Factor Rings and their decompositions in the Eisenstein integers Ring Z [ω]

Armenian Journal of Mathematics Volume 5, Number 1, 013, 58 68 Factor Rings and their decomositions in the Eisenstein integers Ring Z [ω] Manouchehr Misaghian Deartment of Mathematics, Prairie View A&M University Prairie View, TX 77446-USA mamisaghian@vamu.edu Abstract In this aer we will characterize the structure of factor rings for Z [ω] where ω 1+ 3,is a 3rd rimitive root of unity. Consequently, we can recognize rime numbers elements and their ramifications in Z [ω]. Key Words: Euclidean Domain, Unique Factorization Domain, Factor ring, Eisenstein integers Mathematics Subject Classification 000: Primary 13F15, 13F07; Secondary 13F10 Introduction The set of integers, Z {..., n,..., 1, 0, 1,..., n,...}, is the most imortant and simlest Integral Domain. This ring is Euclidean and thus a Unique Factorization Domain UFD. It is also a Princial Ideal Domain PID, thus all ideals of this ring are rincial and are given by: m {km k Z}, for all m Z So the factor rings of Z, are given by Z mz Z m, the ring of integers {0, 1,,, m 1} modulo m. In an attemt to formulate and rove the Recirocity Theorem, one of the most imortant and beautiful Theorems in Number theory arose, Carl Friedrich Gauss realized that he needed to look beyond the set of integers. For this reason Gauss introduced Comlex Integer Numbers [4]. These numbers are now known as Gaussian integers. The Gaussian integers 58

Factor Rings and their decomositions in the Eisenstein integers Ring Z [ω] 59 are sitting in the field of comlex numbers, and by inherited addition and multilication oerations from the field of comlex numbers constitute an integral domain which is a UFD [], [6]. In general we can consider some imaginary extensions of the ring of integers as follows. Let Z, > 1 be a rime number, and let ξ be a rimitive root of the equation x + a 0, where a is an integer; i.e. a number for which ξ a but ξ q a for all q, 0 < q <. Then Z [ξ] { a 0 + a 1 ξ + + a 1 ξ 1 a i Z, 0 i 1 } with aroriate oerations is an extension of Z. In 1847 Gabriel Lame announced that he had a comlete roof of the Last Fermat s Theorem. In his roof, he used the identity x + y x + y x + ξy x + ξ 1 y where and ξ are as above, with the assumtion that all extensions of Z are UFD. Before the Lame s work, Ernst Kummer had already roven that some of these extensions are not UFD. For examle in Z [ 5 ] we have: 6 3 1 + 5 1 5 In connection with this observation, Kummer defined his Ideal Numbers. This led directly to Richard Dedekind s develoment of Algebraic Number Theory in 1870s. Dedekind introduced a form of unique factorization using ideals instead of numbers. Let R be an integral domain for which there exists a subset P such that every non zero element x of R can be written, in a unique way as x ε P vx Where ε is a unit element in R and v x are non-negative integers, all but a finitely many are zero. In the other words the set R P of rincial ideals that coincide with the set of maximal rincial ideals distinct from R, is uniquely determined. A unique factorization domain has also a very simle geometric interretation. In geometry a ring R occurs as a ring of functions defined on some variety V. If n denotes the dimension of V, then R is a UFD means that every subvariety of dimension n 1 can be defined by a single equation. One of the fundamental differences between Z and its extensions is the structure of their Factor Rings. As we know the factor rings of Z, are Z mz Z m, which are isomorhic to the ring {0, 1,,, m 1} modulo m. Even when m is a comosite number like m α 1 1 α αr r 59

60 M. Misaghian we can use The Chinese reminder theorem to factor Z m as Z m Z α 1 1 Z α Z αr. r The structure of factor rings for Z [ξ] is much more comlicated. This structure has been studied for Gaussian integers [3]. In this aer we will characterize the structure of factor rings for Z [ω] where ω 1+ 3 is a rimitive 3rd root of unity. Consequently, we can recognize rime numbers elements and their ramifications in Z [ω]. 1 Notation and Prerequisites Let Z be the ring of integers and let ω 1+ 3 be a rimitive 3rd root of unity; i.e. ω 3 1, and ω + ω + 1 0. Set Z [ω] {a + bω a, b Z} Then Z[ω] with the following oerations is an integral domain a + bω + c + dω a + c + b + d ω a + bω c + dω ac bd + ad + bc bd ω for a + bω, c + dω Z[ω]. For each z a + bω Z[ω], its norm is defined by Then by basic ring theory we have: ν z a + bω a + bω a + b ab Theorem 1 Z[ω] with above norm is an Euclidean Domain and so a Unique Factorization DomainUFD. Definition 1 An element z Z[ω] is called a unit if it has a multilicative inverse in Z[ω]. Lemma 1 The only unit elements in Z[ω] are {±1, ±ω, ±ω }. Proof. A non zero element z a + bω Z[ω] is unit if and only if ν z a + b ab 1. If ab 0, then z ±1 or z ±ω. If ab 0 then a + b and this with a + b ab 1 gives ab 1. on the other hand from a + b ab 1 we have a + b ab 1 ab so ab 1, hence we must have ab 1 that gives us a b ±1, i.e. z ± 1 + ω ω. Definition Legendre s Symbols. Let Z be a rime number. Then the Legendre s symbol, denoted by, for each integer a Z, is defined by a. 1 if there is an integer x Z such that x a mod, 1 if there is no integer x Z such that x a mod, 0 if divides a 60

Factor Rings and their decomositions in the Eisenstein integers Ring Z [ω] 61 The following two theorems can be found in any standard Number Theory book, for examle [], [6] and [5]. Theorem Let Z be a rime number. The Legendre s symbol has the following roerty: i. ab a b ii. a b mod, for all integers a, b Z. a b. Theorem 3 Quadratic Recirocity Theorem. Let Z be a rime number. Then: i. ii. { 1 { 3 1 if 1mod 4, 1 if 3mod 4 1 if 1 or 11mod 1, 1 if 5 or 7mod 1 The following lemmas can be roved in a strait manner. Lemma Let Z be a rime integer. Then 1 mod 3 if and only if 1 mod 6. Lemma 3 Let Z be a rime integer. If 1or 7 mod 1 then 1 mod 6. 3 Theorem 4 Let Z be a rime integer. Then 1 if and only if 1 mod 6. 3 1 3 3 Proof. By art i of the Theorem we have.if 1 then we must have 1or 1. In the first case we have 1mod 4 and 1 3 1 3 1 or 11mod 1. These conditions lead to 1mod 1, so by the Lemma 3 we get 1 mod 6. In the second case we have 3mod 4 and 5 or 7mod 1. These conditions lead to 7mod 1, so by the Lemma 3 we get 1 mod 6. Now suose we have 1 mod 6. So 1 + 6k for some integer k Z. If k is even then 1mod 1, 1 3 so by the Theorem 3 we have 1. If k is odd then 7mod 1, so by the Theorem 3 we have 1. 1 3 Lemma 4 Let Z be a rime integer. Then α + 3β for some integers α, β Z if and only if 3 or 1 mod 6. Proof. Suose α + 3β for some integer α, β Z. If α 0, then we must have 3, because is a rime. If α 0, then α 1 mod 3 or α mod 3. In either case from α + 3β we get 1 mod 3 so by the Lemma we have 1 mod 6. Conversely if 3 or 1 mod 6, then for 3 we have 3 0 + 3 1. For 1 mod 6 by the 61

6 M. Misaghian Theorem 4 there is an integer α such that α + 3 t for some integer t Z, 0 < t <. If t 1 then α + 31. If t 1, then we can choose an integer β Z such that, β α mod t, and - t < β < t. From here and reflexive roerty of congruence relation we get α β mod t and 3 3 mod t, thus α + 3 β + 3 0 mod t. So for some integer λ Z, 1 λ < t, we have β + 3 λt. Since we also have α + 3 t, by multilying these equalities side by side we get α + 3 β + 3 λt. On the other hand we have α + 3 β + 3 αβ + 3 + 3 α β. Thus we have αβ + 3 + 3 α β λt. One can easily show that both αβ + 3 and α β are divisible by t so we have αβ+3 t + 3 α β t λ. This equation shows that a smaller multile of can be written as α + 3β. If λ 1, we are done, if not, we can reeat the above rocedure. After a finite number of stes of reetition we get the result. Theorem 5 Let Z be a rime integer. Then a + b ab for some integers a, b Z if and only if 3 or 1 mod 6. Proof. Let a + b ab for some integer a, b Z, then a and b are relatively rime. so they are not both even. Suose one of them, say a is even, then we have a + b ab a a b + 3 and by the Lemma 4, 3 or 1 mod 6. If both a and b are odd then b a t for some t Z. From here we have b a + t and a + b ab a + b b a a + t a + t a + t + 3t and again by the Lemma 4 we have 3 or 1 mod 6. Conversely, if 3 or 1 mod 6, then for 3 we have 3 + 1. If 1 mod 6, then by the Lemma 4 There are integers α, β Z such that α + 3β. Now set a α + β and b β. Then we have α + 3β a b b + 3 a + b ab 6

Factor Rings and their decomositions in the Eisenstein integers Ring Z [ω] 63 Lemma 5 Let m Z be an integer. Then Z [ω] / m Z m [ω] Proof. For a Z, let [a] m denotes the equivalence class modulo m in Z. Now define f : Z [ω] Z m [ω], by f a + bω [a] m + [b] m ω. One can show that this is a surjective ring homomorhism with the Kernel equal m. Lemma 6 The olynomial q x x + x + 1 is irreducible in Z iff and only if or 5 mod 6. Proof. Let q x x +x+1 be irreducible in Z. If and 5 mod 6, then 3 or 1 mod 6. If 3 then we have q x x + x + 1 x + in Z 3, this is contrary to our assumtion. If 1 mod 6, then by the Theorem 5, a +b ab for some relatively rime integers a, b Z. Since a 1 and b 1 exist in Z and we have a + b ab mod, so we have ab 1 + ba 1 1 mod, Now we have q x x + x + 1 x + ab 1 x + ba 1 in Z and again this is contrary to our assumtion. Conversely, if, then q x x + x + 1 doesn t have any root in Z so it is irreducible. Suose 5 mod 6 and q x x + x + 1 is not irreducible in Z, then there is a Z such that q a a + a + 1 0 in Z, so by the Theorem 5 we must have 3 or 1 mod 6 which is contrary to our assumtion. Theorem 6 Let be a rime integer. Then Z [ω] is a field if and only if or 5 mod 6. Proof. If Z [ω] is a field and and 5 mod 6, then either 3 or 1 mod 6. If 3, then we have + ω 0 in Z 3 [ω]. If 1 mod 6, then by the Lemma 6 there is a Z such that a + a + 1 0 in Z, so we have ω a ω a 0 in Z [ω]. Thus in either case Z [ω] is not a field that is contrary to our assumtion. Now suose or 5 mod 6. For, we have Z [ω] {0, 1, ω, ω } which is a field. For 5 mod 6 consider the following ring homomorhism ϕ : Z [x] Z [ω], ϕ x ω, ϕ m m, for all m Z Obviously it is a surjective homomorhism. Since ϕ x + x + 1 ω + ω + 1 0, so x + x + 1 Ker ϕ. Let a olynomial h x Ker ϕ, then since all coefficients of x + x + 1 are 1 we have h x x + x + 1 g x + ax + b, for some g x Z [x] and a, b Z. From here we get ϕ h x 0, so a b 0 in Z i.e. x + x + 1 Ker ϕ and by the canonical ring Isomorhism Theorem we have Z [x] / x + x + 1 Z [ω]. Since 5 mod 6 so by the Lemma 6, x + x + 1 is irreducible and thus Z [x] / x + x + 1 is a field. 63

64 M. Misaghian Definition 3 An element π Z [ω] is a called a rime element if whenever we have π αβ, for α, β Z [ω], then either π α or π β. Corollary 1 A rime integer Z, is a rime in Z [ω] if and only if or 5 mod 6. Proof. This is a result of the Theorem 6 because Z [ω] is an Euclidean Domain. Theorem 7 If a and b are relatively rime integers, then Z [ω] / a + bω Z a +b ab. Proof. For simlicity, for x Z, we will denote by[x], the equivalence class[x] a +b ab. Since a and b are relatively rime, so [b] 1 exists in Z a +b ab. Now define the following maing: f : Z [ω] Z a +b ab f x + yω [x] [a] [b] 1 [y]. This is a ring homomorhism. To this ends, first note that a +b ab 0 mod a + b ab so from here we get [a] [b] [a] [b] 1 1. Now for every x+yω, and u+vω Z [ω] we have x + yω u + vω xu vy+xv + yu vy ω and [x] [a] [b] 1 [y] [u] [a] [b] 1 [v] [xu vy] [a] [b] 1 [xv + yu vy], so from here we have: Also we have f x + yω u + vω f xu vy + xv + yu vy ω [xu vy] [a] [b] 1 [xv + yu vy] [x] [a] [b] 1 [y] [u] [a] [b] 1 [v] f x + yω f u + vω. f x + yω + u + vω f x + u + y + v ω [x + u] [a] [b] 1 [y + v] [x] [a] [b] 1 [y] + [u] [a] [b] 1 [v] f x + yω + f u + vω Since f a + bω [a] [a] [b] 1 [b] [0], we have a + bω Ker f. Now let x + yω Ker f. In Q [ω] we can write x + yω a + bω ax + by bx ay bx + ω a + b ab a + b ab Since f x + yω [x] [a] [b] 1 [y] [0] we have bx ay λ a + b ab for some λ Z. On the other hand from bx ay 0 mod a + b ab we get ab x a by 0 mod a + b ab which is equivalent to ax a b by 0 mod a + b ab Since a b ab 1 1 mod a + b ab we obtain ax ay+by ax bx+by 0 mod a + b ab, so we have ax bx + by µ a + b ab, for some µ Z. Thus we have x + yω a + bω µ + λω. This shows that Ker f a + bω. So Ker f a + bω. Since f is surjective by standard ring Isomorhism Theorem we have Z [ω] / a + bω Z a +b ab. 64

Factor Rings and their decomositions in the Eisenstein integers Ring Z [ω] 65 Corollary If a and b are relatively rime integers, then z a + bω is rime in Z [ω] if and only if a + b ab is rime in Z. Theorem 8 U to multilication by a unit, the rime elements in Z [ω] are as follows: 1. Every rime integer, that is or 5 mod 6.. Every element ς a + bω such that a + b ab is an integer rime in Z such that 1 mod 6. 3. Element z 1 + ω. Proof. This is an immediate consequence of the Theorems 6, 7 and the Corollaries 1 and.. Remark 1 The rime, z 1 + ω has Ramification roerty; the integer rime 3 is not a rime in Z [ω] and we have 3 1 + ω. 3 is called a ramified rime in Z [ω]. Remark Every element z x + yω in Z [ω] can be written uniquely u to order and multilication by a unit as a roduct of rimes as follows: m x + yω ε α ς βς γ i i 1 + ω n ς where ε is a unit element in Z [ω], α, β ς, γ i, m and n are non-negative integers, ς a+bω are elements in Z [ω] for which a +b ab is an integer rime in Z with 1 mod 6, and i are integer rimes bigger than 3. Since for each rime number such that 1 mod 6 there are only two distinct rime elements, ς 1 a + bω and ς b + aω such that a + b ab, a and b are relatively rime integers, so we can rewrite this factorization as m x + yω ε α ς βς 1 1 ς βς 1 + ω n ς 1 ς i1 i1 γ i i. Factor rings in Z [ω]and their Decomositions Theorem 9 Let k 1 be an integer. Then for n k + 1 we have Z [ω] / 1 + ω n Z [x] / 3 k x, 3 k+1, x + x + 1 Proof. First note that 1 + ω n 1 + ω k+1 1 + ω k 1 + ω 3 k 1 + ω 65

66 M. Misaghian Now define f : Z [x] Z [ω] / 1 + ω n by f x ω 1 mod 1 + ω n This is an onto ring homomorhism. Now note that f 3 k x 3 k ω 1 1 k 1 + ω 1 + ω n 1 + ω n So 3 k+1 x Ker f. Also we have f 3 k+1 3 k+1 1 k+1 1 + ω 1 + ω n 1 + ω n So 3 k+1 Ker f. Obviously we have x + x + 1 Ker f because ω + ω + 1 0, thus 3 k+1 x, 3 k+1, x + x + 1 Ker f. Conversely, let q x Ker f. Then since all coefficients of x +x+1 are 1, we have q x x x + x + 1+ax+b, for some x Z [x], and a, b Z. From here we get f q x a ω 1 + b 0 mod 1 + ω n, so we must have a ω 1 + b c + dω 1 + ω n 3 k c + dω 1 + ω, for some c + dω Z [ω]. Thus we have aω + b a 3 k c d ω + 3 k c d From here we must have { a 3 k c d, b a 3 k c d These give us b 3 k 3c 3d 1 k c d 3 k+1 and a 1 k c d 3 k, so q x x x + x + 1 + 1 k c d 3 k x + 1 k c d 3 k+1 3 k x, 3 k+1, x + x + 1. Now by Fundamental Isomorhism Theorem we have Z [ω] / 1 + ω n Z [x] / 3 k x, 3 k+1, x + x + 1. Lemma 7 Let k 1 be an integer. Then for n k we have Z [ω] / 1 + ω n Z 3 k [ω]. Proof. We have 1 + ω n 1 + ω k 3 k. So 1 + ω n 3 k. Now aly the Lemma 5. Corollary 3 Let R n Z [ω] / 1 + ω n. Then we have Z 3, if n 3 R n Z 3 k [ω], if n k, k 1 Z [x] / 3 k x, 3 k+1, x + x + 1, if n k + 1, k 1 Proof. This follows from the Theorems 7 and 9 and the Lemma7. 66

Factor Rings and their decomositions in the Eisenstein integers Ring Z [ω] 67 Theorem 10 For each element z x + yω in Z [ω] we have Z [ω] / x + yω Z α [ω] Z c +d cd βς 1 Z c +d cd βς Z γ 1 1 where c + dω ς 1 ς βς 1 1, d + cω ς ς βς and R n Z [ω] / 1 + ω n. [ω] Z γm m [ω] R n Proof. First note that, from the Remark, for m x + yω ε α ς βς 1 1 ς βς 1 + ω n ς 1 we have x + yω ς α ς i1 ς βς 1 1 γ i i ς ς βς m Now by standard facts in an Euclidean Domain we have Z [ω] / x + yω Z [ω] / α Z [ω] / Z [ω] / Z [ω] / 1 + ω n. ς 1 ς βς1 1 Now aly the Lemma 5 and the Theorem 7 and this fact that Z [ω] / m i1 γ i i i1 γ i i 1 + ω n ς ς βς Z [ω] / γ 1 1 Z [ω] / γm m Examle 1 Since + 6ω + 3ω 1 + ω we have Z [ω] / m i1 γ i i Examle For z 49 1 + 3ω we have Z [ω] / + 6ω Z [ω] Z 49 Z 3 Z [ω] / 49 1 + 3ω Z 7 [ω] Z 7 Z 49 because we can rewrite z 49 1 + 3ω 7 + 3ω 1 + 3ω which is also isomorhic to Z 49 Z 343 because z 49 1 + 3ω + 3ω 3 + ω 3. Acknowledgements The author would like to thank the anonymous referees for their very careful evaluation of my aer and their meaningful comments. References [1] John A. Beachy and William D. Blair, Abstract Algebra, Third edition, Waveland Press, Inc, 006. [] David M. Burton, Elementary Number Theory, Revised edition, Allyn and Bacon, 1980. 67

68 M. Misaghian [3] Greg Dresden, and Wayne M. Dymacek, Finding Factors of Factor Rings over the Gaussian Integers, the American Mathematical Monthly, Vol. 11, No. 7, August-Setember 005, 60-611. [4] C. F. Gauss, Theoria residuorum biquadraticorum. Rerinted in Werke, George Olms Verlag, Hidelsheim, 1973. [5] Kenneth Ireland and Michael Rosen, A Classical Introduction to Modern Number Theory, second edition, Sringer-Verlag, 1990. [6] Silverman, J. H., A Friendly Introduction to Number Theory, 3rd, Edition, Prentice Hall, 006. [7] E.B. Vinberg, A course in Algebra, American Mathematical Society, 003. 68