The Euclidean Algorithm in Quadratic Number Fields

Transcription

1 The Euclidean Algorithm in Quadratic Number Fields Sarah Mackie Supervised by: Prof. Christopher Smyth September 6, 06 Abstract The purpose of this report is to investigate the Euclidean Algorithm in quadratic number fields. The method originally constructed by Euclid to determine unique prime factorisation of element rings uses the norm function, however, some fields exist which have prime factorisation, but are not norm-euclidean. This report studies different methods used to show a ring is Euclidean. Introduction Anybody who has attained a reasonable level of high school mathematics will have come across Euclid s algorithm in relation to the field of integers. The algorithm is used to find the greatest common divisor of a pair of numbers, break down numbers into their prime factorisation and show that rings are principal ideal domains. This is often done using the norm function. What is much less commonly known, or taught, is that fields can also be Euclidean for other functions. What most students know as a Euclidean ring, is, in fact, a norm-euclidean ring. Definition A domain, R, is said to be a Euclidean domain if there exists a function p(x) defined on R whose values are non-negative integers satisfying the following: if a, b R, b non-zero, then one can find c, r R such that a = bc + r where r = 0 or p(r) < p(b) and p(x) = 0 if and only if x = 0. Theorem If R is euclidean, then every ideal in R is a principal ideal. Where we have K, an algebraic number field, O K denotes the ring of integers of K. When it is stated that K is Euclidean what is meant is that O K is a Euclidean domain. A domain R is said to be norm-euclidean if the above definition holds when p is the absolute value of the norm function for the domain R. When considering the real numbers p(x) is simply x. The norm function in quadratic number

2 fields will be discussed later in definition 3. Definition The class number of a field, K, is the order of the ideal class group of it s ring of integers, O K. Euclidean fields have class number so the hunt for Euclidean fields contributes to the hunt for fields of class number. This result led to the conclusive proof showing which imaginary quadratic fields are Euclidean. Quadratic Fields Lets consider fields of the form Q ( m ) where m is an integer. Theorem The integers of Q ( m ) are of the form if m mod 4 or 3 mod 4 and a + b m a + b( m ) if m mod 4, a and b are rational integers in both cases [3, p. 07]. Definition 3 The norm, Nx, where x = r + s m, is Nx = x x = (r + s m)(r s m) = r ms.. Complex Euclidean Fields A complex field is such that for m < 0 the field Q ( m ) is Euclidean if given any δ Q ( m ), there exists an integer x Q ( m ) such that N(δ x) <. () Theorem 3 There are five complex norm-euclidean quadratic fields: m =,, 3, 7,. Consider m mod 4 or 3 mod 4, then Q ( m ) is norm-euclidean if for each integer x = u+v m there is an ellipse on the complex plane, (x u) m(y v) =. It is easy then to visualise whether a field is norm Euclidean, by plotting the ellipses and determining whether the entire field is covered. For example, consider Q ( ), the fundamental domain to consider is 0 x, 0 y i. From figure it can be deduced that the field is Euclidean since the area inside the red box is covered by the 4 surrounding circles centred at (0, 0), (0, ), (, 0), (, ) which are each integers in the field. Figure shows us that as the magnitude of m increases the spacing of the integers increases, but as we see m = is still Euclidean.

3 To see the role of the integers in a field where m mod 4 m = 3 is plotted in figure 3. This time the fundamental domain is slightly different than in the previous two cases where the rectangle 0 x, 0 y m is used. Here we have the region inside (0, 0) (, 3 ) ( 3, 3 ) (, 0) (0, 0) as the fundamental domain. This region is smaller than if only the integers a + b m were used and so we still have norm-euclidean fields where m >. Finally, an example of a field which is not norm Euclidean is shown in figure 4. Here we have the integers of Q ( 5 ) plotted and the regions around them which contain points that are contained by equation. As we can see, the fundamental domain here is not covered by the circles and, thus, the field is not norm Euclidean. It is proven that the five fields in theorem 3 are the only complex norm-euclidean fields in Hardy and Wright [3, p. 75] and Narkiewicz [7, p. ]. There exist 4 other imaginary fields: m = 9, 43, 67, 63 () which are principal ideal domains, class number, which are not Euclidean for any function, p. This can be shown using Motzkin s characterisation described in section 5. Thus we have counterexamples to the statement: if a domain, K, is a principal ideal domain then K is Euclidean. Thus, it is not enough to show that a domain is a principal ideal domain to prove it is Euclidean. The work considering complex quadratic fields is, therefore, complete and the remainder of this paper will focus on real quadratic fields. 3

4 0 0 Figure : Q(i) Figure : Q ( ) 4

5 Figure 3: Q ( 3 ) Figure 4: Q ( 5 ) 5

6 . Real Euclidean Fields We will now consider when m is a positive integer. Theorem 4 Q ( m ) is norm-euclidean for m =, 3, 5, 6, 7,, 3, 7, 9,, 9, 33, 37, 4, 57, 73. In this case, the area where N(δ x) < covers an area bounded by two hyperbolas. We can only prove Theorem 4 is true for m =, 3, 5, 6, 7,, 3, 7, 9,, 9 which we find in [3, p. 77]. In Cooke [] figures demonstrating this for m =, 3, 6, 7 can be found. 3 k-stage Euclidean Domain There exist fields which are not norm-euclidean for the usual norm function, but are k-stage Euclidean. When refering to R, R is an integral domain. Definition 4 A domain, R, is said to be stage Euclidean if for all pairs (α, β), where α, β R and β 0, there exists q, q R such that where N(r ) < N(β) or r = 0. α = qβ + r and β = q r + r (3) Definition 5 A domain, R, is said to be k-stage Euclidean if for all pairs (α, β), where α, β R and β 0, there exists a sequence of equations α = q β + r β = q r + r r = q 3 r + r 3. r k = q k r k + r k where N(r k ) < N(β) or r k = 0. The division chain is terminating if r k = 0. A division chain is defined by the quotients q,..., q k so that a k-stage division chain for (α, β) is defined. This leads onto the definition of continued fractions with coefficients in R. If N is the norm acting on R where N is multiplicative then N extends to the fraction field K R so that N( α β ) = N(α) N(β). 3. Continued Fractions For r R the process of finding the continued fraction form of r is iterative. r is represented as its integer part, n, and the reciprocal of another number, this number is then written as the sum of its integer part, n and the reciprocal and then again as before, continuing on until an integer, n k, is reached on its own 6

7 or to infinity. The continued fraction is represented as [n,..., n k ]. For example: 37 4 = = = + = + = + = So, 37 4 = [,,,, 4] as a continued fraction Thus, we reach the definition of continued fractions for an integral domain, R. Definition 6 For a sequence q,..., q k of non-zero elements in R, the continued fraction [q,..., q k ] in K is defined by [q ] = q = a b where a = q, b = [q, q ] = q + q = a b where a = q q +, b = q [q,..., q k ] = a k b k where a k = q k a k + a k, b k = q k b k + b k Corollary For (α, β) R, there is a k-stage division chain with N(r k ) < N(β) if and only if there exists a continued fraction a k b k such that N( α β a k b k ) < N(b k ). (4) Understanding this definition is needed to prove that certain fields are n-stage Euclidean. As, if there exists a continued fraction [q,..., q k ] = a k b k of length k n for every α β K then R is n-stage Euclidean. 7

8 3. Q ( 4 ) is -stage Euclidean Let R be the ring of integers of K where K = Q ( 4 ). Embed K in R n so that a + b 4 (a, b n). The fundamental domain, D 4 is 0 x and 0 y < 4. For continued fractions a k b k define the set V ( a { } k ) = x R N( a k x ) < b k b k N(b k ) An integral basis for Q ( 4 ) is {, 4} and = (4 + 4)(4 4) = 4 4 = = = From the definition of a continuous fraction we see that [, 4 4] = 4, thus we have a continuous fraction of length and denominator with norm. This also holds for 4 + therefore we have hyperbolic regions V ( 4 ) and V ( 4 + ). The regions intersect the region 0 x so that all (x, y) where 4 3 < y 4. The lower branch of the boundary V ( 4 ) is x ( y The upper branch of the region V () is 4 ) =. (5) (x ) y =. (6) Since (5) is convex and lies completely beneath (6) which is concave (see figure 5) the points in D 4 K satisfying y < 4 are covered and therefore Q[ 4] is -stage Euclidean. If it is known that a field Q[ n] is k-stage Euclidean then we can begin to investigate whether there exists a different function for which Q[ n] is a Euclidean domain. 8

9 3.5 Equation 6 Equation Figure 5: Showing the coverings of V ( 4 ) and V (). 9

10 4 Z ( 4 ) and a Pairwise Algorithm In [6] Nagata seeks to show Z ( 4 ) is a Euclidean ring by seeking to find a pairwise algorithm. Thus, it is helpful to outline what a pairwise algorithm is: Definition 7 Let E be a ring and U the set of units in E. Let S be an ordered set with minimum condition. Let the class of pairs of elements (a, b) be {(au, bv) u, v U} denoted by (a, b) and let P be the set of all classes. p : P S, x, y E is a pairwise algorithm iff:. If b ae and b / au, then p(a, a) < p(b, b).. If b c ae, then p(a, b) = p(a, c). 3. For each (a, b), there are q, r E so that b = qa + r with either r = a or p(r, a) < p(a, b). Nagata concluded that if a ring has a pairwise algorithm, then E is a principal ideal ring. The proof of this can be found in [6]. It can be tempting to assume that if a ring is a principal ideal ring then it is Euclidean, but this is not be true. Thus, although Nagata found a pairwise algorithm for Z ( 4 ) he could not conclusively say that it is Euclidean. 5 Motzkin s Lemma In [5] a characterisation for Euclidean domains is defined. Its original purpose was to show that the four imaginary quadratic fields listed in section 3, equation, were not Euclidean for any function, p. It was proposed by Samuel [8] that the characterisation could be used to also show that some rings are Euclidean. His suggestion of applying the characterisation to Z ( 4 ) was taken on by Harper [4]. Definition 8 Let R be an integral domain. Define A 0 as the unit group of R, A 0 = R. For n, define A n as the set of all non-zero elements β in R such that every non-zero residue class modulo βr has a representative in A n. A n = {β R \ 0 : A n {0} onto R/(βR)}; A = n 0 A n. Proposition (Motzkin s Lemma) R is Euclidean if and only if every non-zero element of R is in A. The proof of this can be found in Motzkin [5] and Samuel [8]. To understand this definition it is helpful to investigate the case of the integers. We begin with A 0 = {±} and R = Z. Then 0

11 A = {β Z : {±, 0} onto Z/(βZ)} = {±, ±, ±3} A = {β Z : {±, ±, ±3, 0} onto Z/(βZ)}. = {±, ±, ±3 ± 4, ±, 5, ±6, ±7} A n = {β Z : A n {0} onto Z/(βZ)} = {±, ±,..., ±( n+ )} Therefore, we have that A = n 0 A n = Z. We know already that Z is Euclidean by the usual norm so this is not a surprise! The beauty of Motzkin s lemma is that it allows us to show whether a field is Euclidean without having to first find an algorithm. In fact, the construction gives a Euclidean algorithm. For example by defining it as the following: { i if x A i \ A i ψ R (x) = (7) 0 if x A 0 So, ψ R is a function from R \ 0 to N. Harper [4] took Motzkin s Lemma, and constructed a set of elements only containing primes of the integer ring, O K, of K, an algebraic number field. Definition 9 Define B 0 as the monoid generated by the unit group and an admissible set of primes. For n, successively define B n as the set of all primes π of O K such that every non-zero residue class modulo (π) has a representative in B n B 0. B n = {primes π O K : B n B n B = n 0 B n. onto (O K /π) }; This lead on to a variation of Motzkin s Lemma which played a vital role in the proof that Z ( 4 ) is Euclidean. Lemma Let O K be a principal ideal domain. If all primes of O K are in B then O K is Euclidean. The proof of this can be found in Harper [4]. To see that Z ( 4 ) is Euclidean we thus, need to prove that all the primes are in B. The tools used by Harper were the large sieve method and the following lemma: Lemma For β (x) = {(β) : β B and Nβ x}, if #β (x) x log x. The details of the proof that Z ( 4 ) is Euclidean using the above lemmas can be found in Harper [4], but is too vast to detail here.

12 6 Conclusion In this report we have gone in to the details a few different methods of determining whether a field is Euclidean. There is much work to be done in this field. As I have looked in detail at Z ( 4 ) and the two different methods used to show it is Euclidean it may be reasonable to next investigate other -stage Euclidean fields using Harper s method to see whether there also exists an algorithm for those fields. References [] Cooke, George E. A weakening of the Euclidean property for integral domains and applications to algebraic number theory, I. J. Reine Angew. Math. 8, 33 56, 976. [] Graves, Hester K. On Euclidean Ideal Classes, University of Michigan, 009. [3] Hardy, G. H. and Wright, E. M. An Introduction to the Theory of Numbers, Oxford University Press, 008. [4] Harper, Malcolm. Z ( 4 ) is Euclidean, Canad. J. Math. Vol. 56 (), 55-70, 004. [5] Motzkin, Th. The Euclidean Algorithm, Bull. Amer. Math. Soc. 55, 4-6, 949. [6] Nagata, Masayoshi, A Pairwise Algorithm and it s Application to Z ( 4 ), Proceedings of the Algebraic Geometry Seminar, Singapore, 987. [7] Narkiewicz, W. Elementary and Analytic Theory of Algebraic Numbers, Polish Scientific Publishers, Warsaw, 990. [8] Samuel, Pierre, About Euclidean Rings, Journal of Algebra 9, 8-30, 97.