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Finding Factors of Factor Rings Over Eisenstein Integers

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Finding Factors of Factor Rings Over Eisenstein Integers International Mathematical Forum, Vol. 9, 2014, no. 31, 1521 - 1537 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/imf.2014.111121 Finding Factors of Factor Rings over Eisenstein Integers Valmir Bu¸caj Department of Mathematics, Texas Lutheran University 1000 West Court Street, Seguin, TX 78155 Dedicated to my parents, Afrim and Drita Bu¸caj. Copyright c 2014 Valmir Bu¸caj. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this paper we prove a few results related to the factor rings over the Eisenstein integers. In particular we show that the ring Z[!] fac- tored by an ideal generated by any element m + n! of this ring, where g:c:d(m; n) = 1 is isomorphic to the ring Zm2+n2−mn. Next, we find the representatives for the equivalence classes of the ring Z[!] when factor- ing out by a power of a prime ideal, and also identify all the units of these quotient rings respectively. Then, we give a representation for the factor ring Z[!]= hm + n!i in terms of the product of other rings. Fi- nally, we end this paper with a few applications to elementary number theory. Mathematics Subject Classification: Primary 13F15, 13F07, 13B02; Secondary 11D09, 11A51 Keywords: Eisenstein Integers; Eisenstein Primes; Factor Rings; Prime Ideals; Norm; Equivalence Classes; Units 1522 Valmir Bu¸caj 1 Introduction Eisenstein integersp are defined to be the set Z[!] = fa + b! : a; b 2 Zg where ! = (−1 + i 3)=2: This set lies inside the set of complex numbers C and they form a commutative ring in the algebraic number field Q(!): We can define a ring norm N : Z[!] ! Z+ [ f0g by N(m + n!) = (m + n!)(m + n!¯) = m2 + n2 − mn; and it is easy to check that N is multiplicative; that is, N(α · β) = N(α) · N(β), for all α; β 2 Z[!]: p Proposition 1.1. For ! = (−1+i 3)=2; the ring Z[!] is a Euclidean Domain. Proof. Let α = a+b!; β = c+d! be two elements of the ring Z[!] with β 6= 0: Then in the algebraic number field Q(!) we have α ab + cd − ad bc − ad = t + s!; where t = ; and s = : β c2 + d2 − cd c2 + d2 − cd 1 1 Let p; q be two integers such that jt − pj ≤ 2 and js − qj ≤ 2 : Let φ = (t − p) + (s − q)!, and let r = βφ. Then we get r =βφ =β[(t − p) + (s − q)!] =β [(t + s!) − (p + q!)] α =β − (p + q!) β =α − β(p + q!): We notice that r 2 Z[!]: Finally, we have α = β(p + q!) + r: (1) To conclude the proof, we need to show that N(r) < N(β): So, N(r) = N(βφ) = N(β)N(φ) = N(β) (t − p)2 + (s − q)2 − (t − p)(s − q) 1 1 1 ≤ N(β) + + 4 4 4 3 = N(β); 4 Finding factors of factor rings over Eisenstein integers 1523 which is even stronger than necessary. Since Z[!] is a Euclidean Domain, it follows that it is also a Principal Ideal Domain and a Unique Factorization Domain. p Fact 1.2. If ! = (−1 + i 3)=2 then n X i Z[!] = f ai! : ai 2 Zg = fa + b! : a; b 2 Zg: i=0 Proof. We start by observing that ! is a root of the second degree polynomial x2 + x + 1. So, !2 = −! − 1: (2) Therefore, from (2) it is easily seen that Z[!] = fa + b! : a; b 2 Zg: We know that (See [5]) an element α = a + b! is a unit in Z[!] if and only if N(α) = ±1, where in our case N(a + b!) = (a + b!)(a + b!¯) = a2 − ab + b2: Using this fact it readily follows that the only units in Z[!] are {±1; ±!; ±!¯g: Fact 1.3. Let α = m + n! 2 Z[!]: If N(α) = p, where p is a prime in Z; then α is irreducible in Z[!]: Proof. Let α = γ · β where γ; β 2 Z[!]: Then N(α) = m2 + n2 − mn = N(β) · N(γ) = p: Since N(β);N(γ) 2 Z, it follows that N(β) = ±1 and N(γ) = p, or vice-versa. Thus, β is a unit in Z[!]; which concludes the proof. Fact 1.4. Up to associates and an integer multiple, α = 1 − ! is the only element of Z[!] that is associate with its conjugate. Proof. Let α = a+b! and u 2 {±1; ±!; ±!¯g: We want to know when α = u·α;¯ whereα ¯ = a + b!:¯ The proof is straightforward, one merely needs to check the truth of the equation for all such u: 2 Preleminary Results Theorem 2.1. If q is a positive integer larger than 1, then ∼ Z[!]= hqi = Zq[!]: 1524 Valmir Bu¸caj Proof. Define φ : Z[!] ! Zq[!]; by φ(m + n!) = [m]q + [n]q!: First we show that φ is a surjective ring homomorphism. The mapping is clearly surjective, so we need to show only that it is a homomorphism. Let α = a + b!; and β = c + d!: Then φ(α + β) =[a + c]q + [b + d]q! =([a]q + [b]q!) + ([c]q + [d]q!) =φ(α) + φ(β): Next, after a few calculations and using tha fact that !2 = −1 − !; we find that α · β = (ac − bd) + (ad + bc − bd)!: (3) So, φ(α · β) = [ac − bd]q + [ad + bc − bd]q!: In a simmilar fashion we find that φ(α) · φ(β) = [ac − bd]q + [ad + bc − bd]q!; concluding that φ is a ring homomorphism. Now we show that ker(φ) = hqi : Since φ(q) = [q]q = [0]q = 0; it follows that q 2 ker(φ): Since ker(φ) is an ideal we have hqi ⊂ ker(φ): To show the reverse inclusion let α = m + n! 2 ker(φ): Then, φ(α) = φ(m + n!) = [m]q + [n]q! = 0; so both m ≡ 0 (mod q) and n ≡ 0 (mod q): So, there are integers m0; n0 such that m = m0q and n = n0q: Then, m + n! = q(m0 + n0!) which shows that m + nw 2 hqi ; and thus ker(φ) ⊂ hqi : Finally, since ker(φ) = hqi ; by the First Isomorphism Theorem, we have ∼ Z[!]= hqi = Zq[!]; which is what we wanted to show. p Lemma 2.2. Let ! = (−1 + i 3)=2. The following conditions are equivalent for a prime p of Z : (1) p is an irreducible element in Z[!]; 2 (2) x + x + 1 is an irreducible polynomial in Zp[!]; (3) p ≡ 2 (mod 3): Finding factors of factor rings over Eisenstein integers 1525 For a proof see [1], page 426: Proposition 2.3. If p is a positive integer larger than 1, then Zp[!] is a field if and only if p is a prime such that p ≡ 2 (mod 3). Proof. One can show that the following ring isomorphism holds (See [1]): ∼ 2 Z[!]= hpi = Zp[!]= x + x + 1 : (4) Suppose that Zp[!] is a field. Then by Theorem 2:1, Zp[!] is isomorphic to 2 Z[!]= hpi ; thus Z[!]= hpi is a field as well. Then, by (4) Zp[!]= hx + x + 1i is a field, and we know that this happens if and onl if x2 + x + 1 is an irreducible polynomial in Zp[x]: Then the result follows by Lemma 2:2. Lemma 2.4. Let m and n be two relatively prime integers. Then m and m2 + n2 − mn are relatively prime as well. Proof. First we show that m and m2 + n2 are relatively prime. Let (m; m2 + n2) = d, so m = kd and m2 + n2 = hd; for some integers h and k; also since (m; n) = 1 there are some integers x; y such that mx + ny = 1: Then n = n(mx + ny) = mnx + n2y = mnx + (hd − m2)y = mnx + hdy − m2y = m(nx − my) + d(hy) = d(knx − mky) + d(hy) = d[knx − mky + hy]: So, d divides n. Thus, since djm and djn it follows that dj1, thus d = 1: Now let (m; m2 + n2 − mn) = r; so there are some integers h; k such that m = kr and m2 + n2 − mn = hr: Then m2 + n2 = (m2 + n2)(mx + (m2 + n2)y) = (m2 + n2)mx + (m2 + n2)2y = r[kx(m2 + n2)] + (hr + mn)2y = r[kx(m2 + n2)] + (hr + krn)2y = r[kx(m2 + n2) + r(h + kn)2y]: So, since rj(m2 + n2), rjm and the fact that mx + (m2 + n2)y = 1, for some integers x; y, it follows that rj1, which is possible only if r = 1: 1526 Valmir Bu¸caj Theorem 2.5. Let m and n be two integers that are relatively prime, then ∼ Z[!]= hm + n!i = Zm2+n2−mn: Proof. First, we wish to note that it suffices to prove the theorem for the case where both m and n are positive integers. From Lemma 2:4 we know that m and m2 + n2 − mn are relatively prime, so the equivalence class of −1 m has an inverse in Zm2+n2−mn, that is ([m]m2+n2−mn) exists. Using this fact and the fact that m2 + n2 − mn ≡ 0 (mod m2 + n2 − mn), we find that (mn−1)2 ≡ mn−1 − 1 (mod m2 + n2 − mn): Now, let us define a mapping φ : Z[!] ! Zm2+n2−mn, by φ(x + yw) = x − (mn−1)y modulo m2 + n2 − mn: It is not difficult to see that φ is surjective and preserves addition.
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