6 Gaussian Integers and Rings of Algebraic Integers One way that Euler, Lagrange, Jacobi, Kummer and others tackled Fermat’s Last Theorem was to try to show that the equation xn + yn = zn had no non-zero solutions in a ring containing the integers. Ultimately, proofs for a large number of exponents were obtained this way. Many other number theory problems can be approached similarly. Perhaps the simplest example of such a ring is the following: Definition 6.1. The Gaussian integers are the set Z[i] = x + iy : x, y Z of complex numbers f 2 g whose real and imaginary parts are both integers. Z[i] is a ring (really a subring of C) since it is closed under addition and multiplication: (x + iy) + (p + iq) = (x + p) + i(y + q), (x + iy)(p + iq) = (xp yq) + i(xq + yp) − Note how the second follows from the fact that i satisfies the quadratic polynomial i2 + 1 = 0. 6.1 The Division and Euclidean Algorithms for the Gaussian Integers Our first goal is to develop unique factorization in Z[i]. Recall how this works in the integers: every non-zero z Z may be written uniquely as 2 z = upk1 pkn 1 ··· n where k ,..., k N and, more importantly, 1 n 2 • u = 1 is a unit; an element of Z with a multiplicative inverse ( v Z such that uv = 1). ± 9 2 • p ,..., p are distinct irreducibles/primes; a p = a = 1 or p. 1 n j ) ± ± To do the same in Z[i], we first need to identify the ingredients: what are the units and the Gaussian primes? Since primality depends on divisibility, we start by working towards a division algorithm. Definition 6.2 (Divisibility). Givena a, b Z[i], we write a b g such that b = ga. 2 j () 9 aFor the duration of this chapter, Greek letters denote complex numbers and Latin letters integers. Examples 6.3. With the exception of 0 0, testing for divisibility requires the usual ‘multiply by the j conjugate’ trick: 7 + i (7 + i)(1 3i) 10 20i 1. Since = − = − = 1 2i Z[i], we conclude that 1 + 3i 7 + i. 1 + 3i 10 10 − 2 4 + i (4 + i)(1 3i) 7 11i 2. Since = − = − Z[i], we see that 1 + 3i - 4 + i. 1 + 3i 10 10 62 To obtain a version of the division algorithm we need some a notion that the remainder be smaller than a divisor. Thankfully, the modulus provides such for the complex numbers. In part to avoid nasty square roots, we instead use the square of the modulus. 1 Definition 6.4. The norm of a Gaussian integer a = x + iy is N(a) := a 2 = x2 + y2 j j The norm has a couple of useful basic properties analogous to the absolute value in Z: Lemma 6.5. 1. (Multiplicativity) N(ab) = N(a)N(b): this holds for any complex numbers a, b. 2. (Units) a is a unit if and only if N(a) = 1, whence Z[i] has precisely four units: 1, i. ± ± Proof. 1. Just multiply it out. We get almost the same formula as when we discussed sums of squares: if a = x + iy and b = u + iv, then N(ab) = (xu yv)2 + (xv + yu)2 = (x2 + y2)(u2 + v2) = N(a)N(b) − 2. a Z[i] is a unit if and only if b Z[i] such that ab = 1. Taking norms, we see that 2 9 2 N(a)N(b) = N(1) = 1 Since norms are positive integers, this says that N(a) = 1. Conversely, N(a) = 1 = aa = 1 so that a is a unit. ) Theorem 6.6 (Division algorithm). Let a, b Z[i] with b = 0. Then g, r Z[i] for which 2 6 9 2 a = bg + r and N(r) < N(b) The proof is somewhat algorithmic, so we obtain it before computing an example. First observe: • The norm gives the required notion that the remainder r be smaller than the divisor b. • Unlike with the division algorithm in Z, we make no claim that g and r are unique. r N(r) r < Proof. First reframe the goal. Since N( b ) = N(b) , our job is to find r such that N( b ) 1. Since the norm is distance squared, and we want r a = g b b − it is enough for us to find any g Z[i] within distance 1 of a . This is trivial to do, since the Gaussian 2 b integers form an equally-spaced lattice in the complex plane. Let g be any suitable value and define r = a bg, whence − a N(r) = N(a bg) = N g N(b) < N(b) − b − a We could have introduced uniqueness by choosing the closest g to b , preferenced by lexicographic ordering in case of a tie, but this would be unhelpfully messy! 2 a 4+5i Examples 6.7. 1. Let a = 4 + 5i and b = 3. Then b = 3 is iR within unit distance of four Gaussian integers. In this case 3i we have N(b) = 9 and we may write the division algorithm in four different ways 2i 4+5i 4 + 5i = 3(1 + 2i) + (1 i) N(r) = 2 − 3 = 3(2 + 2i) + ( 2 i) = 5 − − i = 3(1 + i) + (1 + 2i) = 5 = 3(2 + i) + ( 2 + 2i) = 8 − 0 R 2. Let a = 2 + 7i and b = 1 + 2i. Then 0123 iR a (2 + 7i)(1 2i) 16 + 3i = − = 4i b 5 5 3i Choose g = 3 + i and write 2i 2 + 7i = (1 + 2i)(3 + i) + 1 where N(r) = N(1) = 1 < 5 = N(b), as required. Two i 16+3i a 5 other Gaussian integers lie within unit distance of b : the corresponding statements of the division algorithm are 0 01234 R 2 + 7i = (1 + 2i) 3 + ( 1 + i) N(r) = 2 · − = (1 + 2i)(4 + i) 2i = 4 − Greatest Common Divisors and the Euclidean Algorithm The discussion of greatest common divisors and the Euclidean Algorithm is almost identical to that in the integers, though there are some subtleties. Definition 6.8. Given a, b Z[i], consider the set 2 S = ka + lb : k, l Z[i] f 2 g A greatest common divisor d of a and b is a non-zero element of S with minimal norm. The definition is algebraically nice, but awkward to compute with directly, except in special cases. Example 6.9. If we take a = 2 + i and b = i, observe that 1 = (1 + i)(2 + i) 3i S − 2 Since no element of S could possibly have smaller positive norm, we conclude that 1 is a greatest common divisor. As previously, we’d say that a, b are relatively prime. Note that there are four distinct greatest common divisors: multiplying the above by any unit shows that 1, i S. Since no other complex numbers have norm 1, these are all the greatest common ± ± 2 divisors. 3 Trying to guess some linear combination of a, b as above is hopeless in general. Thankfully the Eucldiean Algorithm will provide a systematic method. Before seeing this, we check that greatest common divisors really have the properties their name leads us to expect! Theorem 6.10. Let d be any gcd of a, b. Then: 1. (Common divisor) d divides both a and b; 2. (Maximality) Every common divisor of a, b divides d (among all such, d has maximum norm); 3. (Four gcds) S = md : m Z[i] . In particular, d, id are all the greatest common divisors. f 2 g ± ± Since you should have seen a version of this for the integers, we leave the proof to the Exercises. It is legitimate to write d = gcd(a, b) as long as you appreciate (part 3) that the gcd is not unique! It is common to normalize a gcd by insisting that x > 0 and y 0 where d = x + iy: exactly one of the ≥ four choices will satisfy this. Now suppose we have a line from the division algorithm: a = gb + r = ka + lb = (l + k)b + kr ) The sets S generated by the pairs (a, b) and (b, r) are therefore identical, and we conclude that gcd(a, b) = gcd(b, r)! The Eucldiean algorithm now proceeds as in the integers: simply iterate the division algorithm and the last non-zero remainder will be a greatest common divisor. Since division of complex numbers is messier than in the integers, any given example of the Eu- clidean algorithm takes much longer. Examples 6.11. 1. We compute gcd(4 + i,3 ). 4+i Observe that 3 is within unit distance of four Gaussian integers 1, 2, 1 + i, 2 + i. We perform the calculation with the first of these. 4 + i = 1 3 + (1 + i) N(1 + i) = 2 < 9 = N(3) · 3 = (1 i)(1 + i) + 1 N(1) = 1 − The algorithm can be reversed to obtain gcd(4 + i,3 ) = 1 = 3 (1 i)(1 + i) = 3 (1 i) 4 + i 3 − − − − − = (2 i) 3 + (i 1)(4 + i) ( ) − · − ∗ Alternatively, we could have done the following: 4 + i = 2 3 + (i 2) N(i 2) = 5 · − − 3 = ( 1 i)(i 2) + ( i) N( i) = 1 − − − − − from which gcd(4 + i,3 ) = i = 3 + (1 + i)(i 2) = 3 + (1 + i)(4 + i 2 3) − − − · = ( 1 2i) 3 + (1 + i)(4 + i) − − · This is simply ( ) multiplied through by i. ∗ − 4 2. Find gcd(7 + 17i,8 14i). First apply the division algorithm: − a 7 + 17i (7 + 17i)(8 + 14i) 7 + 9i = = = − b 8 14i 82 + 142 10 − so we choose g = 1 + i = 7 + 17i = ( 1 + i)(8 14i) + 1 5i − ) − − − Applying again: 8 14i (8 14i)(1 + 5i) − = − = 3 + i Z[i] = gcd(7 + 17i,8 14i) = 1 5i 1 5i 26 2 ) − − − Indeed we have (3 + i)( 7 + 9i) 8 14i = (3 + i)(1 5i) = 7 + 17i = − (1 5i) = ( 3 + 2i)(1 5i) − − ) 10 − − − Normalizing the gcd, we can write gcd(7 + 17i,8 14i) = 5 + i = i(1 5i) = i(7 + 17i) + (1 + i)(8 14i) − − − Exercises 1.