GAUSSIAN INTEGERS HUNG HO

GAUSSIAN INTEGERS HUNG HO Abstract. We will investigate the ring of Gaussian integers Z[i] = {a + bi a, b Z}. First we will show that this ring shares an imortant roerty with the ring of integers: every element can be factored into a roduct of finitely many rimes. This result is the key to all the remaining concets in this aer, which includes the ring Z[i]/αZ[i], analogous statements of famous theorems in Z, and quadratic recirocity laws. Contents 1. Princial Ideal Domain and Unique Prime Factorization 1 2. The ring Z[i] 6 3. Some Alications of Unique Prime Factorization in Z[i] 8 4. Congruence Classes in Z[i] 11 5. Some imortant theorems and results 13 6. Quadratic Recirocity 18 Acknowledgement 22 References 22 1. Princial Ideal Domain and Unique Prime Factorization Definition 1.1. A ring R is called an integral domain, or domain, if 1 0 and whenever a, b R and ab = 0, then either a = 0 or b = 0. Examle 1.2. Z, Q, R, C are all integral domains. Examle 1.3. The ring Z[i] = {a + bi : a, b Z} is an integral domain. Examle 1.4. The ring Z/nZ is a domain if and only if n is a rime. This is because if n is not a rime then we can write n = ab where a, b Z \ {1} and thus ab = 0 in Z/nZ. Conversely, if n is a rime then n divides ab if and only if n divides either a or b, so a = 0 or b = 0 in Z/nZ. Definition 1.5. An ideal I of a commutative ring R is rincial if it is generated by a single element a of R through multilication by every element of R. In other words, I = Ra = {ra : r R}. It is common to denote the ideal generated by a as (a). Examle 1.6. The set of even integers is a rincial ideal of Z generated by 2. Definition 1.7. A rincial ideal domain is a domain in which every ideal is rincial. 1

2 HUNG HO Definition 1.8. Let a, b be elements of the commutative ring R. If there exists x R such that a = bx then we say that b divides a, or a is divisible by b and write b a. b is called a divisor of a and a is a multile of b. Elements a and b of an integral domain are associates if a b and b a. An element u is a unit if u divides every element of R, or equivalently, u divides 1. We can restate the above claims about divisibility and unit in terms of rincial ideals. From now on, we always assume R to be a commutative ring and an integral domain. Proosition 1.9. Let a, b R, then b divides a if and only if (a) (b). Proof. If b divides a, we write a = bx for some x R. Then for any y (a), we have y = at for some t R, or y = bxt and thus y (b). Conversely, if (a) (b), then obviously a (b) so a = bx for some x R, so b divides a. Corollary 1.10. An element u is a unit if and only if (u) = (1) = R. Corollary 1.11. The following are equivalent: (1) a and b are associates. (2) a = bu for some unit u. (3) (a) = (b). Definition 1.12. An element of the commutative ring R is called rime if {0, 1} and whenever divides ab for a, b R, then either divides a or divides b. Definition 1.13. A non-zero non-unit element in an integral domain is irreducible if it is not the roduct of two non-zero units. In the ring of integers Z, rime and irreducible elements are equivalent and are called interchangeably as rime numbers. In general, however, these two definitions do not coincide. For examle, consider the ring Z 5 = {a + b 5 : a, b Z}. It is easy to check that this ring is an integral domain (because it is a subset of the comlex numbers). The element 2 is irreducible in Z 5 because if 2 = (a + b 5)(c+d 5), taking absolute value of both sides yields 4 = (a 2 +5b 2 )(c 2 +5d 2 ). This is only ossible if b = d = 0, hence ac = 2, where a, c Z, so either a or c must be 1. Therefore either a+b 5 or c+d 5 is a unit in Z 5. On the other hand, 2 is not a rime in Z 5 since 2 divides 4 = ( 1 + 5)(1 + 5) but 2 neither divides 1 + 5 nor 1 + 5 (an integer divides a number a + b 5 in Z 5 if and only if it divides both a and b). The examle above shows that in an integral domain, irreducible elements are not necessarily rimes, but what about the reverse statement? The following theorem addresses this issue. Theorem 1.14. If is a rime element in an integral domain R, then is irreducible. Proof. Assume 0 is a rime but not irreducible in R, then there exists x, y R that are not units such that = xy. Since is a rime element, it follows that x or y. Without loss of generality, suose x, then x = t for some t R. Thus, we can write = ty. Since R is an integral domain and 0, we deduce that ty = 1, or y divides 1, so y is a unit, a contradiction. If R is a rincial ideal domain, then the reverse direction is also true. However, before tackling this roerty, we need some more notions.

GAUSSIAN INTEGERS 3 Definition 1.15. Let a, b be two nonzero elements of R. An element d R is called a greatest common divisor of a and b if d is a divisor of both a and b, and any common divisor of a and b divides d. Later on we will show that any two greatest common divisors of two elements are associates of each other, hence from now on we will use the notation gcd(a, b) and the term the greatest common divisor to denote any of those associates. However, in a general integral domain, two elements need not have a greatest common divisor. In fact, a domain in which every two elements have a greatest common divisor is called a GCD domain. We will show that every rincial ideal domain is also a GCD domain. For elements a 1, a 2,..., a n R, we define (a 1, a 2,..., a n ) = Ra 1 +Ra 2 +...+Ra n = { n r i a i r i R}. It is easy to see that (a 1, a 2,..., a n ) is also an ideal of R. i=1 Theorem 1.16. Let R be a rincial ideal domain and a, b be nonzero elements of R. Then there exists d = gcd(a, b) and (a, b) = (d). Proof. Let I = (a, b) be an ideal of R. Since R is a rincial ideal domain, we have I = (d) for some d R. Because (a), (b) (a, b) = (d), we deduce that d divides both a and b. Now let d be any common divisor of a and b and write a = d a, b = d b. Since d (a, b), we can also write d = au + bv for some u, v R. Thus, we have d = d (a u + b v), so d d. Therefore, d is the greatest common divisor of a and b. Two elements a and b may have more than one greatest common divisor. If d and d are both greatest common divisors of a and b, we can deduce from the definition that d d and d d. Hence, any two greatest common divisors are associates of each other. If gcd(a, b) = u, where u is a unit, then a and b are relatively rime. It follows from theorem 1.16 that if a and b are relatively rime, then (a, b) = R. Corollary 1.17. Let a, b, c R and a bc. If (a, b) = R, then a c. Proof. Since (a, b) = R, there exists x, y R such that ax + by = 1. We have Thus, a divides c. bc = ak ybc = yak c(1 ax) = yak c = a(cx + yk). Corollary 1.18. If is irreducible in a rincial ideal domain R, then is a rime element. Proof. Suose divides ab for some a, b R. Let a = gcd(, a). If a is a unit, then and a are relatively rime and we conclude from corollary 1.17 that divdes b. If a is not a unit, we write = a u and since is irreducible, it follows that u is a unit. So and a are associates, hence divides a. From now on, we will not distinguish irreducible and rime elements. Recall that every ositive integer can be uniquely factored into a roduct of rimes (we only consider ositive rimes). We want to rove a similar result in a

4 HUNG HO rincial ideal domain, that is, every element can be factored into roduct of rime elements. However, this factorization, rovided it exists, may not be unique as we can see via this simle examle in Z: 4 = 2 2 = ( 2) ( 2). But this examle also shows that in Z, if a number can be factored into different roducts of rimes, then we can find corresonding rimes in these roducts that are associates of each other (2 and -2 in the above examle). Therefore, when talking about unique rime factorization in a general rincial ideal domain, we understand that it is unique u to associates. We will now go on to rove that every element in a rincial ideal domain can be factored into irreducible (or rime) elements. The intuition is as followed. Given any element a, if a is irreducible then we are done. If not, then we can write a = bc, where b, c are non-units. Now if either b or c is not irreducible, we roceed similarly to factor that element into another two elements. Eventually, if this rocess terminates after finitely many stes, we have our desired factorization. However, if this rocess goes on forever, then we have a roblem. We will rove that this cannot haen. Proosition 1.19. Let a be a non-zero non-unit element of a rincial ideal domain R. Then a can be factored into a roduct of finitely many irreducible elements. Proof. First we show that every non-zero non-unit element x has an irreducible divisor. Assume otherwise, then obviously x is not irreducible itself, so we can write x = x 1 y 1. Since x has no irreducible divisor, we can again factor x 1 = x 2 y 2, where x 2, y 2 are also not irreducible. Proceed inductively, we have two infinite sequences of (x n ) and (y n ) in R such that all of the terms are not irreducible and x n 1 = x n y n. Thus, (x 1 ) (x 2 )... (the strict subset sign is due to the fact that y n is not irreducible for all n, so a n is not an associate of a n 1 ). Now let I =. We claim that I is an ideal of R. Indeed, it is obvious that (I, +) n=1 is a subgrou of (R, +). For any t I, r R, since t belongs to (x i ) for some i, so does rt, hence rt I. Thus, I is an ideal and we deduce from the fact that R is a rincial ideal domain that I = (b) for some b R. Since x n (b) = I, we have b divides x n for all n. On the other hand, b (x k ) for some k, or x k divides b. Hence, b and x k are associates. But then this imlies that x k divides x k+1, which imlies (x k+1 ) (x k ) (x k+1 ), a contradiction. So every non-zero non-unit element has an irreducible divisor. Now consider an arbitrary non-zero non-unit a R. If a is irreducible, we are done. If not, we can factor a = a 1 b 1, where a 1 is irreducible. Since a is not irreducible, b 1 is not a unit (otherwise a and a 1 are associates), hence b 1 has an irreducible divisor a 2. We write b 1 = a 2 b 2, and aly the same argument for b 2. Proceed inductively, if b n is not a unit, we factor it into a n+1 b n+1, where a n+1 is irreducible. Eventually, if we sto at some N 0 where b N0 is a unit, then b N0 1 is irreducible and a = a 1 a 2... a N0 1b N0 1 is our desired factorization. Otherwise, if we do not sto at some N 0, then we have an infinite sequence of ideals (b 1 ), (b 2 ),... such that (b 1 ) (b 2 )... (a n is non-unit for all n so b n 1 and b n are not associates). Now reeat the roof as above, we have a contradiction, so we must sto after finitely many stes, and thus a can be factored into a roduct of irreducible elements.

GAUSSIAN INTEGERS 5 Now we know that any non-zero non-unit element can be factored into a roduct of irreducibles. To rove that his factorization is unique u to associates, we need the following simle lemma. Lemma 1.20. Let a, b be two irreducible elements of a rincial ideal domain R. Then a and b are relatively rime if and only if they are not associates. Proof. The forward direction is obvious because two associates divide each other. For the reverse direction, assume a and b are not relatively rime, let d be their greatest common divisor. Write a = da and b = db. Since a and b are irreducibles and d is not a unit, it follows that a and b are units. Thus, d is an associate of both a and b, so a and b are associates, a contradiction. From now on, if a and b are not associates, we say that a and b are distinct. Theorem 1.21. Let a be a non-zero element of a PID R. Then we can write k a = u αi i, i=1 where u is a unit, i s are airwise distinct irreducibles that are unique u to associates, and the exonents α i s are uniquely determined. Proof. The existence of such factorization follows from roosition 1.19. Now assume that we have two factorizations a = u k αi i = v l q βi i. For each i = i=1 i=1 l 1, 2,..., k, i divides q βi i, so from lemma 1.20 and corollary 1.18, there must i=1 exists j {1, 2,..., l} such that q j and i are associates. Thus, we can ma each i to its associate q j, and aarently this is a one-to-one ma because two distinct irreducibles have distinct associates. But we can also reeat the argument and deduce that for any q j, there exists a corresonding associate i, hence our ma is an isomorhism. Therefore, k = l, and Without loss of generality, we can assume that i and q i are associates for i = 1, 2,..., k. It remains to show that α i = β i for every i. Assume otherwise and without loss of generality, there exists i 0 such that α i0 > β i0. Since R is an integral domain, we deduce that u i αi 0 βi 0 = v i, so i0 divides q j for some j i 0, a i i 0 αi i 0 i i 0 q βi contradiction. Therefore, α i = β i for all i, and thus the theorem is roven. Definition 1.22. An Euclidean domain R is an integral domain equied with a function f from R \ {0} to {0, 1, 2,..., } such that if a, b R and b is nonzero, then we can write a = bq + r for q R, and either r = 0 or f(r) < f(b). The main reason to define Euclidean domain is the following roosition. Proosition 1.23. Assume R be an Euclidean domain. ideal domain. Then R is a rincial Proof. Let I be an ideal of R, we shall rove that there exists a R such that I = Ra. Indeed, let a be a nonzero element of I such that f(a) is minimum (the existence of such element follows from the fact that the function f takes values on the set of non-negative integers). Now, consider any b I, we can write b = aq + r with q, r R. Since f(a) is minimum, we cannot have f(r) < f(a), so for all b I,

6 HUNG HO b = aq for some q R and thus I Ra. On the other hand, obviously Ra I because I is an ideal. Hence, I = Ra. Proosition 1.23 is imortant because we may show a ring R is a PID by first roving it is an Euclidean domain. We will conclude the first section with an imortant result: the Chinese remainder theorem. Definition 1.24. Two ideals I and J are corime if there exists i I and j J such that i + j = 1. Remarks 1.25. By theorem 1.16, we know that in a rincial ideal domain R, two ideals (a) and (b) are corime if and only if a and b are relatively rime. Theorem 1.26. Let I 1, I 2,..., I k be ideals of a ring R that are airwise corime. Denote I = k I i. We have the isomorhism i=1 R/I R/I 1... R/I k x + I (x + I 1,..., x + I k ). Proof. It suffices to rove for k = 2, as the general case can be roved similarly using induction. We want to show that for any x 1, x 2 R one can find x I such that x x 1 mod I 1 and x x 2 mod I 2. Since I 1 and I 2 are corime, there exists y 1 I 1 and y 2 I 2 such that y 1 + y 2 = x 1 x 2. Let z = y 1 + x 1 = y 2 + x 2, then aarently z R and z x 1 I 1, z x 2 I 2. Thus z x 1 mod I 1 and z x 2 mod I 2. 2. The ring Z[i] Most of this aer is devoted to reroduce many well-known concets and results from Z in the ring Z[i] = {a + bi a, b Z}. First and foremost, we will show that Z[i] is a rincial ideal domain and thus inherits the unique rime factorization roerty. Proosition 2.1. Z[i] is a Euclidean domain. Proof. For each α = a + bi Z[i] \ {0}, we define f(α) = αα = a 2 + b 2. Take an arbitrary β = c + di Z[i], we will show that β = αγ + θ for γ Z[i] and either θ = 0 or f(θ) < f(α). Indeed, let β α = r + si, where r, s Q. Let k be the closest integer to r and l be the closest integer to s, i.e. k r 1 2 and l s 1 2. Let γ = k+li, m = r k, n = s l and θ = (a + bi)(m + ni), we have: c + di = (a + bi)(r + si) = (a + bi)(k + li) + (a + bi)(m + ni) β = αγ + θ. If θ = 0 then β = αγ, as desired. Otherwise, we have f(θ) = θθ = (a 2 + b 2 )(m 2 + n 2 ) (a 2 + b 2 )( 1 4 + 1 4 ) < f(α). Thus, Z[i] is a Euclidean domain with the function f defined as f(α) = αα. Proosition 2.1 is fundamental to the study of the ring Z[i] as we can relate many results and roerties in Z[i] to their counterarts in Z. However, after we roved that Z[i] is a unique rime factorization domain, a natural question arises:

GAUSSIAN INTEGERS 7 What are rimes in Z[i]?. Primes in Z[i] may have some similarities with rimes in Z but aarently they are not the same, as we can see, for instance, that 5 is a rime in Z but not a rime in Z[i] because 5 = (2 i)(2 + i). Lemma 2.2. Let = 4k + 1 be a ositive rime, then ( 1 2 )!2 + 1 is divisible by. Proof. Since for any x {0, 1,..., 2k}, x ( 1)( x) mod, it follows that ( 1 2 )! ( + 1) ( 1)2k ( 1)( 2)... 2 mod. Hence ( 1 2 )!2 ( 1)! 1 mod (the second congruence is due to Wilson s theorem). Lemma 2.3. Let = 4k + 3 be a ositive rime and a, b N such that a 2 + b 2 is divisible by. Then both a and b are divisible by. Proof. We have a 2 b 2 a 2(2k+1) ( b 2 ) (2k+1) mod mod a 1 b 1 mod. Now if a is not divisible by, so is b because a 2 + b 2 is divisible by. But then by the Fermat s little theorem, a 1 b 1 1 mod. Thus, by the above congruence equations, it follows that 2a 1 is divisible by, or a is divisible by, a contradiction. Thus both a and b are divisible by. Corollary 2.4. Let be a rime in Z, then there exists x Z such that x 2 + 1 is divisible by if and only if = 4k + 1. Theorem 2.5. A Gaussian integer α = a + bi is a rime if and only if it falls in one of the following categories: (1) a = 0 and b is a rime number of the form 4k + 3. (2) b = 0 and a is a rime number of the form 4k + 3. (3) a 2 + b 2 is a rime number. Proof. First, we rove that all Gaussian integers described in (1), (2), (3) are rime. (1) We will rove for the case b > 0, the case b < 0 is roved similarly. Let b = 4k + 3 (k 0) be a rime and assume α is not a rime in Z[i]. Then we can write bi = (u + vi)(x + yi), where u + vi and x + yi are not units. We deduce that b 2 = (u 2 + v 2 )(x 2 + y 2 ). Since b is a rime, either u 2 + v 2 or x 2 + y 2 is divisible by b. Without loss of generality, assume u 2 + v 2 is divisible by b. Now because b = 4k + 3, it follows from lemma 2.3 that both u and v are divisible by b. But then both u 2 and v 2 are divisible by b 2, and thus u 2 + v 2 b 2. Now we can conclude from x 2 + y 2 > 1 that (u 2 + v 2 )(x 2 + y 2 ) > b 2, a contradiction. Hence α is a rime. (2) The roof is similar to that of (1).

8 HUNG HO (3) Assume that α is not a rime, then by similar argument we can write a + bi = (u + vi)(x + yi), where u 2 + v 2 and x 2 + y 2 are both greater than 1. We also have a 2 + b 2 = (u 2 + v 2 )(x 2 + y 2 ), which imlies that (u 2 + v 2 )(x 2 + y 2 ) is a rime number. However, this is imossible because the roduct of two numbers greater than 1 can never be a rime number in Z, thus α must be a rime in Z[i]. Conversely, we rove that every rime element of Z[i] belongs to one of the above three categories. Observe that α = a + bi is a rime in Z[i] if and only if α = a bi is also a rime in Z[i] (because a + bi = (u + vi)(x + yi) a bi = (u vi)(x yi), where u 2 + v 2 and x 2 + y 2 are both greater than 1). Also, it is obvious that if both a and b are nonzero, they must be relatively rime in Z[i]. Now we look at q = a 2 + b 2 = (a + bi)(a bi). Consider the following cases: (a) q is even. We have (a + bi)(a bi) is divisible by 2 = (1 + i)(1 i). Note that both 1 + i and 1 i are rimes due to our roof for (3) above. So either a + bi or a bi is divisible by 1 + i, and because they are all rimes in Z[i], we must have a + bi = u(1 + i), where u is a unit. Thus, a bi = u(1 i) and finally a 2 + b 2 = 2, which is a rime. So α falls in category 3. (b) q has a rime divisor with absolute value of the form 4k + 3. Then by lemma 2.3, both a and b are divisible by and thus, a 2 +b 2 is divisible by 2 (in both Z and Z[i]). Since both a + bi and a bi are rimes in Z[i], each of them cannot be divisible by 2, so we deduce that both a + bi and a bi are divisible by. However, this imlies that their sum 2a, and their difference 2bi, are also divisible by in Z[i]. Since 2 and are relatively rime in Z[i] because = 4k + 3, it follows that both a and b are divisible by. This is imossible if a and b are corime, so one of them must be zero. If a = 0, let b = k, then for α = ki to be a rime in Z[i], k must be equal to 1, hence α is in category 1. Similarly, if b = 0 then we can also deduce that α is in category 2. (c) All rime divisors of q have absolute values of the form 4k + 1. First we show that if is a rime in Z of the form 4k + 1, then is not a rime in Z[i]. Indeed, assume othrewise, is also a rime in Z[i]. according to corollary 2.4, there exists x Z such that x 2 + 1 = (x + i)(x i). Since is a rime, it follows that either x + i or x i is divisible by in Z[i]. Without loss of generality, assume x + i = (r + si), then we deduce that 1 = s, a contradiction, hence is not a rime in Z[i]. Now assume for the sake of contradiction that q is not a rime itself, then there exist rimes 1, 2 in Z ( 1 and 2 are not necessarily distinct) such that q is divisible by 1 2. Because both 1 and 2 are of the form 4k + 1 and hence are not rimes in Z[i], we know that 1 2 is a roduct of at least four rimes in Z[i]. This is imossible because 1 2 divides (a + bi)(a bi), which is a roduct of two rimes. Therefore, q = a 2 + b 2 is a rime in Z. 3. Some Alications of Unique Prime Factorization in Z[i] An imortant corollary of theorem 2.5 is the following theorem that characterize the necessary and sufficient condition to exress a rime as a sum of two squares in N.

GAUSSIAN INTEGERS 9 Theorem 3.1. A rime N can be exressed as a sum of two squares if and only if = 2 or = 4k + 1. Proof. By lemma 2.3, we know that if = 4k + 3 then cannot be exressed as a sum of two squares. Now it remains to show that every rime of the form 4k + 1 can be written as = a 2 + b 2. Now from theorem 2.5 we know that is not a rime in Z[i], so has a rime divisor a + bi in Z[i]. Write = (a + bi)(c + di) then we deduce that ac bd = and ad = bc. Since a + bi is a Gaussian rime, we must have gcd(a, b) = 1, so from the latter equation we deduce that a c and b d. Write c = ax and d = by, then because ad = bc we have x = y, so c + di = x(a bi). This imlies that = x(a + bi)(a bi) = x(a 2 + b 2 ), and thus x = 1 because is a rime. We conclude that there exists a, b such that = a 2 + b 2. Proosition 3.2. Let α = a + bi be a Gaussian integer, where a, b Z \ {0} are relatively rime in Z. Suose n is an integer, then n is divisible by α in Z[i] if and only if n is divisible by N(α) in Z. Proof. The reverse direction is obvious because α N(α). For the forward direction, assume n = (a + bi)(c + di), then after exanding we deduce that ad = bc. But since gcd(a, b) = 1, it follows that a c and b d. Write c = ax and d = by, where x, y Z, then from ad = bc we have x = y. Therefore, n = x(a + bi)(a bi) = x(a 2 + b 2 ). From this we conclude that n is divisible by N(α) in Z. Examle 3.3. Solve the equations y 3 1 = x 2 in Z. Solution. First observe that if x is odd then x 2 1 mod 4, so x 2 + 1 is divisible by 2 but not divisible by 4. However, because then y must be even, y 3 = x 2 + 1 is divisible by 8, a contradiction. Therefore, x is even and y is odd. We have y 3 = x 2 + 1 = (x + i)(x i). We claim that x + i and x i are corime. Indeed, if they have a common rime divisor in Z[i], then divides their difference 2i, so is either 1 + i or 1 i. But then because divides y, by roosition 3.2 we deduce that y is divisible by 2, a contradiction. Hence, x+yi and x yi are corime, so each of them must be a cube of a Gaussian integer times a unit. However, observe that Z[i] has four units 1, 1, i, i and each of them itself is a cube of another Gaussian integer. To be secific, 1 = 1 3, 1 = ( 1) 3, i = ( i) 3, ( i) = i 3. Hence we can assume that both x+yi and x yi are cubes of Gaussian integers. We have: x + i = (a + bi) 3 3a 2 b b 3 = 1 b(3a 2 b 2 ) = 1. From this we deduce that b = 1 3a 2 1 = 1 a = 0 b = 1. Thus x + i = i, or x = 0, y = 1. Examle 3.4. Let n be a ositive integer. Find the numbers of solution (x, y) Z 2 of the following equation x 2 + y 2 = n. Solution. For a rime and ositive integers α, n, we write α n if α divides n and α+1 does not divide n. Consider two cases:

10 HUNG HO (a) n is odd. Let n = α1 1... α k k, where j are distinct odd rimes. By lemma 2.3 we can deduce that if α n where is of the form 4k + 3, then in order for the equation to have solution in Z, we must have α = 2β and β x, y. Therefore, we can assume that j 1 mod 4 for all 1 j k. By theorem 3.1, we can write j = (a j + b j i)(a j b j i), where a j + b j i and a j b j i are Gaussian rimes for all j. Now rewrite our equation as (x + yi)(x yi) = k (a j + b j i) αj (a j b j i) αj. j=1 Because a j + b j i and a j b j i are rimes in Z[i], we deduce this reresentation x + yi = i s k j=1 (a j + b j i) βj (a j b j i) γj, where s {0, 1, 2, 3}, β j, γ j {0,..., α j } j = 1,..., k. Taking conjugates of both sides, we have k (x yi) = i 4 s (a j b j i) βj (a j + b j i) γj, j=1 and thus we deduce that β j + γ j = α j for all j. Since x yi is uniquely determined if we know x + yi, it suffices to find all ossible (k + 1)-tules of (s, β 1,..., β k ) such that x + yi = i s k j=1 (a j + b j i) βj (a j b j i) αj βj. However, the only condition required is that s {0, 1, 2, 3} and β j {0,..., α j } j. Thus, the total number of the desired (k + 1)-tules, which is also the number of solutions to the equation, is 4(α 1 + 1)... (α k + 1). (b) n is even. Let n = 2 α α1 1... α k k, where j are distinct odd rimes. By similar argument as in the case n odd, we may assume j are all of the form 4k +1. Denote v 2 (x) as the largest number m such that 2 m divides x, and v 2 (y) is defined similarly. If v 2 (x) v 2 (y) then clearly α = v 2 (n) = min{2v 2 (x), 2v 2 (y)} = 2β. Let x 0 = x, y 2 β 0 = y, n 2 β 0 = n 2 and consider the equation x 2 α 0 + y0 2 = n 0. Since n 0 is odd, we reeat the roof for the first case and conclude that the number of solutions is 4(α 1 + 1)... (α k + 1). If v 2 (x) = v 2 (y) = β then obviously α 2β. Again let x 0 = x, y 2 β 0 = y, n 2 β 0 = n, our equation becomes 2 2β x 2 0 + y 2 0 = 2 α 2β n 0. Since the sum of two odd squares is even but not divisible by 4, we deduce that α 2β = 1. Now roceed similarly as in case 1, write j = (a j + b j i)(a j b j i) and 2 = (1 + i)(1 i), we deduce this reresentation x + yi = i s (1 + i) t (1 i) 1 t k j=1 (a j + b j i) βj (a j b j i) αj βj.

GAUSSIAN INTEGERS 11 where s {0, 1, 2, 3}, t {0, 1} and β j {0, 1,..., α j } j. The number of solutions is therefore 8(α 1 + 1)... (α k + 1). 4. Congruence Classes in Z[i] In the ring of integers Z, the congruence class modulo m of an integer n Z is the set [n] m = {x Z x n mod m}. The set of all congruence classes for a modulus m forms the ring Z/nZ, which is also known as the ring of integers modulo n. In this section, we will study the analogue of Z/nZ in the ring Z[i]. When n 0, Z/nZ is usually defined as Z/nZ = {[0], [1],..., [n 1]}. However, it is not so trivial to define the elements of Z[i]/αZ[i] for an arbitrary α Z[i]. From now on, we shall denote αz[i] by I α, the ideal generated by α. Our first observation is that Z[i]/I α is a finite ring. That is because from the roof of roosition 2.1, we know that for every β Z[i], there exists β such that β < α and β β mod α. Thus, Z[i]/I α {x Z[i] x < α }. Since the latter is a finite set, we deduce that Z[i]/I α is finite for a given α. We will denote the order (the number of elements) of Z[i]/I α by n(i α ). Lemma 4.1. Let α, β be non-zero elements of Z[i], then n(i α I β ) = n(i α )n(i β ). Proof. First note that I α I β = I αβ. That is because for any n 0, αβ n x i y i for x i s I α and y i s I β, so I α I β I αβ. Conversely, any t I αβ can be written as t = αβs, with α I α and βs I β, so I αβ I α I β. Thus I α I β = I αβ. Now assume Z[i]/I α = {[α 1 ],..., [α k ]} and Z[i]/I β = {[β 1 ],..., [β l ]}, where k = n(i α ) and l = n(i β ). Consider the set S consisting of all elements of the form α i β + β j, where 1 i k and 1 j l. We have S = kl, so it suffices to show that S is a comlete residue system modulo αβ. Indeed, for any x Z[i], we can write x = αβq + r. where r < αβ. There exists j 0 {1,..., l} such that r β j0 write r = βr 0 + β j0. Hence. x = αβq + βr 0 + β j0 = β(αq + r 0 ) + β j0. i=0 mod β, so we can Again, there exists i 0 {1,..., k} such that r 0 α i0 mod α, thus r 0 = αr 1 + α i0. Finally, we have this exression x = β(αq + αr 1 + α i0 ) + β j0 = αβ(q + r 1 ) + α i0 β + β j0. From this we deduce that for any x Z[i], there exists α i0 β + β j0 S such that x α i0 β + β j0 mod αβ. It remains to show that for u, v S, we have u v mod αβ if and only if u = v. Assume otherwise, there exists i 1, i 2 {1,..., k} and j 1, j 2 {1,..., l} such that α i1 β + β j1 α i2 β + β j2 mod αβ This imlies that β β j1 β j2, which yields j 1 = j 2. But then, we have αβ β(α i1 α i2 ), which is equivalent to α α i1 α i2, or i 1 = i 2. Hence, S is a comlete residue system modulo αβ, so n(i α I β ) = kl = n(i α )n(i β ).

12 HUNG HO For any ideal I α, denote I α = {a bi a + bi I α }. It is easy to check that I α is also an ideal of Z[i], and we call it the conjugate ideal of I α. Also, for α = a + bi Z[i], we shall denote N(α) as the norm of α, i.e. N(α) = a 2 + b 2. Lemma 4.2. The roduct ideal I = I α I α is the ideal generated by N(α). Proof. We already roved I α I β = I αβ for non-zero α, β in the roof of lemma 4.1. So it suffices to show that I α = I α. We have x I α x I α x = αy x = αy x I α. Thus, x I α if and only if x I α, which shows that I α = I α. Hence, I α I α = I αα = (N(α)). Lemma 4.3. n(i α ) = n(i α ). Proof. Let Z[i]/I α = {[α 1 ],..., [α k ]} and consider S = {α 1,..., α k }, we shall rove that Z[i]/I α = {[α 1 ],..., [α k ]}. Indeed, it suffices to show that S is a comlete residue system modulo α. Clearly, for distinct α i, α j S, we cannot have α i α j mod α, otherwise α α i α j, so α α i α j, or α i α j mod α, a contradiction. Moreover, for any x Z[i], there exists α k such that x α k mod α, hence x α k mod α. Therefore, we conclude that S is a comlete residue system modulo α, so Z[i]/I α = {[α 1 ],..., [α k ]}. Thus, n(i α ) = n(i α ) = n(i α ). Proosition 4.4. n(i α ) = N(α). Before tackling this roblem, we need some additional lemmas. However, first observe that from lemma 4.1, 4.2 and 4.3, we deduce that n(i α ) 2 = n(i α )n(i α ) = n(i αα ). So it remains to show that n(i αα ) = N(α) 2, i.e.we just have to rove roosition 4.4 for the case α Z. Lemma 4.5. Let α Z and β = a + bi Z[i], then α divides β in Z[i] if and only if α divides both a and b in Z. Proof. The reverse direction is trivial. For the forward direction, assume that α β, then we can write β = a + bi = α(c + di), where c, d Z. This is only ossible if a = αc and b = αd, so α divides both a and b in Z. Given α Z, we need to show that n(i α ) = α 2. This can be accomlished by finding α 2 elements of Z[i] that form a comlete residue system modulo α. Assume we have found our desired set S α = {α j = a j +b j i 1 j α 2 }. By lemma 4.5, we have α k α l mod α if and only if a k a l and b k b l mod α. This observation leads to the following result. Lemma 4.6. Let α Z, then S α = {a+bi 0 a, b α 1} is a comlete residue system modulo α. Proof. It follows immediately from lemma 4.5 that for x y S α, we cannot have x y mod α. Now consider any z = + qi Z[i]. Since α Z, there exists 0 and q 0 in {0, 1,..., α 1} such that 0 and q q 0 mod α. Then obviously

GAUSSIAN INTEGERS 13 z 0 + q 0 i mod α, where 0 + q 0 i S α. From this we conclude that S α is a comlete residue system modulo α. Corollary 4.7. Let α Z, then n(i α ) = α 2 = N(α). Proosition 4.4 follows from lemma 4.1, 4.2, 4.3 and corollary 4.7. Lemma 4.6 allows us to recisely define Z[i]/I α for α Z, but what about α Z[i] in general? Assume α = a + bi, can we have a comlete residue system modulo α consisting entirely of integers, namely 0, 1,..., a 2 + b 2 1? Proosition 3.2 and roosition 4.4 answer this question. Corollary 4.8. Let α = a + bi be a Gaussian integer, where a, b Z \ {0} are relatively rime in Z. Then Z[i]/I α = {[0], [1],..., [a 2 + b 2 1]}. Now we are ready to describe Z[i]/I α for an arbitrary α Z[i]. Proosition 4.9. Let α = a + bi Z[i] \ {0}. Assume gcd(a, b) = d in Z and a = da 0, b = db 0, then S α = {x + yi 0 x d(a 2 0 + b 2 0) 1, 0 y d 1} is a comlete residue system modulo α in Z[i]. Proof. Note that S α = a 2 + b 2, so by roosition 4.4 we just need to check that for β, γ S α, β γ mod α if and only if β = γ. Assume there exists β, γ S α such that β γ mod α. Let β = + qi and γ = r + si, we have d β γ, so by lemma 4.5, d r and d q s. However, since 0 q, s d 1, it follows that q = s. Now we also have a 0 + b 0 i β γ = r, and since a 0, b 0 are relatively rime, we deduce from roosition 3.2 that a 2 0 + b 2 0 r. Write r = s(a 2 0 + b 2 0), then we have α r d(a 0 + b 0 i) s(a 2 0 + b 2 0) d s(a 0 b 0 i). Again, by lemma 4.5 we deduce that d divides both sa 0 and sb 0. Suose that s is not divisible by d in Z. Then there must exists a rime in Z and a ositive integer m such that m d but m s. However, since d divides both sa 0 and sb 0, it follows that both a 0 and b 0 is divisible by, a contradiction because a 0 and b 0 are relatively rime. Therefore, d s, so d(a 2 0 + b 2 0) s(a 2 0 + b 2 0) = r. Now use the fact that 0, r d(a 2 0 + b 2 0) 1, we conclude that = r. Thus, β = γ, or S α is a comlete residue system modulo α. 5. Some imortant theorems and results In this section, we shall rove the analogues of some famous theorems and results of Z for the ring Z[i]. Firstly, recall the Euler s totient function ϕ(n) of a ositive integer n is the number of integers in {1, 2,..., n 1} that are relatively rime with n. In other words, ϕ(n) is the number of elements of (Z/nZ). We will now find the formula for the Euler s totient function of a Gaussian integer α, which denotes the number of elements of (Z[i]/I α ). Recall from theorem 2.5 that there are three tyes of rime: the slitting rimes a + bi and a bi where a 2 + b 2 is a rime of the form 4k + 1, the inert rimes where = 4k + 3 and the ramified rimes (1 + i) and (1 i). We will call them tye 1,2,3 resectively.

14 HUNG HO Theorem 5.1. Let α be a Gaussian integer, then ϕ(α) = N(α) η α η rime ( 1 1 N(η) Proof. Observe that from the Chinese remainder theorem, for α, β Z[i] that are relatively rime, we have ϕ(αβ) = ϕ(α)ϕ(β). Therefore, we only need to find the formula for ϕ(α k ), where α is a Gaussian rime. Also note that two associates have the same Euler s totient function. Consider three cases: (a) α = a + bi is a rime of tye 1. Let = a 2 + b 2 be a rime in Z and α k = (a + bi) k = c + di. We claim that gcd(c, d) = 1. Indeed, assume q Z is a common rime divisor of c and d, then q divides α k in Z[i]. Let q Z[i] be a rime divisor of q, then because α is a rime, it follows that q α. This means that q is an associate of α, so by roosition 3.2, we deduce that divides both c and d. But then α k, or a bi (a + bi) k, a contradiction because a bi and a + bi are relatively rime. So gcd(c, d) = 1, hence from corollary 4.8, S α k = {0, 1,..., c 2 + d 2 1} is a comlete residue system modulo α k. We will need to find all elements in S α k that are divisible by α. However, again by roosition 3.2, x S α k is divisible by α if and only if x is divisible by. Hence, we only need to find the number of elements divisible by in {0, 1,..., k 1} (note that c 2 +d 2 = (a 2 +b 2 ) k ). There are k 1 such numbers, so we deduce that ϕ(α k ) = k k 1 = N(α k )(1 1 N(α) ). (b) α is a rime of tye 2. Without loss of generality, assume α Z. By roosition??, we know that S α k = {x + yi 0 x, y α k 1} is a comlete residue system modulo α. We will need to find all elements divisible by α in S α k, which by lemma 4.5 are elements x + yi such that both x and y are divisible by α. Since x, y {0, 1,..., α k 1}, there are α 2(k 1) such elements. Thus, ϕ(α k ) = α 2k α 2(k 1) = N(α k )(1 1 N(α) ). (c) α = 1 + i First we show that β = a + bi Z[i] is divisible by 1 + i if and only if 2 a + b. Indeed, for the forward direction, let a + bi = (1 + i)(c + di), then we deduce that a = c d and b = c + d and aarently a + b = 2c, an even number. Conversely, suose a + b is even, then so is a b. Write α = b(1 + i) + a b, then since 1 + i 2 a b, it follows immediately that 1 + i α. Now a simle induction shows that (1+i) 2k = 2 k i k and (1+i) 2k+1 = 2 k i k (1+i). For the first case, we know that all elements x+yi where 0 x, y 2 k 1 form a comlete residue system modulo (1+i) 2k. There are 2 2 2k 2 = 2 2k 1 elements among these such that x+y is odd, which are also the elements relatively rime to (1 + i). Hence ϕ(α 2k ) = 2 2k 1. For the latter case, the comlete residue system modulo (1 + i) 2k+1 is S = {x + yi 0 x 2 k+1 1, 0 y 2 k 1}. There are 2 2 k 2 k 1 = 2 2k elements x + yi of S such that x + y is odd, hence ϕ(α 2k+1 ) = 2 2k. We conclude that ϕ(α k ) = 2 k 1 = N(α k )(1 1 N(α) ). From the three cases above, we conclude that for any α Z[i], ( ϕ(α) = N(α) 1 1 ). N(η) η α η rime ).

GAUSSIAN INTEGERS 15 We roved that every element of Z[i] can be factored into a roduct of finite irreducible elements and that this factorization is unique u to associates. It is easy to see that every rime α Z[i] has four associates including itself: α, α, iα and iα. For convenience, we introduce the following definition of rimary rime. Definition 5.2. A rime α Z[i] is called rimary if α 1 mod (1 + i) 3. Proosition 5.3. Every rime of tye 1 or tye 2 is associated to a unique rimary rime. Proof. Let α = a + bi be a rime of tye 1 or tye 2. It suffices to show that there exists a unique rimary rime in A α = {a+bi, a bi, b+ai, b ai}. First observe that β = (1+i) 3 = 2(1 i), hence by theorem 5.1, we know that (Z[i]/I β ) = 4. Obviously α is relatively rime to β, so A α (Z[i]/I β ). Therefore, if there does not exist x A α such that x 1 mod β, then since A α = 4, there must be y, z A α such that y z mod β. However, considering all ossible cases, this means that β 2α, or (1 i) α, a contradiction. This roof not only shows that there exists a rimary rime in A α, but this rimary rime is unique because we cannot have y z mod β for y, z A α. Remarks 5.4. If α = a + bi is a rimary rime, we can see that a is odd and b is even. From now on, rimes of tye 1 or tye 2 are assumed to be rimary. Proosition 5.5. Let be a rime of tye 2, then for any β Z[i], we have β β mod. Proof. Assume β = a + bi, then we have ( ) β = (a + bi) = a k b k i k. k k=0 Since ( k) is divisible by for all k 1, we deduce that β a + b i mod. By Fermat s little lheorem for Z, we have a a mod and b b mod. Moreover, since = 4k + 3 is a rime of tye 2, i = 1, and thus β a bi β mod. Proosition 5.6. Let α = a + bi be a rime of tye 1 and = a 2 + b 2 is a rime. Then for any β Z[i] we have β β mod. Proof. The roof is similar to that of roosition 5.5. Note that from roosition 5.5 we can also deduce that for any β relatively rime to : β +1 ββ β 2 1 (ββ) 1 mod β N() β mod. mod The final equation is a deduction from Fermat s little lheorem because ββ Z.

16 HUNG HO Proosition 5.5 and 5.6 imly the analogue statement of Fermat s little theorem in the ring Z[i]. Corollary 5.7. Let α, β Z[i], where α is a rime. Then we have β N(α) β mod α. Note that corollary 5.7 can be roved directly without roosition 5.5 and 5.6. Denote Z[i]/I α \ {0} = T α. We know that for β T α and every η T α, there exists a unique x η T α such that βη x η mod α. From this we deduce that β N(α) 1 η x η mod α η T α η T α β N(α) 1 1 mod α. We will finish this section with the concet of rimitive root modulo α. Definition 5.8. Let α be a non-zero element of Z[i]. An element z Z[i] is called a rimitive root modulo α if for every u such that u and α are relatively rime, there exists an integer n such that z n u mod α. Definition 5.9. Let α, β Z[i] be relatively rime elements. modulo α is the smallest ositive integer k such that β k 1 mod α. The order of β Observe that by corollary 5.7 for every β that is relatively rime to α we always have β N(α) 1 1 mod α, therefore there always exists an order of β modulo α. We denote the order of β modulo α by ord α (β). It is easy to see that if ord α (β) = k then β n 1 mod α if and only if k divides n (if not, then let n = kq + r, where r < k and we deduce that β r 1 mod α, a contradiction). Remarks 5.10. β is a rimitive root modulo α if and only if ord α (β) = N(α). Theorem 5.11. If α is a rime element of Z[i], then there always exists a rimitive root modulo α. Before roving theorem 5.11, we need the following lemmas. Lemma 5.12. Let α Z[i], where α is a rime. Then for any olynomial P (x) Z[x] of degree n such that the highest coefficient of P is relatively rime to α, the following congruence equation has at most n solutions in Z[i]/I α : P (x) 0 mod α. Proof. We will rove by induction on n. The claim is obviously true for n = 0, assume it is true for n = k, consider n = k+1. Assume for the sake of contradiction that there exists P (x), deg P = k + 1 and x 0, x 1,..., x k+1 Z[i]/I α such that P (x i ) 0 mod α i. Since P Z[x], we can write Q(x) = P (x) P (x 0 ) = (x x 0 )R(x), where R Z[x] is a olynomial of degree k. We know that Q(x i ) 0 mod for all i = 1, 2,..., k + 1. However since x i x 0 mod α, it follows that R(x i ) 0 mod α for all i = 1, 2..., k + 1. This is a contradiction because deg R = k, so by induction we cannot have k + 1 solutions for R(x) 0 mod α. Hence the claim is also true for n = k + 1, and thus true for all n.

GAUSSIAN INTEGERS 17 Lemma 5.13. For any ositive integer n, we have ϕ(d) = n. d n Proof. For any d n, let S d = {x x N, x d, gcd(x, d) = 1} and define the following ma: f d : S d {1, 2,..., n} x n d x. Denote f d (S d ) = T d. We wll rove T d T e = for distinct d, e n. Indeed, assume otherwise, there exists x 0 T d T e. Then there exists y 0 S d and z 0 S e such that x 0 = n d y 0 = n e z 0, or dz 0 = ey 0. Since gcd(d, y 0 ) = gcd(e, z 0 ) = 1, it folows that d e and e d, so e = d, a contradiction. Hence T d T e = when d e. Also note that T d {1, 2,..., n} for all d, hence we deduce that S d n. d n ϕ(d) = d n Now we will show that for any m {1, 2,..., n}, there exists m 1 n and n 1 S m1 such that m = n n 1 m 1. Let q = gcd(m, n) and write m = qm 1, n = qn 1, then gcd(m 1, n 1 ) = 1 and m = n n 1 m 1. In other words, for every m n, there exists m 1 n and n 1 S m1 such that f m1 (n 1 ) = m. From this we deduce that n d n S d = d n ϕ(d). Therefore, we can conclude that ϕ(d) = n. d n Lemma 5.14. For every d N(α) 1, the following equation has exactly d solutions in Z[i]/I α : x d 1 mod α. Proof. Let S d = {x d mod α x (Z[i]/I α )\{0}}, then for any y S d, y N(α) 1 d 1 mod α, so by lemma 5.14, S d N(α) 1. d On the other hand, assume S d = {d 1, d 2,..., d k }, for every i we denote [d i ] = {x x Z[i]/I α, x d d i mod α.} We claim that [d i ] = d for all i. Indeed, again by lemma 5.12, we have [d i ] d for all i. Now observe that [d i ] [d j ] = for i j and k [d i ] = (Z[i]/I α ) \ {0}. i=1

18 HUNG HO so we deduce that k [d i ] = N(α) 1 i=1 kd N(α) 1 k N(α) 1. d Thus, we must have equalities everywhere, so S d = N(α) 1 d and [d i ] = d i = 1, 2,..., N(α) 1 d. Obviously there exists i such that d i 1 mod α, so the equation x d 1 mod α has exactly d solutions in Z[i]/I α. Corollary 5.15. Let d N(α) 1, then the number of x Z[i]/I α such that ord α (x) = d is ϕ(d). Proof. For d N(α) 1, let O d = {x Z[i]/I α ord α (x) = d}. Now consider the equation x d 1 mod α with x Z[i]/I α. By lemma 5.14, this equation has exactly d solutions, and each solution x belongs to O e for some e d. Therefore, O e = d. (1) e d This holds for every d N(α) 1. Assume there exists d such that O d ϕ(d), choose the smallest d. Aarently d 1. But then, because every divisor e of d that is not d itself is strictly smaller than d, we have e d e d O e = e d e d ϕ(e). By lemma 5.13, the right hand side is d ϕ(d). Together with the equation at (1), we conclude that O d = d, a contradiction. Hence, O d = d for all d N(α) 1. Theorem 5.11 follows immediately from corollary 5.15. 6. Quadratic Recirocity In this section, we will rove quadratic recirocity laws for the ring Z[i]. First we will recall imortant definitions and results about quadratic recirocity laws for the ring of integers. Definition 6.1. Let be a rime and a be any ositive integer. We define the Legendre s symbol ( ) a = 1 if there exists n Z such that n 2 a mod 0 if divides a 1 otherwise. The Legendre( symbol ) ( indicates ) ( ) whether ( ) a is a quadratic residue modulo. It is easy to see that = and = 1. ab a Theorem 6.2. Let, q be two odd rimes. Then we have ( ) ( ) q = ( 1) 1 q 1 2 2. q b a 2

GAUSSIAN INTEGERS 19 Proosition 6.3 (Sulementary Laws). Suose is an odd rime. Then we have: ( ) 1 (1) = 1 if and only if = 4k + 1. ( ) 1 (2) = ( 1) 2 1 8. Theorem 6.2 is called the Gauss recirocity law and is the most imortant result regarding quadratic recirocity over Z. We will aly the above laws to rove the analogous laws in the ring Z[i]. Definition 6.4. Let α be a Gaussian rime and β be an arbitrary element of Z[i]. We define the following symbol [ ] β = α 1 if there exists η Z[i] such that η 2 β mod α 0 if α divides β 1 otherwise. Similar to the Legendre symbol, the above symbol is also multilicative. Now we introduce the Euler s criterion for the ring Z[i]. Proosition 6.5. Let α be a rime of tye 1 or 2 in Z[i] and β be an arbitrary element of Z[i] not divisible by α. Then we have [ ] β N(α) 1 β 2 mod α. α Proof. Let η be the rimitive root modulo α. Then for every β there ] exists a unique d(β) such that η d(β) β mod α. An easy observation is that = 1 if and only if d(β) is even. Moreover, β N(α) 1 2 1 mod α N(α) 1 d(β) η 2 1 mod α d(β) 0 mod 2. The last equation is due to the fact that η is a rimitive root modulo α, hence η k 1 mod α if and only if k is divisible by N(α) 1. [ ] [ ] Now we conclude that β N(α) 1 2 1 mod α if and only if β α = 1, or β N(α) 1 2 β α mod α. Now observe that if α is a rime of tye 2 then N(α) 1 = α 2 1, which is divisible by 4, so for any integer r, r N(α) 1 2 is a erfect square. Therefore, if α is a rime of tye 2, then any integer is a quadratic residue modulo α. Corollary 6.6. Suose α is a rime of tye 2. Let r Z be relatively rime to α. We have: [ r α] = 1. Proosition 6.7. Suose α is a rime of tye 1 and let r Z be relatively rime to α. Then we have [ ( ) r r =. α] N(α) [ β α

20 HUNG HO [ ( ) r r Proof. We will rove that = 1 if and only if = 1. The reverse α] N(α) direction is obvious since if there exists x Z such that r x 2 mod N(α) then we also have r x 2 mod α because α N(α). For the forward direction, assume η Z[i] such that r η 2 mod α. By corollary 4.8, we know that there exists u Z such that η u mod α, hence r u 2 mod α. This means( that ) α r u 2, so by roosition 3.2, N(α) r u 2. From this we conclude that = 1. r N(α) Proosition 6.7 is very useful in terms of relating quadratic recirocity laws in Z[i] to those of Z. For instance, from corollary 4.8 we know there exists u {0, 1,..., N(α) 1} such that u i mod α, so for any r, s Z, r + si r + su mod α. In terms of quadratic recirocity, we have [ ] r + si = α ( r + su N(α) Proosition 6.8. Suose α = a + bi is a rime of tye 1. Let = N(α). We have: ( ) a = 1. Proof. We know that a is odd and = a 2 = b 2. Assume q 1, q 2,..., q k are all rime divisors of a such that the highest ower of q i that divides a is odd for all i,. In other words, there exists an odd number d i such that q di i a. Aarently, then ( ) ( ) a q1 q 2... q k k ( ) qi = = ). i=1 By theorem 6.2 and the fact that = 4k + 1, we have ( ) ( ) qi =. ( ) However, since b 2 mod q i for all i, we deduce that q i = 1, hence ( ) for all i. From this we conclude that = 1. Corollary 6.9. With the above notations, we also have: ( ) ( ) b 2 =. Moreover, using the same technique, if b = 2 n c, where c is odd then a q i ( ) qi = 1 ( ) c = 1. Theorem 6.10. Let α, β be two different rimes that are not tye 3 in Z[i]. Then we have [ ] [ ] β α =. α β Proof. Consider the following cases: (a) α, β are both rimes of tye 2. It follows easily from corollary 6.6 that [ ] β α = [ α β ] = 1.

GAUSSIAN INTEGERS 21 (b) α, β are both rimes of tye 1. Let α = a + bi and β = c + di. Let u, v Z such that u i mod α and v i mod β. Note that from u i mod α we have bu bi bu a mod α mod α bu a mod a 2 + b 2. The last equation is due to roosition 3.2. Similarly we have dv c mod c 2 + d 2. Now by roosition 6.7 and corollary 6.9, we have: [ ] ( ) β c + du = α a 2 + b [ ] ( ) ( 2 ) β b bc ad α a 2 + b 2 = a 2 + b [ ] ( 2 ) ( ) β 2 bc ad = α a 2 + b 2 a 2 + b 2. Similarly, we can also rove [ ] α = β So it suffices to show ( 2 c 2 + d 2 ( 2 c 2 + d 2 ) ( ) bc ad c 2 + d 2. ) ( ) ( bc ad 2 c 2 + d 2 = a 2 + b 2 Let b = 2b 0 and d = 2d 0. It remains to show that ( ) ( ) b0 c ad 0 b0 c ad 0 c 2 + d 2 = a 2 + b 2. ) ( ) bc ad a 2 + b 2 Let b 0 c ad 0 = 2 k e, where e is odd. Because 4(b 0 c ad 0 ) 2 +(ac+4b 0 d 0 ) 2 = (a 2 + b 2 )(c 2 + d 2 ), by reeating ( the ) same ( argument ) used in the roof of roosition e e 6.7, we know that c 2 + d 2 = a 2 + b 2 = 1. If b 0 +d 0 is odd, then b 0 c ad 0 is also odd because a and c are odd, and we are done. Otherwise, if b 0 + d 0 is even, then 4(b 2 0 d 2 0) is divisible by 8, and since a 2 c 2 is also divisible by 8 because both a and c are odd, we deduce that N(α) N(β) is divisible by 8. Hence, N(α) 2 N(β) 2 By roosition 6.8, it follows that ( ) 2 a 2 + b 2 = mod 16, which means that ( 1) N(α)2 1 8 = ( 1) N(β)2 1 8. ( ) 2 c 2 + d 2. Therefore, either case we will eventually have comletes our roof. (c) α is a rime of tye 1 and β is a rime of tye 2. e have β Z, so by roosition 6.7 [ β α ] = ( β N(α) ) = 1. ( ) ( ) b 0c ad 0 b c 2 +d = 0c ad 0 2 a 2 +b. This 2

22 HUNG HO But by theorem 6.2 and the fact that N(α) is a rime of the form 4k + 1 in Z, we have ( ) ( ) N(α) β = = 1. β N(α) ) [ ] So it suffices to show that = 1 if and only if = 1. The reverse ( N(α) β direction is trivial, because if we assume there exists η Z[i] such that η 2 α mod β, then we also have η 2 α mod β, and thus N(α) N(η) 2 mod β. For the forward direction, assume there exists s Z such that s 2 N(α) mod β. By roosition 5.5, we have α β+1 N(α) mod β. Therefore, α β+1 s 2 mod β α N(β) 1 2 s 1 mod β α N(β) 1 2 1 mod β. The last equation is due to[ Fermat s ] little theorem, and now we can conclude α from Euler s criterion that = 1. β Acknowledgement It is a great leasure to thank my mentor, Tung Nguyen, for heling me throughout the REU rogram. This aer would not have been comleted without his invaluable suort and guidance. I d also want to thank Professor Peter May and the University of Chicago for roviding us with such a great learning and researching exerience. References [1] David Steven Dummit and Richard M. Foote. Abstract Algebra. 3rd ed. Hoboken, NJ : Wiley, c2004. 1991. [2] Kenneth Ireland and Michael Rosen A Classical Introduction to Modern Number Theory Sringer, 1990. α β