Euclidean Domains and the Gaussian Integers : An Application Shreejit Bandyopadhyay July 28, 2013 Abstract Commutative algebra is essentially the study of commutative rings and in this paper, we shed some light on a highly important class of such rings - the Euclidean domain. We then show that the ring of Gaussian integers is an example of such a domain and thereby prove a famous theorem in number theory, due to Fermat, which states that any prime number p, of the form 4n + 1, can be expressed as the sum of squares of two integers while no prime of the form 4n + 3 can be so expressed. 1 Introduction Before we begin our study of Euclidean domains, we'd like to give a few defini- tions to begin our discussion. Definition. A ring is a set R in which are defined two operations, often called the sum and product, and denoted by + and . respectively, which satisfy the following axioms : 1. With respect to the operation +, R is an abelian group. So R has an additive identity, denoted by 0, and for every a 6= 0 2 R, −a 2 R too. The + operation is both associative and commutative. 2. The operation . in R is associative and has the identity element 1. 3. The distributive law a:(b + c) = a:b + a:c holds for all a; b; c 2 R. If, in addition, the operation . in R is commutative, R is said to be a commutative ring. All rings we consider in this paper will be commutative. Definition. An element a 6= 0 in a commutative ring R is said to be a zero divisor if there exists a b 6= 0 in R, such that ab = 0. Definition. A commutative ring R is said to be an integral domain if it has no zero divisors, i.e, if in R, ab = 0 implies either a = 0 or b = 0 (or both). 1 Definition. An element a 6= 0 in a commutative ring R is said to be a unit if it has a multiplicative inverse in R, an element b 6= 0 such that ab = ba = 1, in R. Definition. An integral domain R is said to be an Euclidean domain if in R there is defined, for every a 6= 0, an integer d(a) ≥ 0 satisfying : (i) For all a; b 2 R, d(a) ≤ d(ab), equality holding iff b is a unit in R. (ii) (Division algorithm) For all a; b 2 R, there exist q; r 2 R satisfying a = qb+r with r = 0 or d(r) < d(b). With these definitions in our hand, we now embark on a brief study of Euclidean domains in general and the domain of Gaussian integers in particular, which would eventually enable us to prove the so-called two-square theorem of Fermat. 2 Euclidean Domains In this section, we review some properties of Euclidean domains which will carry over to the ring of Gaussian integers, which, as we shall see, is an example of such a domain. Before we carry on, however, a few more definitions will be necessary. As we've already specified, all rings we discuss will be commutative ones. Definition. An ideal I of a ring R is a subset of R satisfying the following conditions : 1. With respect to the operation +, I is a subgroup of R. 2. For all a 2I and all r 2R, ra 2I (and hence, by commutativity, ar 2I, too). Definition. Given an ideal I of a ring R, a set (a1; a2; :::; an) of elements of I is said to generate I if every a 2I can be written as a linear combination of P these ais with coefficients in R, i.e., a = riai with ri 2R, for some i. In this case, we write I=(a1; a2; :::; an). Definition. An integral domain R is said to be a principal ideal domain (PID) if every ideal of R is generated by one element, i.e, if for any ideal I of R, there is an a 2I such that the elements of I are precisely the elements ra; r 2 R. Theorem. An Euclidean domain is a principal ideal domain. Proof. Let I be an ideal of an Euclidean domain R. If I contains only the element 0, there's nothing to prove. Otherwise, I has an element a0 6= 0. We choose the non-zero element a in I with the least d-value, i.e, with the least value of d(a). Note that since the d-values are non-negative integers, this is always possible. 2 We now show that all elements in I are actually multiples of a. Since the very definition of ideal forces every multiple of a to be in I, it will show that I=(a), proving the theorem. Let b 2I be an arbitrary element. Then b = qa + r, for some q; r 2I, with r = 0 or d(r) < d(a). Since a 2I, so is qa and b 2I also holds. So, by the definition of an ideal, r = b − qa 2I too. But then, by our choice of a, d(r) < d(a) is impossible. So, r = 0 and b = qa is a multiple of a. Definition. Let R be a commutative ring. Given a; b 2R, a greatest common divisor of a and b in R, denoted by (a; b), is an element d 2R such that 1. dja and djb in R. 2. If cja and cjb in R, cjd in R as well. Theorem. Let R be an Euclidean domain. Any two elements a; b 2R has a greatest common divisor in R. Moreover, d = pa + qb, for some p; q 2R. Proof. Let I=(ma + nb : m; n 2R). We claim that I is an ideal of R. This holds because if x = m1a+n1b and y = m2a+n2b are in I, x+y = (m1 +m2)a+(n1 + n2)b 2I since (m1 + m2); (n1 + n2) 2R. Also, if r 2R, rx = (rm1)a + (rn1)b 2I since rm1; rn1 2R. Once I has been shown to be an ideal of R, by the previous theorem, we can conclude that I=(d) for some d 2R. All elements of I are then multiples of this d. Since 0 2R and 1 2R, 1:a + 0:b 2I and 0:a + 1:b 2I both hold, i.e, a; b 2I. So, both a and b must be multiples of d, that is, dja and djb. Also, d being an element of I, d = pa + qb for some p; q 2R. So, if for some c 2R, cja and cjb, then cjpa + qb, i.e., cjd. So d certainly satisfies both conditions for being the greatest common divisor of a and b, proving the assertion of the theorem. Note that in this theoem we make no comments about the uniqueness of the gcd of the two elements. Indeed, it need not be unique at all, but there is a relation between two different gcds of two elements of an Euclidean domain. What this relation is will be seen in the following lemma. Definition. Let R be a commutative ring. Then two elements a and b in R are said to be associates if b = ca and the element c is a unit in R. Lemma 1. In an Euclidean domain R, any two greatest common divisors of two elements a and b are associates of each other. Conversely, any associate of a greatest common divisor of a and b is itself a greatest common divisor of these two elements. Proof. Let c and d be two gcds of a and b in R. Then, by condition 2 for gcd above, cjd and djc. So, c = dk and d = cl for some k; l 2R. So, d = cl = (dk)l = d(kl). Since d 6= 0 and R is an integral domain, kl = 1 and so k is a unit in R. Since c = dk, we see that c and d are indeed associates of each other. 3 Conversely, let d be a gcd of a and b in R and let c be an associate of d, i.e, c = du, where u is a unit in R. Then, d being a gcd of a and b, dja and djb, i.e, a = dp and b = dq for some p; q 2R. But, u being invertible in R, this means that a = c(u−1p) and b = c(u−1q) and (u−1p); (u−1q) 2R. So, cja and cjb in R also hold. Again, if gja and gjb, then since d is a gcd of a and b, djg, i.e, g = dm for some m 2R. But then, g = c(u−1m) and so cjg. So, c satisfies both conditions for being the gcd of a and b. In an Euclidean domain, if the gcd of a and b is a unit, we say that a and b are relatively prime. Since any associate of a gcd is a a gcd and 1 is an associate of any unit in a ring, if a and b are relatively prime, we may safely assume that (a; b)=1 and conversely. Definition. In an Euclidean domain R, an element a 2R is said to be a prime element if it's not a unit in R and if it can't be expressed as a product of two non-units in R. That is, whenever a = bc, either b or c is always a unit in R.