1. Chapter three: Far rings and some results Far rings and some results In This chapter, we introduce some results about far rings and period profiles, and we prove some properties of far rings which are known for the usual rings and some other results. In This chapter, we introduce some results about far rings and period profiles, and we prove some properties of far rings which are known for the usual rings and some other results.

2. Section one: Far rings and period profiles Far rings and period profiles In this section we introduce the meaning of the period profile and we explain the relation between them. This relation is important in the subject of faster way to get fractal images. In this section we introduce the meaning of the period profile and we explain the relation between them. This relation is important in the subject of faster way to get fractal images. First we give the definition of period profile. First we give the definition of period profile.

3. Definition 3.1: Let (Z N,+, ๏) be a far ring; and α 1, …, α k be in Z N. The period profile of α 1,…, α k is a list in descending order of the period lengths of the k-sequences which are obtained by recurrence relation over the far ring: Let (Z N,+, ๏) be a far ring; and α 1, …, α k be in Z N. The period profile of α 1,…, α k is a list in descending order of the period lengths of the k-sequences which are obtained by recurrence relation over the far ring: S n = α 1 ๏S n-1, α 2 ๏ S n-2 +…+ α k ๏S n-k n ≥ k S n = α 1 ๏S n-1, α 2 ๏ S n-2 +…+ α k ๏S n-k n ≥ k

4. Whatever the initial values S 0 S 1 …S k-1 which are elements of Z N. Whatever the initial values S 0 S 1 …S k-1 which are elements of Z N. To illustrate the concept of period profile, we take the simple case k=2 and we consider the far ring of order 4. To illustrate the concept of period profile, we take the simple case k=2 and we consider the far ring of order 4.

5. From examples 2.17 in chapter2, there are four possible far rings of order 4. They are FRA, FRB,FRC, and FRD. From examples 2.17 in chapter2, there are four possible far rings of order 4. They are FRA, FRB,FRC, and FRD. Let α 1, α 2 be in Z 4. Recurrence relations of length k=2 over the far ring (Z 4,+,๏) are given by S n = α 1 ๏S n-1 + α 2 ๏S n-2 n ≥ 2; Let α 1, α 2 be in Z 4. Recurrence relations of length k=2 over the far ring (Z 4,+,๏) are given by S n = α 1 ๏S n-1 + α 2 ๏S n-2 n ≥ 2;

6. Where any two elements S 0 S 1 from Z N can be taken as initial values to obtain each of these recurrence relation. Where any two elements S 0 S 1 from Z N can be taken as initial values to obtain each of these recurrence relation. The period of the sequence S 0 S 1 … cannot exceed 4 2 =16, which is the maximum number of distinct pairs S i S i+1 which can be encountered before repetition. The period of the sequence S 0 S 1 … cannot exceed 4 2 =16, which is the maximum number of distinct pairs S i S i+1 which can be encountered before repetition.

7. Further, any such pair may be taken as initial so is found in some sequence Thus the period lengths sum to 16. Further, any such pair may be taken as initial so is found in some sequence Thus the period lengths sum to 16. The period profile of ( α 1, α 2 ) is a list of period lengths in decending order, and the set of lists for all The period profile of ( α 1, α 2 ) is a list of period lengths in decending order, and the set of lists for all ( α 1, α 2 ) in any order, is the profile of the multiplication table of the far ring itself. ( α 1, α 2 ) in any order, is the profile of the multiplication table of the far ring itself.

8. For example when the coefficients α 1, α 2 =0,1 are taken in FRA, then we get the sequence 120322…, 102300,133…,and 1…, which are periodic of period 6,6,3,and 1 respectively. For example when the coefficients α 1, α 2 =0,1 are taken in FRA, then we get the sequence 120322…, 102300,133…,and 1…, which are periodic of period 6,6,3,and 1 respectively. Therefore the period profile for 0,1 is 6631. Therefore the period profile for 0,1 is 6631.

9. The following tables give the period profile for all coefficients α i, α j in Z 4 and for the four far rings (of order four) FRA,FRB,FRC, and FRD. The following tables give the period profile for all coefficients α i, α j in Z 4 and for the four far rings (of order four) FRA,FRB,FRC, and FRD.

10. coefficients Sequences over far ring FRA (period length) Period profile 11121310212223203132333001020300112310(6);130332(6);220(3);0(1)113120032330102(15);2(1)132230020(9);1103(4);12(2);3(1)122100(6);232030(6);133(3);1(1)113233020121003(15);2(1)1220300231332(13);110(3)123200102213033(15)1(1)131032230(9);11202(5);3(1);0(1)110301312232002(15);3(1)123033100(9);1322(4);20(2);1(1)11203210(8);133023(6);2(1);0(1)113003(6);10220(5);233(3);12(2)120322(6);102300(6);133(3);1(1)13022320(8);112103(6);3(1);0(1)1131003012202(13);233(3)110(3);123(3);132(3);200(3);330(3);2(1)663115,19421663115,113,315,1951115,19421861165326631861113,3333331 Table 3.1

11. coefficients Sequences over far ring FRB (period length) Period profiles 11121310212223203132333001020300112310(6);130332(6);220(3);0(1)120102(6);223003(6);113(3);3(1)110302331320012(15);2(1)122100203(9);1330(4);23(2);1(1)122320(6);100320(6);113(3);3(1)110(3);123(3);132(3);200(3);330(3);2(1)12031022(8);133230(6);1(1);0(1)1300103220233(13);112(3)110203230013312(15);2(1)122103130(9);23320(5);1(1);0(1)1123220100213(13);330(3)113212003022310(15);3(1)120023301031322(15);1(1)133100(6);22030(5);112(3);23(2)113003210(9);1223(4);20(2);3(1)12303320(8);110213(6);2(1);0(1)6631663115,194216631333331861113,315,1951113,315,115,1653294218611 Table 3.2

12. coefficients Sequences over far ring FRC (period length) Period profile 11121310212223203132333001020300112310(6);130332(6);220(3);0(1)120300(6);102322(6);113(3);3(1)110312(6);132330(6);200(3);2(1)122302(6);100320(6);133(3);1(1)120300(6);102322(6);113(3);3(1)110312(6);132330(6);200(3);2(1)122302(6);100320(6);133(3);1(1)112310(6);130332(6);220(3);0(1)110312(6);132330(6);200(3);2(1)122302(6);100320(6);133(3);1(1)112310(6);130332(6);220(3);0(1)120300(6);102322(6);113(3);3(1)122302(6);100320(6);133(3);1(1)112310(6);130332(6);220(3);0(1)120300(6);102322(6);113(3);3(1)110312(6);132330;200(3);2(1)6631663166316631663166316631663166316631663166316631663166316631 Table 3.3

13. coefficients Sequences over far ring FRD(period length) Period profile 11121310212223203132333001020300112310(6);130332(6);220(3);0(1)113(3);120102(6);223003(6);3(1)110312(6);132330(6);200(3);2(1)122100(6);232030(6);133(3);1(1)122320(6);100302(6);113(3);3(1)110(3);123(3);132(3);200(3);330(3);2(1)120322(6);102300(6);133(3);1(1)112(3);130(3);103(3);220(3);233(3);0(1)110312(6);132330(6);200(3);2(1)122100(6);232030(6);133(3);1(1)112310(6);130332(6);220(3);0(1)120102(6);223003(6);113(3);3(1)120322(6);102300(6);133(3);1(1)112(3);130(3);103(3);220(3);233(3);0(1)122320(6);100302(6);113(3);3(1)110(3);123(3);132(3);200(3);330(3);2(1)663166316631663166313333316631333331663166316631663166313333316631333331 Table 3.4

14. Definition 3.2: Let each of (G 1,๏) and (G 2, * ) be a quasigroup. An isomorphism Ψ from G 1 into G 2 is a bijection or a permutation such that for all x,y in G 1 Let each of (G 1,๏) and (G 2, * ) be a quasigroup. An isomorphism Ψ from G 1 into G 2 is a bijection or a permutation such that for all x,y in G 1 Ψ(x ๏ y)=Ψ(x) * Ψ(y). Ψ(x ๏ y)=Ψ(x) * Ψ(y). Then (G 1,๏)and(G 2, * ) are called isomorphic. Then (G 1,๏)and(G 2, * ) are called isomorphic.

15. Our aim is to obtain far rings (of the same order) which have the same period profiles. Our aim is to obtain far rings (of the same order) which have the same period profiles. First the following example shows that the isomorphism of quasigroups does not suffice for equal profiles. First the following example shows that the isomorphism of quasigroups does not suffice for equal profiles.

16. Example 3.3 Consider (Z 4,๏)and (Z 4, * ) such that ๏ and * are two binary operations defined by the following tables Consider (Z 4,๏)and (Z 4, * ) such that ๏ and * are two binary operations defined by the following tables

17. ๏ 1 2 3 0 12301230 2 0 1 3 3 1 0 2 0 3 2 1* 1 2 3 0 12301230 2 3 0 1 3 0 1 2 0 1 2 3

18. Define Ψ from the quasigroup (Z 4,๏) into the quasigroup (Z 4, * )by Define Ψ from the quasigroup (Z 4,๏) into the quasigroup (Z 4, * )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 isomorphism. It is easy to show that Ψ is an isomorphism. But (Z 4,๏)=FRA and (Z 4, * )=FRC have different period profiles ; But (Z 4,๏)=FRA and (Z 4, * )=FRC have different period profiles ;

19. see table 3.1 and table 3.3. see table 3.1 and table 3.3. We shall see that the following definition of isomorphism of far rings is sufficient to make isomorphic far rings of the same period profiles. We shall see that the following definition of isomorphism of far rings is sufficient to make isomorphic far rings of the same period profiles.

20. Definition 3.4: An isomorphism of far rings (Z N,+,๏) → (Z N,+, * ) is a pair of permutations Ψ, ø of Z N such that Ψ(1)=1 and for each α,β,a,b in Z N, An isomorphism of far rings (Z N,+,๏) → (Z N,+, * ) is a pair of permutations Ψ, ø of Z N such that Ψ(1)=1 and for each α,β,a,b in Z N, ø (α ๏ a+ β๏b)= Ψ(α ) * ø(a) + Ψ(β) * ø(b). ø (α ๏ a+ β๏b)= Ψ(α ) * ø(a) + Ψ(β) * ø(b). Then (Z N,+,๏) and (Z N,+, * ) are called isomorphic far rings. Then (Z N,+,๏) and (Z N,+, * ) are called isomorphic far rings.

21. The following theorem proves that any two isomorphic far rings have the same period profiles. The following theorem proves that any two isomorphic far rings have the same period profiles.

22. Theorem 3.5 Isomorphic far rings have the same period profiles.In particular for k=2 If(Ψ,ø) is an isomorphism from the far ring( Z N,+, ๏ ) into the far ring (Z N,+, * ) then the coefficient pairs α,β and Ψ(α), Ψ(β)in Z N have the same period profile, and similarly for k-tuples with k>2. Isomorphic far rings have the same period profiles.In particular for k=2 If(Ψ,ø) is an isomorphism from the far ring( Z N,+, ๏ ) into the far ring (Z N,+, * ) then the coefficient pairs α,β and Ψ(α), Ψ(β)in Z N have the same period profile, and similarly for k-tuples with k>2.

23. Proof First let k=2 and suppose α,β give a sequence {a n } of period t.That is, with addition as usual mod N, we have First let k=2 and suppose α,β give a sequence {a n } of period t.That is, with addition as usual mod N, we have a n =α๏a n-1 + β๏a n-2. a n =α๏a n-1 + β๏a n-2. Using ø(α๏a+β๏b) =Ψ( α ) * ø(a)+Ψ (β)+ø(b)for any α,β, a, b in Z N ….…(1) Using ø(α๏a+β๏b) =Ψ( α ) * ø(a)+Ψ (β)+ø(b)for any α,β, a, b in Z N ….…(1)

24. Then the sequence {a n }={a 0,a 1,…}satisfies that a t =a 0,a t+1 =a 1, and so on. Then the sequence {a n }={a 0,a 1,…}satisfies that a t =a 0,a t+1 =a 1, and so on. Thus, the coefficients Ψ(α),Ψ(β) Thus, the coefficients Ψ(α),Ψ(β) give the sequence {ø(a n )}={ø(a 0 ),ø(a 1 ),…,ø(a n ),…} give the sequence {ø(a n )}={ø(a 0 ),ø(a 1 ),…,ø(a n ),…}

25. Now, we need to prove that {ø(a n )}is periodic of period t; we must prove that Ø(a t )=ø(a 0 ),ø(a t-1 )=ø(a 1 ) and there is no positive integer r such that r<t and Ø(a r )=ø(a 0 ),ø(a r+1 )=ø(a 1 ). Now, we need to prove that {ø(a n )}is periodic of period t; we must prove that Ø(a t )=ø(a 0 ),ø(a t-1 )=ø(a 1 ) and there is no positive integer r such that r<t and Ø(a r )=ø(a 0 ),ø(a r+1 )=ø(a 1 ).

26. It is clear that ø(a t )=ø(a 0 )and ø(a t+1 )=ø(a 1 )…,since {a n } is periodic of period t.If there is r<t such that ø(a r )=ø(a 0 ), ø(a r+1 )=ø(a 1 )and since ø is bijection, then we get that there is r<t such that a r =a 0,a r+1 =a 1, It is clear that ø(a t )=ø(a 0 )and ø(a t+1 )=ø(a 1 )…,since {a n } is periodic of period t.If there is r<t such that ø(a r )=ø(a 0 ), ø(a r+1 )=ø(a 1 )and since ø is bijection, then we get that there is r<t such that a r =a 0,a r+1 =a 1,

27. but this is a contradiction to the fact that the period of {a n } is t and r≠t; and t is the least positive integer making {a n } periodic.Therefore we get that the sequence{ø(an)} is periodic of period t; but this is a contradiction to the fact that the period of {a n } is t and r≠t; and t is the least positive integer making {a n } periodic.Therefore we get that the sequence{ø(an)} is periodic of period t; Which generated by Ψ(α),Ψ(β) with * as multiplication. Which generated by Ψ(α),Ψ(β) with * as multiplication.

28. Finally, we may infer that these two far rings have the same period peofile ;since the bijection Ψ defines a bejiction of pairs (α,β ) → (Ψ(α),Ψ(β)). Finally, we may infer that these two far rings have the same period peofile ;since the bijection Ψ defines a bejiction of pairs (α,β ) → (Ψ(α),Ψ(β)). This complete the proof for the case k=2. This complete the proof for the case k=2.

29. The proof for the case k>2 is by induction. The proof for the case k>2 is by induction. It suffices to illustrate the proof for the case k=3. Let α,β,γ,a,b,c be in Z N and x= α๏a+β๏b+γ๏c, It suffices to illustrate the proof for the case k=3. Let α,β,γ,a,b,c be in Z N and x= α๏a+β๏b+γ๏c, then then

30. Ø(x) =ø(α๏a+β๏b+γ๏c) Ø(x) =ø(α๏a+β๏b+γ๏c) = ø(α๏a+1๏(β๏b+γ๏c)) = ø(α๏a+1๏(β๏b+γ๏c)) Since 1 acts as identify for ๏, Since 1 acts as identify for ๏, =Ψ(α) * ø(a)+Ψ(1) * ø(β๏b+γ๏c) ; since Ψ(1)=1 is an identity for *. =Ψ(α) * ø(a)+Ψ(1) * ø(β๏b+γ๏c) ; since Ψ(1)=1 is an identity for *. =Ψ(α) * ø(α)+Ψ(β) * ø(b)+Ψ(γ) * ø(c); since +is associative and by using (1). =Ψ(α) * ø(α)+Ψ(β) * ø(b)+Ψ(γ) * ø(c); since +is associative and by using (1).

31. By using the same notions in case k=2; it is easy to show that the coefficients Ψ(α),Ψ(β),Ψ(γ)generate the sequence {ø(a n )}of period t,if the coefficients α,β,and γ generate the sequence {a n }of period t. By using the same notions in case k=2; it is easy to show that the coefficients Ψ(α),Ψ(β),Ψ(γ)generate the sequence {ø(a n )}of period t,if the coefficients α,β,and γ generate the sequence {a n }of period t.

32. We get the same result for k>3 by the same method. This means that if the coefficients α 1,α 2,…,α k generate a sequence {a n } of period t. We get the same result for k>3 by the same method. This means that if the coefficients α 1,α 2,…,α k generate a sequence {a n } of period t.

33. under the multiplication๏, then the coefficients Ψ(α 1 ),Ψ(α 2 ),…,Ψ(α k ) generate the sequence {ø(a n )} of period t under multiplication *. Consequently,isomorphic far rings have the same period profiles for all k≥2. under the multiplication๏, then the coefficients Ψ(α 1 ),Ψ(α 2 ),…,Ψ(α k ) generate the sequence {ø(a n )} of period t under multiplication *. Consequently,isomorphic far rings have the same period profiles for all k≥2.

34. Note: The important of theorem 3.5 is the result that if any one of these isomorphic far rings gives an M-sequence, then the others do the same. The important of theorem 3.5 is the result that if any one of these isomorphic far rings gives an M-sequence, then the others do the same. Now we search for what is required to prove the converse of theorem 3.5.We give the following preparatory theorem for k=2. Now we search for what is required to prove the converse of theorem 3.5.We give the following preparatory theorem for k=2.

35. Theorem3.6 let each of (Z N,+,๏) and (Z N,+, * ) be a far ring which are denoted by FRX,FRY respectively. Suppose that Ψ,ø are permutations of Z N,Ψ(1)=1 and ø sends sequences over FRX generated by coefficients α,β to sequences over FRY generated by coefficients Ψ(α), Ψ(β) for any α,β in Z N. let each of (Z N,+,๏) and (Z N,+, * ) be a far ring which are denoted by FRX,FRY respectively. Suppose that Ψ,ø are permutations of Z N,Ψ(1)=1 and ø sends sequences over FRX generated by coefficients α,β to sequences over FRY generated by coefficients Ψ(α), Ψ(β) for any α,β in Z N.

36. Then (Ψ,ø) is an isomorphism from FRX into FRY.

37. Proof: In order to prove this theorem, we must prove the following : In order to prove this theorem, we must prove the following : For any α,β,a,b in Z N ; For any α,β,a,b in Z N ; Ø(α๏a+β๏b)=Ψ(α) * ø(a)+Ψ(β) * ø(b). Ø(α๏a+β๏b)=Ψ(α) * ø(a)+Ψ(β) * ø(b).

38. Now consider α,β,a,b in Z N and take k to be 2 to obtain a k-sequences over FRX particularly when k=2. Then a,b must appear as successive members of some sequence with coefficents α,β (we can take a,b as initial members if necessary). Now consider α,β,a,b in Z N and take k to be 2 to obtain a k-sequences over FRX particularly when k=2. Then a,b must appear as successive members of some sequence with coefficents α,β (we can take a,b as initial members if necessary).

39. Let c=α๏a+β๏b; c will be in Z N. Then ø sends this sequence with coefficients α,β to one over FRY with coefficients Ψ(α),Ψ(β) which results in Let c=α๏a+β๏b; c will be in Z N. Then ø sends this sequence with coefficients α,β to one over FRY with coefficients Ψ(α),Ψ(β) which results in ø ( c) = ø(α๏a+β๏b) ø ( c) = ø(α๏a+β๏b) = Ψ(α) * ø(a)+Ψ(β) * ø(b); As required. = Ψ(α) * ø(a)+Ψ(β) * ø(b); As required.

40. This is true for any four elements of Z N. And since Ψ(1)=1, we get (Ψ,ø) is An isomorphism and the proof is complete. This is true for any four elements of Z N. And since Ψ(1)=1, we get (Ψ,ø) is An isomorphism and the proof is complete. Remark 3.7: let Ψ and ø be permutations of Z N and each of FRX,FRY is a far ring of order N. Remark 3.7: let Ψ and ø be permutations of Z N and each of FRX,FRY is a far ring of order N.

41. The statement that (Ψ,ø) preserves k- profiles from FRX to FRY means that ø sends a sequence defined over FRX by a recurrence with coefficients α i (1≤i≤k) to the sequence over FRY defined by a recurrence with coefficients ø(α i ) (1≤i≤k) The statement that (Ψ,ø) preserves k- profiles from FRX to FRY means that ø sends a sequence defined over FRX by a recurrence with coefficients α i (1≤i≤k) to the sequence over FRY defined by a recurrence with coefficients ø(α i ) (1≤i≤k)

42. The following theorem gives some conditions to get isomorphic far rings The following theorem gives some conditions to get isomorphic far rings

43. Theorem 3.8: Let Ψ,ø be permutations of Z N with Ψ(1)=1 and ø(0)=0, and let FRX, FRY be far rings of order N. Suppose that for some k>2,(Ψ,ø) preserves k-profiles from FRX to FRY then (Ψ,ø) is a far ring isomorphism from FRX to FRY. Let Ψ,ø be permutations of Z N with Ψ(1)=1 and ø(0)=0, and let FRX, FRY be far rings of order N. Suppose that for some k>2,(Ψ,ø) preserves k-profiles from FRX to FRY then (Ψ,ø) is a far ring isomorphism from FRX to FRY.

44. Proof: Let (Ψ,ø) preserves k-profiles from FRX= (Z N,+,๏) to FRY= (Z N,+, * ) for k=2. Then, from Remark 3.7, that means ø Sends sequences over FRX generated by coefficients α,β to sequences over FRY generated by coefficients Ψ(α),Ψ(β); for any α,β in Z N. Let (Ψ,ø) preserves k-profiles from FRX= (Z N,+,๏) to FRY= (Z N,+, * ) for k=2. Then, from Remark 3.7, that means ø Sends sequences over FRX generated by coefficients α,β to sequences over FRY generated by coefficients Ψ(α),Ψ(β); for any α,β in Z N.

45. And since Ψ(1)=1,then we get that (Ψ,ø) is an isomorphism. It is sufficient to show that if for some k>2,(Ψ,ø) preserves k- profiles from FRX to FRY then (Ψ,ø) preserves 2- profiles from FRX to FRY ; and then the theorem is proved by theorem 3.6 And since Ψ(1)=1,then we get that (Ψ,ø) is an isomorphism. It is sufficient to show that if for some k>2,(Ψ,ø) preserves k- profiles from FRX to FRY then (Ψ,ø) preserves 2- profiles from FRX to FRY ; and then the theorem is proved by theorem 3.6

46. Let (Ψ,ø) preserves 3-profiles from FRX to FRY. If c=α๏a+β๏b α,β,a,b€Z N Let (Ψ,ø) preserves 3-profiles from FRX to FRY. If c=α๏a+β๏b α,β,a,b€Z N Then we can put that Then we can put that C=α๏a+β๏b+1๏0 C=α๏a+β๏b+1๏0 And since (Ψ,ø) preserves 3-profiles from FRX to FRY, then we get And since (Ψ,ø) preserves 3-profiles from FRX to FRY, then we get Ø(c) =Ψ(α) * ø(a)+Ψ(β) * ø(b)+Ψ(1) * ø(0). Ø(c) =Ψ(α) * ø(a)+Ψ(β) * ø(b)+Ψ(1) * ø(0).

47. Since Ψ(1)=1 the identity and ø(0)=0, then Since Ψ(1)=1 the identity and ø(0)=0, then Ø(c) =Ψ(α) * ø(a)+Ψ(β) * ø(b) Ø(c) =Ψ(α) * ø(a)+Ψ(β) * ø(b) This shows that (Ψ,ø) preserves 2- profiles from FRX to FRY. This shows that (Ψ,ø) preserves 2- profiles from FRX to FRY. In a similar method if for any k>3,(Ψ,ø) preserves k-profiles from FRX to FRY, In a similar method if for any k>3,(Ψ,ø) preserves k-profiles from FRX to FRY,

48. then, we can show that (Ψ,ø) preserves 2-profiles from FRX to FRY.Thus the theorem is proved. then, we can show that (Ψ,ø) preserves 2-profiles from FRX to FRY.Thus the theorem is proved.

49. Section two Properties of far rings Properties of far rings In this section ; we shall prove some results for the far rings which are knows for the usual rings. In this section ; we shall prove some results for the far rings which are knows for the usual rings. First we give the following theorem. First we give the following theorem.

50. Theorem 3.9: The composition of far ring isomorphisms (Ψ,ø): FRX → FRY and (Ψ’,ø’):FRY → FRZ is an isomorphism The composition of far ring isomorphisms (Ψ,ø): FRX → FRY and (Ψ’,ø’):FRY → FRZ is an isomorphism

51. Proof Let each of (Ψ,ø) and (Ψ’,ø’) be an isomorphism from FRX into FRY and from FRY into FRZ respectively.the composition of (Ψ,ø) and (Ψ’,ø’) consists of two permutations from FRX into FRZ will be denoted by (Ψ’Ψ,ø’ø). Let each of (Ψ,ø) and (Ψ’,ø’) be an isomorphism from FRX into FRY and from FRY into FRZ respectively.the composition of (Ψ,ø) and (Ψ’,ø’) consists of two permutations from FRX into FRZ will be denoted by (Ψ’Ψ,ø’ø).

52. It is clear that each of Ψ’Ψ and ø’ø is a permutation from FRX into FRZ such that Ψ’Ψ is a composition map of Ψ and Ψ’; and ø’ø is a composition map of ø and ø’. It is clear that each of Ψ’Ψ and ø’ø is a permutation from FRX into FRZ such that Ψ’Ψ is a composition map of Ψ and Ψ’; and ø’ø is a composition map of ø and ø’. Since Ψ,Ψ’,ø and ø’ are bijections, then we get that Ψ’Ψ and ø’ø are bijections. We have also that Since Ψ,Ψ’,ø and ø’ are bijections, then we get that Ψ’Ψ and ø’ø are bijections. We have also that

53. Ψ’Ψ (1)= Ψ’(Ψ(1))= Ψ’(1)=1, since Ψ(1)= Ψ’(1)=1 Ψ’Ψ (1)= Ψ’(Ψ(1))= Ψ’(1)=1, since Ψ(1)= Ψ’(1)=1 Now we consider that FRX=(Z N,+, ° ), Now we consider that FRX=(Z N,+, ° ), FRY=(Z N,+, * ), and FRZ=(Z N,+,๏). FRY=(Z N,+, * ), and FRZ=(Z N,+,๏). To prove that (Ψ’Ψ,ø’ø) is an isomorphism from FRX into FRZ, it is sufficient to prove that for any α,β,a,b in Z N, To prove that (Ψ’Ψ,ø’ø) is an isomorphism from FRX into FRZ, it is sufficient to prove that for any α,β,a,b in Z N,

54. ø’ø(α ° a+β ° b)= Ψ’Ψ(α) ๏ ø’ø (a)+ Ψ’Ψ(β) ๏ ø’ø(b). =ø’(ø(α ° a+ β ° b)) =ø’(ø(α ° a+ β ° b)) = ø’(Ψ(α) * ø(a)+Ψ(β) * ø(b)), = ø’(Ψ(α) * ø(a)+Ψ(β) * ø(b)), Since (Ψ,ø) is an isomorphism from FRX into FRY. =Ψ’(Ψ(α))๏ ø’ø(a))+Ψ’(Ψ(β))๏ =Ψ’(Ψ(α))๏ ø’ø(a))+Ψ’(Ψ(β))๏ ø’(ø(b)) ø’(ø(b))

55. Since (Ψ’,ø’) is an isomorphism from FRY into FRZ =Ψ’Ψ(α) ๏ ø’ø(a)+Ψ’Ψ(β) ๏ ø’ø(b) Therefore we get that composition of any two far ring isomorphisms is an isomorphism and the proof is complete. Therefore we get that composition of any two far ring isomorphisms is an isomorphism and the proof is complete. By induction,one can easily prove the following theorem By induction,one can easily prove the following theorem

56. Theorem 3.10: The composition of any finite number of far ring isomorphisms is an isomorphism. The composition of any finite number of far ring isomorphisms is an isomorphism. If (Ψ,ø)is an isomorphism from FRX into FRY then the inverse of(Ψ,ø) is(Ψ,ø) –1. If (Ψ,ø)is an isomorphism from FRX into FRY then the inverse of(Ψ,ø) is(Ψ,ø) –1. We can nominate the inverse of (Ψ,ø) to be (Ψ -1,ø -1 )which consists of two We can nominate the inverse of (Ψ,ø) to be (Ψ -1,ø -1 )which consists of two

57. Permutations ø -1 and Ψ -1 from FRY into FRX. Now we give the following theorem about the inverse of an isomorphism of far rings Permutations ø -1 and Ψ -1 from FRY into FRX. Now we give the following theorem about the inverse of an isomorphism of far rings

58. Theorem 3.11 The inverse of an isomorphism (Ψ,ø) from FRX into FRY is an isomorphism from FRY into FRX. The inverse of an isomorphism (Ψ,ø) from FRX into FRY is an isomorphism from FRY into FRX.

59. Proof Consider (Z N,+,๏) to be the far ring FRX and (Z N,+, * )to be the far ring FRY. Suppose that (Ψ,ø)is an isomorphism from FRX into FRY.To prove that (Ψ -1,ø -1 )is an isomorphism from FRY into FRX. Consider (Z N,+,๏) to be the far ring FRX and (Z N,+, * )to be the far ring FRY. Suppose that (Ψ,ø)is an isomorphism from FRX into FRY.To prove that (Ψ -1,ø -1 )is an isomorphism from FRY into FRX. We have Ψ and ø which are two bijections from Z N into Z N, Ψ(1)=1, and for any α,β,a,b in Z N We have Ψ and ø which are two bijections from Z N into Z N, Ψ(1)=1, and for any α,β,a,b in Z N

60. Ø(α๏a+β๏b)=Ψ(α) * ø(a)+Ψ(β) * ø(b) Ø(α๏a+β๏b)=Ψ(α) * ø(a)+Ψ(β) * ø(b) Since Ψ and ø are bijections from Z N into Z N ; then Ψ -1 and ø -1 are also bijections from Z N into Z N ;and Since Ψ and ø are bijections from Z N into Z N ; then Ψ -1 and ø -1 are also bijections from Z N into Z N ;and ΨΨ -1 = I Z N = Ψ -1 Ψ, øø -1 = I Z N =ø -1 ø, where I Z N is the identity map over Z N. ΨΨ -1 = I Z N = Ψ -1 Ψ, øø -1 = I Z N =ø -1 ø, where I Z N is the identity map over Z N.

61. Since Ψ(1)=1, we get Ψ -1 (1)=1 Since Ψ(1)=1, we get Ψ -1 (1)=1 To prove that (Ψ -1,ø -1 ) is an isomorphism from FRY into FRX, it is sufficient to prove that for any α’,β’,a’,b’ in Z N, To prove that (Ψ -1,ø -1 ) is an isomorphism from FRY into FRX, it is sufficient to prove that for any α’,β’,a’,b’ in Z N, Ø -1 (α’ * a’+β’ * b’)=Ψ -1 (α’)๏ø -1 (a’)+Ψ -1 (β’)๏ø - 1 (b’)………… Ø -1 (α’ * a’+β’ * b’)=Ψ -1 (α’)๏ø -1 (a’)+Ψ -1 (β’)๏ø - 1 (b’)………… 1

62. Since, Ø(Ψ -1 (α’)๏ø -1 (a’)+Ψ -1 (β’)๏ø -1 (b)) = Ψ(Ψ -1 (α’)) * ø(ø -1 (a’))+ Ψ(Ψ -1 (β’)) = Ψ(Ψ -1 (α’)) * ø(ø -1 (a’))+ Ψ(Ψ -1 (β’)) * ø(ø -1 (b’)) * ø(ø -1 (b’)) = ΨΨ -1 (α’) * øø -1 (a’)+ ΨΨ -1 (β’) * øø -1 = ΨΨ -1 (α’) * øø -1 (a’)+ ΨΨ -1 (β’) * øø -1 (b’) (b’) = I Z N (α’) * I Z N (a’)+ I Z N (β’) * I Z N (b’) = I Z N (α’) * I Z N (a’)+ I Z N (β’) * I Z N (b’) = α’ * a’+β’ * b’ = α’ * a’+β’ * b’

63. Thus is satisfied and this completes the proof that the inverse (Ψ -1,ø -1 )=(Ψ,ø) -1 Thus is satisfied and this completes the proof that the inverse (Ψ -1,ø -1 )=(Ψ,ø) -1 of far ring isomorphism (Ψ,ø) is also far ring isomorphism 1

64. Note: The far ring isomorphism (Ψ,ø) from FRX into itself is called an automorphism of FRX The far ring isomorphism (Ψ,ø) from FRX into itself is called an automorphism of FRX Now we prove the following property of the set of automorphisms of far ring Now we prove the following property of the set of automorphisms of far ring

65. Theorem 3.12 The set of automorphisms of the far ring FRX forms a group, under composition 0,denoted by Aut(FRX). The set of automorphisms of the far ring FRX forms a group, under composition 0,denoted by Aut(FRX). Proof : Proof : consider the far ring FRX=(Z N,+,๏). consider the far ring FRX=(Z N,+,๏). To prove that (Aut(FRX), o ) is a group where o is the composition of maps.

66. From theorem 3.9, we get that o is closed binary operation on the set of maps Aut(FRX).It is clear that o is associative If we take ø=Ψ = I Z N the identity over Z N, then it is clear that (I Z N, I Z N ) is an automorphism of FRX.It is easy to show that,for any (Ψ,ø) in aut(FRX), the composition of (I Z N, I Z N ) and (Ψ,ø) is the automorphism (Ψ,ø); From theorem 3.9, we get that o is closed binary operation on the set of maps Aut(FRX).It is clear that o is associative If we take ø=Ψ = I Z N the identity over Z N, then it is clear that (I Z N, I Z N ) is an automorphism of FRX.It is easy to show that,for any (Ψ,ø) in aut(FRX), the composition of (I Z N, I Z N ) and (Ψ,ø) is the automorphism (Ψ,ø);

67. that means there is (I Z N, I Z N ) the identity element of (Aut(FRX), o ).From theorem 3.11 the inverse exists for any element in Aut(FRX). Finally we can say that o is closed binary operation and associative on Aut(FRX);and the identity element exists in Aut(FRX) and each element in Aut(FRX) has an inverse in Aut(FRX)with respect to o. that means there is (I Z N, I Z N ) the identity element of (Aut(FRX), o ).From theorem 3.11 the inverse exists for any element in Aut(FRX). Finally we can say that o is closed binary operation and associative on Aut(FRX);and the identity element exists in Aut(FRX) and each element in Aut(FRX) has an inverse in Aut(FRX)with respect to o.

68. Therefore (Aut(FRX), o ) forms a group and the proof is finished. Therefore (Aut(FRX), o ) forms a group and the proof is finished.

69. Note We denote the number of elements in the set Aut(FRX) by |Aut(FRX)|.And we are going to prove the following theorem We denote the number of elements in the set Aut(FRX) by |Aut(FRX)|.And we are going to prove the following theorem

70. Theorem 3.13: If FRX and FRY are isomorphic far rings, then the number of isomorphisms between them equals both |Aut(FRX)| and by |Aut(FRY)| If FRX and FRY are isomorphic far rings, then the number of isomorphisms between them equals both |Aut(FRX)| and by |Aut(FRY)| Proof: Consider FRX and FRY to be two isomorphic far rings. Then there is an isomorphism (Ψ,ø) denotes by f fom FRX into FRY. Proof: Consider FRX and FRY to be two isomorphic far rings. Then there is an isomorphism (Ψ,ø) denotes by f fom FRX into FRY.

71. If the isomorphism f is the only from FRX into FRY, then the only automorphism of FRX is the identity ; since if there is an automorphism of FRX denoted by (Ψ’,ø’)=k which is different from the identity, then by the above theorems If the isomorphism f is the only from FRX into FRY, then the only automorphism of FRX is the identity ; since if there is an automorphism of FRX denoted by (Ψ’,ø’)=k which is different from the identity, then by the above theorems f o k -1 is an isomorphism from FRX into FRY f o k -1 is an isomorphism from FRX into FRY

72. where k -1 is the inverse of k. where k -1 is the inverse of k. Then f o k -1 must be f. Then f o k -1 must be f. f f FRX → FRY FRX → FRY k ↓↑ k -1 k ↓↑ k -1 FRX FRX

73. Therefore k must be the identity. Therefore k must be the identity. Thus we get that the identity is the only automorphism of FRX. Similarly if f is the only isomorphism from FRX into FRY, then the only automorphism of FRY is the identity. Thus we get that the identity is the only automorphism of FRX. Similarly if f is the only isomorphism from FRX into FRY, then the only automorphism of FRY is the identity.

74. Now if there exists more than one isomorphism from FRX into FRY. Let f be one isomorphism from FRX into FRY and the other is g=(Ψ″,ø″) which is different fromf. Let the set of isomorphisms from FRX into FRY be A. Define a map Ө from A into Aut(FRX) by Ө(g)=g -1 f for any g in A where g -1 is the inverse map of g. Now if there exists more than one isomorphism from FRX into FRY. Let f be one isomorphism from FRX into FRY and the other is g=(Ψ″,ø″) which is different fromf. Let the set of isomorphisms from FRX into FRY be A. Define a map Ө from A into Aut(FRX) by Ө(g)=g -1 f for any g in A where g -1 is the inverse map of g. It is clear that Ө is well defined map. It is clear that Ө is well defined map.

75. 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 1-1 First to prove that Ө is 1-1 Let Ө(g 1 )=Ө(g 2 ), for any g 1,g 2 in A Let Ө(g 1 )=Ө(g 2 ), for any g 1,g 2 in A Then 1 g -1 o f= 2 g -1 o f.this yields to 1 g -1 o f o f -1 = 2 g -1 ๏f๏f -1 Therefore 1 g -1 = 2 g -1 and then we get that 1 g = 2 g This proves that Ө is 1-1.Now to prove that Ө is onto. Therefore 1 g -1 = 2 g -1 and then we get that 1 g = 2 g This proves that Ө is 1-1.Now to prove that Ө is onto.

76. Let h be in Aut(FRX),and h is not the identity over FRX.Then there is g in A,g≠f such that g -1 =hf -1 and Let h be in Aut(FRX),and h is not the identity over FRX.Then there is g in A,g≠f such that g -1 =hf -1 and Ө(g)=g -1 f=hf -1 f=h;(if g=f,then g -1 =f -1 and g -1 =f -1 =hf -1, therefore h is the identity ) this proves that Ө is onto.thus Ө is 1-1 and onto well defined mapping. This proves that the number of 1 to 1 Ө(g)=g -1 f=hf -1 f=h;(if g=f,then g -1 =f -1 and g -1 =f -1 =hf -1, therefore h is the identity ) this proves that Ө is onto.thus Ө is 1-1 and onto well defined mapping. This proves that the number of 1 to 1

77. Isomorphisms from FRX into FRY equals to the number of automorphisms of FRX.one can similarly prove that the number of isomorphisms from FRX into FRY equals to the number of automorphisms of FRY by defining the map Ө from A into Aut(FRY) by Ө(g)=fg -1 for any g in A Thus the theorem is proved. Isomorphisms from FRX into FRY equals to the number of automorphisms of FRX.one can similarly prove that the number of isomorphisms from FRX into FRY equals to the number of automorphisms of FRY by defining the map Ө from A into Aut(FRY) by Ө(g)=fg -1 for any g in A Thus the theorem is proved.

78. Section three Unsolved problems there are several unsolved problems concerning our topics: Unsolved problems there are several unsolved problems concerning our topics: The first problem is to find a test for isomorphic far rings based on inspecting Latin squares. The first problem is to find a test for isomorphic far rings based on inspecting Latin squares. To give something about this for N=1,2,3,4. To give something about this for N=1,2,3,4.

79. First : Let N=1, then the only Latin aquare is 0 which gives only one far ring of order 1 if N=2, then also there is only one Latin square 1 0, which gives only one far ring of order 2 0 1 If N=3, then also there is only one First : Let N=1, then the only Latin aquare is 0 which gives only one far ring of order 1 if N=2, then also there is only one Latin square 1 0, which gives only one far ring of order 2 0 1 If N=3, then also there is only one Latin Square Latin Square

80. 1 2 0 1 2 0 2 0 1 2 0 1 0 1 2 0 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 only one far ring which is isomorphic to itself by the identify isomorphism.If N=4, in this case we have four Latin squares A,B,C, and D as show before in In case N=1 or N=2 or N=3, we have only one far ring which is isomorphic to itself by the identify isomorphism.If N=4, in this case we have four Latin squares A,B,C, and D as show before in

81. Examples: 2.17 and therefore have four far ring FRA, FRB,FRC, and FRD which are of order 4.FRA and FRB are isomorphic by the isomorphism (Ψ,ø) which is defined by Ψ=(1)(234) and ø=(0)(2)(13).But FRC and FRD are not isomorphic ; since if they were, then they have the same period profiles by theorem 3.5 and therefore have four far ring FRA, FRB,FRC, and FRD which are of order 4.FRA and FRB are isomorphic by the isomorphism (Ψ,ø) which is defined by Ψ=(1)(234) and ø=(0)(2)(13).But FRC and FRD are not isomorphic ; since if they were, then they have the same period profiles by theorem 3.5

82. But they have different period profiles (see tables 3.3 and similarly, we obtain that FRA,FRC are not isomorphic,FRA,FRD are also not isomorphic and FRB,FRC are not isomorphic But they have different period profiles (see tables 3.3 and similarly, we obtain that FRA,FRC are not isomorphic,FRA,FRD are also not isomorphic and FRB,FRC are not isomorphic A second problem is to reduce to minimum the number of 4-tuples we must check to verify A second problem is to reduce to minimum the number of 4-tuples we must check to verify

83. Ø(α๏a+β๏b)=Ψ(α) * ø(a)+Ψ(β) * ø(b)………….1 Ø(α๏a+β๏b)=Ψ(α) * ø(a)+Ψ(β) * ø(b)………….1 Where α,β,a,b are in Z N and (Z N,+,๏), (Z N,+, * ) are two far rings. Where α,β,a,b are in Z N and (Z N,+,๏), (Z N,+, * ) are two far rings.

84. We have two far rings of ordered N.(Z N,+, ๏) and (Z N,+, * ). And we have each of Ψ and ø is a permutation of Z N. To prove one, we need to prove that (Ψ,ø) is an isomorphism from (Z N,+,๏) into (Z N,+, * ). In general the number f 4-tuples we must check to verify 1 is N 2.N 2 =N 4. We have two far rings of ordered N.(Z N,+, ๏) and (Z N,+, * ). And we have each of Ψ and ø is a permutation of Z N. To prove one, we need to prove that (Ψ,ø) is an isomorphism from (Z N,+,๏) into (Z N,+, * ). In general the number f 4-tuples we must check to verify 1 is N 2.N 2 =N 4.

85. But if the multiplication of far ring is commutative as for example in case N=1,2,3,4, then the number of 4-tuples that we must check will reduce the number N 2 will reduce toN+(N 2 -N)/2 or less than this number. To show in an explicit manner for cases N≤4. But if the multiplication of far ring is commutative as for example in case N=1,2,3,4, then the number of 4-tuples that we must check will reduce the number N 2 will reduce toN+(N 2 -N)/2 or less than this number. To show in an explicit manner for cases N≤4.

86. In case N=1, it is trivial case we need only one check which is happened by taking α=β=a=b=0 In case N=1, it is trivial case we need only one check which is happened by taking α=β=a=b=0 αβ ab αβ ab 00 00 00 00

87. In case N=2, we have a far ring of order two with elements of Z 2 ={1,0} In case N=2, we have a far ring of order two with elements of Z 2 ={1,0} The following table gives the required cases that we need to prove 1 The following table gives the required cases that we need to prove 1

88. αβ ab ab ab ab αβ ab ab ab ab 11 11 10 01 00 11 11 10 01 00 10 10 00 10 10 00 00 00 00 00 N 2.N 2 =2 2.2 2 =2 4 =16 N 2.N 2 =2 2.2 2 =2 4 =16 16 reduces to 7 16 reduces to 7

89. in case N=3, we have a far ring of order 3 wih elements of Z 3 ={1,2,0} the following table gives the cases that we need to prove 1 in case N=3, we have a far ring of order 3 wih elements of Z 3 ={1,2,0} the following table gives the cases that we need to prove 1

90. αβ ab ab ab ab ab ab αβ ab ab ab ab ab ab 11 11 12 13 22 20 00 11 11 12 13 22 20 00 12 12 10 22 20 02 00 12 12 10 22 20 02 00 10 10 20 01 00 10 10 20 01 00 22 22 20 00 22 22 20 00 20 23 00 20 23 00 00 00 00 00

91. The number N 4 =3 4 =81 reduces to 22 The number N 4 =3 4 =81 reduces to 22 In case N=4, we have a far ring of order 4 with elements of Z 4 ={1,2,3,0} In case N=4, we have a far ring of order 4 with elements of Z 4 ={1,2,3,0} The following tale gives the cases that we need to prove 1 The following tale gives the cases that we need to prove 1

92. αβ ab ab ab ab ab ab ab ab ab ab ab αβ ab ab ab ab ab ab ab ab ab ab ab 11 11 12 13 10 22 23 20 33 30 00 11 11 12 13 10 22 23 20 33 30 00 12 12 13 10 22 23 20 31 32 33 30 00 12 12 13 10 22 23 20 31 32 33 30 00 13 13 10 23 20 33 20 03 00 13 13 10 23 20 33 20 03 00 10 10 20 33 30 00 10 10 20 33 30 00 22 22 23 20 33 30 03 00 22 22 23 20 33 30 03 00 23 23 20 31 33 30 00 23 23 20 31 33 30 00 20 20 30 00 20 20 30 00 33 33 30 00 33 33 30 00 30 30 00 30 30 00 00 00 00 00 The number N 2.N 2 =4 4 =16x16=256reduce to 55 The number N 2.N 2 =4 4 =16x16=256reduce to 55