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

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 Z[ω] fac- tored by an ideal generated by any element m + nω 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[ω] 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[ω]/ hm + nωi 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 integers√ are defined to be the set Z[ω] = {a + bω : a, b ∈ Z} where ω = (−1 + i 3)/2. This set lies inside the set of complex numbers C and they form a in the field Q(ω). We can define a ring norm N : Z[ω] → Z+ ∪ {0} by N(m + nω) = (m + nω)(m + nω¯) = 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 ∈ Z[ω]. √ Proposition 1.1. For ω = (−1+i 3)/2, the ring Z[ω] is a . Proof. Let α = a+bω, β = c+dω be two elements of the ring Z[ω] with β 6= 0. Then in the algebraic number field Q(ω) we have α ab + cd − ad bc − ad = t + sω, where t = , and s = . β c2 + d2 − cd c2 + d2 − cd 1 1 Let p, q be two integers such that |t − p| ≤ 2 and |s − q| ≤ 2 . Let φ = (t − p) + (s − q)ω, and let r = βφ. Then we get

r =βφ =β[(t − p) + (s − q)ω] =β [(t + sω) − (p + qω)] α  =β − (p + qω) β =α − β(p + qω). We notice that r ∈ Z[ω]. Finally, we have α = β(p + qω) + 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[ω] is a Euclidean Domain, it follows that it is also a and a Unique Domain. √ Fact 1.2. If ω = (−1 + i 3)/2 then

n X i Z[ω] = { aiω : ai ∈ Z} = {a + bω : a, b ∈ Z}. i=0

Proof. We start by observing that ω is a root of the second degree polynomial x2 + x + 1. So, ω2 = −ω − 1. (2)

Therefore, from (2) it is easily seen that Z[ω] = {a + bω : a, b ∈ Z}.

We know that (See [5]) an element α = a + bω is a in Z[ω] if and only if N(α) = ±1, where in our case N(a + bω) = (a + bω)(a + bω¯) = a2 − ab + b2. Using this fact it readily follows that the only units in Z[ω] are {±1, ±ω, ±ω¯}.

Fact 1.3. Let α = m + nω ∈ Z[ω]. If N(α) = p, where p is a prime in Z, then α is irreducible in Z[ω].

Proof. Let α = γ · β where γ, β ∈ Z[ω]. Then N(α) = m2 + n2 − mn = N(β) · N(γ) = p. Since N(β),N(γ) ∈ Z, it follows that N(β) = ±1 and N(γ) = p, or vice-versa. Thus, β is a unit in Z[ω], which concludes the proof.

Fact 1.4. Up to associates and an multiple, α = 1 − ω is the only element of Z[ω] that is associate with its conjugate. Proof. Let α = a+bω and u ∈ {±1, ±ω, ±ω¯}. We want to know when α = u·α,¯ whereα ¯ = a + bω.¯ 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[ω]/ hqi = Zq[ω]. 1524 Valmir Bu¸caj

Proof. Define φ : Z[ω] → Zq[ω], by φ(m + nω) = [m]q + [n]qω. 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 + bω, and β = c + dω. Then

φ(α + β) =[a + c]q + [b + d]qω

=([a]q + [b]qω) + ([c]q + [d]qω) =φ(α) + φ(β).

Next, after a few calculations and using tha fact that ω2 = −1 − ω, we find that

α · β = (ac − bd) + (ad + bc − bd)ω. (3)

So, φ(α · β) = [ac − bd]q + [ad + bc − bd]qω. In a simmilar fashion we find that

φ(α) · φ(β) = [ac − bd]q + [ad + bc − bd]qω, 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 ∈ ker(φ). Since ker(φ) is an ideal we have hqi ⊂ ker(φ). To show the reverse inclusion let α = m + nω ∈ ker(φ). Then, φ(α) = φ(m + nω) = [m]q + [n]qω = 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 + nω = q(m0 + n0ω) which shows that m + nw ∈ hqi , and thus ker(φ) ⊂ hqi . Finally, since ker(φ) = hqi , by the First Isomorphism Theorem, we have ∼ Z[ω]/ hqi = Zq[ω], which is what we wanted to show. √ Lemma 2.2. Let ω = (−1 + i 3)/2. The following conditions are equivalent for a prime p of Z : (1) p is an irreducible element in Z[ω]; 2 (2) x + x + 1 is an irreducible polynomial in Zp[ω]; (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[ω] 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[ω]/ hpi = Zp[ω]/ x + x + 1 . (4)

Suppose that Zp[ω] is a field. Then by Theorem 2.1, Zp[ω] is isomorphic to 2 Z[ω]/ hpi , thus Z[ω]/ hpi is a field as well. Then, by (4) Zp[ω]/ 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 d|m and d|n it follows that d|1, 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 r|(m2 + n2), r|m and the fact that mx + (m2 + n2)y = 1, for some integers x, y, it follows that r|1, 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[ω]/ hm + nωi = 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[ω] → 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. Next we show that it preserves multiplication as well. Let α = a + bw, and β = c + dw be two elements from Z[ω]. Then φ(α) · φ(β) = (a − (mn−1)b)(c − (mn−1)d) = ac − (mn−1)ad − bc(mn−1) + (mn−1)2bd = ac − (mn−1)(ad + bc) + (mn−1 − 1)bd = (ac − bd) − (mn−1)(ad + bc − bd) = φ ((ac − bd) + (ad + bc − bd)w) = φ (a + bw)(c + dw)) = φ(α · β).

So, φ is a surjective ring homomorphism. Our next task is to show that ker(φ) = hm + nωi . Since φ(m + nω) = m − (mn−1)n = m − m = 0, it follows that m + nw ∈ ker(φ), and thus hm + nwi ⊂ ker(φ). To show the reverse inclustion, let c + dw ∈ ker(φ), then c + dw cm + dn − cn dm − cn = + w = x + yw. (5) m + nω m2 + n2 − mn m2 + n2 − mn Then, since φ(c + dω) = c − (mn−1)d ≡ 0, multiplying both sides by n we get cn − md ≡ 0, which shows that y is an integer. Now, multiplying the last expression first by n and then by (m−1)2 and using the fact that (nm−1)2 ≡ nm−1 − 1 we get cm + dn − cn ≡ 0, which shows that x is an integer as well. Thus, from the relation in (5) follows that c + dw is a multiple of m + nw, so c + dw ∈ hm + nwi, which shows that ker(φ) ⊂ hm + nωi . Finally, since ker(φ) = hm + nωi , the result follows from the Frist Isomorphism Theorem. Finding factors of factor rings over Eisenstein integers 1527

Next we state an important corollary to the above theorem which carac- terizes primes in Z[ω]. Corollary 2.6. If m and n are relatively prime integers, then α = m + nω is a prime in Z[ω] if and only if N(α) = m2 + n2 − mn is a prime in Z. Proof. The proof follows immediately from the above theorem. Remark 2.7. We wish to note that N(α) ≡ 0, 1 (mod 3).

Theorem 2.8. Up to associates, the primes in Z[ω] are: (1) σ = m + nω and σ0 = n + mω, where N(σ) = m2 + n2 − mn is a prime in Z and N(σ) is congruent to 1 modulo 3; (2) σ = p, where p is a prime in Z and p ≡ 2 (mod 3). (3) σ = 1 − ω. Proof. The proof follows readily from Lemma 2.2 and Corollary 2.6 above. Remark: We will refer to the primes in (1), (2) and (3) as Eisenstein primes of the first, second, and third kind, respectively. Given a nonzero Eisenstein integer α = m + nω ∈ Z[ω], we can factor it into primes as follows:

r Y uk Y 0 vk Y ek t m + nω = ω · σk · σk · pk · (1 − ω) , (6) 0 where r, t, uk, vk, ek are nonnegative integers with uk ≤ vk, N(σk),N(σk) are integers congruent to either 0 or 1 modulo 3, and pk are integers congruent to 2 modulo 3.

3 Main Results

Before we state our main results, we prove the following proposition.

Proposition 3.1. Let p be a prime in Z+. The following conditions are equiv- alent: (1) p ≡ 0 (mod 3) or p ≡ 1 (mod 3), (2) the congruence x2 + x ≡ −1 (mod p) has a solution in Z, (3) p = m2 + n2 − mn. Let us first prove the following two lemmas.

Lemma 3.2. Let p be a prime in Z+. If p is composite in Z[ω], then p must be of the form m2 + n2 − mn, for some m, n ∈ Z. 1528 Valmir Bu¸caj

Proof. Since p is composite in Z[ω], there exist non-units α, β ∈ Z[ω], such that p = α · β. Then, N(p) = p2 = N(α)N(β). Since N(α) 6= 1 and N(β) 6= 1, it follows that N(α) = N(β) = p. So, if α = m+nω, then N(α) = m2+n2−mn = p, which is what we wanted to show.

Lemma 3.3. Let p be a prime in Z+. If p divides (1 − ωx)(1 − ωx¯ ) then p is not a prime in Z[ω]. Proof. To contradiction, suppose that p is a prime in Z[ω]. Then, since p|(1 − ωx)(1 − ωx¯ ), it follows that p|(1 − ωx) or p|(1 − ωx¯ ). If p|(1 − ωx), then (1 − ωx) = p(m + nω), for some m + nω ∈ Z[ω]. But then pm = 1, which is not possible sine p, m ∈ Z. On the other hand, if p|(1 − ωx¯ ), then (1 − ωx¯ ) = p(m + nω). Using the fact thatω ¯ = −ω − 1, it follows that p(m − n) = 1, which is again impossible since p, m, n ∈ Z. This concludes the proof. Now we return to the main proof. Proof. First we show that (1) implies (2). When p = 3 the proof is trivial. So, let p = 3k + 1, for some k ∈ Z, and consider the following polynomial factorization:

3 Xp−1 − 1 = X(p−1)/3 − 1 (7) = X(p−1)/3 − 1 X(2p−2)/3 + X(p−1)/3 + 1 . (8) Next we will count the roots of these polynomials modulo p. By Fermat’s little theorem, the left polynomial in (7) has p − 1 roots modulo p. On the other hand, the first polynomial in (8) has degree (p − 1)/3, so it can have at most (p − 1)/3 roots modulo p. Therefore, the second polynomial in (8) must have some root r modulo p, as well. That is r satisfies r(2p−2)/3 + r(p−1)/3 ≡ −1 (mod p). Therefore, since p = 3k + 1 we have

2 rk + rk ≡ −1 (mod p), which proves the first part. Now we show that (2) implies (3) by showing that (2) implies that p is composite in Z[ω]. Since x2 + x ≡ −1 (mod p), it means that p|(x2 + x + 1), in Z. Since x2 + x + 1 = (1 − ωx)(1 − ωx¯ ), in Z[ω] this relation can be interpreted as p|(1 − ωx)(1 − ωx¯ ). Finding factors of factor rings over Eisenstein integers 1529

Then, by Lemma 3.3, p is composite in Z[ω] and therefore, by Lemma 3.2, it must be of the form m2 + n2 − mn. This proves (3). That (3) implies (1) follows readily by Remark 2.7. This concludes the proof!

Now, we are ready to state and prove the main results of this paper. In the first theorem we find representatives for the equivalence classes for the ring Z[ω] when factored out by some power of a prime ideal, then we identify the units in these quotient rings respectively, and finally in the last theorem we give a decomposition of the ring Z[ω] when factored out by some ideal.

Theorem 3.4. The equivalence classes of Z[ω] modulo a power of a prime are as follows: n n n (1) Z[ω]/ hp i = {[a + bω]pn : 0 ≤ a ≤ p − 1 and 0 ≤ b ≤ p − 1}, where p is a prime in Z such that p ≡ 2 (mod 3); n n (2) Z[ω]/ hπ i = {[a]πn : 0 ≤ a ≤ q − 1}, where q is a prime in Z such that q ≡ 1 (mod 3), and π is a prime factor of q in Z[ω](i.e. q = ππ¯); 2m m m (3) Z[ω]/ hα i = {[a + bω]α2m : 0 ≤ a ≤ 3 − 1, and 0 ≤ b ≤ 3 − 1}, where α = 1 − ω; 2m+1 m+1 m (4) Z[ω]/ hα i = {[a+bω]α2m+1 : 0 ≤ a ≤ 3 −1, and 0 ≤ b ≤ 3 −1}, where α = 1 − ω.

Proof. First let us show that the equivalence classes in (1) are distinct. Sup- n pose that [a + bω]pn = [c + dω]pn . Then p must divide both a − c and b − d. But since 0 ≤ a − c, b − d ≤ pn − 1, it follows that a = c and b = d. Thus, the classes in (1) are distinct. Now, let α = a + bω be some element of Z[ω]. We will show that it belongs to one of the equivalence classes in (1). Reducing both a and b modulo pn, that is a ≡ x (mod pn) and b ≡ y (mod pn) we get

n n n n a + bω = x + t1p + (y + t2p )ω = (x + yω) + (t1 + t2ω)p ∈ Z[ω]/ hp i , which is what we needed to show. Now, as before, we first show that the equivalence classes in (2) are distinct. n If [x]πn = [y]πn , then π divides x − y. Then, there exists some α ∈ Z[ω] such that πn · α = x − y. Taking the conjugates we getπ ¯nα¯ = x − y = x − y (since, x and y are integers), soπ ¯n also divides x − y. But, since qn = πnπ¯n, and π andπ ¯ are not associates, then qn must divide x − y as well. Since, 0 ≤ x − y ≤ qn − 1, this is possible only when x = y, proving that the equivalence classes of Z[ω]/ hπni are distinct. 1530 Valmir Bu¸caj

Next, we want to show that any element α = x + yω belongs to one of the equivalence classes in (2). Observe, that since x and y belong to one of the classes in (2), it suffices to show that ω belongs to one of the equivalence classes of Z[ω]/ hπni. Let πn = c + dω (we wish to note that π is an Eisenstein prime of the first kind). First, we show that c and d are relatively prime. Let p be a prime in Z such that p ≡ 2 (mod 3), then by Theorem 2.8, p is also a prime in Z[ω]. To contradiction, suppose that p divides c + dω. Then, p would have to be an associate to π, that is, π = u · p, for some u ∈ {±1, ±ω, ±ω¯}. But, then we would get for the norm of π, N(π) = p2, which by Corollary 2.6 would contradict the fact that π is a prime in Z[ω]. If p = 3, then since 3 = (1 − ω) · (1 − ω¯), and π 6= (1 − ω), it is clear that 3 cannot divide c + dω. Next, if p is a prime in Z such that p ≡ 1 (mod 3), then by Proposition 3.1, p = σ · σ¯, for some Eisenstein prime, σ, of the first kind. If p divides c + dω, then this would imply that both σ andσ ¯ divide π. So, then for some u1, u2 ∈ {±1, ±ω, ±ω¯} we would have π = u1σ = u2σ.¯ This in turn would imply that σ = uσ¯ for some unit u, which by Fact 1.4 is not possible. The contradiction we arrived at demonstrates that g.c.d(c, d) = 1. So, there are some integers s, t such that cs + td = 1. After a few calculations one finds ω − s(c + dω)(ω + 1) − (c + dω)t to be a real number. From this, it follows that ω is congruent modulo c + dω = πn to a real number, say r, in Z[ω]; that is, ω ≡ r (mod πn). Then, if we reduce r by multiples of qn, we find that ω belongs to one of the equivalence classes of Z[ω]/ hπni , which is what we wanted to show. Again, let us first show that the equivalence classes in (3) are distinct. First, observe that because (1 − ω)2 = −3ω, we have h(1 − ω)2mi = h3mi , and 2m m m thus Z[ω]/ hα i = Z[ω]/ h3 i . Now, if [a + bω]3m = [c + dω]3m , then 3 must divide both a−c and b−d. But, since 0 ≤ a−b, c−d < 3m, then it follows that a = b and c = d. Thus, the equivalence classes in (3) are distinct. Now, let α = x + yω be any element in Z[ω]. Then, as we did in (1) above, by reducing each x and y modulo 3m we get α to be in one of the equivalence classes in (3). Lastly,, as before, we will first show that the equivalence classes in (4) are distinct. First, we notice that hα2m+1i = h(1 − ω)2m(1 − ω)i = h3mαi. Now, if m+1 m [a + bω]α2m+1 = [c + dω]α2m+1 , where 0 ≤ a, c ≤ 3 − 1 and 0 ≤ b, d ≤ 3 − 1, then (a − c) + (b − d)ω ≡ 0 (mod 3mα), in Z[ω]. So,

m (a − c) + (b − d)ω = γ 3 α, for some γ ∈ Z[ω]. Finding factors of factor rings over Eisenstein integers 1531

Then, [2(a − c) − (b − d)] + [(a − c) + (b − d)]ω = 3m+1γ. Thus, 3m+1|[2(a − c) − (b − d)] and 3m+1|[(a − c) + (b − d)], (9) so it must divide also 2[(a − c) + (b − d)] − [2(a − c) − (b − d)] = 3(b − d). Then, 3m divides (b − d), which implies that b = d. Now, from the second expression in (9), it follows that 3m+1 divides (a − c), which in turn implies that a = c. Next, let γ = x + yω. If we reduce x and y by multiples of 3m+1 we get γ ≡ c + dω (mod 3m+1), where c, d < 3m+1. Now, if d < 3m, then γ belongs to one of the classes in (4). Otherwise, suppose d = 3m + k, where k < 3m. Then,

c + dω = (c + 3m) + kω − 3mα.

Then γ ≡ (c + 3m) + kω (mod 3mα), and by reducing c + 3m by powers of 3m+1 we get γ to be in one of the classes in (4). This concludes the proof!

Notice that the above theorem implies that Z[ω]/ hpni has p2n elements, Z[ω]/ hπni has qn, elements and Z[ω]/ hαni has 3n elements. This is a special case of Theorem 2.5 which we have proved in the previous section. The next result helps us identify the units of the quotient rings mentioned in Theorem 3.4. Below we let q, π, p, and α be as in Theorem 3.4.

n Theorem 3.5. Let [a]πn be in Z[ω]/ hπ i. Then [a]πn is a unit if and only if n a and q are relatively prime. Let [a + bω]pn be in Z[ω]/ hp i. Then [a + bω]pn is a unit if and only if a or b is relatively prime with p. Let [a + bω]αn be in n Z[ω]/ hα i . Then [a + bω]αn is a unit if and only if 2a 6≡ b (mod 3).

n Proof. Let [a]πn be in Z[ω]/ hπ i. Then [a]πn is a unit if and only if there n is some γ ∈ Z[ω] such that [a]πn · [γ]πn = [1] in Z[ω]/ hπ i , or equivalently a · γ ≡ 1 (mod πn) in Z[ω]. Then a · γ + β · πn = 1, for some β ∈ Z[ω]. But this means that a and π are relatively prime, and since q = π · π¯, then a and q must be relatively prime as well. Next, let [a + bω]pn be a unit in Z[ω]/ hpni. This is possible if and only if there is some γ ∈ Z[ω]/ hpni such that (a + bω) · γ + (βpn−1) · p = 1, which is true if and only if p and a + bω are relatively prime. But, since p is a prime in Z such that p ≡ 2 (mod 3) and thus is a prime in Z[ω] as well, then p is relatively prime to a + bω if and only if p is relatively prime to a or b. Next, following the same logic as before, 1532 Valmir Bu¸caj

n [a + bω]αn is a unit in Z[ω]/ hα i if and only if (a + bω) and α = 1 − ω are relatively prime; that is if and only if 1 − ω does not divide a + bω. But since

a + bω 2a − b a + b = + ω, (10) 1 − ω 3 3 then (10) is an Eisenstein integer if and only if 2a ≡ b (mod 3). Therefore, α = 1−ω, does not divide a+bω if and only if 2a 6≡ b (mod 3). This concludes the proof!

Example 3.6. Let π = 3 + ω. Observe that π is a prime and π · π¯ = 7. Then by Theorem 3.4 and 3.5 we know that:

2 Z[ω]/ π = {[0], [1], [2], ..., [48]}.

Moreover, an element [a] of this ring is a unit if and only if 7 does not divide a. Next we find the equivalence class to which ω belongs. Since π2 = 8 + 5ω, we have 5ω ≡ −8 (mod π2). Multiplying both sides by 10 we get 50ω ≡ −80 2 2 2 (mod π ), and since π divides 49 we have ω ≡ 18 (mod π ). So, ω ∈ [18]π2 . We can check this result directly as well; that is since ω − 18 = −1 + 2ω, 8 + 5ω then π2 = 8 + 5ω divides ω − 18 in Z[ω], but this in turn means that ω ≡ 18 2 (mod π ), so ω belongs to [18]π2 .

Q uk In what follows, let us adopt the foollowing notation α = N(σk) , β = Q 0 vk Q ek 2 N(σk) , γ = pk . Observe that N(m+nω) = α·β ·γ , and that α divides β.

t Theorem 3.7. If m and n are integers not both zero, and Rt = Z[ω]/ h(1 − ω) i , then the following hold:

∼ Z[ω]/ hm + nωi = Zα ⊕ Zβ ⊕ Zγ[ω] ⊕ Z3t/2 [ω] for t even, and ∼ Z[ω]/ hm + nωi = Zα ⊕ Zβ ⊕ Zγ[ω] ⊕ Rt for t odd. Proof. By the factorization in (6) we can write D E Y uk Y 0 vk Y ek t hm + nωi = σk · σk · pk · (1 − ω) . (11) Finding factors of factor rings over Eisenstein integers 1533

Q uk Q 0 vk t Q ek Since σk , σk , (1−ω) , and pk are pairwise relatively prime and Z[ω] is a Euclidean domain we can appeal to the Chinese Remainder Theorem for rings and use (11) to get

D E D E ∼ Y uk Y 0 vk Z[ω]/ hm + nωi = Z[ω]/ σk ⊕ Z[ω]/ σk D E Y ek t ⊕Z[ω]/ pk ⊕ Z[ω]/ (1 − ω) . (12)

Our next task is to demonstrate that (12) implies our claimed result. Let Q uk σk = c + dω. We want to show that c and d are relatively prime so that we can appeal to Theorem 2.5 to obtain our desired result. Let p be a prime in Z such that p ≡ 2 (mod 3). Then since all such p are also prime in Z[ω], it follows that p cannot divide c + dω. For if p were to divide c + dω, then there would exist some k such that p would divide σk, as well. Then σk = s · p, for 2 some s ∈ {±1, ±ω, ±ω¯}. But then, N(σk) = p , contradicting the fact that σk is a prime in Z[ω]. If p = 3, then p clearly cannot divide c + dω. Next, if 0 p ≡ 1 (mod 3), then by Proposition 3.1, up to a unit we have p = σk · σk, for 0 some k(sinceσ ¯k =ωσ ¯ k), therefore it cannot divide c + dω. So, since c and d are relatively prime integers, by Theorem 2.5, we conclude that D E Y uk ∼ ∼ ∼ Z[ω]/ σk = Z[ω]/ hc + dωi = ZN(c+dω) = Zα. (13)

The last isomorphism in (13) follows from the fact that the norm function N is multiplicative. Similarly we have D E Y 0 vk ∼ Z[ω]/ σk = Zβ. (14) Also, by Theorem 2.1 we have

∼ Z[ω]/ hγi = Zγ[ω]. (15)

The last term by definition is Rt. However, for t even Rt simplifies. That is, since (1 − ω)2 = −3ω, then we have h(1 − ω)ti = (−ω · 3)t/2 = 3t/2 . Thus, for t even we have

t ∼ Z[ω]/ (1 − ω) = Z3t/2 [ω]. (16)

Finally, the result follows by (12), (13), (14), (15), and (16). 1534 Valmir Bu¸caj

3.1 Applications Below we present a few applications to elementary number theory. Proposition 3.8. Let p be a prime that is not congruent to 2 (mod 3). Then, the equation p = m2 + n2 − mn has at most four solutions(not necessarily distinct) for m, n ∈ Z+, and m ≥ n. Proof. By Proposition 3.1, we know there is a solution to p = m2 + n2 − mn. Suppose that there is another solution to this equation, namely there + 2 2 are m1, n1 ∈ Z such that p = m1 + n1 − m1n1. Then, making use of the prime factorization of p in Z[ω] and setting these expressions for p equal to one another we get (m + nω)(m + nω¯) = (m1 + n1ω)(m1 + n1ω¯). Since prime factorization in Z[ω] is unique (remember Z[ω] is a U.F.D) we have one of the following two possibilities: m + nω = u(m1 + n1ω) or m + nω = u(m1 + n1ω¯), where u ∈ {±1, ±ω, ±ω¯}. Case 1. Suppose m + nω = u(m1 + n1ω). (i) If u = ±1, then m1 = ±m and n1 = ±n; (ii) If u = ±ω, then n1 = ∓m and m1 = ±n ∓ m; (iii) If u = ±ω¯, then m1 = ∓n and n1 = ±m ∓ n. Case 2. Suppose m + nω = u(m1 + n1ω¯). (i) If u = ±1, then n1 = ∓n and m1 = ±m ∓ n; (ii) If u = ±ω, then n1 = ±n and m1 = ±m; (iii) If u = ±ω¯, then m1 = ∓m and n1 = ±n ∓ m. So, the only other two solutions of the equation p = m2 + n2 − mn are (m, m − n) and (m − n, m). Notice, that if m = 2n then we get only two distinct solutions, namely (m, n) and (n, m). Proposition 3.9. A positive integer M is of the form M = m2 + n2 − mn, where m, n ∈ Z if and only if any prime p|M such that p ≡ 2 (mod 3) occurs to an even power in the prime factorization of M. Proof. Observe that M = m2 + n2 − mn if and only if N(α) = M for some α ∈ Z[ω]; that is if and only if M is the norm of some element in Z[ω]. Let Y ei Y fj M = pi qj + be the prime factorization for M in Z , where each pi ≡ 2 (mod 3) and qj 6≡ 2 (mod 3). Then, by Proposition 3.1, each qj = πjπ¯j, where πj is an Eisenstein prime of the first or third kind. So,

Y ei Y fj fj M = pi πj π¯j , (17) Finding factors of factor rings over Eisenstein integers 1535 is the prime factorization of M in Z[ω]. Now, suppose that M = m2 + n2 − mn, for some m, n ∈ Z; that is, suppose that N(α) = M, for some α ∈ Z[ω]. Let

Y ui Y vj α = ri sj be the prime factorization of α in Z[ω] where each of ri ≡ 2 (mod 3) and each of sj is an Eisenstein prime of the first or third kind. Then,

Y ui Y vj α¯ = ri s¯j , is the prime factorization ofα. ¯ Then, since N(α) = αα¯ = M, we get

Y 2ui Y vj vj Y ei Y fj fj αα¯ = ri sj s¯j = M = pi πj π¯j .

Since, Z[ω] is a U.F.D we must have ei = 2ui, which is what we wanted to prove. On the other hand, suppose that any prime p|M such that p ≡ 2 (mod 3) occurs to an even power in the prime factorization of M. Then, as in (17) let

Y ei Y fj fj M = pi πj π¯j , be the factorization of M in Z[ω], but now, in addition, suppose that ei = 2ti, for some integer ti; that is

Y 2ti Y fj fj M = pi πj π¯j . Now, let Y ti Y fj Y ti Y fj α = pi πj , thenα ¯ = pi π¯j . Then

Y 2ti Y fj fj αα¯ = pi πj π¯j = M. Let α = m + nω, then since N(α) = αα¯ = m2 + n2 − mn, we conclude that M = m2 + n2 − mn, for some m, n ∈ Z, as we wanted to show. Below we present a different proof of the infinitude of primes of the form 3k + 1. This proof relies on one of our previous results in this paper, but its methodology is similar to one of the methods used to prove the infinitude of primes of the form 4k + 1. To the best of our knowledge, the proof we present here has not appeared in this form elsewhere. 1536 Valmir Bu¸caj

Proposition 3.10. There are infinitely many primes of the form 3k + 1.

Proof. The statement is equivalent to saying that there are infinitely many primes that are congruent to 1 mod 3. Then, by Proposition 3.1 it is sufficient to prove that there are infinitely many primes p that divide the polynomial f(x) = x2 + x + 1, for some value in Z. To contradiction, suppose there are only a finite number of primes p1, p2, ..., pn that divide f(x). Then, consider

2 2 g(y) = f(p1 · p2 · ... · pny) = (p1 · p2 · ... · pn) y + (p1 · p2 · ... · pn)y + 1.

Clearly since all g(y) are contained in f(x), the only primes that divide g(y) are pi for i = 1, 2, .., n. But since

g(y) ≡ 1 (mod pi), for i = 1, 2, ..., n, 2 then g(1) = (p1 · p2 · ... · pn) + (p1 · p2 · ... · pn) + 1 > 1 is not divisible by any of the primes pi, a contradiction, hence the statement of the proposition.

3.2 Conclusion Remarks We would like to end this note with a conjecture and some suggestions for further investigation.

Conjecture 3.11. If m, n, and d are positive integers with m and n relatively prime, then the equivalence classes of Z[ω]/ hdm + dnωi are {[x + yω] : 0 ≤ x < d(m2 + n2 − mn), 0 ≤ y < d}.

It would also be of interest to have a generalization of Theorem 3.5; that is, a result which would characterize all the units in the quotient ring Z[ω]/ hdm + dnωi.

Acknowledgment

The author would like to thank Dr. William Hager of Texas Lutheran University for his guidance, invaluable suggestions, and comments which have increased the quality of this paper. Finding factors of factor rings over Eisenstein integers 1537

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Received: November 1, 2011