Agricultural Journal

Abstract

Let, R be a ring and σ is an endomorphism for R. With multiplication rule xa = σ(a)x for all a∈R, the set of polynomials a₀+a₁x+, ..., +a_nxⁿ where a_i∈R, forms a ring. The ring is called skew polynomial ring. This ring is non-commutative due to the multiplication rule.

INTRODUCTION

Definitions and notations: In mathematics, the quaternions are a real scalar linear combination and three orthogonal imajinary units (denoted by i, j and k) with real coefficients that can be written as:

Where:

For more properties of quaternions, the reader is referred to Shomake and Quaternions. In this study, it is necessary to know the definition of skew polynomial ring and the center of ring. It can be found by Amir (2016, 2017, 2018) and (Amir et al. 2017).

Definition 1.1; McConnel and Robson (1987): Let R be a oring, σ be an endomorphism for R and δ be a σ-derivation. The skew polynomial ring over R is a ring that consists of all polynomials over R with an indeterminate x denoted by:

With multiplication rule, for all, a∈R, then:

For cases where δ = 0, the notations R[x; σ, δ] can be written as R[x; σ.] Moreover, the structure of skew polynomial ring R[x; σ] is different with R[x; σ, δ].

Definition 1.2; McConnel and Robson (1987): Let, R be a ring, the center of R denoted by Z(R) is defined as:

THE MAIN RESULTS

In this part, we give the forms of center of skew polynomial ring over a couple of quaternions. Let:

Where:

It is easy to show that σ is an endomorphism on H×H. So, we have H×H [x; σ] is a skew polynomial ring. Center of ring H×H [x; σ] is stated in the following theorem:

Theorem 2.1:

Proof: We show that:

Where:

The proof will be divided into two parts, i.e., P⊆Z (H×H [x; σ]) and Z(H×H [x; σ])⊆P. First, we show that P⊆Z (H×H [x; σ]):

We will show that p(x) q(x) = q(x) p(x), ∀q(x)∈H×H [x; σ]. Without loss of its generality, the proof will be done for two cases q(x). They are q(x) = (a, b)∈H×H and q(x) = (1, 1)x.

Case q(x) = (a, b). Let:

Then:

So:

(1)

On the other hand:

(2)

From Eq. 1 and 2, we can see that p(x) q(x) = q(x) p(x). Case q(x) = (1, 1) x:

(3)

On the other hand:

(4)

From Eq. 3 and 4, we can see that p(x) q (x) = q(x) p(x.) From the two cases q(x) above it proves that p(x) q(x) = q(x) p(x), then such that p⊆Z (H×H[x; σ]). Second, we show that Z(H×H [x; σ])⊆P. Let, P(x)∈Z (H×H [x; σ]), so, p(x) holds p(x) q(x) = q(x) p(x), ∀q(x) ∈H×H [x; σ] and p(x) By choosing q(x) = (j, i) then: