PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 86(100) (2009), 123–131 DOI: 10.2298/PIM0900123S

FACTORIZATION PROPERTIES OF SUBRINGS IN TRIGONOMETRIC POLYNOMIAL RINGS

Tariq Shah and Ehsan Ullah

Communicated by Žarko Mijajlović

Abstract. We explore the subrings in trigonometric polynomial rings and their factorization properties. Consider the ring S′ of complex trigonometric polynomials over the field ℚ(i) (see [11 ]). We construct the subrings S′_{1}, S′_{0} of S′ such that S′_{1} ⊆ S′_{0} ⊆ S′. Then S′_{1} is a Euclidean domain, whereas S′_{0} is a Noetherian HFD. We also characterize the irreducible elements of S′_{1}, S′_{0} and discuss among these structures the condition: Let A ⊆ B be a unitary (commutative) ring extension. For each x ∈ B there exist x′ ∈ U(B) and x′′∈ A such that x = x′x′′.

1. Introduction

Factorization properties of integral domains have been a common interest of algebraists, particularly for polynomial rings. In this study we investigate the factorization properties of the subrings of S′ (see [11]). The basic concepts, notions and terminology are standard, as in [7].

For the factorization of exponential polynomials, J. F. Ritt developed: “If 1 + a_{1}e^{α_{1} x} +⋯ +a_{n}e^{α_{n} x} is divisible by 1+b_{1}e^{β_{1} x} +⋯ +b_{r}e^{β_{r} x} with no b = 0, then every β is a linear combination of α_{1}, …, α_{n} with rational coefficients” [9, Theorem].

Getting inspired by this, G. Picavet and M. Picavet [7] investigated some factorization properties in trigonometric polynomial rings. Following [7], when we replace all α_{k} above by im, with m ∈ ℤ, we obtain trigonometric polynomials. Whereas

T′= { ∑_{k=0}^{n} (a_{k} Coskx + b_{k} Sinkx) : n ∈ ℕ, a_{k}, b_{k} ∈ ℂ} ,

T = { ∑_{k=0}^{n} (a_{k} Coskx + b_{k} Sinkx) : n ∈ ℕ, a_{k}, b_{k} ∈ ℝ}

2000 Mathematics Subject Classification: 13A05, 13B30, 12D05, 42A05.

Key words and phrases: trigonometric polynomial, HFD, subrings, condition 1, irreducible.

are trigonometric polynomial rings.

Following Cohn [4], an integral domain D is atomic if each nonzero nonunit of D is a product of irreducible elements (atoms) of D, and it is well known that UFDs, PIDs and Noetherian domains are atomic domains. An integral domain D satisfies the ascending chain condition on principal ideals (ACCP) if there does not exist any infinite strictly ascending chain of principal integral ideals of D. Every PID, UFD and Noetherian domain satisfy ACCP and a domain satisfying ACCP is atomic. Grams [6] and Zaks [13] provided examples of atomic domains, which do not satisfy ACCP. Following [12], an integral domain D is said to be a halffactorial domain (HFD) if D is atomic and whenever x_{1} …x_{m} = y_{1} …y_{n}, where x_{1}, x_{2} …x_{m}, y_{1}, y_{2} …y_{n} are irreducibles in D, then m = n. A UFD is obviously an HFD, but the converse fails, since any Krull domain D with CI(D) ≅ ℤ_{2} is an HFD [12], but not a UFD. Moreover a polynomial extension of an HFD is not an HFD, for example, ℤ[\sqrt{−3} ][X] is not an HFD, as ℤ[\sqrt{−3} ] is an HFD but not integrally closed [5]. Following [2], an integral domain D is a finite factorization domain (FFD) if each nonzero nonunit of D has only a finite number of nonassociate divisors and hence only a finite number of factorizations up to order and associates. In general,

UFD ⟹ HFD ⟹ ACCP ⟹ Atomic ,
UFD ⟹ FFD ⟹ ACCP ⟹ Atomic.

But none of the above implications is reversible.

In [7, Theorems 2.1 and 3.1], G. Picavet and M. Picavet demonstrated that T′ is a Euclidean domain and T is a Dedekind half-factorial domain. Moreover, in [11] we extended the study of factorization properties of trigonometric polynomials with coefficients from the field ℚ and its algebraic extension ℚ(i), instead of ℝ and ℂ, that is we study

S′ = { ∑_{k=0}^{n} (a_{k} Coskx + b_{k} Sinkx) : n ∈ ℕ, a_{k}, b_{k} ∈ ℚ(i)} ,

S = { ∑_{k=0}^{n} (a_{k} Coskx + b_{k} Sinkx) : n ∈ ℕ, a_{k}, b_{k} ∈ ℚ} .

where S′ is a Euclidean domain and S is a Dedekind finite factorization domain (see [11, Theorem 1 & Theorem 2]).

Again following [7], Sin^{2} x = (1−Cosx)(1+Cosx) shows that two different nonassociated irreducible factorizations of the same element may appear. Throughout we denote by Coskx and Sinkx the two functions x ↦ Coskx and x ↦ Sinkx (defined over ℝ). Also from basic trigonometric identities, it is obvious that for each n ∈ ℕ ∖ {1}, Cosnx represents a polynomial in Cosx with degree n and Sinnx represents the product of Sinx and a polynomial in Cosx with degree n − 1. Conversely by linearization formulas, it follows that any product Cos^{n}x Sin^{p}x can be written as:

∑_{k=0}^{q} (a_{k} Coskx + b_{k} Sinkx), where q ∈ ℕ and a_{k}, b_{k} ∈ ℚ.

Hence S = ℚ[Cosx, Sinx] ⊆ ℝ[Cosx, Sinx] = T and S′ = ℚ(i)[Cosx, Sinx] ⊆ ℂ[Cosx, Sinx] = T′.

We continue the investigations to find the factorization properties in trigonometric polynomial rings, begun in [7] and extended in [11]. In other words we extend this study towards finding factorization properties of subrings of trigonometric polynomial rings, by establishing S′_{0} and S′_{1} as subrings.

In Section 2 we explore S′_{1} and S′_{0}, and demonstrate that the ring S′_{1} is Euclidean domain (⋍ (ℚ[X])_{X}), whereas S′_{0} is a Notherian HFD (⋍ (ℚ+Xℚ(i)[X])_{X}). In Section 3 we discus Condition 1 (see [8, p. 661]) among the rings S′_{1}, S′_{0} and S′. We also extend the Condition 1, as Condition 2.

2. The Subrings of ℚ(i)[Cosx, Sinx]

A Construction of S′_{1}. We consider

S′_{1} = { ∑_{k=0}^{n} (a_{k} Coskx + ib_{k} Sinkx), n ∈ ℕ, a_{k}, b_{k} ∈ ℚ} .

Let z = ∑ _{k=0}^{n} (a_{k} Coskx + ib_{k} Sinkx) ∈ S′_{1}. As Cosx = \frac{1}{2}(e^{ix} + e^{−ix} ) and Sinx = \frac{1}{2i} (e^{ix} − e^{−ix} ), so

z = ∑_{k=0}^{n} \{( \frac{a_{k} + b_{k}}{2} ) e^{ikx} + ( \frac{a_{k} − b_{k}}{2} ) e^{−ikx} \}

= e^{−inx} [ ∑_{k=0}^{n} \{( \frac{a_{k} + b_{k}}{2} ) e^{i(n+k)x} + ( \frac{a_{k} − b_{k}}{2} ) e^{i(n−k)x} \}] ,

where (a_{k} + b_{k})/2, (a_{k} − b_{k})/2 ∈ ℚ. Therefore any element z is of the form e^{−inx} P (e^{ix}), n ∈ ℕ, where P (X) ∈ ℚ[X] and deg(P) ⩽ 2n.

Conversely, for α_{k} ∈ ℚ, 0 ⩽ k ⩽ 2n, we have

e^{−inx} P (e^{ix}) = e^{−inx} ( ∑_{k=0}^{2n} α_{k}e^{ikx} ) = ∑_{k=0}^{n−1} (α_{k}e^{−i(n−k)x} + α_{2n−k} e^{i(n−k)x} ) + α_{n}.

As e^{ix} = Cosx + i Sinx, so

e^{−inx} P (e^{ix}) = ∑_{k=0}^{n−1} { α_{k}(Cos(n − k)x − i Sin(n − k)x)
+ α_{2n−k} (Cos(n − k)x + i Sin(n − k)x)} + α_{n}
= ∑_{k=0}^{n−1} { (α_{k} + α_{2n−k} ) Cos(n − k)x
+ i(α_{2n−k} − α_{k})) Sin(n − k)x} + α_{n},

where α_{k} + α_{2n−k} , α_{2n−k} − α_{k} ∈ ℚ. Therefore S′_{1} contains all the elements that are of the form e^{−inx} P (e^{ix}), n ∈ ℕ, where P (X) ∈ ℚ[X] has degree at most 2n.

Conclusion 1. A consequence of the above construction is : S′_{1}={e^{−inx} P (e^{ix}), n ∈ ℕ, where P (X) ∈ ℚ[X] and deg(P) ⩽ 2n}. So we have an isomorphism f : (ℚ[X])_{X} → S′_{1} through the substitution morphism X → e^{ix}. Therefore S′_{1} ⋍ (ℚ[X])_{X}.

Theorem 2.1. S′_{1} is a Euclidean domain having nonzero elements of ℚ as units and irreducible elements, up to units, trigonometric polynomials of the form Cosx + i Sinx − a, where a ∈ ℚ ∖ {0}.

Proof. (ℚ[X])_{X} is a localization of ℚ[X] by a multiplicative system generated by a prime because X is a prime in ℚ[X] [1, Example 1.8 (b)]. Also ℚ[X] is a Euclidean domain. Therefore (ℚ[X])_{X} is a Eucledean domain [10, Proposition 7]. Now use the isomorphism S′_{1} ⋍ (ℚ[X])_{X} in Conclusion 1. □

A Construction of S′_{0}. Let z = ∑ _{k=0}^{n} (a_{k} Coskx + b_{k} Sinkx), n ∈ ℕ, a_{k}, b_{k} ∈

ℚ(i), such that a_{n} = α + γ + iβ and b_{n} = −β + i(α − γ), where α, β, γ ∈ ℚ; obviously z ∈ S′. We define S′_{0} to be the set of all the polynomials of the form ∑ _{k=0}^{n} (a_{k} Coskx + b_{k} Sinkx), n ∈ ℕ, a_{k}, b_{k} ∈ ℚ(i) and a_{n} = α + γ + iβ, b_{n} = −β + i(α − γ). Let z be a polynomial from S′_{0}. We may write

z = a_{0} + ∑_{k=1}^{n−1} (a_{k} Coskx+b_{k} Sinkx) + {(α+γ+iβ) Cosnx+(−β+i(α−γ))  Sinnx}

= a_{0} + ∑_{k=1}^{n−1} \{( \frac{a′_{k} + b′′_{k}+ i(a′′_{k}− b′_{k})}{2} ) e^{ikx} + ( \frac{a′_{k} − b′′_{k}+ i(a′′_{k}+ b′_{k})}{2} ) e^{−ikx} \}
+ (α + iβ)e^{inx} + γ e^{−inx} ,

where a_{k} = a′_{k} + ia′′_{k}, b_{k} = b′_{k} + ib′′_{k} and a′_{k}, a′′_{k}, b′_{k}, b′′_{k} ∈ ℚ, a_{0} ∈ ℚ(i). Setting α′_{k} = \frac{1}{2}(a′_{k} + b′′_{k}+ i(a′′_{k}− b′_{k})) and β′_{k} = \frac{1}{2}(a′_{k} − b′′_{k}+ i(a′′_{k}+ b′_{k})), we have

z = e^{−inx} [a_{0}e^{inx} + ∑_{k=1}^{n−1} {α′_{k}e^{i(n+k)x} + β′_{k}e^{i(n−k)x} } + (α + iβ)e^{i2nx} + γ],

where α′_{k}, β′_{k}, a_{0} ∈ ℚ(i) and α, β, γ ∈ ℚ. So z is of the form e^{−inx} P (e^{ix}), n ∈ ℕ, where P (X) ∈ ℚ + Xℚ(i)[X] and deg(P) ⩽ 2n.

Conversely, for α_{0} ∈ ℚ, and α_{k} ∈ ℚ(i), 1 ⩽ k ⩽ 2n, we have

e^{−inx} P (e^{ix}) = e^{−inx} (α_{0} + α_{1}e^{ix} + ⋯+ α_{2n} e^{i2nx} )

= α_{0}e^{−inx} + ∑_{k=1}^{2n−1} α_{k}e^{−i(n−k)x} + α_{2n} e^{inx}

= α_{0}e^{−inx} + α_{2n} e^{inx} + ∑_{k=1}^{n−1} (α_{k}e^{−i(n−k)x} + α_{2n−k} e^{i(n−k)x} ) + α_{n}

= α_{0}(Cosnx − i Sinnx) + α_{2n} (Cosnx + i Sinnx)

+ ∑_{k=1}^{n−1} {α_{k}( Cos(n − k)x − i Sin(n − k)x)

+ α_{2n−k} (Cos(n − k)x + i Sin(n − k)x)} + α_{n}.

Take α_{k} = α′_{k} + iα′′_{k}, α_{2n−k} = α′_{2n−k} + iα′′_{2n−k} and α_{2n} = α′_{2n} + iα′′_{2n}. Thus

e^{−inx} P (e^{ix}) = (α_{0} + α′_{2n} + iα′′_{2n}) Cosnx + (−α′′_{2n} + i(α′_{2n} − α_{0})) Sinnx

+ ∑_{k=1}^{n−1} {(α′_{k} + α′_{2n−k} + i(α′′_{k}+ α′′_{2n−k} )) Cos(n − k)x

+ (α′′_{k}− α′′_{2n−k} + i(α′_{2n−k} − α′_{k})) Sin(n − k)x} + α_{n}
= a_{n} Cosnx + b_{n} Sinnx + ∑_{k=1}^{n−1} {a_{k} Cos(n − k)x + b_{k} Sin(n − k)x} + α_{n},

where

a_{n} = α_{0} + α′_{2n} + iα′′_{2n} ,  a_{k} = α′_{k} + α′_{2n−k} + i(α′′_{k}+ α′′_{2n−k} ),
b_{n} = −α′′_{2n} + i(α′_{2n} − α_{0}),  b_{k} = α′′_{k}− α′′_{2n−k} + i(α′_{2n−k} − α′_{k}).

So, every element of the form e^{−inx} P (e^{ix}), n ∈ ℕ, where P (X) ∈ ℚ + Xℚ(i)[X] and deg(P) ⩽ 2n is in S′_{0}.

Conclusion 2. A consequence of above construction is: S′_{0} = {e^{−inx} P (e^{ix}), n ∈ ℕ, where P (X) ∈ ℚ+Xℚ(i)[X] and deg(P) ⩽ 2n}. So we have an isomorphism f : (ℚ+Xℚ(i)[X])_{X} → S′_{0} through the substitution morphism X → e^{ix}. Therefore S′_{0} ⋍ (ℚ + Xℚ(i)[X])_{X}.

Theorem 2.2. The integral domain S′_{0} is a Noetherian HFD having nonzero elements of ℚ(i) as units and trigonometric polynomials Cosx + i Sinx − a, where a ∈ ℚ(i) ∖ {0} are irreducible elements, up to units.

Proof. Since X is a prime in ℚ + Xℚ(i)[X] [1, Example 1.8(b)], we have that (ℚ + Xℚ(i)[X])_{X} is a localization of ℚ + Xℚ(i)[X] by a multiplicative system generated by a prime. Also ℚ + Xℚ(i)[X] is a Notherian HFD [3, Theorem 4], [2, Proposition 3.1]. Therefore (ℚ + Xℚ(i)[X])_{X} is an HFD [1, Corollary 2.5] and Notherian [14, Corollary 1, p. 224]. Hence the isomorphism S′_{0} ⋍ (ℚ+Xℚ(i)[X])_{X} in Conclusion 2 gives the result. □

The following is an analogue of [11, Corollary 1] and gives a factorization in S′_{0} instead of S′.

Corollary 2.1. Let z = ∑ _{k=0}^{n} (a_{k} Coskx + b_{k} Sinkx), n ∈ ℕ ∖ {1}, a_{k}, b_{k} ∈

ℚ(i) with (a_{n}, b_{n}) ≠ (0, 0), such that a_{n} = α+γ +iβ and b_{n} = −β +i(α−γ), where α, β, γ ∈ ℚ. Let d be a common divisor of the integers k such that (a_{k}, b_{k}) ≠ (0, 0). Then z has a unique factorization

λ(Cosnx − i Sinnx) ∏_{j=1}^{2n/d} (Cosdx + i Sindx − α_{j}), where λ, α_{j} ∈ ℚ(i) ∖ {0}.

Proof. Since S′_{0} ⊂ S′, the proof follows by [11, Corollary 1]. □

Remark 2.1. The factorization in S′_{1} is an analogue of Corollary 2.1.

Now onwards the symbol ∩ in all diagrams will represent the inclusion ⊆.

Remark 2.2. ℚ + Xℚ(i)[X] is a Noetherian HFD wedged between two Euclidean domains ℚ[X] and ℚ(i)[X], that is ℚ[X] ⊆ Q + Xℚ(i)[X] ⊆ ℚ(i)[X] and the localization of all these by a multiplicative system generated by X preserves their factorization properties as

ℚ[X] ⊆  ℚ + Xℚ(i)[X]  ⊆  ℚ(i)[X]
∩  ∩  ∩
(ℚ[X])_{X} ⊆ (ℚ + Xℚ(i)[X])_{X} ⊆ (ℚ(i)[X])_{X} .

Using Conclusion 1, Conclusion 2 and [11, Theorem 1], we have

ℚ[X] ⊆ ℚ + Xℚ(i)[X] ⊆ ℚ(i)[X]
∩  ∩  ∩
S′_{1} ⊆  S′_{0}  ⊆  S′,

where S′_{0} is a Noetherian HFD wedged between two Euclidean domains S′_{1} and S′.

Remark 2.3. (a) Consider the domain extension ℚ[X] ⊆ (Q[X])_{X}. As Xℚ[X] is a maximal ideal of ℚ[X] and Xℚ[X] ∩ (X) ≠ ϕ. Therefore the extended ideal (Xℚ[X])^{e} = (ℚ[X])_{X} [14, Corollary 2]. Hence (Xℚ[X])^{e} ⋍ S′_{1} by Conclusion 1.

(b) If we consider the domain extension ℚ+Xℚ(i)[X] ⊆ (ℚ+Xℚ(i)[X])_{X}. We observe that Xℚ(i)[X] is a maximal ideal of ℚ+Xℚ(i)[X] and Xℚ(i)[X]∩(X) ≠ ϕ. Therefore the extended ideal (Xℚ(i)[X])^{e} = (ℚ + Xℚ(i)[X])_{X} [14, Corollary 2]. Hence (Xℚ(i)[X])^{e} ⋍ S′_{0} by Conclusion 2.

(c) On the same lines we can apply the same result to the domain extension ℚ(i)[X] ⊆ (Q(i)[X])_{X}. In this case Xℚ(i)[X] is a maximal ideal of ℚ(i)[X] and Xℚ(i)[X] ∩ (X) ≠ ϕ. Therefore the extended ideal (Xℚ(i)[X])^{e} = (ℚ(i)[X])_{X} [14, Corollary 2]. Hence (Xℚ(i)[X])^{e} ⋍ S′ by [11, Theorem 1].

Definition 2.1. Let J′ be the subset of S′_{1} defined by

J′ = { ∑_{k=0}^{n} (a_{k} Coskx + ib_{k} Sinkx), n ∈ ℕ, a_{k}, b_{k} ∈ ℚ and a_{n} = b_{n}} .

Definition 2.2. Let I′ be the subset of S′_{0} defined by

I′={∑_{k=0}^{n} (a_{k} Coskx+b_{k} Sinkx) : n∈ℕ, a_{k}, b_{k} ∈ ℚ(i) and a_{n} = α+iβ, b_{n} = −β+iα} .

Lemma 2.1. For the maximal ideal Xℚ[X] (respectively Xℚ(i)[X]) of ℚ[X] (respectively ℚ + Xℚ(i)[X]) we have (Xℚ[X])_{X} ⋍ J′ (respectively (Xℚ(i)[X])_{X} ⋍ I′).

Proof. Follows by Conclusion 1 (respectively Conclusion 2). □

3. Conditions satisfied by ring extensions

In this section we discuss two special conditions. First one, known as Condition 1, is borrowed from [8] and the second one is derived from Condition 1. Moreover, we study a few interesting results about these conditions and trigonometric polynomial ring extensions satisfying them.

Condition 1. Let A ⊆ B be a unitary (commutative) ring extension. For every x ∈ B there exist x′ ∈ U(B) and x′′∈ A such that x = x′x′′[8, page 661].

Example 3.1. Following [8, Example 1.1 ]; (a) If the ring extension A ⊆ B satisfies Condition 1, then the ring extension A+XB[X] ⊆ B[X] (or A+XB[[X]] ⊆ B[[X]]) also satisfies Condition 1.

(b) If the ring extensions A ⊆ B and B ⊆ C satisfy Condition 1, then so does the ring extension A ⊆ C.

(c) If B is a fraction ring of A, then the ring extension A ⊆ B satisfies Condition 1. Hence the ring extension A ⊆ B satisfies Condition 1 is the generalization of localization.

(d) If B is a field, then the ring extension A ⊆ B satisfies Condition 1.

Condition 2. Let A, A_{1}, B and B_{1} be unitary (commutative) rings such that

\begin{aligned}A & ⊆ & B \\ ∩ & & ∩ \\ A_{1} & ⊆ & B_{1} \end{aligned} .

Then for each x ∈ B_{1} there exist x′ ∈ U(B) and x′′∈ A_{1} such that x = x′x′′.

Lemma 3.1. Let A ⊆ B be a unitary (commutative) ring extension which satisfies Condition 1. If N is a multiplicative system in A, then the ring extension N^{−1} A ⊆ N^{−1} B satisfies Condition 2.

Proof. Since the ring extension A ⊆ B satisfies Condition 1. Therefore for each a ∈ B there exist b ∈ U(B) and c ∈ A such that a = bc. Obviously N^{−1} A ⊆ N^{−1} B. Let x = \frac{a}{s} ∈ N^{−1} B, where a ∈ B, s ∈ N. This implies x = \frac{bc}{s} = b\frac{c}{s}, where b ∈ U(B) and \frac{c}{s} ∈ N^{−1} A. □

Example 3.2. (a) If the ring extensions A ⊆ B and B ⊆ C satisfy Condition 2, then so does the ring extension A ⊆ C.

(b) By Lemma 3.1 the ring extensions S′_{1} ⊆ S′_{0} and S′_{0} ⊆ S′ satisfy Condition 2 so does the ring extension S′_{1} ⊆ S′.

(c) If the ring extension A ⊆ B satisfies Condition 1, then obviously it satisfies Condition 2.

Proposition 3.1. Let A ⊆ B and A_{1} ⊆ B_{1} be unitary (commutative) ring extensions, where A ⊆ A_{1} and B ⊆ B_{1}. Let M be a commen ideal of A, B, A_{1} and B_{1} for which the extension A_{1}/M ⊆ B_{1}/M satisfies Condition 2. Assume for each α ∈ U(B_{1}/M) there exists a ∈ U(B) such that p(a) = α, where p : B_{1} → B_{1}/M is the canonical surjection; then A_{1} ⊆ B_{1} satisfies Condition 2.

Proof. Let b ∈ B_{1}. We represent the class of b by b̂ in B_{1}/M. Using Condition 2, we have b̂= b̂′b̂′′, with b̂′∈ U(B/M), b̂′′∈ A_{1}/M. By hypothesis b′ ∈ U(B), since b̂′′∈ A_{1}/M, for b′′∈ A_{1}, we have b = b′b′′+ m = b′(b′′+ b^{′−1} m) with m ∈ M. Thus b′′+ b^{′−1} m ∈ A_{1}. □

Lemma 3.2. Let A ⊆ B and A_{1} ⊆ B_{1} be unitary (commutative) ring extensions, where A ⊆ A_{1} and B ⊆ B_{1}. Let M be an ideal of A_{1} that is also an ideal in B_{1}. If for each b ∈ B_{1} ∖ M there exists m ∈ M such that b + m ∈ U(B), then the extension A ⊆ B satisfies Condition 2.

Proof. If b ∈ M, then b = 1.b. Let b ∈ B_{1} ∖ M, then there exists m ∈ M with b + m ∈ U(B). So we can write, b = (b + m)(b + m)^{−1} b and (b + m)^{−1} b ∈ A_{1}, because (b + m)^{−1} b = 1 + m′ with m′ ∈ M. □

Proposition 3.2. Let A ⊆ B_{1} ⊆ B_{2} be a unitary (commutative) ring extension such that A ⊆ B_{2} satisfies Condition 2. If for each x ∈ U(B_{1}), we have x ∈ A or x^{−1} ∈ A then B_{1} = N^{−1} A, where N = U(B_{1}) ∩ A.

Proof. The inclusion N^{−1} A ⊆ B_{1} is obvious. Let x ∈ B_{2}. We can write x = x′x′′, where x′ ∈ U(B_{1}), x′′∈ A. If x′ ∈ A then x′ ∈ A ∩ U(B_{1}) = N and x ∈ A. If x^{′−1} ∈ A then x^{′−1} ∈ A ∩U(B_{1}) = N and x = \frac{x′′}{x^{′−1}} ∈ N^{−1} A. □

Remark 3.1. Consider the following commutative inclusion diagram which follows from Remark 2.2.

\begin{aligned}ℚ[X] & ⊆ & ℚ + Xℚ(i)[X] & ⊆ & ℚ(i)[X] \\ ∩ & ↘ & ∩ & ↘ & ∩ \\ S′_{1} & ⊆ & S′_{0} & ⊆ & S′. \end{aligned}

Now the following table concludes our discussion on Condition 1 and Condition 2 among trigonometric polynomial ring extensioins.

Ring Extension  Condition 1 Condition 2
ℚ[X] ⊆ ℚ + Xℚ(i)[X]  No  No
ℚ + Xℚ(i)[X] ⊆ ℚ(i)[X]  Yes  Yes
ℚ[X] ⊆ S′_{1}  Yes  Yes
ℚ + Xℚ(i)[X] ⊆ S′_{0}  Yes  Yes
ℚ(i)[X] ⊆ S′  Yes  Yes
S′_{1} ⊆ S′_{0}  No  Yes
S′_{0} ⊆ S′  No  Yes

By transitivity the domain extensions ℚ[X] ⊆ S′_{0}, ℚ + Xℚ(i)[X] ⊆ S′ and S′_{1} ⊆ S′ also satisfy Condition 2.

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Department of Mathematics (Received 29 02 2008) Quaid-i-Azam University (Revised 30 11 2008) Islamabad, Pakistan

stariqshah@gmail.com

Fakultät für Informatik und Mathematik Passau Universität Passau, Germany

ehsanu01@stud.uni-passau.de