Let ⟨P,⊕,⊙⟩ be a Pythagorean ring. For distinct prime integers pi,1⩽i⩽m , the set of zero divisors of P including ⟨0,0,0⟩ is

Z(P)∪(0)=⟨μ,0,μ⟩ μ=0,1orℤ\2ℤ∋μ=p1…pm ∪{⟨0,μ,μ⟩}∪{⟨0,−μ,μ⟩}∪{⟨0,μ,−μ⟩} μ∈ℤ∪{±kpi,±k(1−p2i)2,±k(1+p2i)2foroddintegerkandoddprime integerp}

From Theorem 2.3 in,⁵ φ:P⟶ℤ×ℤ given by φ(⟨α,β,γ⟩)=(γ−β,d′(⟨α,β,γ⟩)) is bijective. Since Z(ℤ×ℤ)={(0,μ)|μ∈ℤ}∪{(μ,0)|μ∈ℤ} , it suffices to determine φ−1((0,μ)) and φ−1((μ,0)) . To determine φ−1((μ,0)) , we consider μ=γ−β and d′(⟨α,β,γ⟩)=0 from φ defined above. Using Definition 2.2 in,⁵ we consider the following two cases:

Now, if γ−β is even and γ−β≠0 , then r′(γ−β) is even and non-zero, implying that α=0 . Consequently ⟨0,β,γ⟩∈P contingent upon β=−γ or γ=−β . If γ−β is odd, then r′(γ−β) is odd and αr′(γ−β)=1 in order that α=r′(γ−β) . Suppose α=γ−β=p1α1…pmαm for some αi>1 and pi is odd prime. With a constraining assumption, let α1=2 then αr′(γ−β)=p1>1 , a contradiction. Therefore α1=α2=α3=…=αm=1 .

Now, let α≠γ−β and αr′(γ−β)=1 . Then by definition of r′,γ−β is a prime multiple of α . Let p be an odd prime integer and i∈ℕ . If α is an even multiple of pi then γ−β is even. Now, let α=mpi where m=±k is odd, then

(mpi)2+β2=γ2⇒m2p2i=(γ−β)(γ+β)

So γ−β=mp2i,γ+β=m .

Therefore

β=m−γ=m−β−mp2i⟹2β=m(1−p2i)⟹β=m(1−p2i)2

But

γ=m−β=m−m(1−p2i)2=m(1+p2i)2

Therefore

⟨α,β,γ⟩=mpi,m(1−p2i)2,m(1+p2i)2,

where m=±k is odd.

Similarly, to determine φ−1((0,μ)) , we consider γ−β=0 and d′(⟨α,β,γ⟩)=μ . If γ−β=0 then d′(⟨α,β,γ⟩)=β . So, β=μ=γ so that α2=0 , implying that α=0 . Therefore φ−1((0,μ))=⟨0,μ,μ⟩ .

From Theorem 3.1, it is of interest to identify ideal forming subsets of P because of their natural role in the construction of new (quotient) rings. In the sequel, the subsets of Z(P)∪(0) shall be expressed as follows:

S(P)={⟨μ,0,μ⟩ μ=0,1orℤ\2ℤ∋μ=p1…pm};R1(P)={⟨0,−μ, μ⟩ μ∈ℤ};R2(P)={⟨0, μ,−μ⟩ μ∈ ℤ};T(P)={⟨0, μ,  μ⟩ μ∈ℤ}

and

Q(P)={±kpi,±k(1−p2i)2,±k(1+p2i)2 kis anoddinteger,pis anoddprime integer}.

From the given expressions, it can easily be verified that R1(P),R2(P) and T(P) are ideals of P . It is well known that the union of any pair of ideals of a ring is not always an ideal. However, R1(P)∪R2(P) is an ideal.

Let Z(P) be the set of zero divisors of the Pythagorean ring ⟨P,⊕,⊙⟩ . Then Z(P) is not an ideal of ⟨P,⊕,⊙⟩ . For instance, if u∈T(P) and v∈S(P),u+v∉Z(P) .

The set I(P) of the units in the Pythagorean ring ⟨P,⊕,⊙⟩ is the Klein 4 -group.

Obviously, the unit group U(ℤ×ℤ)={(1,1),(−1,1),(−1,−1),(1,−1)} . Using Theorem 2.3 in,⁵ there exists an an isomorphism φ:P⟶ℤ×ℤ given by,

φ(⟨α,β,γ⟩)=(γ−β,d′(⟨α,β,γ⟩)).

So, by direct calculation, we obtain the units of P ,

I(P)=φ−1(U(ℤ×ℤ))={⟨3,4,5⟩,⟨−3,−4,−5⟩,⟨1,0,−1⟩,⟨−1,0,1⟩}

For all x∈I(P) , we have that: x⊙x=⟨3,4,5⟩ , where ⟨3,4,5⟩ is the identity element in I(P) . Furthermore,

I(P)={⟨1,0,−1⟩,⟨−3,−4,−5⟩,⟨−1,0,1⟩,⟨3,4,5⟩}={⟨1,0,−1⟩,⟨3,4,5⟩}×{⟨−3,−4,−5⟩,⟨3,4,5⟩}=C2×C2

Let P be the Pythagorean ring. There exists a ring isomorphism T(P)→(0)×ℤ .

Let μ∈ℤ and suppose that φ:T(P)→(0)×ℤ is defined by φ(⟨0,μ,μ⟩)=(0,μ) . Then

φ(⟨0,μ,μ⟩⊕⟨0,μ′,μ′⟩)=φ(φ−1(φ(⟨0,μ,μ⟩)+φ(⟨0,μ′,μ′⟩)))=φ(φ−1((0,μ)+(0,μ′)))=φ(φ−1(0,μ+μ′))=(0,μ+μ′)=(0,μ)+(0,μ′)=φ(⟨0,μ,μ⟩)+φ(⟨0,μ′,μ′⟩)

and

φ(⟨0,μ,μ⟩⊙⟨0,μ′,μ′⟩)=φ(φ−1(φ(⟨0,μ,μ⟩)⋅φ(⟨0,μ′,μ′⟩)))=φ(φ−1((0,μ)⋅(0,μ′)))=φ(φ−1(0,μ⋅μ′))=(0,μ⋅μ′)=(0,μ)⋅(0,μ′)=φ(⟨0,μ,μ⟩)⋅φ(⟨0,μ′,μ′⟩).

The kernel of φ is given by:

Kerφ={s∈T(P)|φ(s)=(0,0)}={⟨0,μ,μ⟩∈T(P)|φ(⟨0,μ,μ⟩)=(0,0)}={⟨0,μ,μ⟩∈T(P)|(0,μ)=(0,0)}={⟨0,0,0⟩}.

Thus φ is injective. It can be seen that φ is surjective because for any (0,μ)∈(0)×ℤ there exists ⟨0,μ,μ⟩∈T(P) so that φ(⟨0,μ,μ⟩)=(0,μ) . Therefore φ is an isomorphism.

Next, we describe the structure of the zero divisors of the Pythagorean ring P via the extended zero divisor graph of P denoted by Γ‾(P) and the extended zero divisor graph of P determined by equivalence classes and denoted by Γ‾E(P) .

⁷ Let P be a Pythagorean ring. The extended zero divisor graph of P is the simple graph with vertex set being the set of zero divisors of P and with ( a,b ) an edge if and only if a≠b and ambn=0 , (m,n=1,2,3…) , with am≠0 and bn≠0 .

Let P be the Pythagorean ring. The extended zero divisor graph of P symbolized by Γ‾(P) is a complete bipartite graph K∞,∞ , diam(Γ‾(P))=2 and gr(Γ‾(P))=4 .

Let A(P)=S(P)∪R1(P)∪R2(P)∪Q(P) . Then Z(P)=T(P)∪A(P) and the vertex set Z(P)∗=(T(P)∪A(P))∗ . Since T(P)∗ and A(P)∗ are disjoint and no pair of vertices in either T(P)∗ or A(P)∗ are adjacent while every vertex in T(P)∗ is adjacent to all the other vertices in A(P)∗ and vice versa, Γ‾(P) is bipartite. The completeness and the diameter of the graph Γ‾(P) are obvious from Figure 1 below.

Finally, let x1,x2,x3,…∈T(P)∗ and y1,y2,y3,…∈A(P)∗ . Then, for distinct vertices xi and yj,i,j=1,2,3,… , the shortest cycle in Γ‾(P) is as referenced in Figure 2 below:

Let u∈P . The annihilator of u,Ann(u)={v∈P|u⊙v=⟨0,0,0⟩} .

It is well known that two vertices u,v∈Z(P)∗ are equivalent if AnnP(u)=AnnP(v) and an extended zero divisor graph determined by the equivalence classes in P is denoted by Γ‾E(P) .

Γ‾E(P) is a complete bipartite graph K1,1 of diam (Γ‾E(P))=1 and gr(Γ‾E(P))=∞ .

The result can easily be seen, from the fact that AnnP(T(P))=A(P) and AnnP(A(P))=T(P) . So Γ‾E(P) is A(P)∗ ____ T(P)∗

In the next section, we construct a new class of Pythagorean rings, by taking the quotients P/I , where I is an ideal of P .

P/T(P) is a commutative ring with identity ⟨3,4,5⟩⊕T(P) .

First, we show that addition and multiplication on P/T(P) are well defined. Let a,a′,b and b∈P such that

a⊕T(P)=a′⊕T(P)

and

b⊕T(P)=b′⊕T(P)

so that r=a′⊖a∈T(P) and s=b′⊖b∈T(P) where ⊖a denotes the additive inverse of a with respect to ⊕ defined above.

Then

a′⊕b′=(r⊕a)⊕(s⊕b)=φ−1(φ(r)+φ(a))⊕φ−1(φ(s)+φ(b))=φ−1(φ(φ−1(φ(r)+φ(a)))+φ(φ−1(φ(s)+φ(b))))=φ−1(φ(r)+φ(a)+φ(s)+φ(b))∈(a⊕b)⊕T(P)

a′⊙b′=(r⊕a)⊙(s⊕b)=φ−1(φ(r)+φ(a))⊙φ−1(φ(s)+φ(b))=φ−1(φ(φ−1(φ(r)+φ(a)))⋅φ(φ−1(φ(s)+φ(b))))=φ−1((φ(r)+φ(a))⋅(φ(s)+φ(b)))=φ−1(φ(r)φ(s)+φ(r)φ(b)+φ(a)φ(s)+φ(a)φ(b))∈(a⊙b)⊕T(P)

Let a=⟨α,β,γ⟩∈P . From,⁵ the additive inverse

⊖(a⊕T(P))={⟨α,−β,−γ⟩⊕T(P)ifγ−βis even,γ−β≠0e,e2−f22f,e2+f22f⊕T(P)ifγ−βisodd⟨0,−β,−γ⟩⊕T(P)ifγ−β=0

where e=α−2r′(γ−β) and f=β−γ .

The commutativity of P implies that P/T(P) is commutative and for a,b,c∈P ,

(a⊕T(P))⊙((b⊕T(P))⊕(c⊕T(P)))=(a⊕T(P))⊙(b⊕T(P))⊕(a⊕T(P))⊙(c⊕T(P))

It is easy to verify the other axioms of a ring.

Finally, let ⟨α,β,γ⟩∈P . Then

(⟨α,β,γ⟩⊕T(P))⊙(⟨3,4,5⟩⊕T(P))=(⟨α,β,γ⟩⊙⟨3,4,5⟩)⊕T(P)=φ−1((φ⟨α,β,γ⟩⋅φ⟨3,4,5⟩))⊕T(P)=φ−1((γ−β,d′(⟨α,β,γ⟩))⋅(1,1))⊕T(P)=φ−1(γ−β,d′(⟨α,β,γ⟩))⊕T(P)=⟨α,β,γ⟩⊕T(P)

So P/T(P) forms a commutative ring which has the identity ⟨3,4,5⟩⊕T(P) .

P/T(P) forms an integral domain.

Suppose on the contrary that P/T(P) has a zero divisor a⊕T(P) where a∈P . Then there exists a nonzero b⊕T(P)∈P/T(P) so that

(a⊕T(P))⊙(b⊕T(P))=0⊕T(P),a≠0,b≠0

Now (a⊙b)⊕T(P)=0⊙T(P) . Let a=⟨α,β,γ⟩ and b=⟨α′,β′,γ′⟩ . Since a⊙b∈T(P) , let a⊙b=⟨0,μ,μ⟩ . So

φ−1(φ(α)⋅φ(β))=⟨0,μ,μ⟩.

Applying φ on both sides, we have

φ(a)⋅φ(b)=φ(⟨0,μ,μ⟩)=(0,μ)⟹φ(⟨α,β,γ⟩)⋅φ(⟨α′,β′,γ′⟩)=(0,μ)⟹(γ−β,d′⟨α,β,γ⟩)⋅(γ′−β′,d′⟨α′,β′,γ′⟩)=(0,μ)⟹((γ−β)(γ′−β′),d′⟨α,β,γ⟩d′⟨α′,β′,γ′⟩)=(0,μ)

This implies that either γ−β=0 or γ′−β′=0 . Now, if γ−β=0 , then γ=β,α=0 , so that ⟨α,β,γ⟩=⟨0,β,β⟩ . Consequently, if γ′−β′=0 , then γ′=β′,α′=0 , so that ⟨α′,β′,γ′⟩=⟨0,β′,β′⟩ . In either case, a⊕T(P) or b⊕T(P) is a zero element in P/T(P) , a contradiction. Therefore, P/T(P) has no zero divisor. □

From the above result, it is clear that T(P) forms a prime ideal of P . But T(P) forms a principal ideal since it is generated by the triple ⟨0,1,1⟩ . Similarly, S(P)∪R1(P)∪R2(P)∪Q(P) is generated by ⟨1,0,1⟩ while R1(P)∪R2(P) is generated by ⟨0,−1,1⟩ .

Let P be the Pythagorean ring. Suppose U is a P -module generated by {u1,…,uh} and {σ1,…,σh} is a set of automorphisms of P such that R=P(+)U is an additive abelian group. On R , one defines multiplication by

where σi is an automorphism of P .

It is easy to see that this multiplication turns R into a commutative ring with identity (⟨3,4,5⟩,0,…,0) . Indeed, R is an additive Abelian group, and it is elementary to show that multiplication is associative and distributive over addition. Identifying U with 0(+)U , we can think of U as a subset of R . Clearly P⊆R is a subring with identity ⟨3,4,5⟩ . Therefore, with the usual identifications, P and R have the same identity.

Now, from the given multiplication, the set of zero divisors Z(R)=0(+)U≅U and (Z(R))2=(0) (the product of any two zero divisors is zero).

From the construction in Section 5.1, R is an infinite ring with identity ( ⟨3,4,5⟩,0,…,0 ) and is commutative if and only if σi=idP . In this case, (r0)σi⊙ui=ui⊙r0 in equation (4.1). The infiniteness of R implies that it contains elements which are neither zero divisors nor units. Furthermore, the ring is generated by {⟨3,4,5⟩,u1,…,uh} and any element of R can be uniquely expressed as r0⊕u with r0∈P and u∈Z(R) , and therefore any element of R is of the form r0⊕∑i=1hri⊙ui for r0,ri∈P .

Let Aut(R) denote the group of automorphisms of R . We show that the ring P is invariant under any automorphism θ∈Aut(R) and then establish the necessary and sufficient conditions for a map ϕ to be an automorphism of R .

We refer to Lemma 3.1 in⁴ and Proposition 2 in¹³ and state the following result.

No data was used to support this research as this article does not require data.

The authors Raymond Calvin Ochieng, Maurice Owino Oduor and Vitalis Onyango- Otieno appreciate Strathmore University for financially supporting this research.

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