈ C_{0} = B + C_{0}x implies v = v_{1} + v_{2}x + u_{2}x^{2} with v_{2} ∈ B and u_{2} ∈ C_{0}. Continuing in this way, we get

v = v_{1} + v_{2}x + v_{3}x^{2} + ⋯ + v_{k}x^{k−1} + u_{k}x^{k}, v_{i} ∈ B, u_{k} ∈ C_{0}.

Substitute this expression for v in v ∘ (w + ux^{k}) − v ∘ w and after simplification, we will get a sum with terms of the form

v_{1} ∘ (w + ux^{k}) − v_{1} ∘ w
v_{i+1}x^{i} ∘ (w + ux^{k}) − v_{i+1}x^{i} ∘ w,  i = 1, 2, ⋯, k − 1, and
u_{k}x^{k} ∘ (w + ux^{k}) − u_{k}x^{k} ∘ w.

We consider these, each in turn:

v_{1} ∘ (w + ux^{k}) − v_{1} ∘ w = v_{1} ∘ w + v_{1} ∘ ux^{k} − v_{1} ∘ w
= (v_{1} ∘ u)x^{k} ∈ C_{0}x^{k};
v_{i+1}x^{i} ∘ (w + ux^{k}) − v_{i+1}x^{i} ∘ w
= v_{i+1} ∘ (w + ux^{k}) · (w + ux^{k})^{i} − v_{i+1} ∘ w · w^{i}
= (v_{i+1} ∘ w + v_{i+1} ∘ ux^{k})(w^{i} + d_{i}x^{k}) − v_{i+1} ∘ w · w^{i} for some d_{i} ∈ C_{0}
= v_{i+1} ∘ w · d_{i}x^{k} + (v_{i+1} ∘ ux^{k})w^{i} + (v_{i+1} ∘ ux^{k})d_{i}x^{k}
= (v_{i+1} ∘ w · d_{i})x^{k} + (v_{i+1} ∘ u)x^{k}w^{i} + (v_{i+1} ∘ ux^{k})d_{i}x^{k} ∈ C_{0}x^{k}

and finally,

u_{k}x^{k} ∘ (w + ux^{k}) − u_{k}x^{k} ∘ w
= u_{k} ∘ (w + ux^{k})^{k} · (w + ux^{k}) − u_{k} ∘ w · w^{k} ∈ C_{0}^{k+1} = C_{0}x^{k}. □

There is much more to say about and to do with composition p-rings. To make them even more like polynomial and power series composition rings, one could add the requirement (0 : x)_{C_{1}} = 0. We have refrained from doing so, since this requirement was not generally required in what we want to present here. Futhermore, if c ∈ C we may define the degree of c by using the decomposition c = n + v_{0} + v_{1}x + ⋯ + v_{k} + u_{k}x^{k}, n ∈ F, v_{i} ∈ B, u_{k} ∈ C_{0}. For example, if v_{0} = v_{1} = ⋯ = v_{k} = u_{k} = 0, deg(c) = 0, if v_{0} ≠ 0 and v_{1} = ⋯ = v_{k} = u_{k} = 0, deg(c) = 1 and if u_{k} ≠ 0 for all k, then deg(c) = ∞, etc. One may also investigate the composition rings C_{0}/C_{0}x^{k}, k ≥ 2. But these issues are not our main concern here and we will not pursue them any further. We much rather want to proceed with some examples of composition p-rings.

3.3. Examples. (1) Any one of the composition rings R[x], R_{0}[x] and R_{N}[[x]] is a composition p-ring. For the associated composition rings of polynomial functions \overline{R[x]} and \overline{R_{0}[x]} things are not so clear. For example, if R is an infinite field, then \overline{R[x]} ≅ R[x] is a composition p-ring, but if R is a finite field, then \overline{R[x]} = M(R) which is not a composition p-ring (if C = M(R), then 0 ≠ Base(C) ⊆ C_{0}x; cf Example 2.6(2)). In fact, we have: C = \overline{R[x]}, R a commutative ring with 1 ∈ R, is not a composition p-ring if and only if there exists an f = f_{1}x + f_{2}x^{2} + ⋯ + f_{k}x^{k} in R[x] with f_{1} ≠ 0 and f(t) = 0 for all t ∈ R. Indeed, if such an f exists, then (−f_{1})t = f_{2}t^{2} + ⋯ + f_{k}t^{k} for all t ∈ R and so 0 ≠ (−f_{1})x = f_{2}x^{2}+⋯+f_{k}x^{k} ∈ B∩C_{0}x. Conversely, if B∩C_{0}x ≠ 0, then there are f_{1}, f_{2}, …, f_{k} ∈ R with f_{1} ≠ 0 and f_{1}x = f_{2}x^{2}+⋯+f_{k}x^{k} ∈ B ∩C_{0}x. Then f := (−f_{1})x+f_{2}x^{2}+⋯ f_{k}x^{k} is the desired element of R[x]. As examples of such elements, one may consider 2x + 2x^{2} in ℤ_{4}[x] or 3x + 3x^{2} in ℤ_{6}.

(2) Let I ⊲ R, R a commutative ring with 1 ∈ R. Then C := {c ∈ R[x] | c ∘ 0 ∈ I} is a subcomposition ring of R[x]. It contains x and it is a composition p-ring.

(3) Let R be a commutative ring with 1 ∈ R and char R ≠ 2. Suppose further that R has an ideal I which satisfies I^{2} ≠ 0, I^{3} = 0 and for a, b ∈ R, if ai = bi^{2} for all i ∈ I, then b ∈ I (below we will give an example of such a ring). Let C = {f ∈ M(I) | f is of the form f = f_{0} + f_{1}x + f_{2}x^{2} where f_{0} ∈ I and f_{1}, f_{2} ∈ R}. Here, as usual x = 1_{I} (identity function on I) and f(t) = f_{0} + f_{1}t + f_{2}t^{2} ∈ I for all t ∈ I. It can be verified that C is a subcomposition ring of M(I) and x = 0 + 1x + 0x^{2} ∈ C is the commuting composition identity. We show B = Rx. The inclusion Rx ⊆ B is clear, so let v = v_{1}x+v_{2}x^{2} ∈ B. For any a, b ∈ C, v∘(a+b) = v∘a+v∘b implies 2v_{2}ab = 0. Thus v_{2}ab = 0. Let i ∈ I and let a(t) = b(t) = i for all t ∈ I. Then a, b ∈ C and v_{2}i^{2} = v_{2}a(t)b(t) = 0 = 0i. By our assumption on R, we get v_{2} ∈ I. This means v_{2}x^{2} = 0 (since (v_{2}x^{2})(i) = v_{2}i^{2} ∈ I^{3} = 0) and so v = v_{1}x ∈ Rx. Thus B = Rx. Note also, since I^{3} = 0, C_{0}x = Rx^{2}. Next we show B ∩ C_{0}x = 0. Let ax = bx^{2} ∈ B ∩ C_{0}x, a, b ∈ R. For any i ∈ I, ai = bi^{2} from which b ∈ I follows. Thus bx^{2} = 0 and so B ∩ C_{0}x = 0. We conclude that (C_{0}, +) = (B, +) ⊕ (C_{0}x, +) and so C is a composition p-ring. We note that (0 : x)_{C_{1}} ≠ 0. For example, choose 0 ≠ b ∈ I^{2}. Then bx = 0 but b ≠ 0.

Finally we show the existence of a ring R with the above claimed properties. Let ℤ be the

ring of integers and let I be the commutative ring of all 3 x 3 matrices of the form[ \begin{matrix}0 & a & b \\ 0 & 0 & a \\ 0 & 0 & 0 \end{matrix} ]

where a, b ∈ ℤ.

Then I^{2} ≠ 0 but I^{3} = 0 and char I ≠ 2. Let R be the Dorroh extension of I (i.e. the standard unital extension). This means (R, +) = (I, +) ⊕ (ℤ, +) and (a, n)(b, m) = (ab + nb + ma, nm) and I ≅ (I, 0) ⊲ R. R is a commutative ring with identity (0, 1) and char R = 0. Finally, let (a, n), (b, m) ∈ R such that (a, n)(i, 0) = (b, m)(i, 0)^{2} for all i ∈ I. From this we get ai + ni = bi^{2} = mi^{2} for all i ∈ I (where a, b ∈ I, n, m ∈ ℤ). Suppose

a = [ \begin{matrix}0 & α & α′ \\ 0 & 0 & α \\ 0 & 0 & 0 \end{matrix} ] , b = [ \begin{matrix}0 & β & β′ \\ 0 & 0 & β \\ 0 & 0 & 0 \end{matrix} ] and i = [ \begin{matrix}0 & j & j′ \\ 0 & 0 & j \\ 0 & 0 & 0 \end{matrix} ] where α, α′, β, β′, j, j′ ∈ ℤ.

From ai + ni = bi^{2} + mi^{2} we get αj + nj′ = mj^{2} and nj = 0 for all j, j′ ∈ ℤ. For j = 1 we get n = 0 and so αj = mj^{2} for all j ∈ ℤ. From this we get m = α = 0 and we conclude with (b, m) = (b, 0) ∈ I.

(4) Let C = (R, +, ·, ∘) where (R, +, ·) is a zero ring and (R, +, ∘) is a ring with identity x. Then F = F ound(C) = 0, B = Base(C) = C and C_{0}x = Cx = 0. Hence C is (trivially) a composition p-ring.

(5) Let (R[x, y], +, ·) be the ring as in Example 2.6 (7). On R[x, y] × R[x, y] define ∘ by: (f(x, y), g(x, y)) ∘ (h(x, y), k(x, y)) = (f(h(x, y), y), g(x, k(x, y))). Then C = (R[x, y] × R[x, y], +, ·, ∘) is a composition ring with commuting composition identity (x, y) and constant multiplicative identity (1, 1). Here F = F ound(C) = {(f(x, y), g(x, y)) ∈ C | f(x, y) is a polynomial in y and g(x, y) is a polynomial in x, i.e. f(x, y) = f(r, y) and g(x, y) = g(x, r) for all r ∈ R}. By Proposition 2.3(4) we get B = Base(C) = {(f(x, y)x, g(x, y)y) | (f(x, y), g(x, y)) ∈ F }. We note that C_{0} = {(f(x, y), g(x, y)) | every term of f(x, y) contains at least one x, and every term of g(x, y) contains at least one y}. Then it can be verified that (C_{0}, +) = (B, +) ⊕ (C_{0}(x, y), +), hence C is a composition p-ring.

We should point out that the composition ring on R[x, y]×R[x, y], as defined in Example 2.6(9), is not a composition p-ring. Even though it may have some other useful applications (in connection to abstract affine near-rings), the above shows that the composition on R[x, y] × R[x, y] as defined in the present example, is more ‘polynomial like’.

It should be clear that this example can be generalized to any finite number of commuting indeterminates x_{1}, x_{2}, x_{3}, ..., x_{n}: On n copies of the ring R[x_{1}, x_{2}, x_{3}, ..., x_{n}], define the composition by:

(f_{1}, f_{2}, …, f_{n}) ∘ (g_{1}, g_{2}, …, g_{n}) =
(f_{1}(g_{1}, x_{2}, x_{3}, …, x_{n}), f_{2}(x_{1}, g_{2}, x_{3}, ..., x_{n}), …, f_{n}(x_{1}, x_{2}, x_{3}, ..., g_{n}))

where f_{i} = f_{i}(x_{1}, x_{2}, x_{3}, ..., x_{n}) and g_{i} = g_{i}(x_{1}, x_{2}, x_{3}, ..., x_{n}) are from R[x_{1}, x_{2}, x_{3}, ..., x_{n}].

3.4. Remark. In our definition of a composition p-ring, we have opted for clarity of presentation following our motivating examples as closely as possible. If more generality is required, one need not require the existence of the commuting composition identity. In such a case, one could replace all occurences of F x and C_{0}x with \overline{F B} and \overline{C_{0}B} respectively. A composition ring C can then be called a composition p-ring if BF = F B and (C_{0}, +) = (B, +) ⊕ (\overline{C_{0}B}, +). In this case, the definition of β stays the same, but then one can show that β(cd) ∈ \overline{F B} and with slightly more effort that β(c ∘ d) − β(c) ∘ β(d) ∈ \overline{F B}. In case C does have a commuting composition identity x, then we know from Proposition 2.2(4) that \overline{BF} = \overline{F B} = F x. But also \overline{C_{0}B} = C_{0}x: Let u ∈ C_{0} and v ∈ B. Since C_{0} = B + \overline{C_{0}B}, u = w + ∑ u_{i}v_{i} for some u_{i} ∈ C_{0} and w, v_{i} ∈ B. Using the fact that B^{2} = Bx ⊆ C_{0}x, we will get uv ∈ C_{0}x. Thus C_{0}x ⊆ C_{0}B ⊆ C_{0}x and so \overline{C_{0}B} = C_{0}x.

3.5. Example. Let R be the ring of even integers, i.e. R = (2ℤ, +, ·). Then C := (R[x], +, ·, ∘) is a composition ring which does not have a multiplicative identity nor a composition identity. Here F = F ound(C) = R and B = F x = Rx with BF = F B = \overline{F B} ⊂ F x (for example, 2x ∈ F x, but 2x ∉ F B). Futhermore, B ∩ \overline{C_{0}B} = B ∩ C_{0}B = 0 and (C_{0}, +) = (B, +) ⊕ (\overline{C_{0}B}, +).

4. Semi-constants

We start with 4.1. Proposition. Let C be a composition ring with x ∈ C a commuting composition identity. Let s ∈ C be such that

(1) c ∘ s ∘ a · a = c ∘ s · a for all c, a ∈ C.

(2) c ∘ s · s = c ∘ s for all c ∈ C.

(3) (0 : s)_{C_{1}} ∩ B = 0 (i.e. vs = 0, v ∈ B, implies v = 0 ).

Then (C, +) = (B, +) ⊕ (D_{s}, +) where D_{s} = {c ∈ C | (c ∘ s)x = 0} ⊲ C_{1}.

Proof. For any c ∈ C, c = (c ∘ s)x + (c − (c ∘ s)x). We show (c ∘ s)x ∈ B. For any a, b ∈ C we have:

(c ∘ s)x ∘ (a + b) = c ∘ s ∘ (a + b) · (a + b)
= c ∘ s · (a + b) by (1)
= (c ∘ s)a + (c ∘ s)b
= c ∘ s ∘ a · a + c ∘ s ∘ b · b by (1)
= (c ∘ s)x ∘ a + (c ∘ s)x ∘ b

and

(c ∘ s)x ∘ ab = c ∘ s ∘ ab · ab = c ∘ s · ab = ((c ∘ s)a)b = (c ∘ s ∘ a · a)b = ((c ∘ s)x ∘ a)b. Next we show c − (c ∘ s)x ∈ D_{s} :

((c − (c ∘ s)x) ∘ s)x
= (c ∘ s)x − (c ∘ s ∘ s · s)x
= (c ∘ s)x − (c ∘ s · s)x by (1)
= (c ∘ s)x − (c ∘ s)x by (2)
= 0.

If v ∈ B ∩ D_{s}, then 0 = (v ∘ s)x = vs. Hence v ∈ (0 : s)_{C_{1}} ∩ B = 0 by (3) and we conclude that (C, +) = (B, +) ⊕ (D_{s}, +). Finally we show D_{s}⊲C_{1}. Let c_{1}, c_{2} ∈ D_{s} and c ∈ C. Then ((c_{1}−c_{2})∘s)x = (c_{1}∘s)x−(c_{2}∘s)x = 0, (c_{1}c ∘ s)x = c_{1} ∘ s · c ∘ s · x = ((c_{1} ∘ s)x)c ∘ s = 0 and (cc_{1} ∘ s)x = c ∘ s · (c_{1} ∘ s)x = 0. □

4.2. Corollary. Let x ∈ C be a commuting composition identity and let s ∈ C be such that:

(1) c ∘ s ∘ a · a = c ∘ s · a for all c, a ∈ C and

(2) x = sx. Then (0 : s)_{C_{1}} ∩ B = 0 and c ∘ s · s = c ∘ s for all c ∈ C; hence (C, +) = (B, +) ⊕ (D_{s}, +) and D_{s} ⊲ C_{1}.

Proof. We only have to verify that conditions (2) and (3) of Proposition 4.1 are fulfilled. For any c ∈ C, c ∘ s = x ∘ (c ∘ s) = sx ∘ (c ∘ s) = s ∘ c ∘ s · c ∘ s

= x ∘ s ∘ (c ∘ s) · c ∘ s = x ∘ s · c ∘ s by (1)

= s · c ∘ s

= c ∘ s · s by Proposition 2.2(1)

If v ∈ B, then vs = (v ∘ s)x = v ∘ sx = v ∘ x = v; hence vs = 0 if and only if v = 0. Thus (0 : s)_{C_{1}} ∩ B = 0. □

4.3. Definition. Let C be a composition ring. Then s ∈ C is called a semi-constant if:

(1) s ∘ s = s,

(2) (c ∘ s)b ∘ a = c ∘ s · b ∘ a for all c ∈ C and a, b ∈ C_{0}. A minimal semi-constant s is a semi-constant which satisfies: c ∘ s ∘ t · a = c ∘ s · a for all c ∈ C, a ∈ C_{0} and non-zero semi-constants t.

Any constant s (i.e. s ∈ F) is a minimal semi-constant. We now describe all the minimal semi-constants of M_{0}(R) and M(R). In case C has a commuting composition identity x and s ∈ C is a semi-constant, then c ∘ s ∘ a · a = c ∘ s · a for all c ∈ C and a ∈ C_{0} ( just let b = x).

4.4. Proposition. Let R be a commutative ring which satisfies Ry = 0 implies y = 0 (e.g., if 1 ∈ R). Let C = M_{0}(R). Then s ∈ M_{0}(R) is a semi-constant if and only if s = r̃ for some r ∈ R (for the definition of r̃, see Example 2.6(2)).

Proof. If S is of the form s = r̃ for some r ∈ R, then it is straightforward to verify that S is a semi-constant. Conversely, suppose s ∈ C is a semi-constant. If s = 0 we are done, so suppose s(t_{0}) = r_{0} ≠ 0 for some 0 ≠ t_{0} ∈ R. Choose y ∈ R with yt_{0} ≠ 0. We firstly show that t ∈ R, t ≠ 0 implies s(t) ≠ 0. To the contrary, if s(t) = 0 for some t ≠ 0, let

a(r) = \{ \begin{matrix}t_{0} & if & r = t \\ 0 & if & r ≠ t. \end{matrix}

Then a ∈ C = C_{0} and from (ỹ ∘ s)b ∘ a = ỹ ∘ s · b ∘ a, with b = 1_{R}, we get 0 ≠ yt_{0} = ỹ(r_{0})t_{0} = ŷ(s(t_{0})) · t_{0} = ỹ(s(a(t))a(t)) = ỹ(s(t)) · a(t) = ỹ(0) · t_{0} = 0; a contradiction. Thus t ≠ 0 implies s(t) ≠ 0. Suppose for t_{1} ∈ R, t_{1} ≠ 0, t_{1} ≠ t_{0}, we have s(t_{1}) = r_{1} ≠ r_{0}. Then r_{1} ≠ 0. Define functions d and b by

d(t) = \{ \begin{matrix}y & if & t = r_{0} \\ 0 & if & t ≠ r_{0} \end{matrix} and b(t) = \{ \begin{matrix}t_{0} & if & t = t_{1} \\ 0 & if & t ≠ t_{1} \end{matrix}

Then d ∈ C, b ∈ C = C_{0} and from (d ∘ s)1_{R} ∘ b = d ∘ s · 1_{R} ∘ b we get 0 ≠ yt_{0} = d(r_{0})b(t_{1}) = d(s(t_{0})) · b(t_{1}) = d(s(b(t_{1})))b(t_{1}) = d(s(t_{1})b(t_{1}) = d(r_{1})b(t_{1}) = 0; a contradiction. Hence r_{1} = r_{0} and we conclude that s = r̃_{0}. □

4.5. Corollary. Let R be a commutative ring which satisfies Ry = 0 implies y = 0. Let C = M_{0}(R). Then the following are equivalent:

(1) s ∈ C is a semi-constant.

(2) s = r̃ for some r ∈ R.

(3) s is a minimal semi-constant. □

There is a similar characterization of the semi-constants in M(R) but, since the mappings are not necessarily 0-symmetric, the proof is slightly different.

4.6. Proposition. Let R be a commutative ring which satisfies Ry = 0 implies y = 0. Let C = M(R). Then the following are equivalent:

(1) s ∈ C is a semi-constant.

(2) There exists an r ∈ R such that s = r̂ (i.e. s is constant) or s = r̃.

(3) s is a minimal semi-constant.

Proof. We will only provide the verification of (1) ⇒ (2); the other implications being straightforward. Suppose s ∈ C is a semi-constant. If s is not a constant, we are done, so suppose s is a constant. Then there are t_{1} ≠ t_{2} in R such that s(t_{1}) = r_{1} ≠ r_{2} = s(t_{2}). At least one of t_{1} or t_{2} is non-zero, say t_{2} ≠ 0. We will show that t_{1} = 0. Suppose this is not the case, i.e. t_{1} ≠ 0. We also know that at least one of r_{1} or r_{2} must be non-zero. We will consider both these cases. Suppose r_{1} ≠ 0 and choose y ∈ R with yt_{1} ≠ 0. Define a and c in C by:

a(t) = \{ \begin{matrix}t_{1} & if & t = t_{2} \\ 0 & if & t ≠ t_{2} \end{matrix} and c(t) = \{ \begin{matrix}y & if & t = r_{1} \\ 0 & if & t ≠ r_{1} \end{matrix}

Then a ∈ C_{0} and from (c ∘ s)1_{R} ∘ a = c ∘ s · 1_{R} ∘ a we get 0 ≠ yt_{1} = c(r_{1})t_{1} = c(s(t_{1}))t_{1} = c(s(a(t_{2})))a(t_{2}) = c(s(t_{2}))a(t_{2}) = c(r_{2})a(t_{2}) = 0; a contradiction.

If r_{2} ≠ 0, choose z ∈ R such that zt_{2} ≠ 0 and let

b(t) = \{ \begin{matrix}t_{2} & if & t = t_{1} \\ 0 & if & t ≠ t_{1} \end{matrix} and d(t) = \{ \begin{matrix}z & if & t = t_{2} \\ 0 & if & t ≠ t_{2}. \end{matrix}

Then d ∈ C, b ∈ C_{0} and from d ∘ s ∘ b · b = d ∘ s · b we will get the contradiction zt_{2} = 0.

Thus we must have t_{1} = 0. But this means that s is of the form s(t) = \{ \begin{matrix}r_{2} & if & t ≠ 0 \\ r_{1} & if & t = 0. \end{matrix}

Since s ∘ s = s, for any 0 ≠ t ∈ R, we then have r_{2} = s(t) = s(s(t)) = s(r_{2}). This means r_{2} ≠ 0 and r_{1} = s(0) = s(s(0)) = s(r_{1}). Thus r_{1} = 0 and so s = r̃_{2}. □

4.7. Example. In C = \overline{ℤ_{4}[x]}, s = x + x^{3} is a non-zero semi-constant (which is not a constant). □

Let 0 ≠ s ∈ C be a minimal semi-constant, C any composition ring. Let

k = \{ \begin{matrix}s & if & F = 0 \\ s ∘ n & for & 0 ≠ n ∈ F. \end{matrix}

We note that in general, if F ≠ 0, then k is not uniquely defined. One would be tempted to argue if x ∈ C_{2} is a commuting composition identity and if i ∈ C_{1} is a multiplicative identity, then s ∘ n = s since s ∘ n = x ∘ s ∘ n · i = x ∘ s · i = si = s (using the fact that s is a minimal semi-constant and n ≠ 0 is a semi-constant). But this is not a valid argument since i ∈ C_{0} need not hold (as is required in the definition of a minimal semi-constant). If i ∈ C_{0}, then C = C_{0} and F = 0 in which case s ∘ n is of no concern. In any case, k itself is a minimal semi-constant.

Let v ∈ B. Define a mapping β : (C, +, ·) → (B, +, ∘) by β(c) = (c ∘ k)v for all c ∈ C. This mapping depends only on s and v (and not on the n chosen) since s is a minimal semiconstant and v ∈ B ⊆ C_{0}. It is easy to see that β is a group homomorphism, but first priority should be to confirm that it is well-defined, i.e. (c ∘ k)v ∈ B for all c ∈ C. We distinguish two cases.

If F = 0, then for all a, b ∈ C = C_{0}: (c∘k)v∘(a+b) = (c∘s)v∘(a+b) = c∘s·v∘(a+b) = c ∘ s · (v ∘ a + v ∘ b) = c ∘ s · v ∘ a + c ∘ s · v ∘ b = (c ∘ s)v ∘ a + (c ∘ s)v ∘ b. Futhermore, (c ∘ k)v ∘ ab = (c ∘ s)v ∘ ab = c ∘ s ∘ ab · v ∘ ab = c ∘ s · v ∘ ab = c ∘ s · (v ∘ a)b = (c ∘ s ∘ a · v ∘ a)b = ((c ∘ s)v ∘ a)b = ((c ∘ k)v ∘ a)b. Thus if F = 0, then (c ∘ k)v ∈ B for all c ∈ C. If F ≠ 0, then (c ∘ k)v ∈ F x ⊆ B. Thus β(c) ∈ B for all c ∈ C. We record this (and more), for the special case v = x, in

4.8. Proposition. Let C be a composition ring with commuting composition identity x and non-zero semi-constant s. Let k = s if F = 0 and k = s ∘ n for 0 ≠ n ∈ F. Then β_{s} : (C, +, ·) → (B, +, ∘) defined by β_{s}(c) = (c ∘ k)x = (c ∘ s)x is a ring homomorphism. If F ≠ 0, then β_{s}(C) ⊆ F x. The homomorphism β_{s} is surjective if and only if there is a d ∈ C with d ∘ s · x = x. If β_{s} is surjective and F ≠ 0, then (C ∘ k)x = β_{s}(C) = B = F x.

Proof. We start by showing that β_{s} is a ring homomorphism. For c, d ∈ C, and since k is a semi-constant with (d ∘ k)x ∈ C_{0}, we get

β_{s}(cd) = (cd ∘ k)x = c ∘ k · d ∘ k · x = c ∘ k ∘ (d ∘ k)x · (d ∘ k)x
= (c ∘ k)x ∘ ((d ∘ k)x) = β_{s}(c) ∘ β_{s}(d).

If β_{s} is surjective, then x ∈ B = β_{s}(C) implies there is a d ∈ C with x = (d ∘ k)x = d ∘ s · x. Conversely, if such a d ∈ C exists, then for any v ∈ B, v ∘d ∈ C and β_{s}(v ∘d) = (v ∘d∘k)x = (v ∘ d ∘ s)x = (v ∘ (d ∘ s))x = v ∘ (d ∘ s)x = v ∘ x = v. Finally, if β_{s} is surjective and F ≠ 0, then F x ⊆ B = β_{s}(C) = (C ∘ k)x ⊆ F x. □

For two non-zero minimal semi-constants s_{1} and s_{2} in C, with associated ring homomorphisms β_{1} and β_{2} respectively (i.e. β_{i}(c) = (c∘s_{i})x ), we always have β_{1}(C) = β_{2}(C), but β_{1} need not coincide with β_{2}. Indeed, for any c ∈ C, β_{1}(c) = (c ∘ s_{1})x = (c ∘ s_{1} ∘ s_{2})x ∈ (C ∘ s_{2})x = β_{2}(C) and β_{2}(C) ⊆ β_{1}(C) follows similarly. The next example will show that in general β_{1} ≠ β_{2} and in Proposition 5.1(9) we will show under which circumstances they do coincide.

4.9. Example. Let R = ℤ_{4} and let C = M_{0}(R). Then s_{1} = ˜1 and s_{2} = ˜2 are two nonzero minimal semi-constants in C. Let β_{1} and β_{2} be the associated ring homomorphism respectively. Then β_{1}(x) = (x ∘ s_{1})x = s_{1}x = ˜1 while β_{2}(x) = (x ∘ s_{2})x = s_{2}x = ˜2 ≠ ˜1. □

Worthwhile to point out is that if β_{s} is surjective for any one non-zero minimal semi-constant s, then β_{s′} is surjective for all non-zero minimal semi-constants s′. Indeed, since β_{s} is surjective, there is a d ∈ C with d ∘ s · x = x ( by Proposition 4.8). Since s′ is a non-zero semi-constant and s is minimal (d ∘ s) ∘ s′ · x = d ∘ s ∘ s′ · x = d ∘ s · x = x; hence β_{s′} is surjective.

5. Composition t-rings

A composition ring C is called a composition t-ring if C has a commuting composition identity x and a non-zero minimal semi-constant s with x = sx. If we use this s to define k as in the previous section, there are many useful properties that we list in:

5.1. Proposition. Let C be a composition t-ring and let

k = \{ \begin{matrix}s & if & F = 0 \\ s ∘ n & for & 0 ≠ n ∈ F \end{matrix} where 0 ≠ s is a minimal semi-constant in C with x = sx.

Let β_{s} : C → B be the mapping defined by β_{s}(c) = (c ∘ s)x. Then:

(1) k is a non-zero minimal semi-constant and kx = x = xk.

(2) c ∘ k ∘ a · a = c ∘ k · a for all a, c ∈ C.

(3) a = ka for all a ∈ C and vk = v for all v ∈ B.

(4) If F ≠ 0, then k ∈ F is an identity for the ring (F, +, ·) and B = F x.

(5) k ∘ k = k, k ∘ s = k = s ∘ k and kk = k.

(6) (0 : x)_{C_{1}} ∩ F = 0

(7) c ∘ k = k · c ∘ k = c ∘ k · k

(8) If s′ is any other non-zero minimal semi-constant in C and k′ is defined analogously to k, then k = k′ and (c ∘ k)x = (c ∘ k′)x for all c ∈ C (i.e. β_{s} = β_{s′})

(9) β_{s} is surjective.

Proof. (1) If k = s ∘ n = 0 (0 ≠ n ∈ F), then n = x ∘ n = sx ∘ n = s ∘ n · n = 0; a contradiction. That k is a semi-constant is clear, since either k = s or k ∈ F. If k = s, then kx = x = xk. If k = s ∘ n, then kx = (x ∘ s ∘ n)x = (x ∘ s)x = sx = x since s is a minimal semi-constant.

(2) We distinguish two cases: If F = 0, then c ∘ k ∘ a · a = c ∘ s ∘ a · a = c ∘ s · a = c ∘ k · a for all c ∈ C, a ∈ C = C_{0}. If F ≠ 0, then c ∘ k ∘ a · a = c ∘ k · a for all c, a ∈ C (since k ∈ F).

(3) For any a ∈ C, a = x ∘ a = kx ∘ a = k ∘ a · a = x ∘ k ∘ a · a = x ∘ k · a = ka (by (2) above) and v = v ∘ x = v ∘ xk = (v ∘ x)k = vk.

(4) If F ≠ 0, then k ∈ F and the result follows from (3) above and Proposition 2.2(1). For any v ∈ B, v = v ∘ x = v ∘ kx = (v ∘ k)x ∈ F x.

(5) Straightforward.

(6) If F = 0 we are done. If F ≠ 0 and m ∈ F with mx = 0, then m = mk = mx ∘ k = 0.

(7) By (3) above and Proposition 2.2(1).

(8) Let k′ = s′ if F = 0 and let k′ = s′ ∘ n for 0 ≠ n ∈ F. Then k′ has all the properties that k does. From (4) above we get k′ = k.

(9) β_{s} surjective follows from Proposition 4.8 since x = kx = (x ∘ k)x. □

From Propositions 5.1, 4.8 and 4.1 we get most of :

5.2. Proposition. Let C be a composition t-ring. Then there exists a non-zero minimal semi-constant k such that (C, +) = (B, +) ⊕ (D, +) where D = {c ∈ C | (c ∘ k)x = 0} ⊲ C_{1} and β : (C, +, ·) → (B, +, ∘) defined by β(c) = (c ∘ k)x for all c ∈ C is a surjective ring homomorphism with kerβ = D and β(v) = v for all v ∈ B. The mapping β is independent of which minimal semi-constant k ≠ 0 is chosen. If F ≠ 0, then B = F x. □

We note that if C is a composition ring with F = 0, then it cannot be both a composition p-ring as well as a composition t-ring. The reason being that any 0-symmetric composition t-ring has 0 ≠ x ∈ B ⊆ C_{0}x (cf. Proposition 2.4). But if F ≠ 0, a composition ring can be of both types. For example, let C = R[x]. Then C is a composition p-ring as well as a composition t-ring (s = 1 ∈ F is a non-zero minimal (semi-) constant with x = sx). We note further that the two associated mappings, say β_{p} and β_{t} respectively, are not the same. For c = c_{0} + c_{1}x + c_{2}x^{2} + ⋯ + c_{k}x^{k} ∈ C, β_{p}(c) = c_{1}x and β_{t}(c) = (c_{0} + c_{1} + ⋯ + c_{k})x. Of course, both have image B = F x = Rx, but kerβ_{p} = {c ∈ C | c_{1} = 0} and ker β_{t} = {c ∈ C | c∘1 = 0} are not comparable: For any 0 ≠ r ∈ R, −r + rx ∈ kerβ_{t} but −r + rx ∉ kerβ_{p} and for a, b ∈ R with a + b ≠ 0, a + bx^{2} ∈ kerβ_{p} but a + bx^{2} ∉ ker β_{t}.

5.3. Examples. (1) Let R be a commutative ring with 1 ∈ R and let C be a subcomposition ring of M(R) such that x = 1_{R} ∈ C. If F ound(C) ≠ 0, then C is a composition t-ring if and only if ˆ1 ∈ C. Indeed, if ˆ1 ∈ C, then ˆ1 is a non-zero minimal (semi-) constant in C with x = ˆ1x. Conversely, if C is a composition t-ring, let 0 ≠ k ∈ F ound(C) be such that x = kx. Then t = k(t)t for all t ∈ R and so k(1) = 1. Since k is constant, we conclude that ˆ1 = k ∈ C. If F ound(C) = 0 and ˜1 = k ∈ C, then C is a composition t-ring. If R has no non-zero zero divisors, then also the converse is true, i.e. if C is a composition t-ring, then ˜1 ∈ C.

(2) \overline{R[x]} is always a composition t-ring since ˆ1 ∈ \overline{R[x]}. \overline{R_{0}[x]} need not be a composition t-ring. For example, suppose R has a non-zero nilpotent element b, say b^{n} = 0 but b^{n−1} ≠ 0. Then \overline{R_{0}[x]} is not a composition t-ring. To the contrary, if it is, then x = sx for some 0 ≠ s ∈ \overline{R_{0}[x]}. Thus, if s = s_{1}x+s_{2}x^{2} +···+s_{k}x^{k}, we get t = s(t)t = s_{1}t^{2} +s_{2}t^{3} +···+s_{k}t^{k+1} for all t ∈ R. Hence t^{n−1} = s_{1}t^{n−1} + s_{2}t^{n+2} + ⋯ s_{k}t^{n+k−1} for all t ∈ R. For t = b, we get 0 ≠ b^{n−1} = s_{1}b^{n+1} + s_{2}b^{n+2} + ⋯ + s_{k}b^{n+k−1} = 0, a contradiction.

6. Ideals in composition rings

At first we will consider the relationship between the ideals of C and those of the foundation F of C. Most of these are known results (often with their origins in the more general composition near-rings) and are, of course, only interesting if F ≠ 0.

Let ψ : C → M(F) be the composition ring homomorphism defined by ψ(c) = ψ_{c} : F → F (F = F ound(C)), ψ_{c}(n) = c∘n for all n ∈ F, c ∈ C. Kerψ = (0 : F)_{C_{2}} = {c ∈ C | c∘F = 0}⊲C and if Kerψ = 0, then C can be regarded as a subcomposition ring of M(F). In particular, if C is simple and F ≠ 0, then C is a subcomposition ring of M(F). But one can say more:

6.1. Proposition. (Adler [1]) Let R be a ring. If every subcomposition ring C of M(R) with F ound (C) = R is simple, then R is simple. Conversely, if R is a simple ring with identity, then every subcomposition ring C of M(R) with F ound(C) = R is simple. □

The foundation F can always be considered as a C-module in the following sense: Let C be a composition ring and let S = (S, +, ·) be a ring. Then S is called a C-module if there is a mapping C × S → S, (c, s) ⟼ cs which satisfies (c + d)s = cs + ds, (cd)s = (cs)(ds) and (c ∘ d)s = c(ds) for all c, d ∈ C, s ∈ S. A subset J of S is an C-ideal of S if J is an ideal of S and c(s + j) − cs ∈ J for all c ∈ C, s ∈ S and j ∈ J. S is C-simple if it has no non-trivial proper C-ideals. Clearly, if S is simple, then it is C-simple. The foundation F of C is always an C-module in the cannonical sense: (c, n) := c ∘ n. The composition rings C for which the notions “C-ideal” and “ideal” of F coincide, are called compatible. Any composition ring with F = 0 is trivially compatible. Not all composition p-rings are compatible. For example, let R be a commutative ring with 1 ∈ R which contains a nil ideal N and suppose N has a non-zero ideal J which is not an ideal in R (an example of such a ring R is given below). Then there are elements j ∈ J and r ∈ R such that rj ∉ J. Let C = R_{N}[[x]]. Then F ound(C) = N and J is an ideal of N which is not a C-ideal of N: rx ∘ (n + j) − rx ∘ n = rj ∉ J (rx ∈ C, n ∈ N). We now give an example of the ring R with the properties as claimed. Let A be a commutative ring with 1 ∈ R and which contains an ideal B ≠ 0 with B^{2} = 0 (for example, take B any ring with zero multiplication and let A be any unital ideal extension of B). Let R be the subring of the full 3 × 3 matrix ring over A

consisting of all matrices of the form[ \begin{matrix}a & b & c \\ 0 & a & b \\ 0 & 0 & a \end{matrix} ] . It can be verified that R is a commutative

ring with identity (the usual matrix ring identity).

Let J be the set of all matrices of the form

[ \begin{matrix}0 & b & 0 \\ 0 & 0 & b \\ 0 & 0 & 0 \end{matrix} ] with b ∈ B and let N be the set of all matrices of the form [ \begin{matrix}0 & b & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{matrix} ]

with b ∈ B and c ∈ A. Then J ⊲ N ⊲ R, N^{2} = 0 and J is not an ideal of R: For 0 ≠ b ∈ B,

[ \begin{matrix}0 & b & 0 \\ 0 & 0 & b \\ 0 & 0 & 0 \end{matrix} ] ∈ J, but [ \begin{matrix}0 & b & 0 \\ 0 & 0 & b \\ 0 & 0 & 0 \end{matrix} ] [ \begin{matrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} ] = [ \begin{matrix}0 & b & b \\ 0 & 0 & b \\ 0 & 0 & 0 \end{matrix} ] ∉ J.

6.2. Proposition. Let C be a composition ring with commuting composition identity x. If every c ∈ C_{0} is of the form c = v_{0} + v_{1}x + v_{2}x^{2} + ⋯ + v_{k}x^{k} for some k ≥ 0, v_{i} ∈ B for all i = 0, 1, 2, ⋯, k, then C is compatible.

Proof. Let j ∈ J ⊲ F = F ound(C), c ∈ C with c = m + u ∈ F + C_{0} and u = Σ_{i=0}^{k}v_{i}x^{i}, k ≥ 0, v_{i} ∈ B. Let n ∈ F. Then (n + j)^{i} = n^{i} + d_{i}j for some d_{i} ∈ F and we get:

c ∘ (n + j) − c ∘ n
= m ∘ (n + j) + Σv_{i}x^{i} ∘ (n + j) − m ∘ n − Σv_{i}x^{i} ∘ n
= Σv_{i} ∘ (n + j) · (n + j)^{i} − Σv_{i} ∘ n · n^{i}
= Σv_{i} ∘ n · (n + j)^{i} + Σv_{i} ∘ j · (n + j)^{i} − Σv_{i} ∘ n · n^{i}
= Σv_{i} ∘ n · d_{i}j + Σv_{i} ∘ j · n^{i} + Σv_{i} ∘ j · d_{i}j
= Σv_{i} ∘ (nd_{i}j) + Σv_{i} ∘ (jn^{i}) + Σv_{i} ∘ (jd_{i}j).

Now v_{i}∘(nd_{i}j) = (v_{i}∘nd_{i})j ∈ F J ⊆ J, v_{i}∘(jn^{i}) = (v_{i}∘n^{i})j ∈ F J ⊆ J (cf. Proposition 2.2(1)) and v_{i} ∘ (jd_{i}j) = (v_{i} ∘ jd_{i})j ∈ F J ⊆ J. Hence c ∘ (n + j) − c ∘ n ∈ J. □

For a composition ring C with F = F ound(C) we will use the following notation: Let A ⊆ C and let S ⊆ C be either a subring of C_{1} or a subnear-ring of C_{2} or a subcomposition ring of C.

⟨A⟩_{s} denotes the ideal in S generated by A ∩ S. For example, if S = C_{1} then ⟨A⟩_{C_{1}} denotes the ideal in C_{1} generated by A. If S = (C, +, ·, ∘), then we usually write ⟨A⟩_{C} just as ⟨A⟩. We will also use

(A ∩ F : F)_{C_{2}} := {c ∈ C | c ∘ F ⊆ A ∩ F }.

We recall two further results; the latter already part of the folklore of the theory of polynomial near-rings.

6.3. Proposition. [4] Let J ⊲F, F = F ound(C), C a composition ring. Then the following is equivalent:

(1) J is an C-ideal of F.

(2) (J : F)_{C_{2}} ⊲ C.

(3) J = ⟨J⟩ ∩ F. □

6.4. Proposition. Let I ⊲ C. Then there is a unique C-ideal J of F (namely J = F ∩ I) such that ⟨J⟩ ⊆ I ⊆ (J : F)_{C_{2}}. □

6.5. Proposition. The mapping θ : {I | I ⊲ C} → {J | J an C-ideal of F } defined by θ(I) = I∩F is a surjective function which retains sums of ideals, inclusions and intersections. In general, θ need not be a bijection.

Proof. Let I = Σ_{α}I_{α}, I_{α} ⊲C. Then c ∈ I ∩F if and only if c ∈ F and c = ∑_{i=1}^{k} s_{α_{i}}, s_{α_{i}} ∈ I_{α_{i}}. For such a c, c = c ∘ 0 = Σs_{α_{i}} ∘ 0 and s_{α_{i}} ∘ 0 ∈ I_{α_{i}} ∩ F. Thus θ(I) = ∑_{α} θ(I_{α}) and so θ retains sums of ideals. θ is surjective: Let J be an C-ideal of F. Then ⟨J⟩ ⊆ (J : F)_{C_{2}} and J ⊆ ⟨J⟩∩F = θ(⟨J⟩) ⊆ θ((J : F)_{C_{2}}) = (J : F)_{C_{2}} ∩F = J. θ need not be a bijection: For a Cideal J of F, θ(⟨J⟩) = θ((J : F)_{C_{2}}) = J, but ⟨J⟩ and (J : F)_{C_{2}} need not coincide. For example if C = C_{0}, let J = 0. Then (0 : F)_{C_{2}} = (0 : 0)_{C_{2}} = C and θ(0) = 0 = θ(C) = θ((0 : F)_{C_{2}}). The remaining assertions are clear. □

6.6. Proposition. (Peterson and Veldsman [3]) Let C be a composition ring with F ound(C) = F ≠ 0. Then (0 : F)_{C_{2}} is a maximal ideal of C if and only if F is C-simple and (0 : F)_{C_{2}} + ⟨F ⟩ = C. □

6.7. Corollary. (Peterson and Veldsman [3]) Let C be a composition ring with F ound(C) = F ≠ 0 and ⟨F ⟩ = C. Then:

(1) F is C-simple if and only if (0 : F)_{C_{2}} is the unique maximal ideal of C.

(2) C is simple if and only if F is C-simple and (0 : F)_{C_{2}} = 0. □

Let us mention that if C is infinite and 0 ≠ F ound(C) = F is finite, then C is not simple. Indeed, (0 : F)_{C_{2}} is non-zero (otherwise ψ : C → M(F) defined by ψ(c) = ψ_{c} : F → F, ψ_{c}(n) = c ∘ n, is an injective mapping from the infinite set C into the finite set M(F)) and (0 : F)_{C_{2}} ≠ C (otherwise F = 0). Note also that if the composition ring is compatible, then all occurrences of “C-ideal” can be replaced by “ideal”. We next look at the relationship between the ideals of F and those of B.

6.8. Proposition. Let C be a composition ring with commuting composition identity x, F ound(C) = F and Base(C) = B.

(1) If J ⊲ F, then Jx ⊲ (F x, +, ∘).

(2) If J is an C-ideal of F, then B ∘ J + J ∘ B ⊆ J and Jx ⊲ (B, +, ∘).

(3) Let K ⊲ B. Let J = {n ∈ F | nx ∈ K ∩ F }. Then Jx = K ∩ F x and J ⊲ F. If B = F x, then K ⊲ B if and only if K = Jx for some J ⊲ F.

Proof. (1) For j ∈ J and n ∈ F, jx ∘ nx = (jn)x ∈ Jx and nx ∘ jx = (nj)x ∈ Jx.

(2) J ∘ B = J (since J ⊆ F). For v ∈ B, j ∈ J, v ∘ j = v ∘ (0 + j) − v ∘ 0 ∈ J since J is an C-ideal of F. Hence J ∘ B + B ∘ J ⊆ J. Furthermore,

jx ∘ v = jv

= vj since j ∈ J ⊆ F, cf. Proposition 2.2(1) = (v ∘ j)x since v ∈ B ∈ (B ∘ j)x ⊆ Jx

and v ∘ jx = (v ∘ j)x ∈ (B ∘ J)x ⊆ Jx.

(3) Clearly Jx = K ∩ F x. Let j ∈ J, n ∈ F. Then (jn)x = (jx)n ∈ (K ∩ F x)F ⊆ K ∩ F x (cf. Proposition 2.2 (6)). Hence jn ∈ J and thus also nj = jn ∈ J. □

Next we will look at the transfer of ideals between C and its base B. For this we will make the following

Assumption. In the sequel we suppose that β : (C, +) → (B, +) is a surjective group homomorphism.

Let A ⊆ B. We define

[A] := \{c ∈ C | β(c) ∈ A + \overline{F B}\}.

If C is a composition p-ring, we take β to be the mapping as defined in §3 and if C is a composition t-ring we take β as defined in Proposition 5.2. For both these two cases, \overline{F B} = F x. In case C is both a composition p-ring as well as a composition t-ring (cf §5), then it doesn’t matter how we choose β, at least for the purpose we have in mind here. This follows, since in this case, F ≠ 0, B = F x and hence [A] = {c ∈ C | β(c) ∈ A + B} = C.

We will at times impose further conditions on β. The first of these are:

(B_{1}) For all a, b ∈ C, β(ab) ∈ \overline{F B} or for all a, b ∈ C, β(ab) = β(a) ∘ β(b). (B_{2}) For all a, b ∈ C, β(a ∘ b) − β(a) ∘ β(b) ∈ \overline{F B}.

For a composition p-ring, both these conditions are satisfied. For a composition t-ring condition (B_{1}) holds and if F ≠ 0, then also (B_{2}) is satisfied. More generally, any composition ring C with β surjective and B = \overline{F B} satisfies (B_{1}) and (B_{2}) (trivially) but in such a case [A] = C for any A ⊆ B.

6.9. Proposition. Let J be a subgroup of B such that J ∘ B + B ∘ J ⊆ J + \overline{F B} (for example, if J ⊲ B or if J ⊆ \overline{F B}). Then β([J]) = J + \overline{F B} and:

(1) If β satisfies (B_{1}), then [J] ⊲ C_{1}.

(2) If β satisfies (B_{1}) and (B_{2}), then [J] ⊲ C.

Proof. β([J]) = J + \overline{F B} follows from the definition and since β is assumed to be surjective. Let c, d ∈ [J] and p, q ∈ C. Then β(c − d) = β(c) − β(d) ∈ J + \overline{F B}.

(1) Suppose β satisfies condition (B_{1}). Either β(cp) ∈ \overline{F B} ⊆ J +\overline{F B} or β(cp) = β(c)∘β(p) ∈ (J + \overline{F B}) ∘ B ⊆ J + \overline{F B} and similarly either β(pc) ∈ \overline{F B} ⊆ J + \overline{F B} or β(pc) = β(p) ∘ β(c) ∈ B ∘ (J + \overline{F B}) ⊆ J + \overline{F B}. Thus [J] ⊲ C_{1}.

(2) Suppose β also satisfies condition (B_{2}). Then β(c ∘ p) = β(c) ∘ β(p) + u ∈ (J + \overline{F B}) ∘ B + \overline{F B} ⊆ J + \overline{F B} for some u ∈ \overline{F B}. Lastly, β(p ∘ (q + c) − p ∘ q) = β(p) ∘ β(c) + w ∈ B ∘ (J + \overline{F B}) + \overline{F B} ⊆ J + \overline{F B} for some w ∈ \overline{F B}. Hence [J] ⊲ C. □

6.10 Proposition. Suppose β satisfies condition (B_{2}). Let I ⊲ C. Then J := β(I) is a subgroup of B which satisfies J ∘ B + B ∘ J ⊆ J + \overline{F B} and I ⊆ [J]. If J′ is any other subgroup of B with I ⊆ [J′], then J + \overline{F B} ⊆ J′ + \overline{F B}.

Proof. Let a, d ∈ I and b ∈ B = β(C), say b = β(c), c ∈ C. Then β(a) − β(d) = β(a − d) ∈ β(I). Also, β(a) ∘ β(c) = β(a ∘ c) + g ∈ β(I) + \overline{F B} = J + \overline{F B} for some g ∈ \overline{F B} and β(c) ∘ β(a) = β(c ∘ a) + w for some w ∈ \overline{F B}. If c = k + u with k ∈ F and u ∈ C_{0}, then β(c ∘ a) = β(u ∘ a) ∈ β(I) + \overline{F B} since C_{0} ∘ I ⊆ I and β(k) = β(k ∘ 0) = β(k) ∘ β(0) + h = h for some h ∈ \overline{F B}. Thus β(c) ∘ β(a) ∈ β(I) + \overline{F B}. Clearly I ⊆ [J]. Let J′ be a subgroup of B such that I ⊆ [J′]. Let j + d ∈ J + \overline{F B}, say j = β(i), i ∈ I. Since I ⊆ [J′], β(i) ∈ J′ + \overline{F B} and so j + d ∈ J′ + \overline{F B}. □

6.11. Corollary. Suppose β satisfies conditions (B_{1}) and (B_{2}). The mapping Φ : {I | I ⊲ C\} ⟶ \{J + \overline{F B} | J a subgroup of B with J ∘ B + B ∘ J ⊆ J + \overline{F B}\} defined by Φ(I) = β(I) + \overline{F B} is a well-defined surjection which preserves inclusions and sums.

Proof. Most of the proof follows from Propositions 6.9 and 6.10. The remainder is straightforward. □

6.12. Proposition. Suppose β satisfies conditions (B_{1}) and (B_{2}). Let I be a maximal ideal of C. Then one and only one of the following three cases holds:

(1) F ⊈ I and I = (I ∩ F : F)_{C_{2}} with I ∩ F a maximal C-ideal of F.

(2) F ⊆ I and I = [β(I)] with β(I) a maximal ideal of B.

(3) F ⊆ I and β(I) + \overline{F B} = B.

Proof. By Proposition 6.4 and the fact that I is maximal, we have I = (I ∩ F : F)_{C_{2}} or (I ∩ F : F)_{C_{2}} = C. Suppose firstly I = (I ∩ F : F)_{C_{2}}. Then F ⊈ I for if F ⊆ I then I = (I ∩ F : F)_{C_{2}} = (F : F)_{C_{2}} = C; a contradiction. We show I ∩ F is a maximal C-ideal of F. Let K be a C-ideal of F with I ∩ F ⊂ K ⊂ F. Then I = (I ∩ F : F)_{C_{2}} ⊆ (K : F)_{C_{2}} and so I = (K : F)_{C_{2}} or (K : F)_{C_{2}} = C. If the latter equality holds, then F = C ∘ F = K, a contradiction. Thus I = (K : F)_{C_{2}} and from this we get K ⊆ I ∩ F, also a contradiction. Hence I ∩ F is a maximal C-ideal of F. Suppose now (I ∩ F : F)_{C_{2}} = C. Then F ⊆ I. By Proposition 6.10 J := β(I) is a subgroup of B which satisfies J ∘B +B ∘J ⊆ J +\overline{F B} and I ⊆ [J]. From Proposition 6.9 we get [J]⊲C and so I = [J] or [J] = C.

If [J] = C, then B = β(C) = β([J]) = J + \overline{F B} = β(I) + \overline{F B}. Suppose thus [J] = I. Then J ⊲ B: Note firstly that J = β(I) = β([J]) = J + \overline{F B}; hence J ⊆ \overline{F B}. Thus J ∘ B + B ∘ J ⊆ J + \overline{F B} = J. Finally we show that J is maximal in B. Let K ⊲ B such that J ⊂ K ⊂ B. By Proposition 6.9 we have [K] ⊲ C and I = [J] ⊆ [K]. Thus [J] = [K] or [K] = C. If [J] = [K], then J = J + \overline{F B} = β([J]) = β([K]) = K + \overline{F B} = K; a contradiction. If [K] = C, then B = β(C) = β([K]) = K + \overline{F B} = K; also a contradiction. Hence β(I) = J is a maximal ideal of B. □

The next two results are immediate special cases of this result:

6.13. Corollary. Suppose β satisfies conditions (B_{1}) and (B_{2}) and that F ound(C) = 0. If I is a maximal ideal of C, then only case (2) or (3) of Proposition 6.12 can occur, i.e.: I = [β(I)] and β(I) is a maximal ideal of B or β(I) = B. □

6.14. Corollary. Suppose C = ⟨F ⟩ and β satisfies conditions (B_{1}) and (B_{2}). If I is a maximal ideal of C, then I = (I ∩ F : F)_{C_{2}} and I ∩ F is a maximal C-ideal of F. □

The third case in Proposition 6.12 can occur (cf. Example 6.16 below) and in the known cases with a stronger conclusion:

6.15. Proposition. Let C be a composition ring with commuting composition identity x, let I ⊲ C such that F ⊆ I and β(I) + F x = B. Suppose C and β satisfies: (B_{3}) For all m ≥ 0 (m ∈ ℤ) there exists a function β_{m} : C → B with β_{0}(c) := (c ∘ 0)x for all c ∈ C and β_{1} = β, such that for all m ≥ 0 and for all c ∈ C, there exists a c′ ∈ ⟨c⟩ such that β_{m}(c) = β_{m+1}(c′).

Then β_{m}(I) = B for all m ≥ 1.

Proof. We start by showing β_{1}(I) = β(I) = B, i.e. F x ⊆ β(I). Let k ∈ F. Then kx = (k ∘ 0)x = β_{0}(k) = β_{1}(k′) for some k′ ∈ ⟨k⟩. From F ⊆ I we get k′ ∈ I and thus kx ∈ β_{1}(I) = β(I). Hence β(I) = B. We proceed by induction and suppose B = β_{m}(I) for some m ≥ 1. Let b ∈ B = β_{m}(I), say b = β_{m}(a), a ∈ I. By (B_{3}) there is an a′ ∈ ⟨a⟩ ⊆ I such that b = β_{m}(a) = β_{m+1}(a′) ∈ β_{m+1}(I); hence B = β_{m+1}(I). □

6.16. Example. Let C be a composition p-ring. As we have seen before, for any k ≥ 0 and any c ∈ C, there is a unique representation.

c = n_{c} + v_{0} + v_{1}x + v_{2}x + ⋯ + v_{k}x^{k−1} + u_{k}x^{k}

where n_{c} ∈ F, v_{i} ∈ B and u_{k} ∈ C_{0}. Define β_{m} : C → B as follows: β_{0}(c) = n_{c}x = (c∘0)x; β_{1}(c) = β(c) = v_{0}, β_{2}(c) = v_{1}, …, β_{m+1}(c) = v_{m}, . . .. Let c ∈ C. Then cx ∈ ⟨c⟩ and since cx = n_{c}x + v_{0}x + v_{1}x^{2} + ⋯ + v_{k}x^{k+1} + u_{k}x^{k+1} (and remember B ∩ Bx ⊆ B ∩ C_{0}x = 0), we get β_{m}(c) = v_{m−1} = β_{m+1}(cx) for m ≥ 2, β_{0}(c) = n_{c}x = β_{1}(cx) and β_{1}(c) = v_{0} = β_{2}(cx).

Thus β_{m}(c) = β_{m+1}(cx) and cx ∈ ⟨c⟩ for all m ≥ 0. □

We know exactly when only cases (1) and (2) of Proposition 6.12 can occur:

6.17. Proposition.. Suppose β satisfies conditions (B_{1}) and (B_{2}). Let I be a maximal ideal of C. Then only cases (1) and (2) of 6.12 can occur if and only if there is a c ∈ C such that β(c + I) ∩ \overline{F B} = ϕ.

Proof. Suppose only cases (1) and (2) can occur. From the proof of Proposition 6.12 we then know that I = [β(I)] ⊂ C. If β(c + I) ∩ \overline{F B} ≠ ϕ for all c ∈ C, we will derive a contradiction. Let c ∈ C and i ∈ I such that β(c + i) = w for some w ∈ \overline{F B}. Then β(c) = −β(i) + w ∈ β(I) + \overline{F B}. Thus c ∈ [β(I)]. Hence [β(I)] = C; a contradiction. Conversely, suppose β(c + I) ∩ \overline{F B} = ϕ for some c ∈ C. If [β(I)] = C, then c ∈ [β(I)] implies β(c) = β(i) + u for some i ∈ I, u ∈ \overline{F B}. Thus β(c − i) ∈ \overline{F B} which contradicts our assumption. Hence β(I) ⊂ C. □

6.18. Corollary. Suppose F ound(C) = 0 and β satisfies conditions (B_{1}) and (B_{2}). Let I be a maximal ideal of C. Then I = [β(I)] and β(I) is a maximal ideal of B if and only if there is a c ∈ C such that β(c + i) ≠ 0 for all i ∈ I. □

References

[1] Adler, I.: Composition Rings. Duke J. Math. 29 (1962), 607–623.

[2] Kautschitsch, H.; Mlitz, R.: Maximal ideals in composition-rings of formal power series. Contr. to General Algebra 6. Verlag Holder - Pichler - Tempsky, Wien 1988.

[3] Petersen, Q. N.; Veldsman, S.: Composition near-rings. Nearrings, Nearfields and Kloops, Proc. Conf. Nearrings and Nearfields, Hamburg 1995, Editors: G. Saad and M. J. Thomsen), Kluwer Academic Publishers, The Netherlands, 1997, 357–372.

[4] Veldsman, S.: On the radical of composition near-rings. Proc. Conf. Near-Rings and Near-fields, Stellenbosch 1997, to appear.

Received September 13, 1999