Chapter three: Chapter three: Far rings and some results Far rings and some results In This chapter , we introduce In This chapter , we introduce some results about far rings some results about far rings and period profiles, and we and period profiles, and we prove some properties of far prove some properties of far rings which are known for the rings which are known for the usual rings and some other usual rings and some other results. results.
Transcript
Chapter three:Chapter three:
Far rings and some results Far rings and some results In This chapter , we introduce some In This chapter , we introduce some
results about far rings and period results about far rings and period profiles, and we prove some profiles, and we prove some properties of far rings which are properties of far rings which are known for the usual rings and some known for the usual rings and some other results.other results.
Section one:Section one:
Far rings and period profilesFar rings and period profiles In this section we introduce the In this section we introduce the
meaning of the period profile and we meaning of the period profile and we explain the relation between them . explain the relation between them . This relation is important in the This relation is important in the subject of faster way to get fractal subject of faster way to get fractal images.images.
First we give the definition of period First we give the definition of period profile.profile.
Definition 3.1:Definition 3.1:
Let (ZLet (ZNN,+,,+, ๏๏) be a far ring; and) be a far ring; and αα11,,…, …, ααkk be in Z be in ZNN. The period profile of . The period profile of αα11,…, ,…, ααkk is a list in descending order of the is a list in descending order of the period lengths of the k-sequences period lengths of the k-sequences which are obtained by recurrence which are obtained by recurrence relation over the far ring:relation over the far ring:
SSnn= = αα11๏๏SSn-1n-1, , αα22๏๏ SSn-2n-2+…+ +…+ ααkk๏๏SSn-kn-k n n ≥ ≥
k k
Whatever the initial values SWhatever the initial values S00SS11…S…Sk-1k-1 which are elements of Zwhich are elements of ZNN..
To illustrate the concept of period To illustrate the concept of period profile, we take the simple case k=2 profile, we take the simple case k=2 and we consider the far ring of order and we consider the far ring of order 4.4.
From examples 2.17 in chapter2 , From examples 2.17 in chapter2 , there are four possible far rings of there are four possible far rings of order 4. They are FRA, FRB,FRC, and order 4. They are FRA, FRB,FRC, and FRD.FRD.
Let Let αα11, , αα22 be in Z be in Z44. Recurrence . Recurrence relations of length k=2 over the far relations of length k=2 over the far ring (Zring (Z44,+,,+,๏๏) are given by S) are given by Snn= = αα
11๏๏SSn-1n-1+ +
αα22๏๏SSn-2n-2 n ≥ 2; n ≥ 2;
Where any two elements SWhere any two elements S00SS11from Zfrom ZNN can be taken as initial values to obtain can be taken as initial values to obtain each of these recurrence relation .each of these recurrence relation .
The period of the sequence SThe period of the sequence S00SS11 … … cannot exceed 4cannot exceed 422=16 , which is the =16 , which is the maximum number of distinct pairs maximum number of distinct pairs SSiiSSi+1i+1 which can be encountered before which can be encountered before repetition. repetition.
Further, any such pair may be taken Further, any such pair may be taken as initial so is found in some sequence as initial so is found in some sequence Thus the period lengths sum to 16. Thus the period lengths sum to 16.
The period profile of (The period profile of (αα11, α, α22)) is a list of is a list of period lengths in decending order, period lengths in decending order, and the set of lists for all and the set of lists for all
((αα11, α, α22)) in any order , is the profile of in any order , is the profile of the multiplication table of the far ring the multiplication table of the far ring itself .itself .
For example when the coefficients For example when the coefficients αα11, α, α22 =0,1 are taken in FRA, then we get the =0,1 are taken in FRA, then we get the sequence 120322…, 102300 ,133…,and sequence 120322…, 102300 ,133…,and 1…, which are periodic of period 1…, which are periodic of period 6,6,3,and 1 respectively.6,6,3,and 1 respectively.
Therefore the period profile for 0,1 is Therefore the period profile for 0,1 is 6631. 6631.
The following tables give the The following tables give the period profile for all period profile for all coefficients coefficients ααii, α, αjj in Z in Z44and for and for the four far rings (of order the four far rings (of order four) FRA,FRB,FRC, and FRD.four) FRA,FRB,FRC, and FRD.
coefficientscoefficients Sequences over far ring FRA (period Sequences over far ring FRA (period length)length)
Let each of (GLet each of (G11,,๏๏) and (G) and (G22,,**) be a ) be a
quasigroup. An isomorphism Ψquasigroup. An isomorphism Ψ from from GG11 into G into G22 is a bijection or a is a bijection or a permutation such that for all x,y in Gpermutation such that for all x,y in G11
Ψ(x Ψ(x ๏๏ y)=Ψ(x) y)=Ψ(x) **Ψ(y).Ψ(y).
Then (GThen (G11,,๏๏)and(G)and(G22, , **) are called ) are called
isomorphic.isomorphic.
Our aim is to obtain far rings (of the Our aim is to obtain far rings (of the same order) which have the same same order) which have the same period profiles.period profiles.
First the following example shows First the following example shows that the isomorphism of quasigroups that the isomorphism of quasigroups does not suffice for equal profiles.does not suffice for equal profiles.
Example 3.3Example 3.3
Consider (ZConsider (Z44,,๏๏)and (Z)and (Z44, , **) such that ) such that ๏๏
and and ** are two binary operations are two binary operations
defined by the following tablesdefined by the following tables
๏ 1 2 3 0
1230
1 2 3 02 0 1 33 1 0 20 3 2 1
** 1 2 3 0
1230
1 2 3 02 3 0 13 0 1 20 1 2 3
Define Ψ from the quasigroup (ZDefine Ψ from the quasigroup (Z44,,๏๏) ) into the quasigroup (Zinto the quasigroup (Z44, , **)by )by
Ψ(0)=3, Ψ(1)=1,Ψ(2)=2, and Ψ(3)=0.Ψ(0)=3, Ψ(1)=1,Ψ(2)=2, and Ψ(3)=0. It is easy to show that Ψ is an It is easy to show that Ψ is an
isomorphism. isomorphism. But (ZBut (Z44,,๏๏)=FRA and (Z)=FRA and (Z44, , **)=FRC have )=FRC have
different period profiles ; different period profiles ;
see table 3.1 and table 3.3.see table 3.1 and table 3.3. We shall see that the following We shall see that the following
definition of isomorphism of far rings definition of isomorphism of far rings is sufficient to make isomorphic far is sufficient to make isomorphic far rings of the same period profiles.rings of the same period profiles.
Definition 3.4:Definition 3.4:
An isomorphism of far rings (ZAn isomorphism of far rings (ZNN,,+,+,๏๏)→(Z)→(ZNN,+, ,+, **) is a pair of ) is a pair of permutations Ψ , ø of Zpermutations Ψ , ø of ZNN such that such that Ψ(1)=1 and for each α,β,a,b in ZΨ(1)=1 and for each α,β,a,b in ZNN, ,
Then (ZThen (ZNN,+,,+,๏๏) and (Z) and (ZNN,+, ,+, **) are called ) are called isomorphic far rings .isomorphic far rings .
The following theorem proves that The following theorem proves that any two isomorphic far rings have any two isomorphic far rings have the same period profiles.the same period profiles.
Theorem 3.5Theorem 3.5
Isomorphic far rings have the same Isomorphic far rings have the same period profiles .In particular for k=2 period profiles .In particular for k=2 If(Ψ,ø) is an isomorphism from the far If(Ψ,ø) is an isomorphism from the far ring( Zring( ZNN,+, ,+, ๏๏ ) into the far ring (Z) into the far ring (ZNN,+, ,+,
**) then the coefficient pairs α,β and ) then the coefficient pairs α,β and
Ψ(α), Ψ(β)in ZΨ(α), Ψ(β)in ZNN have the same period have the same period profile, and similarly for k-tuples with profile, and similarly for k-tuples with k>2.k>2.
ProofProof
First let k=2 and suppose α,β give a First let k=2 and suppose α,β give a sequence {asequence {ann} of period t.That is , } of period t.That is , with addition as usual mod N, we with addition as usual mod N, we havehave
aann=α=α๏๏aan-1n-1+ β+ β๏๏aan-2n-2. .
Using ø(αUsing ø(α๏๏a+βa+β๏๏b) =Ψ(b) =Ψ(αα ) )**ø(a)+Ψ (β)ø(a)+Ψ (β)
+ø(b)for any α,β, a, b in Z+ø(b)for any α,β, a, b in ZNN ….…(1) ….…(1)
Then the sequence {aThen the sequence {ann}={a}={a00,a,a11,,…}satisfies that a…}satisfies that att=a=a00,a,at+1t+1=a=a11, and so , and so on.on.
Thus , the coefficients Ψ(α),Ψ(β)Thus , the coefficients Ψ(α),Ψ(β) give the sequence give the sequence
Now, we need to prove that {ø(aNow, we need to prove that {ø(ann)}is )}is periodic of period t; we must prove periodic of period t; we must prove that that Ø(aØ(att)=ø(a)=ø(a00),ø(a),ø(at-1t-1)=ø(a)=ø(a11) and there is ) and there is no positive integer r such that r<t no positive integer r such that r<t and and Ø(aØ(arr)=ø(a)=ø(a00),ø(a),ø(ar+1r+1)=ø(a)=ø(a11).).
It is clear that ø(aIt is clear that ø(att)=ø(a)=ø(a00)and )and ø(aø(at+1t+1)=ø(a)=ø(a11)…,since {a)…,since {ann} is periodic } is periodic of period t.If there is r<t such that of period t.If there is r<t such that øø(a(arr)=ø(a)=ø(a00), ø(a), ø(ar+1r+1)=ø(a)=ø(a11)and since ø )and since ø is bijection, then we get that there is is bijection, then we get that there is r<t such that ar<t such that arr=a=a00,a,ar+1r+1=a=a11,,
but this is a contradiction to the fact but this is a contradiction to the fact that the period of {athat the period of {ann} is t and r≠t;} is t and r≠t; and t is the least positive integer and t is the least positive integer making {amaking {ann} periodic .Therefore we } periodic .Therefore we get that the sequence{get that the sequence{ø(an)} is ø(an)} is periodic of period t; periodic of period t;
Which generated by Which generated by ΨΨ((αα),),ΨΨ((ββ) with * ) with * as multiplication. as multiplication.
Finally , we may infer that these two far Finally , we may infer that these two far rings have the same period peofile ;since rings have the same period peofile ;since the bijection the bijection ΨΨ defines defines a bejiction of pairs a bejiction of pairs (α,β )→(Ψ(α),Ψ(β)).(α,β )→(Ψ(α),Ψ(β)).
This complete the proof for the case k=2 .This complete the proof for the case k=2 .
The proof for the case k>2 is by The proof for the case k>2 is by induction .induction .
It suffices to illustrate the proof for It suffices to illustrate the proof for the case k=3. Let α,β,γ,a,b,c be in the case k=3. Let α,β,γ,a,b,c be in ZZNN and x= α and x= α๏๏a+βa+β๏๏b+γb+γ๏๏c, c,
then then
Ø(x) Ø(x) ==ø(αø(α๏๏a+βa+β๏๏b+γb+γ๏๏c)c) = ø(α= ø(α๏๏a+1a+1๏๏(β(β๏๏b+γb+γ๏๏c))c)) Since 1 acts as identify for Since 1 acts as identify for ๏๏,, =Ψ(α) =Ψ(α) **ø(a)+Ψ(1) ø(a)+Ψ(1) **ø(βø(β๏๏b+γb+γ๏๏c) ; since c) ; since
Ψ(1)=1 is an identity for Ψ(1)=1 is an identity for **.. =Ψ(α) =Ψ(α) **ø(α)+Ψ(β) ø(α)+Ψ(β) **ø(b)+Ψ(γ) ø(b)+Ψ(γ) **ø(c); ø(c);
since +is associative and by using (1).since +is associative and by using (1).
By using the same notions in case By using the same notions in case k=2; it is easy to show that the k=2; it is easy to show that the coefficients Ψ(α),Ψ(β),Ψ(γ)generate coefficients Ψ(α),Ψ(β),Ψ(γ)generate the sequence {ø(athe sequence {ø(ann)}of period t,if the )}of period t,if the coefficients α,β,and γ generate the coefficients α,β,and γ generate the sequence {asequence {ann}of period t.}of period t.
We get the same result for k>3 by We get the same result for k>3 by the same method.the same method. This means that if This means that if the coefficients αthe coefficients α11,α,α22,…,α,…,αkk generate a generate a sequence {asequence {ann} of period t.} of period t.
under the multiplicationunder the multiplication๏๏,, then the then the coefficients Ψ(αcoefficients Ψ(α11),Ψ(α),Ψ(α22),…,Ψ(α),…,Ψ(αkk) ) generate the sequence {ø(agenerate the sequence {ø(ann)} of )} of period t under multiplication period t under multiplication *.*.Consequently ,isomorphic far rings Consequently ,isomorphic far rings have the same period profiles for all have the same period profiles for all k≥2.k≥2.
Note:Note:
The important of theorem 3.5 is the The important of theorem 3.5 is the result that if any one of these result that if any one of these isomorphic far rings gives an M-isomorphic far rings gives an M-sequence,sequence, then the others do the same.then the others do the same.
Now we search for what is required to Now we search for what is required to prove the converse of theorem 3.5 .We prove the converse of theorem 3.5 .We give the following preparatory theorem give the following preparatory theorem for k=2.for k=2.
Theorem3.6Theorem3.6
let each of (Zlet each of (ZNN,+,,+,๏๏) and (Z) and (ZNN,+, ,+, **) )
be a far ring which are denoted by be a far ring which are denoted by FRX,FRY respectively . Suppose that FRX,FRY respectively . Suppose that Ψ,ø are permutations of ZΨ,ø are permutations of ZNN,Ψ(1)=1 and ,Ψ(1)=1 and ø sends sequences over FRX generated ø sends sequences over FRX generated by coefficients α,β to sequences over by coefficients α,β to sequences over FRY generated by coefficients Ψ(α), FRY generated by coefficients Ψ(α), Ψ(β) for any α,βΨ(β) for any α,β in Zin ZNN . .
Then (Ψ,ø) is an isomorphism from FRX Then (Ψ,ø) is an isomorphism from FRX into FRY. into FRY.
Proof:Proof:
In order to prove this theorem, we In order to prove this theorem, we must prove the following :must prove the following :
Now consider α,β,a,b in ZNow consider α,β,a,b in ZNN and take k and take k to be 2 to obtain a k-sequences over to be 2 to obtain a k-sequences over FRX particularly when k=2. Then a,b FRX particularly when k=2. Then a,b must appear as successive members must appear as successive members of some sequence with coefficents of some sequence with coefficents α,β (we can take a,b as initial α,β (we can take a,b as initial members if necessary).members if necessary).
Let c=αLet c=α๏๏a+βa+β๏๏b; c will be in Zb; c will be in ZNN. Then ø . Then ø sends this sequence with coefficients sends this sequence with coefficients α,β to one over FRY with coefficients α,β to one over FRY with coefficients Ψ(α),Ψ(β) which results inΨ(α),Ψ(β) which results in
ø ( c) ø ( c) == ø(α ø(α๏๏a+βa+β๏๏b)b) == Ψ(α) Ψ(α) **ø(a)+Ψ(β) ø(a)+Ψ(β) **ø(b); As ø(b); As
required.required.
This is true for any four elements of This is true for any four elements of ZZNN . And since Ψ(1)=1, we get (Ψ,ø) is . And since Ψ(1)=1, we get (Ψ,ø) is An isomorphism and the proof is An isomorphism and the proof is complete . complete .
Remark 3.7: let Ψ and ø be Remark 3.7: let Ψ and ø be permutations of Zpermutations of ZNNand each of and each of FRX,FRY is a far ring of order N.FRX,FRY is a far ring of order N.
The statement that (Ψ,ø) preserves The statement that (Ψ,ø) preserves k-profiles from FRX to FRY means k-profiles from FRX to FRY means that ø sends a sequence defined over that ø sends a sequence defined over FRX by a recurrence with coefficients FRX by a recurrence with coefficients ααii(1≤i≤k) to the sequence over FRY (1≤i≤k) to the sequence over FRY defined by a recurrence with defined by a recurrence with coefficients ø(αcoefficients ø(αii) (1≤i≤k)) (1≤i≤k)
The following theorem gives some The following theorem gives some conditions to get isomorphic far ringsconditions to get isomorphic far rings
Theorem 3.8:Theorem 3.8:
Let Ψ,ø be permutations of ZLet Ψ,ø be permutations of ZNN with with Ψ(1)=1 and ø(0)=0, and let FRX, FRY Ψ(1)=1 and ø(0)=0, and let FRX, FRY be far rings of order N. Suppose that be far rings of order N. Suppose that for some k>2,(Ψ,ø) preserves k-for some k>2,(Ψ,ø) preserves k-profiles from FRX to FRY then (Ψ,ø) is profiles from FRX to FRY then (Ψ,ø) is a far ring isomorphism from FRX to a far ring isomorphism from FRX to FRY.FRY.
Proof:Proof:
Let (Ψ,ø) preserves k-profiles from Let (Ψ,ø) preserves k-profiles from FRX= (ZFRX= (ZNN,+,,+,๏๏) to FRY= (Z) to FRY= (ZNN,+, ,+, **) for ) for
k=2 . Then , from Remark 3.7 , that k=2 . Then , from Remark 3.7 , that means ø Sends sequences over FRX means ø Sends sequences over FRX generated by coefficients α,β to generated by coefficients α,β to sequences over FRY generated by sequences over FRY generated by coefficients Ψ(α),Ψ(β); for any α,β in coefficients Ψ(α),Ψ(β); for any α,β in ZZNN..
And since Ψ(1)=1,then we get that And since Ψ(1)=1,then we get that (Ψ,ø) is an isomorphism. It is (Ψ,ø) is an isomorphism. It is sufficient to show that if for some sufficient to show that if for some k>2 ,(Ψ,ø) preserves k-profiles from k>2 ,(Ψ,ø) preserves k-profiles from FRX to FRY then (Ψ,ø) preserves 2- FRX to FRY then (Ψ,ø) preserves 2- profiles from FRX to FRY ; and then profiles from FRX to FRY ; and then the theorem is proved by theorem the theorem is proved by theorem 3.6 3.6
Let (Ψ,ø) preserves 3-profiles from FRX Let (Ψ,ø) preserves 3-profiles from FRX to FRY. If c=αto FRY. If c=α๏๏a+βa+β๏๏b α,β,a,b€Zb α,β,a,b€ZNN
Then we can put that Then we can put that C=αC=α๏๏a+βa+β๏๏b+1b+1๏๏00 And since (Ψ,ø) preserves 3-profiles And since (Ψ,ø) preserves 3-profiles
from FRX to FRY, then we get from FRX to FRY, then we get Ø(c) =Ψ(α) Ø(c) =Ψ(α) ** ø(a)+Ψ(β) ø(a)+Ψ(β) ** ø(b)+Ψ(1) ø(b)+Ψ(1) **
ø(0). ø(0).
Since Ψ(1)=1 the identity and ø(0)=0, Since Ψ(1)=1 the identity and ø(0)=0, then then
Ø(c) =Ψ(α) Ø(c) =Ψ(α) ** ø(a)+Ψ(β) ø(a)+Ψ(β) ** ø(b) ø(b) This shows that (Ψ,ø) preserves 2- This shows that (Ψ,ø) preserves 2-
profiles from FRX to FRY.profiles from FRX to FRY. In a similar method if for any k>3,In a similar method if for any k>3,
(Ψ,ø) preserves k-profiles from FRX to (Ψ,ø) preserves k-profiles from FRX to FRY, FRY,
then , we can show that (Ψ,ø) then , we can show that (Ψ,ø) preserves 2-profiles from FRX to preserves 2-profiles from FRX to FRY .Thus the theorem is FRY .Thus the theorem is proved .proved .
Section twoSection two
Properties of far rings Properties of far rings In this section ; we shall prove some In this section ; we shall prove some
results for the far rings which are results for the far rings which are knows for the usual rings .knows for the usual rings .
First we give the following First we give the following theorem.theorem.
Theorem 3.9:Theorem 3.9:
The composition of far ring The composition of far ring isomorphisms (Ψ,ø): FRX→FRY and isomorphisms (Ψ,ø): FRX→FRY and (Ψ’,ø’):FRY→FRZ is an isomorphism(Ψ’,ø’):FRY→FRZ is an isomorphism
ProofProof
Let each of (Ψ,ø) and (Ψ’,ø’) be an Let each of (Ψ,ø) and (Ψ’,ø’) be an isomorphism from FRX into FRY and isomorphism from FRX into FRY and from FRY into FRZ respectively .the from FRY into FRZ respectively .the composition of (Ψ,ø) and (Ψ’,ø’) composition of (Ψ,ø) and (Ψ’,ø’) consists of two permutations from consists of two permutations from FRX into FRZ will be denoted by FRX into FRZ will be denoted by (Ψ’Ψ,ø’ø).(Ψ’Ψ,ø’ø).
It is clear that each of Ψ’Ψ and ø’ø is It is clear that each of Ψ’Ψ and ø’ø is a permutation from FRX into FRZ a permutation from FRX into FRZ such that Ψ’Ψ is a composition map such that Ψ’Ψ is a composition map of Ψ and Ψ’; and ø’ø is a composition of Ψ and Ψ’; and ø’ø is a composition map of ø and ø’.map of ø and ø’.
Since Ψ,Ψ’,ø and ø’ are bijections , Since Ψ,Ψ’,ø and ø’ are bijections , then we get that Ψ’Ψ and ø’ø are then we get that Ψ’Ψ and ø’ø are bijections. We have also thatbijections. We have also that
Ψ’Ψ (1)= Ψ’(Ψ(1))= Ψ’(1)=1, since Ψ’Ψ (1)= Ψ’(Ψ(1))= Ψ’(1)=1, since Ψ(1)= Ψ’(1)=1 Ψ(1)= Ψ’(1)=1
Now we consider that FRX=(ZNow we consider that FRX=(ZNN,+,,+,°°), ), FRY=(ZFRY=(ZNN,+, ,+, **), and FRZ=(Z), and FRZ=(ZNN,+,,+,๏๏).). To prove that (Ψ’Ψ,ø’ø) is an To prove that (Ψ’Ψ,ø’ø) is an
isomorphism from FRX into FRZ, it is isomorphism from FRX into FRZ, it is sufficient to prove that for any α,β,a,b sufficient to prove that for any α,β,a,b in Zin ZNN,,
= ø’(Ψ(α) = ø’(Ψ(α) ** ø(a)+Ψ(β) ø(a)+Ψ(β) ** ø(b)), ø(b)),Since (Ψ,ø) is an isomorphism from FRX intoSince (Ψ,ø) is an isomorphism from FRX intoFRY.FRY. =Ψ’(Ψ(=Ψ’(Ψ(αα))))๏๏ ø’ø(a))+Ψ’(Ψ(ø’ø(a))+Ψ’(Ψ(ββ))))๏๏ ø’(ø(b))ø’(ø(b))
Since (Ψ’,ø’) is an isomorphism from Since (Ψ’,ø’) is an isomorphism from FRY into FRZFRY into FRZ
=Ψ’Ψ(=Ψ’Ψ(αα) ) ๏๏ ø’ø(a)+Ψ’Ψ(ø’ø(a)+Ψ’Ψ(ββ) ) ๏๏ ø’ø(b)ø’ø(b) Therefore we get that composition of Therefore we get that composition of
any two far ring isomorphisms is an any two far ring isomorphisms is an isomorphism and the proof is complete. isomorphism and the proof is complete.
By induction ,one can easily prove By induction ,one can easily prove the following theoremthe following theorem
Theorem 3.10:Theorem 3.10:
The composition of any finite number The composition of any finite number of far ring isomorphisms is an of far ring isomorphisms is an isomorphism .isomorphism .
If (Ψ,ø)is an isomorphism from FRX If (Ψ,ø)is an isomorphism from FRX into FRY then the inverse of(Ψ,ø) into FRY then the inverse of(Ψ,ø) is(Ψ,ø)is(Ψ,ø)–1–1..
We can nominate the inverse of (Ψ,ø) We can nominate the inverse of (Ψ,ø) to be (Ψto be (Ψ-1-1,ø,ø-1-1)which consists of two)which consists of two
Permutations øPermutations ø-1-1 and Ψ and Ψ-1-1 from from FRY into FRX . Now we give the FRY into FRX . Now we give the following theorem about the following theorem about the inverse of an isomorphism of far inverse of an isomorphism of far ringsrings
Theorem 3.11Theorem 3.11
The inverse of an isomorphism (Ψ,ø) The inverse of an isomorphism (Ψ,ø) from FRX into FRY is an isomorphism from FRX into FRY is an isomorphism from FRY into FRX.from FRY into FRX.
ProofProof
Consider (ZConsider (ZNN,+,,+,๏๏) to be the far ring FRX ) to be the far ring FRX and (Zand (ZNN,+, ,+, **)to be the far ring FRY. )to be the far ring FRY.
Suppose that (Ψ,ø)is an isomorphism Suppose that (Ψ,ø)is an isomorphism from FRX into FRY .To prove that (Ψfrom FRX into FRY .To prove that (Ψ-1-1,ø,ø--
11)is an isomorphism from FRY into FRX.)is an isomorphism from FRY into FRX. We have Ψ and ø which are two We have Ψ and ø which are two
bijections from Zbijections from ZNN into Z into ZNN, Ψ(1)=1, and , Ψ(1)=1, and for any α,β,a,b in Zfor any α,β,a,b in ZNN
Since Ψ and ø are bijections from ZSince Ψ and ø are bijections from ZNN
into Zinto ZNN; then Ψ; then Ψ-1 -1 and ø and ø-1-1 are also are also
bijections from Zbijections from ZNN into Z into ZN N ;and;and
ΨΨΨΨ-1-1= I= IZZNN = Ψ = Ψ-1-1Ψ, øøΨ, øø-1-1= I= IZZNN =ø =ø-1-1ø, ø, where Iwhere IZZNN is the identity map over Z is the identity map over ZNN..
Since Ψ(1)=1, we get ΨSince Ψ(1)=1, we get Ψ-1-1 (1)=1(1)=1 To prove that (ΨTo prove that (Ψ-1-1,ø,ø-1-1) is an ) is an
isomorphism from FRY into FRX , it is isomorphism from FRY into FRX , it is sufficient to prove that for any sufficient to prove that for any α’,β’,a’,b’ in Zα’,β’,a’,b’ in ZNN, ,
= I= IZZNN(α’) (α’) ** I IZZNN(a’)+ I(a’)+ IZZNN(β’) (β’) ** I IZZNN(b’)(b’)
= α’ = α’ **a’+β’ a’+β’ **b’b’
Thus is satisfied and this completes Thus is satisfied and this completes the proof that the inverse (Ψthe proof that the inverse (Ψ-1-1,ø,ø--
11)=(Ψ,ø) )=(Ψ,ø) -1-1
of far ring isomorphism (Ψ,ø) is also far of far ring isomorphism (Ψ,ø) is also far ring isomorphismring isomorphism
1
Note:Note:
The far ring isomorphism (Ψ,ø) from The far ring isomorphism (Ψ,ø) from FRX into itself is called an FRX into itself is called an automorphism of FRXautomorphism of FRX
Now we prove the following property Now we prove the following property of the set of automorphisms of far of the set of automorphisms of far ring ring
Theorem 3.12Theorem 3.12
The set of automorphisms of the far The set of automorphisms of the far ring FRX forms a group, under ring FRX forms a group, under composition 0,denoted by Aut(FRX).composition 0,denoted by Aut(FRX).
Proof :Proof :
consider the far ring FRX=(Zconsider the far ring FRX=(ZNN,,+,+,๏๏).).
To prove that (Aut(FRX), To prove that (Aut(FRX), oo) is a group ) is a group where where oo is the composition of maps. is the composition of maps.
From theorem 3.9 , we get that From theorem 3.9 , we get that oo is closed is closed binary operation on the set of maps binary operation on the set of maps Aut(FRX).It is clear that Aut(FRX).It is clear that oo is associative If is associative If we take ø=Ψ = Iwe take ø=Ψ = IZZNN the identity over Z the identity over ZNN, , then it is clear that (Ithen it is clear that (IZZNN, I, IZZNN) is an ) is an automorphism of FRX .It is easy to show automorphism of FRX .It is easy to show that ,for any (Ψ,ø) in aut(FRX), the that ,for any (Ψ,ø) in aut(FRX), the composition of (Icomposition of (IZZNN, I, IZZNN) and (Ψ,ø) is the ) and (Ψ,ø) is the automorphism (Ψ,ø);automorphism (Ψ,ø);
that means there is (Ithat means there is (IZZNN, I, IZZNN) the identity ) the identity element of (Aut(FRX),element of (Aut(FRX),oo).From theorem ).From theorem 3.11 the inverse exists for any element in 3.11 the inverse exists for any element in Aut(FRX) . Finally we can say that Aut(FRX) . Finally we can say that oo is is closed binary operation and associative closed binary operation and associative on Aut(FRX);and the identity element on Aut(FRX);and the identity element exists in Aut(FRX) and each element in exists in Aut(FRX) and each element in Aut(FRX) has an inverse in Aut(FRX)with Aut(FRX) has an inverse in Aut(FRX)with respect to respect to oo . .
Therefore (Aut(FRX), Therefore (Aut(FRX), oo) forms a ) forms a group and the proof is finished.group and the proof is finished.
NoteNote
We denote the number of elements We denote the number of elements in the set Aut(FRX) by |in the set Aut(FRX) by |Aut(FRX)| .And we are going to prove Aut(FRX)| .And we are going to prove the following theoremthe following theorem
Theorem 3.13:Theorem 3.13:
If FRX and FRY are isomorphic far If FRX and FRY are isomorphic far rings, then the number of rings, then the number of isomorphisms between them equals isomorphisms between them equals both |Aut(FRX)| and by |Aut(FRY)|both |Aut(FRX)| and by |Aut(FRY)|
Proof: Consider FRX and FRY to be Proof: Consider FRX and FRY to be two isomorphic far rings. Then there two isomorphic far rings. Then there is an isomorphism (Ψ,ø) denotes by f is an isomorphism (Ψ,ø) denotes by f fom FRX into FRY.fom FRX into FRY.
If the isomorphism f is the only from FRX If the isomorphism f is the only from FRX into FRY, then the only automorphism of into FRY, then the only automorphism of FRX is the identity ; since if there is an FRX is the identity ; since if there is an automorphism of FRX denoted by automorphism of FRX denoted by (Ψ’,ø’)=k which is different from the (Ψ’,ø’)=k which is different from the identity , then by the above theoremsidentity , then by the above theorems
ffookk-1-1 is an isomorphism from FRX into FRY is an isomorphism from FRX into FRY
where kwhere k-1-1 is the inverse of k. is the inverse of k. Then f Then f ookk-1-1 must be f. must be f. ff FRXFRX→→FRYFRY
kk↓↑↓↑kk-1-1
FRXFRX
Therefore k must be the identity .Therefore k must be the identity . Thus we get that the identity is the Thus we get that the identity is the
only automorphism of FRX . Similarly only automorphism of FRX . Similarly if f is the only isomorphism from FRX if f is the only isomorphism from FRX into FRY , then the only into FRY , then the only automorphism of FRY is the identity .automorphism of FRY is the identity .
Now if there exists more than one Now if there exists more than one isomorphism from FRX into FRY. Let f be isomorphism from FRX into FRY. Let f be one isomorphism from FRX into FRY and one isomorphism from FRX into FRY and the other is g=(Ψ″,ø″) which is different the other is g=(Ψ″,ø″) which is different fromf. Let the set of isomorphisms from fromf. Let the set of isomorphisms from FRX into FRY be A. Define a map Ө from A FRX into FRY be A. Define a map Ө from A into Aut(FRX) by Ө(g)=ginto Aut(FRX) by Ө(g)=g-1-1f for any g in A f for any g in A where gwhere g-1-1 is the inverse map of g. is the inverse map of g.
It is clear that Ө is well defined map.It is clear that Ө is well defined map.
To prove that Ө is bijection (1-1 and onto) To prove that Ө is bijection (1-1 and onto) First to prove that Ө is 1-1First to prove that Ө is 1-1
Let Ө(gLet Ө(g11)=Ө(g)=Ө(g22), for any g), for any g11,g,g22 in A in A
Then Then 11gg-1-1
oof=f=22gg-1-1
oof.this yields to f.this yields to 11gg-1-1
ooffooff-1-1==22gg-1-1๏๏ff
๏๏ff-1-1
Therefore Therefore 11gg-1-1==22
gg-1-1 and then we get that and then we get that 11gg==22
g g This proves that Ө is 1-1.Now to This proves that Ө is 1-1.Now to prove that Ө is onto.prove that Ө is onto.
Let h be in Aut(FRX),and h is not the Let h be in Aut(FRX),and h is not the identity over FRX .Then there is g in identity over FRX .Then there is g in A ,g≠f such that gA ,g≠f such that g-1-1=hf=hf-1-1 and and
Ө(g)=gӨ(g)=g-1-1f=hff=hf-1-1f=h;(if g=f,then gf=h;(if g=f,then g-1-1=f=f-1-1 and and gg-1-1=f=f-1-1=hf=hf-1-1, therefore h is the identity ) , therefore h is the identity ) this proves that Ө is onto .thus Ө is 1-1 this proves that Ө is onto .thus Ө is 1-1 and onto well defined mapping. This and onto well defined mapping. This proves that the number of 1 to 1proves that the number of 1 to 1
Isomorphisms from FRX into FRY equals Isomorphisms from FRX into FRY equals to the number of automorphisms of to the number of automorphisms of FRX .one can similarly prove that the FRX .one can similarly prove that the number of isomorphisms from FRX into number of isomorphisms from FRX into FRY equals to the number of FRY equals to the number of automorphisms of FRY by defining the automorphisms of FRY by defining the map Ө from A into Aut(FRY) by Ө(g)=fgmap Ө from A into Aut(FRY) by Ө(g)=fg-1-1 for any g in A Thus the theorem is for any g in A Thus the theorem is proved.proved.
Section threeSection three
Unsolved problems there are several Unsolved problems there are several unsolved problems concerning our unsolved problems concerning our topics:topics:
The first problem is to find a test for The first problem is to find a test for isomorphic far rings based on isomorphic far rings based on inspecting Latin squares .inspecting Latin squares .
To give something about this for To give something about this for N=1,2,3,4.N=1,2,3,4.
First : Let N=1, then the only Latin First : Let N=1, then the only Latin aquare is 0 which gives only one far aquare is 0 which gives only one far ring of order 1 if N=2, then also there ring of order 1 if N=2, then also there is only one Latin square 1 0, which is only one Latin square 1 0, which gives only one far ring of order 2 gives only one far ring of order 2 0 1 If N=3, then also there is only one0 1 If N=3, then also there is only one
Latin Square Latin Square
1 2 01 2 0 2 0 12 0 1 0 1 20 1 2 which gives only one far ring of order 3 which gives only one far ring of order 3 In case N=1 or N=2 or N=3 , we have In case N=1 or N=2 or N=3 , we have
only one far ring which is isomorphic to only one far ring which is isomorphic to itself byitself by the identify isomorphism .If the identify isomorphism .If N=4, in this case we have four Latin N=4, in this case we have four Latin squares A,B,C, and D as show before in squares A,B,C, and D as show before in
Examples: 2.17Examples: 2.17
and therefore have four far ring FRA , and therefore have four far ring FRA , FRB,FRC,FRB,FRC, and FRD which are of order and FRD which are of order 4.FRA and FRB are isomorphic by the 4.FRA and FRB are isomorphic by the isomorphism (Ψ,ø) which is defined by isomorphism (Ψ,ø) which is defined by Ψ=(1)(234) and ø=(0)(2)(13).But FRC Ψ=(1)(234) and ø=(0)(2)(13).But FRC and FRD are not isomorphic ; since if and FRD are not isomorphic ; since if they were , then they have the same they were , then they have the same period profiles by theorem 3.5period profiles by theorem 3.5
But they have different period But they have different period profilesprofiles (see tables 3.3 and similarly, (see tables 3.3 and similarly, we obtain that FRA,FRC are not we obtain that FRA,FRC are not isomorphic ,FRA,FRD are also not isomorphic ,FRA,FRD are also not isomorphic and FRB ,FRC are not isomorphic and FRB ,FRC are not isomorphicisomorphic
A second problem is to reduce to A second problem is to reduce to minimum the number of 4-tuples we minimum the number of 4-tuples we must check to verifymust check to verify
**ø(b)………….1ø(b)………….1 Where α,β,a,b are in ZWhere α,β,a,b are in ZNN and (Z and (ZNN,,
+,+,๏๏) , (Z) , (ZNN,+, ,+, **) are two far rings .) are two far rings .
We have two far rings of ordered N.(ZWe have two far rings of ordered N.(ZNN,,+,+,๏๏) and (Z) and (ZNN,+, ,+, **). And we have each of ). And we have each of
Ψ and ø is a permutation of ZΨ and ø is a permutation of ZNN. To . To prove one , we need to prove that (Ψ,ø) prove one , we need to prove that (Ψ,ø) is an isomorphism from (Zis an isomorphism from (ZNN,+,,+,๏๏) into ) into (Z(ZNN,+, ,+, **). In general the number f 4-). In general the number f 4-
tuples we must check to verify 1 is tuples we must check to verify 1 is NN22.N.N22=N=N44..
But if the multiplication of far ring is But if the multiplication of far ring is commutative as for example in case commutative as for example in case N=1,2,3,4, then the number of 4-N=1,2,3,4, then the number of 4-tuples that we must check will tuples that we must check will reduce the number Nreduce the number N22 will reduce will reduce toN+(NtoN+(N22-N)/2 or less than this -N)/2 or less than this number . To show in an explicit number . To show in an explicit manner for cases N≤4.manner for cases N≤4.
In case N=1, it is trivial case we need In case N=1, it is trivial case we need only one check which is happened by only one check which is happened by taking α=β=a=b=0taking α=β=a=b=0
αβ abαβ ab 00 0000 00
In case N=2, we have a far ring of In case N=2, we have a far ring of order two with elements of Zorder two with elements of Z22={1,0}={1,0}
The following table gives the The following table gives the required cases that we need to prove required cases that we need to prove 1 1
αβ ab ab ab abαβ ab ab ab ab 11 11 10 01 0011 11 10 01 00 10 10 00 10 10 00 00 0000 00 NN22.N.N22=2=222.2.222=2=244=16=16 16 reduces to 716 reduces to 7
in case N=3, we have a far ring of in case N=3, we have a far ring of order 3 wih elements of Zorder 3 wih elements of Z33={1,2,0} ={1,2,0} the following table gives the cases the following table gives the cases that we need to prove 1that we need to prove 1
αβ ab αβ ab ab ab ab ab abab ab ab ab ab 11 11 12 13 22 20 0011 11 12 13 22 20 00 12 12 10 22 20 02 0012 12 10 22 20 02 00 10 10 10 20 01 00 10 20 01 00 22 22 20 22 22 20 00 00 20 23 0020 23 00 00 0000 00
The number NThe number N44=3=344=81 =81 reduces to 22 reduces to 22
In case N=4, we have a far In case N=4, we have a far ring of order 4 with elements ring of order 4 with elements of Zof Z44={1,2,3,0}={1,2,3,0}
The following tale gives the The following tale gives the cases that we need to prove 1cases that we need to prove 1
αβ ab ab ab ab ab ab ab ab ab ab abαβ ab ab ab ab ab ab ab ab ab ab ab 11 11 12 13 10 22 23 20 33 30 0011 11 12 13 10 22 23 20 33 30 00 12 12 13 10 22 23 20 31 32 33 30 0012 12 13 10 22 23 20 31 32 33 30 00 13 13 10 23 20 33 20 03 0013 13 10 23 20 33 20 03 00 10 10 20 33 30 0010 10 20 33 30 00 22 22 23 20 33 30 03 0022 22 23 20 33 30 03 00 23 23 20 31 33 30 0023 23 20 31 33 30 00 20 20 30 0020 20 30 00 33 33 30 0033 33 30 00 30 30 0030 30 00 00 0000 00
The number NThe number N22.N.N22=4=444=16x16=256reduce to 55=16x16=256reduce to 55
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