Computation of Units in Number Fields

Transcription

1 UTRECHT UNIVERSITY MASTER THESIS Computaton of Unts n Number Felds Author: Bas JACOBS Supervsor: Prof. dr. Frts BEUKERS Second revewer: Prof. dr. Gunther CORNELISSEN June 14, 2016

2 Abstract We dscuss three algorthms to fnd small norm elements n number felds. One of these algorthms s a contnued fracton-lke algorthm based on the LLL-reducton of postve defnte quadratc forms as suggested by Beukers. The other two algorthms are adaptatons of that algorthm. We dscuss how to fnd unts from these small norm elements and how to extract a system of ndependent unts from that. We dscuss propertes of these algorthms and compare them to algorthms by Cohen, Daz y Daz, Olver, by Buchmann, Pethő and by Pohst, Zassenhaus. We run tests on an mplementaton of these algorthms n Mathematca. Acknowledgements I would lke to thank Frts Beukers for helpng me pck the subject and for hs help and support along the way. I also thank Gunther Cornelssen for beng the second revewer.

3 Contents 1 Introducton 1 2 Quadratc Forms Lattces Reduced lattces and forms Mnkowsk reduced HKZ reduced LLL reducton LLL reduced lattce bases LLL reduced forms Implemented unt algorthms Geodesc algorthm Another quadratc form Reducton n drectons Exstng unt algorthms Cohen, Daz y Daz, Olver Buchmann, Pethő Pohst, Zassenhaus Computng the unt group Computng the roots of unty Computng fundamental unts Generatng unts Constructng a set of ndependent unts Implementaton and results Descrpton of code Tests and comparsons Choosng the parameters Statstcs Hgh degree number felds A Mathematca Code 42

4 Chapter 1 Introducton In ths thess, K = Q(α) wll denote an algebrac number feld obtaned by adjonng to Q a root α of an rreducble polynomal f of degree n. Let σ 1,..., σ n be the embeddngs of K n C of whch r 1 are real and 2r 2 are complex. Then the degree of K s [K : Q = n = r 1 + 2r 2. We wll wrte O K for the rng of ntegers of K, whch has ntegral bass α 1,..., α n O K. The dscrmnant of K, denoted K, s the square of the determnant of the n n matrx wth entres σ (α j ). An element x O K can be wrtten unquely as x = n x α, =1 wth x Z. We wll wrte x () = σ (x) for ts th conjugate. By the norm of x, denoted N(x), we mean N(x) = N K/Q (x) = n x (). We are nterested n the elements x of O K wth N(x) = ±1, whch are called unts. The unts of O K form a group, whch we wll denote by OK. In 1846, Drchlet [9 showed that ths group s fntely generated of rank r 1 + r 2 1. Theorem 1.1 (Drchlet s Unt Theorem). Let O K be the rng of ntegers of a feld K, admttng r 1 real and 2r 2 complex embeddngs, and wrte µ for the group of roots of unty n O K. Then µ s fnte, and O K /µ s a free abelan group of rank r 1 + r 2 1. A proof can be found n [26. The theorem mples that we can wrte =1 O K = µ η 1 η r1 +r 2 1, where µ s the group of roots of unty n O K and η 1,..., η r1 +r 2 1 are so-called fundamental unts. Ths system of fundamental unts s unque up to coordnate transformatons and multplcaton by roots of unty. We wll gve a small example to llustrate ths theorem. Example 1.2. Consder f(x) = X 5 19, and let α = 19 1/5 be a root of f. Then K = Q(α) has degree 5 over Q. Now O K has 1 real and 4 complex embeddngs, so Drchlet s Unt Theorem tells us that O K = µ η 1 η 2. It turns out that µ = {±1} and that η 1 = 1 + α + α 3 and η 2 = 4 + 2α + α 4 are fundamental unts. In general, f r 1 > 0, then we have µ = {±1}, snce R contans no other roots of unty. In the totally complex case r 1 = 0, the roots of unty are also easly calculated. In Chapter 5, we wll dscuss how that could be done. 1

5 Chapter 1. Introducton 2 We wll gve another example, whch shows that the ablty of fndng unts can solve certan Dophantne equatons. Example 1.3. The Pell equaton s the equaton x 2 dy 2 = 1, to be solved for postve ntegers x, y for a gven nonzero nteger d. Ths equaton has a rch hstory, and solvng t bols down to fndng unts n a number feld. To see ths, note that we can rewrte the equaton as (x+y d)(x y d) = 1. Hence, a soluton to the Pell equaton corresponds to a unt n the rng Z[ d. On the other hand, f we can fnd a unt of norm 1 n Z[ d, we have found a soluton to the Pell equaton. Here the unts ±1 of Z[ d are consdered trval. If d = 19, then (x, y) = (170, 39) found n the example above solve the Pell equaton. The other solutons are gven by powers of Note that the above example can be solved usng a smple contnued fracton algorthm lke n [16, p. 11. However, for larger degree number felds, fndng a system of fundamental unts s a more dffcult task. In ths thess, we wll focus on fndng elements of small norm, whch we can then combne to create unts. From these unts, we then try to fnd a system of fundamental unts. In Chapter 3 we dscuss three algorthms that we have also mplemented. These algorthms are desgned to return many elements of bounded norm usng quadratc form reductons. We wll also dscuss some exstng algorthms developed by others n Chapter 4. We compare these algorthms and dscuss dfferences and smlartes. In Chapter 5, we explan how to obtan a system of fundamental unts from elements of small norm. Ths s mplemented together wth the algorthms of Chapter 3 n Chapter 6. Before we do all that, we need to ntroduce the notons of quadratc forms and LLL reducton. Ths wll be the topc of Chapter 2.

6 Chapter 2 Quadratc Forms A quadratc form s a polynomal of the form Q(x) = n q j x x j,j=1 n the varables x 1, x 2,..., x n and wth q j = q j R for 1, j n. To such a quadratc form we assocate a matrx Q = (q j ) 1,j n. The absolute value of the determnant of Q s called the determnant of the form, whch we wll denote by D(Q). We call a quadratc form postve defnte f Q(x) 0 for all x R n and Q(x) = 0 f and only f x = 0. From now on, wth form we wll mean a postve defnte quadratc form. Two forms Q, Q wll be called equvalent f there exsts P GL n (Z) such that Q = P t QP. Note that D(Q ) = D(Q), snce det P = ±1. A central task wth quadratc forms s fndng mnma. Ths s also our goal, because, as we wll show later, by fndng mnma of forms we are able to fnd elements of small norm n number felds. By µ(q) we wll denote the mnmal non-zero value of the set {Q(x) x Z n }. The followng theorem by Hermte gves a bound on ths mnmum n terms of the determnant of Q. Theorem 2.1 (Hermte). For every n 2 there exsts γ n such that µ(q) γ n D(Q) 1/n for all postve defnte quadratc forms Q n n varables. The smallest possble values of γ n are called Hermte s constants. We have n general γ n 2n/3. We wll show below that quadratc forms are related to lattces. 2.1 Lattces A lattce L n a vector space V s a subgroup of V of the form L = Z b 1 + Z b Z b n for lnearly ndependent b 1, b 2,..., b n V. The nteger n s called the rank of L. We wll look at lattces L of maxmal rank, meanng that the vector space V has dmenson n. Because of our nterests, from now on we wll look at lattces n V = R n. Let b 1, b 2,..., b n V be a Z-bass of L. We defne the matrx Q = (b b j ) 1,j n, whch we wll call the Gram matrx of the b. We defne the determnant d(l) of L to be d(l) = det Q. A geometrc nterpretaton s that d(l) s the volume of the parallelepped spanned by the bass vectors. Let A be the matrx wth columns b 1, b 2,..., b n. Then Q = A T A, so det Q = (det A) 2. Hence f b 1, b 2,..., b n V form a Z-bass of L, then det Q > 0. Moreover, Q s postve defnte and symmetrc, hence t corresponds to a 3

7 Chapter 2. Quadratc Forms 4 postve defnte quadratc form. In partcular, Q(x 1,..., x n ) = x 1 b x n b n 2. If we change the Z-bass, ths amounts to replacng A wth AP for a P GL n (Z), hence we get Q = P T QP. Hence equvalence classes of lattces correspond to equvalence classes of forms. Conversely, f Q s the n n matrx of a postve defnte quadratc form, we can decompose Q as Q = F 1 DF where F s the matrx of egenvectors and D the dagonal matrx whose dagonal elements are the correspondng egenvalues, whch are real and postve. Let E be the matrx wth entres E = D, so E 2 = D. If we let b 1, b 2,..., b n be the columns of A = EF, then Q s ther Gram matrx. Indeed, A T A = F T E T EF = F 1 EEF = F 1 DF = Q Moreover, the vectors b are lnearly ndependent, so they span a lattce n R n. Smlar to before, f we consder a form Q equvalent to Q, then we have Q = P T QP = P T A T AP for P GL n (Z), so we have replaced A by AP and obtaned an equvalent lattce. Hence we have shown that there s a natural correspondence between forms and lattces. There are many reducton algorthms for lattces, whch we can now translate to reductons of forms. 2.2 Reduced lattces and forms We saw before that there are equvalence classes of lattces and forms. For determnaton of µ(q) we want to be able to fnd "nce" representatves for an equvalence class, whch are called reduced. The process of fndng such representatves s called reducton. There are dfferent defntons of reduced, among whch there are Hermte, Mnkowsk, Hermte-Korkn-Zolotarev (HKZ), Venkov and Lenstra-Lenstra-Lovasz (LLL) reduced. We wll brefly dscuss the notons of Mnkowsk and HKZ reducedness here Mnkowsk reduced In 1891, Mnkowsk [22 came up wth a noton of reduced bases, now known as Mnkowsk reduced bases. We wll start by statng the defnton n terms of lattce bases. Defnton 2.2. A lattce bass b 1,..., b n of a lattce L s called Mnkowsk reduced f for all we have b m = n m j b j for all m L wth gcd(m,..., m n ) = 1. j=1 Another way of puttng t, s that for all, b s a shortest lattce element that can be extended to a bass wth (b 1,..., b 1 ). For forms, as one mght expect, we have the followng defnton. Defnton 2.3. A form Q(x) s called Mnkowsk reduced f for all we have Q(e ) Q(m) for all m Z n wth gcd(m,..., m n ) = 1. We can see that f we have a Mnkowsk reduced form Q = (q j ) j, then q 11 s the smallest non-zero value of Q restrcted to Z n. Hence f we are able to calculate a Mnkowsk reduced form equvalent to a gven form Q, we know µ(q). In [15, an algorthm to construct a Mnkowsk reduced lattce bass s dscussed. However, the number of nequaltes that characterse Mnkowsk reducedness grows very quckly, makng the

8 Chapter 2. Quadratc Forms 5 algorthm run n exponental tme wth respect to n. Hence the algorthm s mpractcal for our purposes HKZ reduced The noton of HKZ reducedness was ntated n 1850 by Hermte [17 and n 1873 by Korkn and Zolotareff [19. Let µ j be as defned n (2.2). We have the followng defnton. Defnton 2.4. A lattce bass b 1,..., b n of a lattce L s called HKZ reduced f the followng condtons hold: 1. µ j 1 2 for < j, 2. b 1 s the smallest non-zero vector of L, 3. the orthogonal projecton of the vectors b 2,..., b n to b 1 s HKZ reduced. For forms, we have the followng defnton, where µ j s as defned n the recursve form of Q. Defnton 2.5. A form Q(x) s called HKZ reduced f the followng condtons hold: 1. µ j 1 2 for < j, 2. b 1 = µ(q), 3. the form Q b 1 (x 1 + µ 12 x µ 1n x n ) 2 n x 2,..., x n s HKZ reduced. Agan, t s easly seen that the two defntons are equvalent. It turns out that when a form s HKZ reduced, then b b +1 + µ 2,+1 b, whch s a stronger verson of (2.7) whch we wll see n the next secton. Lke wth a Mnkowsk reduced form, f we have a HKZ reduced form, we mmedately have µ(q). However, agan lke wth Mnkowsk reducedness, t s very expensve to compute an HKZ reduced bass. Several algorthms exst, for example the algorthm by Kannan [13. However, all known algorthms run n exponental tme. 2.3 LLL reducton LLL reducton s named after Lovasz, Lenstra and Lenstra, who proposed t n 1982 [20. Even though t does not gve optmal output, t has proven to be fast and yeldng good results. It wll prove useful for fndng unts, whch we wll see n the next chapters. Frst, we wll look at the noton of LLL reduced lattces, then at LLL reduced forms LLL reduced lattce bases Gven a bass b 1, b 2,..., b n of a lattce L, defne the vectors b numbers µ j (1 < j n) recursvely by (1 n) and the 1 b = b µ j b j, (2.1) j=1 µ j = (b b j)/(b j b j). (2.2)

9 Chapter 2. Quadratc Forms 6 Ths s the well-known Gram-Schmdt process and we know that the b form an orthogonal bass of L. Note that f we defne A to be the matrx wth columns b and A the matrx wth columns b, then we have 1 µ 12 µ µ 1n 0 1 µ µ 2n A = A µ 3n = A P Hence we can see that that d(l) 2 = 1 n b 2, where denotes the Eucldean dstance n R n. Now, we can defne the noton of LLL reducedness of a lattce base. Defnton 2.6. The bass b 1, b 2,..., b n s called LLL reduced f µ j 1 for < j, (2.3) b 2 b +1 + µ,+1 b 2 for < n. (2.4) We have the followng theorem whch gves bounds for the b. Theorem 2.7. Let b 1, b 2,..., b n be an LLL reduced bass of a lattce L. Then 1. d(l) n =1 b 2 n(n 1)/4 d(l). 2. b 1 2 (n 1)/4 d(l) 1/n. 3. For every x L wth x 0, we have b 1 2 (n 1)/2 x. Proof. We start by provng part 1. We know that d(l) 2 = 1 n b 2. We can rearrange and square (2.1), to get 1 b 2 = b 2 + µ 2 j b j 2, by orthogonalty of the b. Hence, we have d(l)2 1 n b 2 and the frst nequalty follows. We have b +1 2 (3/4 µ 2,+1) b 2 b 2 /2, where the frst nequalty follows from (2.4) and the orthogonalty of b, and the second nequalty follows from (2.3). By nducton, we now have for j. Ths gves j=1 b j 2 2 j b 2 (2.5) 1 1 b 2 = b 2 + µ 2 j b j 2 b j b 2 /4 ( ) b 2 /2, (2.6) j=1 j=1 gvng the second nequalty and provng 1. For part 2, note that we can combne (2.5) and (2.6) to get that for j, we have b j 2 ( j 1 ) b 2. Settng j = 1 and multplyng the nequaltes for 1 n,

10 Chapter 2. Quadratc Forms 7 we get part 2. For part 3, note that there exsts an such that x = 1 j r jb j = 1 j s jb j, for r Z, s j R and wth r 0. From (2.1), we see that r = s, so we have x 2 s 2 b 2 = r 2 b 2 b 2, snce r 0 s an nteger. By (2.5), we have x j b n b 1 2, from whch part 3 follows. We can see that even though the vector b 1 n a reduced bass of L does not have to be the shortest non-zero vector n L, t s not too far from t. Hence, ndeed the results are n that sense not optmal, but the algorthmc advantage over other types of reducedness n combnaton wth the above bounds make t a powerful noton nevertheless. Gven a bass of a lattce, the LLL Algorthm transforms the bass vectors so that they form an LLL reduced bass. The algorthm s smple and very effcent and s gven below. It s mplemented n most mathematcs software packages, such as Mathematca, PARI/GP and SageMath. The algorthm requres a lattce bass b 1, b 2,..., b n and returns an LLL reduced bass. For more detals, we refer to [6. Algorthm Start 2. Reducton Compute b 1,..., b n. (a) For = 2 to n (b) For j = 1 to 1 (c) 3. Swap b b µ j b j (a) If such that 3 4 b 2 > b +1 + µ,+1b 2, b b +1. Go to 1. The noton of LLL reduced forms s very smlar to that of LLL reduced lattces, whch s not a surprse, gven the correspondence between lattces and forms dscussed before. We wll treat t for completeness and to be able to compare the theorems and results LLL reduced forms We can wrte a form Q(x) n the so-called recursve form Q(x) = b 1 (x 1 + µ 12 x µ 1n x n ) 2 +b 2 (x 2 + µ 23 x µ 2n x n ) 2. +b n 1 (x n 1 + µ n 1,n x n ) 2 + b n x 2 n. We can wrte the correspondng matrx as Q = P T Q P, wth 1 µ 12 µ µ 1n 0 1 µ µ 2n P = µ 3n b b and Q = 0 0 b b n

11 Chapter 2. Quadratc Forms 8 Hence we see that f we let A and A be as n Secton and Q = (A ) T A, then Q = P T (A ) T A P = A T A. Therefore, f we defne b = b 2 and let the µ j defned above correspond to those n Secton 2.3.1, then Q s Gram matrx of the lattce spanned by b and Q s Gram matrx of the lattce spanned by b. We can now defne LLL reducedness of a form. Defnton 2.9. We call the form Q LLL reduced f µ j 1 for < j, (2.7) b b +1 + µ 2,+1b for < n. (2.8) As one would expect, for quadratc forms, we have a theorem smlar to Theorem 2.7. Theorem Let Q be an LLL reduced form n n varables and denote by e the th bass vector of R n. Then 1. D(Q) n =1 Q(e ) 2 n(n 1)/2 D(Q). 2. Q(e 1 ) 2 (n 1)/2 D(Q) 1/n 3. Q(e 1 ) 2 n 1 µ(q). Proof. By the dscusson above, t s clear that Q s the gram matrx of the bass b. Hence we have D(Q) = d(l) 2 and Q(e ) = b 2. Now t should be clear that ths theorem s a drect translaton of Theorem 2.7 to forms. A proof for forms specfcally can be found n [3. Ths theorem wll prove useful for provng the results of the algorthms of the next chapter. Lke wth lattces, the process of fndng a LLL reduced form s called LLL reducton of the form. An mplementaton as gven n [3 uses shft and swap operatons. A shft s a substtuton of the form x r x r + ax s for 1 r < s n and a Z such that µ rs 1/2. A swap for 1 r n nterchanges the varables x r, x r+1. In [3 and [21, explct formulae for b and µ j n terms of the coeffcents of Q are gven, whch allow for LLL reducton to be appled drectly on the form Q. For certan quadratc forms, ths gves a verson of LLL reducton that can be used for fndng unts n a number feld. Ths s explaned n the next chapter and mplemented later.

12 Chapter 3 Implemented unt algorthms We ntroduce a notaton whch allows us to rewrte the LLL reducedness condtons n a for us more convenent way. For proofs of the statements, we refer to [3 and [21. Theorem 3.1. Let Q be a form n n varables wth matrx (q j ),j=1,...,n and let b and µ j be the coeffcents of the recursve form of Q. For, j wth 1 j n, we defne q q 1, 1 q 1,j q q 2, 1 q 2,j B j =....,.. q 1... q, 1 q,j wth B 00 = 1. Then b = B, /B 1, 1 for all 1 n. Also, µ j 1 < j n. If we also defne q q 1, 1 q 1,+1 q q 2, 1 q 2,+1 C =......, q 1,1... q 1, 1 q 1,+1 q +1,1... q +1, 1 q +1,+1 then we have C /B 1, 1 = b +1 + µ 2,+1 b. = B j /B for all Wth ths notaton, we can rewrte the LLL reducedness condtons as follows. Corollary 3.2. Let Q be a quadratc form and let B,j and C be as defned above. We call the form Q LLL reduced f 2 B j B for < j, 3 4 B C for < n. The shfts and swaps defned n the prevous chapter can now be expressed n terms of the determnants B j and C. In the next sectons, we wll use quadratc forms and these determnants to fnd unts. 3.1 Geodesc algorthm Let K = Q(α) be a number feld of degree n + 1 wth a real embeddng wth ntegral bass 1, α 1, α 2,..., α n. In [3 and [21, the authors consder the quadratc form gven by Q t = x x x 2 n + t(x 0 + x 1 α x n α n ) 2 9

13 Chapter 3. Implemented unt algorthms 10 n n + 1 varables x for 0 n and a parameter t. Its matrx has the form t tα 1 tα 2... tα n tα tα1 2 tα 1 α 2... tα 1 α n Q t = tα 2 tα 1 α tα tα 2 α n tα n tα 1 α n tα 2 α n tαn 2 By Gaussan elmnaton, ths can be transformed to t tα 1 tα 2... tα n From ths we see that det(q t ) = t. Assume that ths form has a mnmum at y = (y 0 + y y n ). Then by 2.1, we have Q t (y) = y y y 2 n + t(y 0 + y 1 α y n α n ) 2 γ n+1 t 1/(n+1). Ths gves t(y 0 +y 1 α 1 + +y n α n ) 2 γ n+1 t 1/(n+1) and y 2 1 +y y2 n γ n+1 t 1/(n+1). Multplyng the two nequaltes gves hence (y 0 + y 1 α y n α n ) 2 (y y y 2 n) n γ n+1 n+1, y 0 + y 1 α y n α n (y y y 2 n) n/2 γ (n+1)/2 n+1 (n + 1) n/2. (3.1) Ths makes t temptng to hope that by fndng mnma of quadratc forms of the above type, we may fnd elements of small norm. Usng LLL reducton, we can fnd elements close to mnma relatvely easly. The followng geodesc algorthm from [3 and [21 does just that. Algorthm Intalzaton 2. Repeat Q (0) t = x x x2 n + t(x 0 + x 1 α x n α n ) 2, P (0) = I n+1, k 0. (a) Determne the maxmum of the set {t Q (k) t s LLL reduced} and call ths t k. (b) Perform LLL reducton on Q (k) t k +ɛ for nfntesmal ɛ > 0 and let A k GL n+1 (Z) be such that x A k x s the correspondng change of varables. (c) Defne Q (k+1) t (x) = Q (k) t (A k x) and P (k+1) = P (k) A k. (d) k k + 1

14 Chapter 3. Implemented unt algorthms 11 Snce LLL reducton does not gve a mnmum of a quadratc form, we have a weaker verson of 3.1. Proposton 3.4. Let y = (y 0, y 1,..., y n ) be the frst column of P (k). Then y 0 + y 1 α y n α n (y y y 2 n) n/2 2 n(n+1)/4. Proof. Frst, note that det(q (k) t ) = t, snce A k GL n+1 (Z). We have Q t (P (k) x) = (x) for every k. Hence Q t (y) = Q (k) (e 1 ) and by part 2 of Theorem 2.10, we have Q (k) t t y y y 2 n + t(y 0 + y 1 α y n α n ) 2 2 n/2 t 1/(n+1). Hence we have t(y 0 + y 1 α y n α n ) 2 2 n/2 t 1/(n+1) and y y y2 n 2 n/2 t 1/(n+1). Multplyng these, we get y 0 + y 1 α y n α n (y y y 2 n) n/2 2 n(n+1)/4. From ths proposton, we can say somethng about the norm of y 0 + y 1 α y n α n. Note that we have y 0 + y 1 α y n α n 2 n/4 t n/(n+1). As we let t get bg, ths gves y 0 + y 1 α y n α n 1, so y 0 y 1 α y n α n. If we wrte α (j) = σ j (α ), wth α (1) = α, we have n+1 y 0 +y 1 α () 1 + +y nα n () y 0 +y 1 α 1 + +y n α n =1 Call the rght-hand sde S. We know that =2 n+1 =2 y 1 (α () 1 α(1) 1 )+ +y n(α n () α n (1) ). n+1 y 1 (α () 1 α(1) 1 ) + + y n(α n () α n (1) ) 2 c (y yn) 2 for a constant c R +. Hence for all > 1, we have y 1 (α () 1 α(1) 1 ) + + y n(α () n α (1) n ) ( c (y y 2 n) ) 1/2. If we multply these nequaltes for all > 1, S y 0 + y 1 α y n α n (c (y y y 2 n) ) n/2. By the proposton, we now have N(y 0 + y 1 α y n α n ) S 2 n(n+1)/4 c n/2. Hence the absolute norms of the elements correspondng to the frst column of P (k) are bounded. We have the followng proposton, showng a practcal aspect of the algorthm. Proposton 3.5. Let Q t = x x x2 n + t(x 0 + x 1 α x n α n ) 2. Then the determnants B j and C as defned before are of the form ut + v where v Z and where u s quadratc n α.

15 Chapter 3. Implemented unt algorthms 12 For a proof, we refer to [21. Ths proposton shows that the values of t 1 for whch Q t (x) s LLL reduced are closed ntervals n R 1. Hence t s easy to fnd t k whch break the LLL condtons, allowng us to reduce agan to fnd elements that may have small norm. Moreover, t s shown n [3 that even though the rules for updatng the B j and C are not lnear, the only non-lnear part conssts of dvson by an nteger. These two observatons allow ths algorthm to be very practcal and easy to mplement. In order to fnd unts, we perform the above algorthm, and n each teraton we look at the frst column of P (k), whch we denote by (y 0,..., y n ). By the dscusson after Proposton 3.4, we see that N(y 0 + y 1 α y n α n ) s bounded. We can calculate ths norm, and check f t s a unt. If t s not, we may be able to dvde t by another element of the same norm to get a unt. Ths s further explaned n Chapter 6. Note, however, that we requred the feld K to have at least one real embeddng. Otherwse, the quadratc form s not necessarly real valued and our theory does not apply. If we change the quadratc form to Q t = x x x2 n + t x 0 + x 1 α x n α n 2, the quadratc form stays real valued. However, Proposton 3.5 no longer holds and the determnants can become quadratc n B j and C. Ths results n forms that cannot be reduced by changng t. Moreover, the results of the algorthm can n theory (and do n practce) depend on the choce of the real embeddng. In the next secton, we consder another quadratc form and dscuss ts propertes. 3.2 Another quadratc form Let K be a number feld of degree n wth ntegral bass α 1,..., α n, let r 1 be the number of real embeddngs of K, and r 2 be the number of complex embeddngs. We wrte α (j) = σ j (α ), wth α (1) 1 j r 1 and α (j) = α. Index the conjugates such that we have α (j) R for = α (j+r 2) for r j r 1 + r 2, where x denotes the complex conjugate of x. In the quadratc form of the prevous secton, we use one embeddng of the algebrac nteger x 0 + x 1 α x n α n. Now, we want to look at all embeddngs of such an nteger, and use them all n some sense. To that end, we defne Q t = t x 1 α (1) x n α (1) n 2 + x 1 α (2) f σ 1 s a real embeddng, and Q t = t x 1 α (1) 1 + +x nα (1) n 2 + +t x 1 α (1+r 2) x n α n (2) 1 + +x n α (1+r 2) n x 1 α (n) x n α (n) n x 1 α (n) 1 + +x nα (n) n 2 otherwse. We want to use Algorthm 3.3 wth ths quadratc form. We wll frst show some propertes of ths form. Frst, note that n both cases ths quadratc form s real-valued. Next, we can see that n the real case, ts matrx s gven by tre(α (1) 1 α(1) 1 ) + n k=2 Re(α(k) 1 α(k) 1 )... tre(α(1) tre(α (1) 2 Q t = α(1) 1 ) + n k=2 Re(α(k) 2 α(k) 1 )... tre(α(1).... tre(α n (1) α (1) 1 ) + n k=2 Re(α(k) n α (k) 1 )... tre(α(1) 1 α(1) n 2 α(1) n n α n (1) ) + n k=2 Re(α(k) 1 α(k) n ) ) + n k=2 Re(α(k) 2 α(k). ) + n k=2 Re(α(k) n n ) α n (k) ).

16 Chapter 3. Implemented unt algorthms 13 If K has a real embeddng, then det(q t ) = K t. In the complex case, we have an other matrx, wth det(q t ) = K t 2. We wll prove ths n the next secton n a more general settng. The followng algorthm s smply a verson of Algorthm 3.3 wth the new quadratc form. Algorthm Intalzaton 2. Repeat Q (0) t = t x 1 α (1) x n α n (1) 2 + x 1 α (2) + x n α n (n) 2, P (0) = I n, k x n α n (2) x 1 α (n) 1 + (a) Determne the maxmum of the set {t Q (k) t s LLL reduced} and call ths t k. (b) Perform LLL reducton on Q (k) t k +ɛ for nfntesmal ɛ > 0 and let A k GL n (Z) be such that x A k x s the correspondng change of varables. (c) Defne Q (k+1) t (x) = Q (k) t (A k x) and P (k+1) = P (k) A k. (d) k k + 1 We would lke to have Propostons smlar to 3.4 and 3.5. Indeed, Proposton 3.4 translates to the followng. Proposton 3.7. Let y = (y 1,..., y n ) be the frst column of P (k). Then, N(y 1 α y n α n ) 2 n(n 1)/4 K 1/2. Proof. We have Q t (P (k) x) = Q (k) t (x) for every k. Hence Q t (y) = Q (k) t (e 1 ) and by part 2 of 2.10, we have Q(y) 2 (n 1)/2 det(q t ) 1/n. Frst, assume that K has a real embeddng. Then, we have t y 1 α (1) y n α n (1) y 1 α (n) y n α n (n) 2 2 (n 1)/2 K 1/n t 1/n. Ths gves y 1 α (1) y n α n (1) 2 2 (n 1)/2 K 1/n t (1 n)/n for the frst embeddng, and for > 1, If we multply these together, we get y 1 α () y nα () n 2 2 (n 1)/2 K 1/n t 1/n. y 1 α () y nα n () 2 2 n(n 1)/2 K. If K does not have a real embeddng, then we have y 1 α (1) y n α n (1) 2 2 (n 1)/2 K 1/n t 2/n t 1

17 Chapter 3. Implemented unt algorthms 14 for embeddng 1 and 1 + r 2. For > 1, Multplyng these, we get whch proves the statement. y 1 α () y nα () n 2 2 (n 1)/2 K 1/n t 2/n. y 1 α () y nα n () 2 2 n(n 1)/2 K, Ths proposton gves a bound on the norms of the elements obtaned by the algorthm. Ths bound depends on the degree of the feld and ts dscrmnant. For felds K wth a real embeddng, we have the followng proposton, whch s a weaker verson of Proposton 3.5. Proposton 3.8. Let K be a number feld wth a real embeddng and let Q t = t x 1 α (1) x n α n (1) x 1 α (n) x n α n (n) 2. Then the determnants B j and C as defned before are lnear n t. Proof. Snce α (1) tα (1) R for all, we can wrte 1 α(1) 1 + n k=2 Re(α(k) 1 α(k) 1 )... tα(1) tα (1) 2 Q t = α(1) 1 + n k=2 Re(α(k) 2 α(k) 1 )... tα(1).... tα n (1) α (1) 1 + n k=2 Re(α(k) n 1 α(1) n 2 α(1) n α (k) 1 )... tα(1) n α n (1) + n + n k=2 Re(α(k) 1 α(k) n ) + n k=2 Re(α(k) 2 α(k). k=2 Re(α(k) n n ) α n (k) ) We can subtract α (1) /α (1) 1 tmes the frst row from the th row to obtan a matrx where t only occurs n the frst row. Snce n the defnton of B j and C, these stay n the frst row, the determnants are lnear n t. However, when K s totally magnary, Proposton 3.5 does not translate well to our form, as we can see n the followng example. Example 3.9. Consder the totally magnary feld K = Q(α), where α s a root of the polynomal X Let B j be as defned n Theorem 3.1. Then, we have B 22 = ( t + t 2 )/ 2 and C 2 = (3 + t) 2, so the condton 3 4 B 22 C 2 becomes quadratc. We see that n those cases t becomes a lot harder to determne t k from Algorthm 3.6. The fact that the condtons can be quadratc, also leads to losng the lnear aspect of the algorthm. Moreover, t turns out that n certan cases wth totally magnary K, at a certan pont the LLL condtons cannot be satsfed at all. Therefore, we do not use ths algorthm on totally complex felds n our mplementaton and n the tests. We wll now dscuss yet another quadratc form that does work for totally complex felds Reducton n drectons Usng the number feld and the ndexng from before, consder the quadratc form Q t1,t 2,...,t n = n =1 t x 1 α () x nα () n 2

18 Chapter 3. Implemented unt algorthms 15 n n varables x and n parameters t. Note that the quadratc form from the prevous secton s a specal case of ths one, wth t = 1 for = 2,..., n. In ths more general settng, by adjustng the parameters, we can now change the nfluence each embeddng has. Proposton The quadratc form Q t1,t 2,...,t n = n =1 t x 1 α () x nα () n 2 has determnant Proof. We can rewrte entry (, j) as n k=1 t k Re(α (k) α (k) j ) = det(q t1 t n ) = K r 1 k=1 t k α (k) α (k) j + n k=1 t k n k=r 1 +1 t k Re(α (k) α (k) j ). Snce Re(z) = Re(z) = (z + z)/2 for complex z, for every k r we have so t k Re(α (k) α (k) j ) + t k+r2 Re(α (k+r 2) t k Re(α (k) α (k) j ) + t k+r2 Re(α (k+r 2) Hence we can wrte entry (, j) as n k=1 α (k+r 2) j α (k+r 2) j τ k α (k) α (k) j, wth τ k = ) = t k Re(α (k) α (k) j ) = (t k + t k+r2 )(α (k) ) + t k+r2 Re(α (k) α (k) j ), α (k) j + α (k+r 2) α (k+r 2) j )/2. t k f 1 k r 1 ; (t k + t k+r2 )/2 f r 1 < k r 1 + r 2 ; (t k + t k r2 )/2 f r 1 + r 2 < k r 1 + 2r 2. Hence we can see that τ τ Q t1,t 2,...,t n = A AT, τ n wth A = 1... α (n) α (n) α n (1) α n (2)... α n (n) α (1) 1 α (2) α (1) 2 α (2) Snce det(aa T ) = K, we have det(q t1 t n ) = K n k=1 τ k. The determnant of the form Q t from Secton 3.2 follows after settng t = 1 for = 2,..., n.

19 Chapter 3. Implemented unt algorthms 16 From now on, we wll set t +r2 = t for > r 1 to make sure that the conjugate embeddngs get the same parameter. In that case, the determnant becomes smply K n k=1 t k. Wth ths new quadratc form, we wll reduce n "drectons", whch means that we wll choose t and LLL reduce the obtaned form. If we repeat ths, ths leads to the followng algorthm. Algorthm Intalzaton 2. Repeat Q t1,t 2,...,t n n =1 t x 1 α () x nα () n 2. k 1. (a) Choose t 1, t 2,..., t n R. (b) Perform LLL reducton on Q t 1,t 2,...,t n and let A k GL n (Z) be such that x A k x s the correspondng change of varables. (c) k k + 1. There s a freedom of choosng t n step (a). One can do ths n a systematc fashon, smlar to the geodesc algorthms. Ths could be done to hope that no unts are mssed n the drecton we are lookng at. Another opton s to choose random t and hope that we wll fnd unts. Ths way, we can look at very dfferent drectons, but we can mss unts that the systematc algorthm does fnd. Both optons are mplemented n Chapter 6. The followng proposton gves a bound for the norms of the elements found. Proposton Let y = (y 1,..., y n ) be the frst column of A k. Then, N(y 1 α y n α n ) 2 n(n 1)/2 K 1/2. Proof. The proof s the same as the proofs of Propostons 3.4 and 3.7 but wth another determnant and s therefore not ncluded. For a dscusson on the results of the three algorthms ntroduced above, we refer to Chapter 6.

20 Chapter 4 Exstng unt algorthms In ths chapter, we wll have a look at a couple of unt algorthms that have been developed by others. There exst more algorthms, but most of them share deas wth the algorthms we chose to dscuss. A lot of these algorthms started as deas by some researchers, and have then been changed and adapted by others. Most of the algorthms have ther roots n the 1980s and have been mplemented by the authors. In ths chapter, we wll dscuss the algorthms brefly and show the smlartes and dfferences they have to each other and to the algorthms of Chapter Cohen, Daz y Daz, Olver In 1997, Cohen, Daz y Daz and Olver [7 descrbe the mplementaton of an algorthm whch computes the class group and the unt group of a number feld. It s based on deas of Buchmann [4 and ts mplementaton s used n software packages PARI/GP and SageMath. In Cohen s "A Course n Computatonal Algebrac Number Theory" [6, the algorthm s explaned more thoroughly. The algorthm actually computes the class group of a number feld, gettng the fundamental unts "for free". The algorthm s the fastest algorthm known asymptotcally, but one needs to accept the Generalzed Remann Hypothess (GRH) for ts correctness. The GRH s assumed n step 1 and 4 of the algorthm. For more detals, we refer to remark below the algorthm. We wll start explanng some number theoretc defntons needed for ths algorthm. Denote by I and J two nonzero fractonal deals of O K. We say that I and J are related, denoted I J, whenever there exst nonzero elements a and b of O K such that (a)i = (b)j. Ths s an equvalence relaton and ts equvalence classes are called the deal classes of O K. Ideal classes can be multpled and the class of prncpal deals s an dentty element for ths multplcaton. Wth ths multplcaton, the set of fractonal deal classes are an abelan group, called the deal class group of O K, denoted Cl(K). The sze of ths group s called the class number of K, and s denoted by h(k). Let η 1,..., η r1 +r 2 1 be a system of fundamental unts of O K. Number the embeddngs nto C that are not conjugates by 1,..., r 1 + r 2. Let c j be 1 f embeddng j s real and 2 otherwse. Then the determnant of the (r 1 + r 2 1) (r 1 + r 2 ) matrx wth entres c j log η j s called the regulator of K, denoted by R(K). If O K s fnte, we put R(K) = 1. Unlke the dscrmnant, the regulator s a postve real number whch s usually expected to be transcendental. In the algorthm, the authors work wth deals rather than elements of the feld. The authors cut the algorthm n the followng 5 steps, whch are explaned n more detal below. 17

21 Chapter 4. Exstng unt algorthms 18 Algorthm Calculate the Mnkowsk bound and consder prme deals wth norms up to ths bound. Denote these by g 1,..., g k. It s known that they generate Cl(K). 2. Fnd many (say l) relatons n the class group among the g. Wrte these relatons as k g m,j = α j Z K (1 j l), =1 where m,j Z and α j K. 3. Let M = (m,j ) 1 k,1 j l be the k l matrx of exponents, and let V = (α j ) 1 j l be the vector of the α j. Perform Hermte and Smth normal form reductons on M, dong the same operatons on V. Ths way, we obtan a tentatve class group and unt group. Let h (K) and R (K) be the correspondng tentatve class number and regulator. 4. By usng the analytc class number and regulator formula, check that the product h (K)R (K) s correct up to a factor of 2. If t s not, fnd a few more relatons and go back to step Now h (K) = h(k) and R (K) = R(K). From the data obtaned above, compute a system of fundamental unts, and output t as well as the class group. Remark. If we are wllng to accept the GRH, then n step 1 we do not need to go as far as the Mnkowsk bound. There are other bounds that are usually much smaller than the Mnkowsk bound, for example as dscussed n [2. The bound proven there s 12 ln 2 D. In step 4, GRH s assumed to be able to prove that f h (K)R (K) s correct up to a factor of 2, then h (K) = h(k) and R (K) = R(K). In ths proof, the authors rely on a truncaton of the Euler product of the analytc class number and regulator formula whch can only be done assumng GRH. In step 1, we can calculate the Mnkowsk bound M K = ( ) 4 r2 n! D π n n, and consder the prme deals wth norms up to ths bound. It s known that these generate Cl(K), and they can be found usng the Kummer-Dedeknd theorem [8. Note that although these deals generate Cl(K), there can be deals amongst these that correspond to the same class or that are prncpal. Snce a pror we do not know Cl(K) (ths s one of the thngs the algorthm gves us), we just work wth all prme deals g we found. In step 2, the relatons are found n three ways. Frstly, there are the trval relatons obtaned by factorng prme numbers of Z nto deals of the number feld. Ths s already done n order to fnd the generators n step 1. Secondly, the authors use algorthms as n [10 to fnd elements α O K of small norm. Ths s done by enumeratng elements of a lattce that le n sutable ellpsods, smlar to Secton 4.3. They hope to be able to factor these n terms of the deals generatng Cl(K), obtanng new relatons. Fnally, the authors generate random exponents u and consder the deal I = k =1 g u.

22 Chapter 4. Exstng unt algorthms 19 Let α 1,..., α n be a Z-bass of ths deal. Choose a vector v = (v ) 1 n of real numbers such that v r2 + = v for r 1 < r 1 + r 2. Consder the form Q = n =1 e v x 1 α () x nα () n 2. Perform LLL reducton on ths form and let β 1,..., β n be the LLL reduced bass of I obtaned. Let α = β 1, whch s small by Theorem Then, β /α form a Z-bass of the fractonal deal I/α whch we wll call J. Next, the authors try to factor J to get so k =1 J = k =1 g v, g u v = αo K. Note that J does not necessarly factor n g. If t does, the authors obtan a new relaton. Otherwse, the authors try other random u, obtanng another deal I. Remark. The quadratc form Q s smply Q t1,...,t n from Secton 3.3 wth t = e v. Hence the reducton of the form Q we performed n Secton 3.3 s used n ths algorthm, too. Moreover, ths step of the algorthm just bols down to fndng small α O K and tryng to factor an deal above t (namely IJ 1 ) t n terms of g. The method of fndng these small α s the same as Algorthm In step 3, Hermte and Smth normal form reductons on M (dong the same operatons on V ) are used to dentfy certan subdetermnants of the matrces wth multples of the class number and regulator. In partcular, certan columns of an obtaned matrx correspond to elements α K such that αo K = k g 0 = O K. =1 Hence these columns corresponds to unts. The detals of these matrx reductons are tedous and wll not be treated here. They can be found n [6. For fndng unts, t bols down to tryng to dvde elements such that the correspondng row n the matrx of exponents M only contans zeros, such that the above relaton holds. Usng a relaton between the class number and regulator, t can be shown that f h (K)R (K) s correct up to a factor of 2, then we have h (K) = h(k) and R (K) = R(K). In partcular, truncate the Euler product of the analytc class number and regulator formula s at a carefully chosen place (smaller than 12 log 2 D(K) ). Denote by z the quantty obtaned. Then, h(k)r(k) < z < 2h(K)R(K). 2 To be able to show the nequaltes, GRH must be assumed. Now f h (K)R (K) < z 2, the authors are able to show that h (K) = h(k) and R (K) = R(K). Ths s checked n step 4, and n step 5, the matrces of step 3 are processed to get Cl(K) and the fundamental unts. For more detals, we refer to [6 and [7. To summarse, ths algorthm generates many relatons between prme deals of the

23 Chapter 4. Exstng unt algorthms 20 feld. Then, usng matrx reductons, from these relatons a tentatve class group and regulator are obtaned. These are checked to be correct and can then be used to compute the class group and a system of fundamental unts. Although ths algorthm s substantally dfferent from the algorthms of Chapter 3, a part of Step 2 concdes wth a part of Secton Buchmann, Pethő Among the many unt group algorthms Johannes Buchmann ntroduced, there s one algorthm he created together wth Atlla Pethő [5 that we wll dscuss here. The algorthm uses LLL reducton of lattces, whch we wll translate to quadratc forms. For more detals, we refer to the artcle. The algorthm fnds a system of ndependent unts rather than fundamental unts. A system {ɛ 1,..., ɛ u } OK s called ndependent f ɛm ɛ mu u = 1 mples m 1 = = m u = 0 for every system of ntegers {m 1,..., m u }. In Chapter 5, we wll explan how from such a system we can fnd fundamental unts. Agan, let the number of real embeddngs be r 1 and the number of complex embeddngs r 2. We number them lke before, such that σ j s a real embeddng for 1 j r 1 and that σ j and σ j+r2 are conjugate embeddngs for r 1 < j r 1 + r 2. We wrte σ (α) = α (). The authors construct a sequence of numbers (γ k ) k N n O K for every {1,..., r 1 + r 2 }. Ths sequence satsfes γ () k < γ() k 1 for k 2, (4.1) γ (j) k > γ(j) k 1 for k 2, j {1,..., r 1 + r 2 }, j. (4.2) Hence the absolute value of the chosen embeddng becomes smaller and the rest become larger. Moreover, the numbers γ k are constructed n such a way that they are of bounded norm. The numbers γ k are parwse dstnct. Hence after a fnte amount of steps, two of them are assocated to a unt, n the sense that γ k2 /γ k1 = ɛ for a certan k 1, k 2 N. Ths unt satsfes ɛ () < 1 and ɛ (j) > 1 for j. (4.3) To construct the sequence (γ k ) k N, start wth γ 1 = 1. For every k we compute a number β k such that β () k < 1 and c 1 > β (j) k > 1 for j. (4.4) Here c 1 s a constant dependng on the degree and the dscrmnant of K. We can now set γ k+1 = γ k β k, so (γ k ) k N satsfes 4.1 and 4.2. We let β k be an element of the module R k = 1 γ k O K. We also want N(β k ) c 2 / N(γ k ), where c 2 s a constant dependng on the degree and the dscrmnant of K. Ths way, the conjugates of β k are small for all k, whch makes computatons nexpensve. Moreover, ths s used to guarantee that γ k s of bounded norm. In fact, we only need to keep track of the small β k and not the γ k. Two elements γ k1 and γ k2 are assocated f the modules R k1 and R k2 are the same. Then, the unt can be computed by ɛ = k 2 1 l=k 1 β l.

24 Chapter 4. Exstng unt algorthms 21 We wll now explan how we can calculate β k. Frst, the authors fnd a reduced bass of the module R k. A module bass s called reduced f ts mage under the mappng K R n α (α (1),..., α (r 1), Re α (r 1+1),..., Re α (r 1+r 2 ), Im α (r 1+1),..., Im α (r 1+r 2 ) ) s an LLL reduced lattce bass. Ths way, the authors can assure that the bass elements are bounded. Specfcally, they are able to bound α () 1 from above, whch wll prove useful later. Secondly, the authors LLL reduce the columns of a well-chosen matrx. If the embeddng we are consderng s a real embeddng, the authors defne U = δ δ δ α () 1 α () 2 α () 3... α n (). (4.5) Here, α 1,..., α n s an LLL reduced bass of R k n the above sense. The postve constant δ s defned as δ = 2 n/4 α () 1 κ n, (4.6) where κ 1 s a constant that s chosen n the begnnng of the algorthm. Although there are some restrctons on ths choce of κ (for whch we refer to the artcle), the choce s rather arbtrary. The Gram matrx of the columns of U s G = whch corresponds to the quadratc form (α () 1 )2 α () 1 α() 2... α () 1 α() n α () 2 α() 1 δ 2 + (α () 2 )2... α () 2 α() n α n () α () 1 α n () α () 2... δ 2 + (α n () ) 2, Q δ = δ 2 x δ 2 x δ 2 x 2 n + (α () 1 x 1 + α () 2 x α () n x n ) 2. Ths corresponds to the form of Algorthm 3.3 wth t = 1/δ 2. The determnant of Q s D(Q δ ) = α () 1 δn 1. Let y = (y 1,..., y n ) be the frst column of the transformaton matrx correspondng to the reducton. Let β be β = α 1 y 1 + α 2 y α n y n. Then, by part 2 of Theorem 2.7, we see that β () 2 (n 1)/4 α () 1 1/n δ n 1. Snce α 1,..., α n form an LLL reduced bass of R k, the authors can bound α () 1 from above. Therefore, they are able to bound β () n terms of only constants, n and N(γ k ). We can also bound N(β) smlar to the dscusson after Proposton 3.4. The authors fnd N(β) c 3 N(γ k ) 1. Ths means that we can bound the absolute values of the embeddngs β (j) for j from below. Hence f we carefully choose δ, we can force 4.4 to hold, hence satsfyng 4.1 and 4.2.

25 Chapter 4. Exstng unt algorthms 22 If corresponds to a complex embedddng, the authors proceed n the same way, but wth another matrx δ U = δ (4.7) Re α () 1 Re α () 2 Re α () 3... Re α n () Im α () 1 Im α () 2 Im α () 3... Im α n () The Gram matrx of the columns of U s Re α () 1 α() 1 Re α () 1 α() 2 Re α () 1 α() 3... Re α () 1 α() n Re α () 2 α() 1 Re α () 2 α() 2 Re α () 2 α() 3... Re α () 2 α() n G = Re α () 3 α() 1 Re α () 3 α() 2 δ 2 + Re α () 3 α() 3... Re α () 3 α() n Re α n () α () 1 Re α n () α () 2 Re α n () α () 3... δ 2 + Re α n () α n () The correspondng quadratc form s Q δ = δ 2 x δ 2 x 2 n + α () 1 x 1 + α () 2 x α () n x n 2, whch s smlar to the form n the real case. Agan, we let y = (y 1,..., y n ) be the frst column of the transformaton matrx of the reducton. Usng arguments smlar to the real case, the authors guarantee that β = α 1 y α n y n satsfes 4.4. For a more detaled explanaton, we agan refer to [5. Now we know how to fnd γ () k for all k, gvng a unt ɛ for k large enough. Hence after applyng ths for every, we fnd r 1 + r 2 unts ɛ satsfyng 4.3. We then check f among ths set of r 1 + r 2 unts there are r 1 + r 2 1 ndependent unts. If ths does not hold, we apply the algorthm agan but wth bgger κ n 4.6. As we saw, ths algorthm s rather dfferent from the algorthms we mplemented. The LLL reducton of lattce bases the authors perform corresponds to the LLL reducton of quadratc forms we also consdered. However, n ths algorthm the author also reduces the bases of the modules, hence resultng n dfferent lattces. We consdered geodesc algorthms and algorthms that have a certan randomness. Buchmann and Pethő chose to use a chan of modules R k and guarantee that the unts satsfy certan propertes. Therefore, they are able to fnd an ndependent system of unts. We wll dscuss how to get from an ndependent system to fundamental unts n Chapter 5. We conclude by summarsng the algorthm for a drecton {1,..., r 1 + r 2 }. Ths algorthm should be run for = 1,..., r 1 + r 2. Ths s a very hgh-level summary, more detals can be found n [5.

26 Chapter 4. Exstng unt algorthms 23 Algorthm Intalzaton 2. Repeat k 1, R 1 = O K. (a) Fnd an LLL reduced bass of R k and call t α 1,..., α n. (b) Defne U as n 4.5 or 4.7 f corresponds to a real or a complex embeddng, respectvely. (c) LLL reduce the columns of U. Let y = (y 1,..., y n ) be the frst column of the transformaton matrx correspondng to the reducton. (d) Let β k = α 1 y α n y n and R k+1 = (1/β k )R k. (e) If R k = R l for l k, return ɛ = k l=l β l. (f) k k Pohst, Zassenhaus The thrd and fnal algorthm we wll dscuss s due to Pohst and Zassenhaus and s explaned n ther book Algorthmc Algebrac Number Theory [25. It uses Mnkowsk s Convex Body Theorem to fnd non-trval lattce ponts n certan lattces. Those ponts correspond to elements of bounded norm, whch can be used to fnd unts. In Chapter 5, we wll dscuss how we can fnd unts usng those elements. We wll start by statng Mnkowsk s Convex Body Theorem. A proof can for example be found n [25. Theorem 4.3 (Mnkowsk s Convex Body Theorem). Let C R n be a convex, 0-symmetrc set and L a lattce. If V (C) > 2 n d(l) or f V (C) = 2 n d(l) and C s compact, then C contans a lattce pont x L wth x 0. Here, V (C) s the volume of the set C, whch s Jordan-measurable. By 0-symmetrc, we mean that x L mples x L. The authors create these types of sets C such that they can fnd non-trval lattce ponts. The authors call these sets parallelotopes. The parallelotopes are constructed n a specal way whch we wll explan below. Let K be a feld wth [K : Q = n and let α 1,..., α n be a bass of O K. There s a bjectve mappng φ : O K Z n : x 1 α x n α n (x 1,..., x n ). The authors use ths mappng to go back and forth from the rng of ntegers to Z n R n, and we wll follow the same notaton. Consder the set Π = {(x 1,..., x n ) T R n 1 x 1, 1 n}. Ths set s convex, compact and 0-symmetrc, and has volume V (Π) = 2 n. If we consder the lattce L = Z n, then we see that Π satsfes the premses of Mnkowsk s Convex Body Theorem. Obvously, we could have seen a pror that Π contans non-trval ponts of Z n. However, we wll now transform Π n such a way that these premses stll hold, but such that we can fnd other lattce ponts that correspond to other elements

27 Chapter 4. Exstng unt algorthms 24 of small norm. If we defne n B = max y 1 α (j) y n α n (j) y Π, (4.8) j=1 then we have N(φ 1 (x)) B for all x Z n Π. Choose an element ω O K \Z. More about the choce of ω can be found n the end of ths secton. Then ts rght regular representaton M ω Z n n s defned by (α 1,..., α n )ω = (α 1,..., α n )M ω. We have N(ω) = det M ω, and we can defne a lnear transformaton Ψ ω by Ψ ω = N(ω) 1/n M ω. Now, Ψ ω s lnear and has determnant ±1. Therefore, Ψ ω (Π) := { N(ω) 1/n M ω (x 1,..., x n ) T R n 1 x 1, 1 n} s a convex, 0-symmetrc parallelotope of volume 2 n. For 1 x 1,..., x n 1, we have n N(ω) 1/n (α (j) 1,..., α(j) n )M ω x = N(ω) 1 j=1 n ω (j) (α (j) 1,..., α(j) n )x B. Hence the absolute norms of elements of φ 1 (Ψ ω (Π) Z n ) are bounded by B. Ths means that by usng these knds of transformatons, we can fnd new elements of bounded norm. The only thng left to show s how to compute Ψ ω (Π) Z n. The authors show that there exst unmodular matrces U ω, Uω 1 such that Mω T U ω s n Hermte normal form. We wll call ths matrx N ω. It s a lower trangular matrx, and snce det M ω = ± det N ω, the product of the dagonal elements of N ω s up to sgn N(ω). The authors now create a lower trangular matrx B ω Z n n such that j=1 N(ω) Mω T 0 N(ω)... 0 U ω B ω = N(ω) They create B ω by recursvely computng ts elements from top left to bottom rght. Consder a lattce pont c = (c 1,..., c n ) T of Ψ ω (Π). Hence c = N(ω) 1/n M ω x for x R n wth 1 x 1 for all. We defne d = U T ω c. Then d Z n, snce c Z n. We have B T ω d = N(ω) 1/n B T ω U T ω M ω x = N(ω) (n 1)/n x. Hence we can fnd d Z n by solvng N(ω) (n 1)/n ( B T ω d ) N(ω) (n 1)/n for {1,..., n}. Snce B T ω s an upper trangular matrx ths can be done recursvely, startng from d n and workng to d 1. When we have d, we calculate c = (U 1 ω ) T d,