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SIM PLIFYING NESTED RADICALS AND SOLVING POLYNOMIALS BY NESTED RADICALS IN MINIMUM D EPTH by Gwoboa Horng A D issertation Presented to the FACULTY OF TH E GRADUATE SCHOOL UNIVERSITY OF SOUTHERN CALIFORNIA In P artial Fulfillment of the Requirem ents for the Degree D O C TO R O F PHILOSOPHY (Com puter Science) May 1992 i Copyright 1992 Gwoboa Horng UMI Number: DP22848 All rights reserved INFORMATION TO ALL USERS The quality of this reproduction is dependent upon the quality of the copy submitted. In the unlikely event that the author did not send a complete manuscript and there are missing pages, these will be noted. Also, if material had to be removed, a note will indicate the deletion. UMI DP22848 Published by ProQuest LLC (2014). Copyright in the Dissertation held by the Author. Dissertation Publishing Microform Edition © ProQuest LLC. All rights reserved. This work is protected against unauthorized copying under Title 17, United States Code ProQuest LLC. 789 East Eisenhower Parkway P.O. Box 1346 Ann Arbor, Ml 4 8 106- 1346 UNIVERSITY OF SOUTHERN CALIFORNIA TH E GRADUATE SCHOOL UNIVERSITY PARK LOS ANGELES, CALIFORNIA 90007 This dissertation, w ritten by Gwoboa # Horng.................................................................. under the direction of h i s D issertation Committee, and approved by all its members, has been presented to and accepted by The Graduate School, in partial fulfillm ent of re quirem ents for the degree of Ph.ix CpS ) 92 H S I 6 D O C T O R OF PH ILO SOPH Y Dean of Graduate Studies Date ..Marchit2 fct,1992 DISSERTATION COMMITTEE Chairperson D ed ica tio n This dissertation is dedicated to my parents, M a-tsao Horng and Sui-sheng Huang, and my wife, Chiung-yi Huang. i ii A ck n ow led gem en ts I would like to thank my advisor, Professor Ming-Deh Huang, for his advice, m oti vation and support throughout the years of research and writing of this dissertation. I would like to thank the other members of my dissertation com m ittee, Professor Leonard Adlem an and Professor Solomon Golomb, for their insights, comments and their valuable tim e. I would like to thank Professor Seymour Ginsburg and Professor Douglas Ierardi for serving as m em bers of my guidance com m ittee and providing valuable suggestions for my research. I would like to thank my wife, Chiung-yi, for her support and encouragement. j This research has been supported partially by M inistry of Education, R. 0 . C. and , NSF through grant CCR 8957317. j I I ! t iii | C o n ten ts D ed ica tio n ii A ck n o w led g em en ts • • • m A b stra ct vi 1 In tro d u ctio n 1 1.1 Problem s and M o tiv a tio n s ................................................. ............................ 1 1.2 Overview of the Approach and R e s u l t s ........................ ......................... 6 1.3 O utline of the D isse rta tio n ................................................. ......................... 8 2 P relim in a ries on G alois T h eo ry and C o m p u ta tio n s in A lgeb raic N u m b er F ield s 1 0 2.1 Elem entary Galois T h e o r y ................................................. ......................... 10 2.2 Com putations in Algebraic Num ber F i e l d s ................. ......................... 14 3 N e ste d R ad icals - D efin itio n s and B a sic Facts 18 3.1 D epth and Value of Nested R ad icals............................... ........................ 18 3.2 Nested Radicals and Root T o w e rs ................................... ......................... 21 3.3 Basic Facts ............................................................................. ......................... 23 3.4 Field Theoretical P ro b le m s................................................. ......................... 27 4 S im p lify in g N e ste d R ad icals 30 4.1 In tro d u c tio n ............................................................................. ........................ 30 4.2 Conductor-D iscrim inant F o r m u l a ................................... ......................... 32 4.3 Proof of Theorem 4 . 1 ........................................................... ......................... 33 5 R o o ts o f U n ity 38 5.1 In tro d u c tio n ............................................................................. ......................... 38 5.2 Lagrange R e so lv e n t............................................................... ......................... 39 5.3 Two Useful T h eo rem s........................................................... ......................... 41 5.4 Proof of Theorem 5 . 1 ........................................................... ........................ 46 IV 6 S o lv in g P o ly n o m ia l E q u ation s 51 6.1 In tro d u c tio n ........................................................................................................ 51 6.2 Proof of Theorem 6 . 1 ........................................................................................ 53 6.3 Proof of Theorem 6 . 2 ........................................................................................ 53 6.4 Proof of Theorem 6 . 3 ........................ 62 6.5 A lgorithm s............................................................................................................ 74 7 C o n stru ctib le N u m b ers 78 7.1 In tro d u c tio n ........................................................................................................ 78 7.2 Proof of Theorem 7.1 and 7 . 2 ...................................................................... 81 7.3 Proof of Theorem 7.3 and 7 . 4 ...................................................................... 86 8 C on clu d in g R em ark s 89 8.1 Sum m ary of the D issertatio n .......................................................................... 89 8.2 Directions for Further R e s e a r c h ..................................................................... 90 j I I I 1 v A b stra ct This dissertation addresses the issue of algebraic simplification in com puter alge bra, particularly the problem of simplifying expressions involving nested radicals. The problem of solving polynomials by nested radicals in m inim um depth, is also studied. A theoretical framework is developed which provides: form al definitions of nested radicals, their depth and value; th e notion of pure nested radicals and their field theoretical properties; and a field theoretical translation of the problems m entioned above. Then an algorithm is developed for denesting nested radicals over algebraic num ber fields in the case where roots of unity are treated as symbols of no nesting depth. ] N ext, the m inim um depth of nested radicals for roots of unity is studied. A j polynom ial tim e algorithm is developed to find a nested radical of m inim um depth ' over Q which represents exactly all th e roots of a cyclotomic polynomial. j For the general problem of solving polynomials by nested radicals in m inim um depth, it is shown th at the m inim um possible depth of a pure nested radical rep resenting the set of roots of an irreducible polynomial is at m ost one more than th e m inim um possible depth of nested radical representing the same set. It is also shown th a t there exists a polynomial tim e algorithm to solve irreducible, abelian polynomials over Q by nested radicals in m inim um depth. Finally, the special class of nested radicals involving only square roots is studied. This class of nested radicals represents exactly the constructible num bers. It is shown th a t exactly n square root extractions are needed if the m inim um polynomial of the constructible num ber is of degree 2n. Furtherm ore, there is a polynomial tim e algorithm to determ ine if the roots of an irreducible polynom ial f{ x ) over an algebraic num ber field are constructible, and, if it is the case, return an expression VI involving m inim um num ber of square root for representing all the roots of f( x ) . The m inim um nesting depth of square roots for representing constructible num bers is also studied. It is shown th a t there is an algorithm to' express constructible num bers in nested radicals of m inim um nesting depth of square roots. C h ap ter 1 In tro d u ctio n 1.1 P ro b lem s and M o tiv a tio n s Com puter algebra concerns th e use of com puters for perform ing symbolic m athe m atics such as algebra and calculus. M any com puter systems exist for performing some form of com puter algebra [43, 39]. The application areas of com puter algebra and com puter algebra systems are numerous. For examples, com puter algebra sys tem s can be used as research tools in Physics such as quantum mechanics [9] and com puter algebra is useful in chemical inform ation systems and the enum eration and structural representation of chemical isomers [2]. A survey of applications based either on th e fundam ental algorithm s in com puter algebra or on the use of a com puter algebra system can be found in [5]. A m ajor issue in the design and im plem entation of com puter algebra systems is the simplification of expressions. Most com puter algebra systems, for examples M ACSYMA, RED U CE, SYMBAL, provide various simplification procedures for classes of expressions [12, 13, 14, 18, 31, 11]. Sometimes, users m ay teach the system new simplification rules. As observed in [14], there are at least three im p o rtan t reasons for simplifying expressions in these systems. The first reason is to m ake an expression smaller in store and to speed up subsequent calculation. The second reason is to put an expression into an intelligible form. The th ird reason is to see if an expression is identically zero. For a survey of some basic techniques for th e simplification of various classes of expressions, we refer to th e paper by Buchberger and Loos [4]. 1 The problem of simplifying expressions is in general undecidable. In fact the following result shows th a t for expressions from a sufficiently rich class of func tions even th e question: w hether an expression can be simplified to 0, is already undecidable [34, 7, 28]. T h e o re m (Richardson, Caviness, M atijasevic) Let R be the class o f expressions generated by (i) the rational numbers and the real number ir, (ii) the variable x, (Hi) the operations o f additions, multiplications and compositions and (iv) the sine and absolute value functions. Then the set { E \E (E R ,E — 0} is undecidable. In this dissertation, we consider th e class of expressions involving only radical signs ™ /~ , constants (elements in a field K ) and arithm etical operation symbols , x , /, These expressions are called nested radicals defined over th e ground field K . They are im portant objects in algebra since they represent roots of solvable polynomials. As m entioned in [43], it is difficult to give a rigorous definition of the simplest j fo rm of an expression because it depends on the intended use and even the per- ! sonal taste of th e users. However, for nested radicals, it is n atural to consider sim plification w ith respect to the nesting depth. We give two examples of this kind of simplification. The first example is taken from th e work of Ram anujan [32]. Consider the following identity ^ ^ 3 2 / 5 - ^ 7 7 5 = ^1/25(1 + \/3 - (v^ )2). We see th a t the nested radical of depth 2 at the left hand side is simplified to one of depth 1 at the right hand side. The second exam ple is more com plicated and is originally due to Shanks [36]: \ ] l l + 2 V 2 9 + \ / l 6 - 2V^29 + 2\/55 - 10^ = y/22 + 2y/5 + VE. 2 Again, the nested radical of depth 3 at the left hand side is simplified to one of depth 2 at the right hand side. O ur first research problem is to study how to simplify a given nested radical to one of m inim um depth. It is called the denesting problem. A nested radical over a ground field K represents a set of values in the algebraic closure K of K . It is said to be exact if it represents exactly an element in K and all its conjugates over the ground field. For example, \J§ + 2\/6 is exact while ■\/—8 is not. W hen denesting an exact nested radical, one expects the simplified nested radical to be exact as well; th a t is, representing the same set of values. For exam ple, \J§ + 2 \/6 can be simplified to y/2 + \/3 as an exact nested radical. However, y \ / — 3 -f 1 can not be simplified to v^—8 as an exact nested radical. Since ; a nested radical represents roots of a solvable polynomial, an exact nested radical t represents exactly all the roots of an irreducible solvable polynomial. Therefore, denesting in this context is reduced to solving irreducible, solvable polynomials by exact nested radicals in m inim um depth which is our second research problem. We call it the min~depth exact-solution problem. For exam ple, consider the polynomial x 6 — 2a;3 -J- 4. It can be solved exactly by first solving u2 = — 3, then for all such u solve v3 = 1 + u. Symbolically, the set of roots of the polynomial can be exactly represented by the nested radical yja /— 3 + T . However, the same polynomial can be solved, though not exactly, in one step by solving the binom ial x 9 + 8, since x6 — 2a:3 + 4 is an irreducible factor ! of x9 + 8 . j i An im portant consideration when dealing w ith the problems of denesting and m in-depth exact-solution is th e depth of roots of unity. As it turns out, these two problems become easier when roots of unity are taken as symbols of no nesting depth. In this case, we will offer a complete algorithm ic solution for the problems. I W hen roots of unity are not taken as symbols of no nesting depth, the denesting j and m in-depth exact-solution problems are much m ore difficult. In fact, a basic | I problem is to study nested radicals of m inim um possible depth representing roots of unity; or equivalently, to solve the cyclotomic polynomials exactly by nested radicals in m inim um depth. 3 There are m ore th an one m ethods to find a nested radical which represents exactly all th e roots of a cyclotomic polynomial over Q. For example, we give two m ethods for solving < & 7(x) = x6 + x s + -- - + x + l = 0 exactly by nested radical. T he first m ethod is to reduce th e problem to solving a lower degree polynomial by m eans of transform ations. Let y — x + x~x. Then the equation becomes y3 + y 2 - 2y - 1 = 0. I I Let 7 1 Then the equation becomes 3 343 7 z A ------------------------ = 0 . ^ 729z3 27 I t’s easy to find * = 5 # 1 + s ^ ' ' Hence, » = | ( - 1 + # 1 + 3^ ) + _ _ I It can be shown th a t | (f/ + y /y 2 — 4) has depth 3 and represents all the prim itive I seventh root of unity. The second m ethod is to use Lagrange resolvents. Let t(a , /?) = a + a 3(3 + a 2(32 + a 6j33 + a A (34 -f a 5/35 where $ 7(0:) = 0 and f3 is a prim itive 6-th root of unity. Let ti = t(a , (31 ) for i = 1,..., 6 . Then a = (ix + 12 + • • • + te)/6. It can be shown j th a t and tQ are in Q(/?). Therefore, they can be expressed in ; term s of (3. Let = 7 € Q (/?). Then ti = t f y and and be expressed in I term s of and some elements in Q(/?). Since (3 can be expressed as (1 + y /—3)/2, 4 i we see th a t a can be expressed as a nested radical of depth 2 . (See C hapter 5 for details.) The second m ethod is actually quite general, it can be used to solve any p-th cyclotomic polynom ial equation. If the second m ethod does give us a m inim um depth nested radical for a prim itive 7-th root of unity then an interesting question is: Do we always get a m inim um depth nested radical which solves 3 > p(x) = 0 exactly by using Lagrange resolvents? We will provide an affirmative answer to this question. In fact, the theoretical m achinery developed to prove this result will be generalized to solve th e m in-depth 1 exact-solution problem for abelian polynomials over Q. For the last research problem, we consider a very special type of nested radicals where the radical signs involved are only square root They represent the so called constructible numbers. A complex num ber is called constructible if we can locate it in the complex plane by using ruler and compass only. Each step of a construction calls for finding a : l point of intersection either of two lines, a line and a circle, or two circles. T hat is, finding the sim ultaneous solution of either two linear equations, a linear equation and a quadratic equation, or two quadratic equations. Therefore, in any case, the worst th a t is involved algebraically is a square root. Two kinds of simplification are considered for this type of nested radicals. ; One is to simplify w ith respect to the num ber of square roots and the other is to simplify w ith respect to the nesting depth of square roots. T he former corresponds to minim izing the num ber of square root extractions when we evaluate a nested radical sequentially. T he later corresponds to m inim izing the num ber of parallel square root extractions when we evaluate a nested radical w ith m any processors in parallel. Consider the exam ple given by Shanks. Let A = y ^ T + 2 V ^ + y i 6 ^ 2 v i 9 > 2 \ / ^ 5 ^ 1 0 ^ and let b = \ j n + 2 \/5 + VE. 5 Let a be the value represented by A , B . Then a can be constructed by taking 4 square roots if we use the nested radical A. However, it takes only 2 square roots to construct a using th e nested radical B . T he last research problem is to study the num ber (length) as well as the nested depth of square-root extractions needed to construct a constructible num ber. 1.2 O verview o f th e A p p roach and R esu lts Consider the exam ple given by Shanks. The nested radical y/ll + 2V29 + ^ 1 6 - 2V 29 + 2V55 - lO v ^ is defined over Q. Let a , /?, 7 , S be positive real num bers such th a t a 2 — 29, /?2 = 11 + 2 a, 7 2 = 55 — 10a and S2 = 16 — 2 a + 2 7 . T h at is, a is th e positive root of x 2 — 29, is the positive root of x 2 — (11 + 2 a), 7 is the positive root of x 2 — (55 — 10a ), and < 5 is th e positive root of x 2 — (16 — 2a + 2 7 ). Let K 0 = Q, K \ = K 0(a), K 2 = K \{(3,7 ) and I( 3 = 1( 2(8 ). Then, for i = 1,2,3, I(.t is generated by some roots of binomials over I ( i ~ \ and I({ is said to be a root extension of K i- 1. The sequence of extension fields I( 0 C I(\ C K 2 C I(z is called a root tower of depth 3. Hence there is a natural way to associate nested radicals w ith root towers. Conversely, there is also a natural way to associate elem ents in a root tower w ith nested radicals. For examples, a can be represented by \/29 and /3 can be represented by ^ 1 1 + 2a/29 etc. (See C hapter 3 for a m ore detailed discussion.) Therefore, the denesting problem is reduced to finding a root tower of m inim um depth containing a value of the nested radical. In [22], Landau showed th a t when the ground field contains all roots of unity the m inim um depth of root towers containing a value a is equal to the m inim um depth of root towers containing the ' splitting field of the m inim al polynomial of a. For a nested radical over Q or more generally over a num ber field, the problem rem ains as to w hat roots of unity are sufficient in order to build such a tower. It is also shown in [22] th a t, given a nested radical representing a value a , one can find a nested radical w ith depth at most one m ore th an the m inim um depth over the ground field adjoining a prim itive I-th root of unity where / is the degree of the splitting field of a. Though this gives a 6 near optim al solution, it rem ains open to decide, when roots of unity are taken as symbols of no nested depth, w hether a nested radical can be denested and to find a denested form when it exists. We solve this question when the ground field is an algebraic num ber field. We show th at, given a nested radical A over an algebraic num ber field K and a value a of A , there is an algorithm to find a positive integer n and a nested radical B over K (£ n), where is a prim itive ra-th root of unity, such th a t the m inim um depth of nested radicals representing a over K(fJ,< x > ), the ground field adjoining all roots of unity, is equal to the depth of B . For th e m in-depth exact solution problem, the situation is more complicated. We w ant to find a m inim um depth root tower which not only contains a root of the given polynomial b u t also has an exact nested radical associated w ith th e root. O ur approach is to work on a special kind of towers (called pure root towers) having the property th a t the associated nested radicals (called pure nested radicals) for elem ents in such a tower are exact. Then we relate the m inim um depth of pure nested radicals for th e given polynom ial and the m inim um depth of exact nested radicals for the same polynomial. W hen th e ground field contains enough roots of unity (or equivalently roots j of unity are treated as symbols of no nesting depth) we show th a t the m inim um depth of nested radicals solving a given irreducible polynom ial exactly is equal to th e m inim um depth of (pure) root towers containing the splitting field of the given polynomial. Moreover, there is an algorithm which finds a nested radical of m inim um depth over an algebraic num ber field adjoining all roots of unity which solves th e given polynom ial exactly. W hen roots of unity are not present in the ground field, th e problem becomes much m ore difficult and rem ains largely open. We show th a t th e m inim um depth t of pure nested radicals for a given irreducible polynom ial is at m ost one m ore than the m inim um depth of exact nested radicals for the polynomial. Furtherm ore, for irreducible polynomials f(x) over Q such th at the splitting field of f(x) is abelian over Q we show th a t the m inim um depth of a pure nested radical for f(x) is equal to th e m inim um depth of the exact nested radicals for the same polynom ial. In particular, we obtain a recursive form ula for the m inim um depth of exact nested radicals for cyclotomic polynomials. Let d(n) be the m inim um depth of pure nested radical for the n-th cyclotomic polynom ial over Q . Then 7 ___l d{ 1) = d{2) = 0 ; d(p) = d(p — 1) -j- 1 and d(pe) = d(p) + 1 for prim e p and e > 1; d{mn) — m ax{d(m ), d(n)} for m,n w ith (to, n) = 1. In addition, such a m inim um depth nested radical can be obtained by using Lagrange resolvents through a specific recursion. Polynomial tim e algorithm s for solving this kind of polynom ials exactly by nested radicals of m inim um depth are also given. For constructible num bers, we show th a t exactly n square root extractions are needed if th e m inim um polynom ial of the constructible num ber is of degree 2n. M oreover, we show th at there is a polynomial tim e algorithm which on input an irreducible polynom ial / over an algebraic num ber field, determ ines if the roots of / are constructible, and if it is the case, returns a nested radical involving m inim um num ber of square roots for the roots of / . We also show th a t there is an algorithm I to find a nested radical of m inim um possible depth representing a constructible num ber a, subject to th e condition th at only square roots are used in the nested radical. The running tim e is polynomial in th e length of the nested radicals and the degree of the splitting field of a. ] 1.3 O u tlin e o f th e D isserta tio n The rest of this dissertation is organized as follows. C hapter 2 consists of two sections. The first section contains a brief overview of the Galois theory. The second section contains some Galois theoretical compu- I tations over algebraic num ber fields. C hapter 3 begins w ith the formal definitions of nested radicals, their depth and value. Then (pure) root extensions and (pure) root towers are defined, the correspondence betw een nested radicals and root towers is established, and basic properties of nested radicals are studied. In the end, our research problems are 1 restated in field theoretical term s. The next four chapters are devoted to studying the four research problems. The denesting problem is tackled in chapter 4. The pure nested radical depth of the root of unity is treated in chapter 5. C hapter 6 contains some results of the m in-depth exact-solution problem. C hapter 7 is devoted to the study of the length and depth of the constructible num bers. The organization of these four chapters is | similar. They all begin w ith an introductory section containing some related work j 8 and sum m ary of our results. The proofs of these results are contained in the rest sections of each chapter. C hapter 8 , the last chapter, summarizes the contributions of this dissertation and suggests directions for further research. C h ap ter 2 P relim in a ries on G alois T h eory and C o m p u ta tio n s in A lgeb raic N u m b er F ield s 2.1 E lem en ta ry G alois T h eo ry j Galois theory is generally regarded as one of the central parts of algebra. Its cre ation resolved one of the oldest problems in algebra, the solvability of polynomials j by radicals. The purpose of this section is to review some fundam ental theorem s of Galois theory and to define some symbols th a t will be used throughout the dissertation. For details and proofs of th e theorem s in this section, we refer to [26, 20]. Let K be a field. An extension field L of K is a field having K as a subfield. We w rite L j K for th e field L considered as an extension field of K . Let M / K and N / K be two extension fields. If M and N are contained in j some field L , then we denote by M N the smallest subfield of L containing both ' i M and N , and called it the composite field of M and N in L. Thus whenever we ' w rite M N we autom atically assume th at M and N are contained in some field L. j Similarly, the com posite field of any finite num ber of subfields £ ] r . . , £ n ofa field j L is defined to be the smallest subfield of L containing E x ,..., E n. If S C L, we denote by K(S) the subfield of L generated by K U S, and say th a t K ( S ) is the subfield o f L generated by S over K . We say th a t L is finitely generated over K if there exists a finite subset S of L such th at L = K{S). If S — {a} then we use K(a) to denote AT({a}). An elem ent a € L is said to be a | prim itive element o f L over K if L = K(a). L is said to be a simple extension j of i f if L adm its a prim itive element over if . T he dimension of the if-space L, denoted by [L : if], is called the degree of the extension. We say th a t L is a finite extension o f if when [L : if] is finite. Let E be an interm ediate field between i f and L. T h at is, i f C E C L. It is easy to see th a t [L : i f ] = [L : E][E : if]. Also, [E : if] = [L : if] implies E = L and [L : E] = [L : if] implies E = if . A nonconstant polynom ial f ( x ) in K[x] is said to split in i f [a ? ] when each of its irreducible factors in i f [a:] has degree 1. By a splitting field of f(x) over i f we shall m ean an extension field L of i f generated by all the roots of f{x) and f(x) splits in L[x\. All splitting fields of a given polynomial / over a given ground field i f are isom orphic over i f , so th at it is legitim ate to speak of the splitting field of / over if. Let S be a finite set of non constant polynomials in i f [a:]. It can be shown th at there exists an extension field L of i f such th a t every polynom ial in S splits in L[x]. A nonconstant polynomial / 6 K[x\ is said to be separable in K[x] if / and / ' are relatively prime. Let i f be a field and let L be an extension field of if . An elem ent a £ L is said to be algebraic over i f if it is a zero of a nonzero polynom ial in K[x\. The monic polynom ial in i f [a;] th a t adm its a as a zero and divides every polynomial in i f [a:] adm itting a as a zero is called the m inim al polynomial o f a over if . We say L is algebraic over if , or th a t L is an algebraic extension o f i f if every elem ent of L is algebraic over i f . It can be shown th a t L is finite over i f if and only if L i is algebraic and finitely generated over if. I Let i f be a field, i f is called algebraically closed if every nonconstant polynom ial in i f [a:] adm its a zero in if . An algebraic closure o f i f , denoted by K , is an algebraically closed algebraic extension of if . Let L be an extension field of if . L is said to be a norm al extension o f i f , or to be norm al over i f , if L is algebraic over i f and th e m inim al polynom ial over i f of every elem ent of L splits in L[x], An elem ent of L is said to be separable over K if it is algebraic over i f and its 1 i m inim al polynom ial over i f is separable. L is said to be a separable extension o f i f , or to be separable over if , if every elem ent of L is separable over If. L is said to be a Galois extension o f K , or to be Galois over i f , if L is norm al and separable over i f . 11 T h eo re m 2.1 Let K be a field. Then every finite separable extension o f K is a simple extension o f K . N orm ality and separability are independent properties. However, for field of characteristic 0, every algebraic extension is separable. Therefore, the Galois ex tensions of such a field are ju st its norm al extensions. A nd every finite extension ! of such a field is a simple extension. Throughout this dissertation, we will assume th a t all fields have characteristic 0. We will let Z denote th e ring of integers, let Q denote the field of rational num bers, let R denote the field of real num bers and let C denote the field of complex numbers. Let K be a field and let L be an extension field of K . The set of all autom or phisms < f > : L y-y L such th a t f>(a) = a for all a € K is a group under composition of functions. It is called the Galois group of L over K , denoted by GiLfK). Let f(x) € K[x\ and let I be a splitting field for f(x) over K. Then G(L/K) is called the Galois group of f(x) over K. Any element of G(L/K) defines a perm utation of roots of f(x). Also \G(L/K)\ = [L : K). Let H C G(L/K). We define L H by : L H = { a \a € L , er(a) = a for all a H }. It is called the fixed field of H. I T h eo rem 2.2 (Fundamental Theorem o f Finite Galois Theory) Let K be a field and let L be a finite Galois extension over K . (i) The mapping M y — y G{L/M) from the set o f all intermediate fields between L and K to the set o f all subgroups o f G (L / K ) and the mapping H y-y L H from the j set o f all subgroups o f G(L/K) to the set o f all intermediate fields between L and ! K are m utually inverse inclusion-reversing bijections. ' (ii) I f K C M C L then M /K is normal if and only if G(L/M) is a normal j subgroup o f G(L/K), in this case G (M /K ) = G(L/K)/G(L/M). The usual Galois correspondence between subgroups of Galois groups of finite Galois extensions and interm ediate fields is not valid for infinite Galois extensions. However, it can -be corrected by the consideration of a n atu ral topology w ith the Galois group G of a Galois extension L/K. This topology is called K rull topology and is obtained as follows. For each crgG w e take the cosets crG (L/M ) as a basis I of neighborhoods of a where M / K runs through all finite Galois subextensions | of L/K. The m ultiplication m ap G x G y-y G, (<r, 7 ) y-y < J ~ f and the inverse m ap 1 G y-y G, cr y — y cr~x are continuous, so G becomes a topological group. ! 12 ; T h eo rem 2.3 (Fundamental Theorem o f Infinite Galois Theory) Let L f K be a (finite or infinite) Galois extension with Galois group G. (i) The map M ► G(L/M) is an inclusion-reversing bijection between the inter mediate fields M o f L / K and the closed subgroups H o f G with the inverse map given by H t — > L H . Furthermore, the open subgroups o f G correspond to the finite subextensions o f LfK. (ii) I f K C M C L and M / K is norm al then the restriction map G iL fK ) i — * • G(M/K) gives rise to the following isom orphism o f topological groups: G(M/K) = G(L/K)/G(L/M). T he following theorem about the degree of composite field is very useful. T h eo rem 2.4 Let M / K and N / K be extension fields o f finite degree where M is Galois over K . Then M N / N and M /M (IN are Galois extensions, G (M N /N ) = j G(M/MON), [MN : N] = [M : Mf)N] and [MN : K] = [M : I<][N : K]/[MDN : . K}. This theorem can be illustrated by the following diagram: M N M, N Lines slanting up indicate an inclusion relation between fields. This kind of lattice ! I diagram s of fields is extrem ely suggestive in handling extension fields. Let K be a field and let K be an algebraic closure of K. Let L be an extension field of K such th a t K C L C K and let < r € G(K/K). We shall use a* to denote th e elem ent a (a ) € K and use L a to denote the field { o f € K\a € L }. Let f ( x ) = anx n H b a \x + ao € K[x\. We shall use / C T (a;) to denote th e polynomial a ° nx n H b a{x + ag 6 K a[x]. 13 A finite extension L / K is said to be abelian (cyclic) if L / K is Galois and G(L/K) is abelian (cyclic). The following two theorem s are about abelian exten sions. T h eo rem 2.5 A n y finite abelian extension is a composite o f cyclic extensions o f prim e power degree. T h eo re m 2.6 (i) I f M / K and N / K are abelian extensions then so is M N /K . (ii) I f L / K is abelian and M /K is any extension then L M /M is abelian. | (Hi) I f L / K is abelian and M is a field such that K C M C L, then L /M and | M / K are abelian. J A polynom ial equation is said to be solvable by radicals if its solutions can be obtained from the coefficients in a finite sequence of steps using exclusively the field operations and the extraction of roots. A group G is called solvable if there is a sequence G — Go D Gi Z) • • ■ D Gr = {1) such th a t G{/G{+i is abelian for all i,0 < i < r — 1. An extension field L / K is said to solvable if L / K is Galois and G(L/K) is solvable. T h eo re m 2 .7 (Galois) Let K be a field o f characteristic 0, and let f(x) € K [x\ be nonconstant. Then the equation f(x) = 0 is solvable by radicals over K if and only if the Galois group o f f(x) over K is solvable. I I 1 2.2 C o m p u ta tio n s in A lgeb raic N u m b er F ield s j All the algorithm s for denesting and solving polynom ial equations given in this i dissertation are over algebraic num ber fields. In, this section, we introduce some Galois theoretical algorithm s over algebraic num ber fields. Let A be a subfield of C such th a t [K : Q] is finite. Then K is said to be an algebraic number field. An algebraic integer is a complex num ber th a t is a root of a monic polynom ial in 7i[x\. The subset of K consisting of all algebraic integers in K forms a ring, denoted by O k , called the ring o f algebraic integers in K . Let f(x) € Z[.t] be monic and irreducible of degree n and let a be a root of f(x) in C. Let K — Q (a ) and Ok th e ring of algebraic integers in K. Let 14 <Ti,..., < rn be the embeddings of K in C. If /? £ K then the norm of (3 is defined by N(/3) = n t i K a n} C 0 K and 0 K = n * ,Q !<K G z ) then {«i,..., a n} is said to be an integral basis for Ok - T he discrim inant of Ok , disc Ok , is defined to be the square of the determ inant of [0 »(aj)]i<»,j<»i. T hat is, disc Ok — |«r*(otj) |2 £ Z. Since a is a root of f(x), the set { 1, a , a2, ..., a n~x } is ; a basis for K as a vector space over Q but not necessary an integral basis for Ok - T he discrim inant of f(x), disc f , and the discrim inant of a , disc a , are defined to be disc f = disc a = |<7,-(a^-1 )|2. It is well known th a t disc a = N(f'(a )) and disc a = k 2 disc Ok for some k £ Z . The norm can be extended in a natural m anner to polynomials g{x) £ K[x\. Each em bedding crj can act on g{x) by replacing coefficients of g(x) by theirs conjugates by th e action of c r,-. By using this action, the norm N(g) is defined by N(g) = Y\^=1g,Ti- Let A be a com m utative ring w ith identity. Let f(x) = amx m + oT O _ ix m-1 + ■ • • + a0 £ A[x] and g(x) = bnx n + bn^ 1x n~1 + ■ ■ ■ + b0 £ A[x\. M ultiply f(x) successively by 1, x , ..., x n~x and g(x) by 1, x , ..., x m~x. We have m + n linear forms in l,a:,... ,x m+n_1. Their determ inant is th e resultant Rx(f,g ) of f(x) and g(x). T h at is, Rx(f,g ) = det(M) where ^ 7 T i Q " n n— 1 &m— l . . . . . . i & m — 1 ' * ' « 0 j T he discrim inant of a polynomial can be com puted easily as the resultant of i the polynom ial and its derivative. The norm defined for polynomials can be used to com pute the m inim al polynomials of elements in K . Let /? £ K . Then /? can be expressed as a polynom ial, say /3(a), in a. Let fp(x) be the m inim al polynomial of f3 over Q. Then the norm of x — /3(a), N (x — /3(a)), is a power of fp(x) and it can be com puted by N(x — /3(a)) — Rx(f(y),x — /3(y)). \ i 15 ! M bn— i In th e following, we define th e size of polynomials over algebraic num ber fields. Let f ( x ) = anx n + an-\X n~x + . . . + a \x + «o € Z[a;] be irreducible. T he elements j3 of Q (a ) are represented w ith rational coefficients 6, in the basis 1, a , ...,, a n~x of the vector space Q (« ) of dimension n. T h at is, 0 — YfiZo he? where bi G Q for 0 < i < n — 1. We use th e isomorphism Q (a ) = Q [x ]/(/(x )). T he operations of negation, addition, subtraction and m ultiplication in Q(ct) are operations in Q[x] w ith a reduction m od f(x). The inverse is com puted by the extended Euclidean algorithm . Algebraically, a does not have to be distinguished from its conjugates. Following [41, 23], the size of /( x ) , |/( x ) |, is defined to be (XTiLo al T he size j of 0, ||/?||, is defined to be th e m axim um of the absolute values of th e conjugates I of /3. Let g(x) = 0mx m + • • • + 0 \x + 0O w ith 0i = J2]Zo h jed. Then th e size of g{x), ||flr(*)||, is defined to be m a x ^ J Z o T h eo rem 2.8 [29] Let /( x ) = anx n + an^ i x n~x + . . . + a,\x + a0 G Z[x] and let I g{x) — bmx m + 6m_i x m~x + . . . + b\x + bo be a divisor o f f(x). Then |^*| < \bm/an\2m\f(x)\. T h eo re m 2.9 [41, 23] Let a be an algebraic integer with m inim al polynomial f[t) = t n + an_it n_1 + ••• + a it + a 0 G Z[<]. Let 0 be a root o f g(x), a poly nom ial in 0 Q(a)[®]. Then ||/?|| < ||^(a:)|| + 1 . Assum e g(x) is monic and let h(x) be a factor o f g{x) in OQ(a)[a:]- Then ||/i(a;)|| < n||<5 r(x )||(rc|/(i)|)n. For th e rest of this section, we quote some im portant results on polynomial factorization and Galois theoretical com putations in algebraic num ber fields. We will not give th e description of the algorithm s and their tim e com plexity analysis. Instead, we state them as theorem s. In [27], Lenstra, Lenstra and Lovasz announced a polynomial tim e algorithm to factor polynom ials over th e field of rational numbers. T h eo rem 2.10 [27] Let f(x) = anx n + - • - + ao G Z [x ]. Then there is an algorithm to factor f(x) into irreducible factors over Z[x] in 0 ( n 9+e + n r+ £ log |/( x ) |) steps fo r all e > 0 . In [23], Landau showed th a t there is a polynomial tim e algorithm to factor polynomials over algebraic num ber fields. 16 T h eo re m 2.11 [23] Let g(t) be a monic irreducible polynomial o f degree m over 7i and let a be such that g(a) = 0. Let K = Q(o;) and let f(x) be a polynomial o f degree n over Ok- Then there is an algorithm to factor f(x) into irreducible factors over O k in 0 ( m 9+cn 7+ e log2 |^(<)| log2+t(||/(a;)J|(m |5f(t)|)r ! '(m n)T l)) steps fo r all e > 0 . It is also shown in [23] th a t the splitting field and its Galois group of a poly nom ial over an algebraic num ber field K can be com puted. ; j T h eo rem 2.12 [23] Let f{x) be an irreducible polynomial over an algebraic n u m -1 her field K . Then the Galois group o f f(x) and the m inim al polynomial o f a prim itive element o f the splitting field L [ K o f f(x) can be computed in tim e polynomial in [L : K] and log ||/(a;)||. To find the fixed field is also im portant in the Galois theoretical com putations. T h eo re m 2.1 3 [22] Let f(x) be an irreducible polynomial over an algebraic num ber field K . Let L / K be the splitting field o f degree r. Let H = {<Xi,..., a a} be a subgroup o f G(L/K) = {ai, ..., ar}. Then a set o f algebraic numbers fdi , ..., fdt such that L H = K{jd1, ..., /?<) can be computed in tim e polynomial in [L : K] and \ log 1 1 / 0*0 II- ' i The following theorem can be used to find the m inim al polynom ial of a prim itive \ elem ent of th e composite field of a finite num ber of fields. T h eo rem 2.1 4 [42] Let a , fd € C. Let f(x) be the m inim al polynomial polynomial o f a over Q and let g(x) be the m inim al polynomial o f (i over Q (o ). Let N = [Q (a, /3) : Q]. Then there is at least one integer s among N distinct integers such that e x . -f* sf3 is a prim itive element o f Q(a, /?) and its m inim al polynomial over Q can be computed in tim e polynomial in n, m , log |/(a:)| and log Ug^aO H where rn and n are the degrees o f polynomials f and g. 17 C h ap ter 3 N e s te d R a d ica ls — D efin itio n s an d B a sic F acts 3.1 D e p th an d V alue o f N e s te d R ad icals T he symbol first appeared in print in Rudolff’s Coss (1525). Using th e symbol ■s/a to denote a rc-th root of a was becoming fairly well standardized by the end of the 17th century [37]. This section presents definitions concerning nested radicals. They are defined as symbolic objects taking values in th e algebraic closure of a ground field. Their m ulti-valued nature and their associated field theoretic con structs are discussed and clarified through various exam ples, rem arks, lemmas, and theorem s. T he proofs in this chapter are Galois theoretic and elem entary in nature. Let K be a field and let K be an algebraic closure of K . In the following definitions, let * £ {+>—, j | I I D e fin itio n 3.1 T he nested radicals and their depths over K are defined as follows (1) An elem ent a of K is a nested radical of depth 0 over K . We w rite depthk {a) = 0 for all a £ K . (2) Let A, B be nested radicals over K . Then A * B is a nested radical of depth m ax(depthj((A),depthK (B)). i (3) Let A be a nested radical and n > 1. Then y/~A is a nested radical of depth equal J to d e p th i/A )- f 1. We shall call y/~A a simple nested radical of depth depth k {A) + 1 . ! We shall use depth(A) instead of d epthx(A ) when there is no ambiguity. | 18 D efin itio n 3.2 Let A be a nested radical over K . T he associated simple nested radicals of A , 5(A ), is defined as follows. (1) If A = a £ K then 5 (A ) = {a}. I ( 2 ) U A = B * C then 5(A ) = S ( B ) U 5 (C ). (3) If A = V B then 5(A ) = S ( B ) U { \fB } . Let A be a nested radical over a field K and let 5(A ) = {Ai,..., A n }. Then it is easy to see th a t there exists a f ( x ) 6 K ( xi ,..., x n) such th a t A = / ( A 1}..., An). D efin itio n 3.3 Let k be a field, a 6 k and n > 1 b e an integer. The radical tfa is an irreducible radical of degree n over k if xn — a is an irreducible polynom ial over | k E x a m p le 3.1 Let A — y \ / 2 A 1 A v^2 A v ^ . Then depf/iQ(A) — depth(A) = 3 and 5(A ) = {y] # 2 A 1 A ^ A 1, </2,2,1}. ^ 2 is an irreducible radical I of degree 3 over Q. D e fin itio n 3 .4 Let R be the set of all nested radicals over K . A radical valuation j v over K is a m ap v : R — * K such th a t ! ( 1) u(A) = A if A € K , and (2) v(A * B ) = v(A) * v(B ) for nested radicals A, B over K , and (3) (u(v^A))n = u(A) where A is a nested radical over K . Let R be the set of all nested radicals over a field K and let 5 C R- Let v be a radical valuation over K . We shall use v (S ) to denote the set {u(o)|o: € 5}. j D e fin itio n 3.5 Let A be a nested radical over K . Then value(A) = { u(A )|v is a ! radical valuation over K }. j < i i E x a m p le 3.2 (i) Let a € Q. Then value{tfa) = {aC N = 1, • • •, w) where a € C, olu — a and (n is a prim itive n -th root of unity. 19 (ii) Let Vi = value(\/2) = { \ i e C|A? = 2}. Then value{\J</2 + 1) = {A2 < E C IA ® = A j + l,Ai € Fi}. I t R e m a r k 3.1 The conventional use of identity between two nested radicals should be taken as identity under certain radical valuation. For exam ples, \/4 and (a/2)2 are distinct radicals th a t are equal under some valuations and unequal under others. The same rem ark can be m ade about y/2y/3 and \ / 6 , y/8 and 2 \/2 for examples. R e m a r k 3 .2 Consider th e exam ple by Ram anujan in C hapter 1. Let A = ^ 3 2 / 5 - ^ 2 7 /5 and let B = ^ 1 /2 5 (1 + ^ 3 - ( v ^ ) 2). To be precise, the equality A — B holds in the sense th at A and B are equal under some radical i valuations. In particular, there is a radical valuation v such th a t t>(^/32/5) = a x > 0, = a 2 > 0, v(^j-^32/E — ^ 2 7 /5 ) = a 3 > 0, = a 4 > 0, v m ) — «5 > 0, and — ol2. U nder the valuation v, A and B are equal. | On th e other hand, \value{A)\ = 75 and \value(B)\ = 25, so there are radical j valuations under which A and B are not equal. D e fin itio n 3.6 Let A be a nested radical over a field K . Then A is said to be exact if there is an irreducible polynom ial f( x ) € K [x ] such th a t value(A) = { o ja e K , f ( a ) = 0 }. i In general, a nested radical over a ground field K is not exact. T h at is, it does ; not represent an elem ent in K up to conjugacy. This is the case in particular for j reducible radicals. For exam ple, \/4 has values 2 and -2 which are not conjugate over Q. R e m a r k 3.3 Let v be a radical valuation over a field K . Let a € G (K /K ). Then a • v is a radical valuation over K . Consequently, for all nested radicals A over K , value(A) is closed under the action of G (K /K ). On the other hand, it is clear from R em ark 3.2 th a t two radical valuations over K are in general not related by an elem ent in G (K /K ). This is the case even if we restrict to the valuations of irreducible radicals. For exam ple, there are radical valuations tq and v2 such t i 2 0 | th a t u i( \/ 8) = vi(2y/2) > 0, and 0 > u2( \/ 8 ) = — u2(2\/2)- Hence there is no cr E C (Q /Q ) such th a t er • v\ = v2. D e fin itio n 3 .7 Let L f K be a field extension. Let R k , resp. R l , be the set of all nested radicals over K , resp. L. Let Vk , resp. Vl , be the set of all radical valuations over K , resp. L. Then R k C R l . Let A E R k and let v E Vk - A nested radical B E R l is said to be a denested form of A over L w ith respect to v if the following conditions hold: j (i) there is a v' E Vl such th a t v'{B) = v(A) and VC E •S'(A), v'(C ) = v(C ); j (ii) VD E R l such th at (i) holds, we have depth,L{D) > depth,L(B). We denote by dv(A /L ) the depth of a denested form of A over L w ith respect to j v. D e fin itio n 3.8 Given a nested radical A over K , a radical valuation v over K and a field extension L f K , the denesting problem is to find a denested form of A over L w ith respect to v. We are m ainly interested in the case when L = K (p <») or L — K . From Definition 3.7, if a nested radical B over a field K is a denested form of a nested radical A w ith respect to some radical valuation v over K , then there is a radical valuation v' over K such th at v'(B) = v(A). Since VC E S(A ) we have v'(C ) = ! v(C ), by Definition 3.7, v'(A — B ) = 0. T h at is, A — B represents 0 under the valuation v'. So for example, y/2 + y/3 is not a denested form of \/2 — v/3 under any radical valuation. For the exam ple considered in R em ark 3.2, B is a denested form of A w ith respect to some valuations, but not all. 3.2 N e s te d R ad icals and R o o t Tow ers D e fin itio n 3.9 Let k be a field. A field extension K over & is a root extension if K = k(ax, ..., a m) where, for all 1 < i < m , a f x = a, E k for some integer n,- > 1. It is a pure root extension if in addition, for all 1 < i < m , x ni — a,i is an irreducible binom ial over fc(o:i,..., a,-_i, Oj+i,..., a m). Each a * - is called a generating root of degree r^. 2 1 A root tower1, resp. pure root tower, of depth n over k is a tower of extensions k — k0 C A ? i C • • ■ C kn such th a t ki/ki-i is a root extension, resp. pure root extension, for 1 < i < n. If a is a generating root of degree m for ki/ki-x then it is called a generating root of degree m for the tower at level i. For exam ple, Q(Cp) is a root extension for all p > 2 since Q = 1. However, it is not a pure root extension for prim e p > 3 though it is always contained in a pure root tower. R e m a r k 3.4 There is a n atural way to associate an elem ent in a root tower w ith a nested radical. Let k = k0 C k\ C • • • C kn be a root tower. Let a be a generating root at level 1 of degree m such th a t a m = a £ k0. Then yfa is the nested radical associated w ith a . Inductively, let a be a generating root at level i of degree m. Then a™ = g(/31, ..., /?/) where /?, are the generating roots | at level < i and g € k { x \,... ,x{). Let B{ be the nested radical associated w ith j Pi. Then ^ g ( B 1,... , B t) is th e nested radical associated w ith a. Similarly, let j a = < * 1,2, • • •, " n .s j € K - for some g € k (x ltl, x lf2, ■■■, x n< S n) where j cxi'i, ..., cti> S i are all the generating roots at level i. Then the nested radical as sociated w ith a is A = g (A \ti, A 1>2, ..., AniS?l) where A{j is th e nested radical associated w ith the generating root a ,j . A is of depth n and there is a radical valuation v such th a t v(Aitj) — o;4 -j for all i,j, 1 < * < n, 1 < j < S{. Conversely, there is a natural way to associate a root tower w ith a nested radical over a field k w ith respect to a given radical valuation over k. Let A be a nested radical over a field k w ith depth(A) = d and let v be a radical valuation over k. Let Si C S (A ) be the set of all the depth i simple nested radicals associated w ith J A. Let ki = )b,_x(n(5’ {)). Then it is easy to see k = k0 C kx C • • ■ C kj is a root ' tower. It is called the root tower for A w ith respect to v. \ We are interested in a subclass of nested radicals, the pure nested radicals, defined in the following. x An extension Kfk is sometimes called a radical extension if there is a root tower from k to K. j 2 2 ! D efin itio n 3.1 0 Let vl be a nested radical over a field k. A is called a pure nested radical over k if for all radical valuations v over k the root tower for A over k w ith ! respect to v is a pure root tower. In the next section, we will show th at a pure nested radical is always exact. Let K be a field and let a E K . Then a is said to be solvable if the minim al polynom ial f ( x ) of a over K is solvable by radicals. Let a be solvable over a field K . Then a (pure) nested radical for a is a (pure) nested radical A over K such th a t value(A) = { a^lcr E G (K /K ) }. Note th a t such a nested radical m ust be exact. T he pure depth of a over K , denoted by d^-(ct), is defined to be the m inim um possible depth of a pure nested radical for a over K and the optimal depth of a. over K , denoted by Dk (&), is defined to be the m inim um possible depth j of a nested radical for a over K . Let K be a field and let a be solvable over K . Let fi E K (a ). Then it is easy to see th a t D k (o c ) < d/<:(a), D k {@) < D k (c h ) and dfc(P) < dfc(cn). 3.3 B a sic F acts j Irreducible radicals distinguish themselves from other radicals in th a t they repre sent elem ents in the algebraic closure of a ground field uniquely up to conjugacy. The next theorem shows th a t this is also tru e for pure nested radicals. T h eo rem 3.1 Let A be a pure nested radical over a field K . Then A is exact. To prove Theorem 3.1, we first observe the following simple lemma. s I L em m a 3.1 Let k be a field and let K / k be a finite extension. Let a be a root of an irreducible binomial x n — ft E K\x]. Let a E , G (k /k ) and let aq E k such that a " = . Then there exists a r E G (k /k ) such that t \ k = ct\k and atT = c*i. Proof: Since a™ = — (a n)( T = (aa)n, a° is a root of th e irreducible binomial x n — j3a E K a[x\. Hence, a a and c*i are conjugates over K a. Therefore, there exists a < 7i E G (k )K a) such th a t aq = a aai. Let r = a < 7 \ E G (k /k ). Then a T = c*i. 23 1 For every fii G K , we have fil = (fif)ai since /3f € K a and a\ G G { k /K a). Hence /?f = = {3{. Therefore, t \k = cr|j^. □ i L e m m a 3.2 Let K be a field. Let R be the set of all nested radicals over K and i let V be the set of all radical valuations over K . Let G = G ( K /K ) . Let A G R be a pure nested radical. Let T = 5(A ) and let v G V . Then W G V,3er G G such that V B G T v'(B ) — v (B )a, and v \ A ) = v(A)*. Proof: Let A = g (C i, ..., Cs) where g G K ( x\ , ..., x s) and Ci G R are simple nested radicals. Then T = 5(A ) = {Ci,..., Cs}. Let Si = {B \B G T, depth(B) < *}, for 0 < i < depth(A). We prove by induction on i th a t W G V, 3a G G such th a t v '(B ) = v ( B y , VL? G S{. Then th e lem m a will follow since W G V, 3a G such th a t v'(Ci) = for 1 < i < s and hence u'(A) = v'{g{C \,..., (7S )) = g (v'(C i),..., v'(Ca)) = 9 ( t i ( C , 17(0.)') = s ( v ( C ,) v (C .)Y = v(s(C 1, . . . , C . ) Y = v (A )° . W hen * = 0, the assertion is trivial since 5o C K . Assume the assertion holds up to n. Let S n = {B \,..., B i} and let Sn+i — Sn = { A%,..., Am}. Let ui be a radical valuation over K . Let L = K ( v i( B i) ,... ,vi(Bi)). By induction hypothesis, there exists a d G G such th at Vi(Bi) = v(Bi)a for i = 1,...,Z. Let /3S - = v(B,) for j i = 1,...,/. By definition, A, = n y/gi(Bi, ..., 5 ;) for some n; > 1 and < / ,- G I K ( x i , ... ,xi). Let q :,- = u(A„) and let a* = Vi(Ai). Then a ™ * = gi(/3\ , ..., fii) and = gi(fii, • • • ,/?f) = (< 7 i(/?i, • - •, fii)Y • By Lemm a 3.1, there is a a\ G G such th a t < 7i |l = &\l and a *1 = di. Since a 2 is a root of an irreducible binom ial over L(o;i), applying Lemm a 3.1 to a 2 over L{oc\), it follows th a t there is a cr2 G G such th a t cr2|£(ai) = <7\\l(<xi) and = d 2. Therefore, there is a ct2 G G such th at < 7 2|l = o-i\l, « i 2 = « i and a^ 2 =: < * 2- In this m anner, we see th a t there is a u 6 G such th a t o-\l = d \l and a f = di for 1 < i < m. □ P r o o f o f T h e o re m 3.1: Let 5(A ) = { A i, ..., A m} be the set of simple nested radicals associated w ith A. Then A = g(At,..., Am) for some g G K { xi , ..., xm). By Lemm a 3.2, V radical val uation vi over K , 3a G G (K /K ) such th a t vi(g (A i, ..., Am)) = v (g (A i, ..., A m) Y . \ i I 1 And by Rem ark 3.3, V <r £ G (K /K ), there is a radical valuation vx over K such th a t vi(g(A%,..., Am)) = w (< 7(A i, ..., Am))< '. The theorem follows. □ T h e o re m 3.2 Let k be a field. Let k = ko C k% C • • ■ C kn be a pure root tower, i Let A be the associated nested radical of some element a £ kn — kn- \ . Then A is a pure nested radical over k. Proof: Let k = (o;,i,... where is a root of an irreducible binom ial of degree m j over k - i( a i,i, ..., ..., ar*,s< ) and a f'f = aitj £ Let kitl = A;,-ijS ,._1(q!8i1 ) and let for j > 1. Then k = k0t0 C ki,i C A qj2 C ••• C kiiSl C A j2,x C ••• C & n > S n is a pure root tower. Renum bering the indices and rew rite the tower as k = K q C K \ C • • • C = kniSn where Ki = Ki-i(fii) and fii = a U jW for some u,w . Let Bi be th e associated nested radical for / ? ,■ for 1 < i < m. Let V be the set of all radical valuations over k. It is easy to see = "i^/oTJ is a pure nested radical since Vu € V, v{B\) is a root of the irreducible binom ial ar™ 1 -1 — alf\ over Ko- Inductively, we show th a t all the associated nested radicals for elem ents in Ki, i > 1, are pure nested radicals by showing th a t Ko = Kq C K \ C • • • C K™ is a pure root tower for all v 6 V where K j = K j_ 1(v(Bj)) for 1 < j < i. By induction hypothesis, Kq C K \ C * • • C K^_ 1 is a pure root tower. We show in the following th a t K^ is a pure root extension over K\’ _1. Let Bi = n '$fAi where Ai is a pure nested radical associated w ith the elem ent a,- £ K {- 1. Since k — K 0 C K i C • • • C K m is a pure root tower, x ni — ai is ! irreducible over K i- 1. Hence x n' — a? is irreducible over Kf_x for all a £ G (k/k). j By Lem m a 3.2, v(Ai) = aJ for some < r £ G (k/k). Therefore, v(B i) satisfies the | irreducible binom ial x ni — u(Ai) = x n' — a? £ Hence K " = A']L1('u(i?i)) is I a pure root extension over Kf_x. □ i i l Let a be solvable over a field K . Then, by Theorem 3.1 and 3.2, dx(oi) is equal ! to m inim um depth of a pure root tower containing a. Furtherm ore, the associated j 1 pure nested radical for a is a pure nested radical of m inim um depth for a. j T h e o re m 3.3 Let k be a field. Let f be an irreducible polynomial over k. Then j the following conditions are equivalent: ' i l (i) f is solvable by radicals over k; (ii) f is solvable by pure nested radicals; (Hi) f is solvable by exact nested radicals. Theorem 3.3 will follow from a sequence of simple lemmas. D e fin itio n 3.11 Let G b e a group. The exponent of G is th e least positive integer n such th a t a n = 1 for all a 6 G. L e m m a 3.3 Let L / K be an abelian extension with Galois group G (L jK ) of expo nent n. Assume p,n C K where p n is the set of all n-th roots of unity. Then L / K is a pure root extension. j Proof: Since L / K is abelian, there is an isomorphism cr from G (L /K ) to ®\-\H i for some cyclic groups Hi, 1 < i < I. Let Gi is th e subgroup of G (L fK ) isomorphic to (B j^H j under cr. Let Li be th e fixed field of Gi. Then G (L i/K ) = H Hence L ijK is cyclic and as [Hi : l]|n and p n C K , there is an a , over K such th a t L, = K(a.i) and is a root of an irreducible binom ial over K . Let Mi be the com posite field of L \ , . . . , Li and let M 0 = K . Then [L : K] — n to t-^ i+ i : Since [L : K] = n L i^ * ' : 1] and [M + i : M(\ < [Hi+i : 1], we see th a t [Mi+i : M(\ = : 1]. It follows L — K {o i\,..., on) is a pure root extension \ over K . □ L e m m a 3.4 Let n be a positive integer and let K be a field. Then there is a pure root tower over K which contains (n where (n is a primitive n-th root of unity. Proof: We use induction to prove this lemma. For n = 2, it is trivial th a t £2 G K . Assume for all 2 < n < I there is a pure root tower K = Ko C K \ C • • • C K m such th a t C n € K m. Now consider n = I + 1. Let d = [Ff(Cn) : K ]. Then d < I. By induction hypothesis, there is a pure root tower K ~ K 0 C K \ C • • • C K m such th a t (d G K m. Since K m((n)/ K m is abelian and Q G K m, by Lemm a 3.3, L = K { (n) is a pure root extension of K m. Therefore, K q C K i C • • • C K m C L is a pure root tower. □ 26 L em m a 3.5 Let k be a field and let K be contained in a root tower over k. Then K is contained in a pure root tower over k. Proof: By assum ption, K is contained in a root tower k = ko C k\ C • • • C km where & ,•+1 = kfioti) w ith a ”’ £ ki for some n,-. Let n be th e l.c.m. of th e n% . By Lem m a 3.4, £n is contained in a pure root tower k = Lo C L \ C • • • C Li = L. Let Ki = Lk{. Since (n £ Ki and [Ki+1 : Ki]\n, K i+ i/K i is abelian. By Lemm a 3.3, K i/K i-i is a pure root extension for all 1 < i < m. Therefore, k = Lq C Li C • • • C Li = L C K i C • ■ • C K m is a pure root tower containing K . □ P r o o f o f T h eo re m 3.3: It is easy to see (iii) — * (i). (ii) — ► (iii) follows from the definition of pure nested radical and Theorem 3.1. (i) — * • (ii) follows from Lemm a 3.5 and Theorem 3.2. □ 3.4 F ield T h eo retica l P ro b lem s In this section we translate our research problems into field theoretical problems. < t For th e first problem (the denesting problem ), we are given a nested radical A ; and a radical valuation v over a field k, and an extension field L over k. We want j to find a nested radical w ith m inim um possible depth which has value t>(.A) over L. By Rem ark 3.4, this problem can be translated into the following field theoretical problem. Given k, a field; A, a nested radical over k; v; a radical valuation v over k; L, a field extension of k. Find a root tower L = L q C L\ C • • • C L& of minimum possible depth such that v{A ) £ Lj,. For the second problem (the m in-depth exact-solution problem ), we are given an irreducible, solvable polynom ial over a field k. We w ant to find a nested radical 27 of m inim um possible depth which represents exactly all th e roots of the polyno m ial. This problem can be restated as follows: Given k, a field; f ( x ) £ k\x], irreducible, solvable. Find a root tower k — ko C ki C • • • C kd of minimum possible depth such that there is an a € kd with f( a ) = 0 and the associated nested radical for a has no other value than the conjugates of a. i If we restrict ourself to pure nested radicals then this problem can be trans lated into th e following field theoretical problem by Theorem 3.1 and Theorem 3.2. j i Given k, a fietd; f( x ) € k[x\, irreducible, solvable. Find a pure root tower k = ko C k\ C ■ • • C kd of minimum possible depth such 1 I that there is an a € kd with f ( a ) = 0 . j 1 T he th ird research problem (depth of roots of unity) is a special case of th e j second problem w ith k = Q and the n -th cyclotomic polynom ial over Q. j T he fourth research problem can be divided into two part. In both parts, we are given a constructive num ber a. For the first p art of the problem, We want to express a by a nested radical w ith m inim um num ber of square roots. The corre sponding field theoretic problem is the following. Given k, a field; a, a constructible number over k. Find a tower of root extensions k — ko C k\ C • • • C kd of minimum possible depth d such that, for all 1 < i < d, k{/ki-\ is generated by a single generating root of degree 2 , and a £ kd. For th e second p art of th e problem, we want to express a by a nested radical of m inim um nesting depth of square roots. The corresponding field theoretic problem is the following. 28 Given k, a field; a, a constructible Find a tower of root extensions k depth d such that, for all 1 < i < d, roots of degree 2 , and a € k^. i I number over k. = k0 C kx C • • ■ C kd of minimum possible ki/ki-1 is generated by a set of generating I ! 29 C h ap ter 4 S im p lify in g N e s te d R ad icals 4.1 In tro d u ctio n Borodin, Fagin, Hopcroft and Tompa [3] considered how to simplify nested radicals involving only square roots. Polynom ial tim e algorithm s were given for denesting the following two special forms of nested radicals A in this class: (1) V B £ S(A), depth(B) = and (2) depth(A) = 2. They also showed the following T h eo re m (Borodin, Fagin, Hopcroft and Tom pa1) Let K be a real extension of Q , that is, Q C K C R , and let a ,b ,r £ K with y/r £ K . Let n i , . . . , n k > 1 and ai,. . . ,a k £ K be positive. If \Ja + by/r £ K ( ..., then \ja + by/r £ K ( ^ r , yjp{,..., ^/pf) for some positive p i , . . . , p k £ K . \ Therefore, when we restrict our attention to denesting over a real extension of I Q , the above theorem shows th a t no roots other th an square roots or fourth roots j are ever helpful for denesting \ja + b^/r, thus characterizing th e denestable nested i radicals in this case. I ! Zippel [44] produced th e following structure theorem th a t indicates when a field ] < generated by a nested radical of depth n can be m apped isom orphically to a field | generated by nested radicals of depth n — 1. j I 1 All roots are positive real roots. That is, only the radical valuation v such that v{A) > 0 for all simple nested radicals A are considered. i I I 30 | T h eo re m (Zippel) Let k be a field containing a primitive r~th root of unity. Let K be an extension field of k. Let L = K ( l/a ) for some ot E K . I f there exists a field F such that the intersection of K and F is k and L = K F then there exists a (3 E k such that a/3 E K*r . Furthermore, F = k(y/fi). \ I I Recently, Landau [24, 22] obtained the following general result. j i I T h eo rem (Landau) Let A be a nested radical and let v be a radical valuation over k where k is a field o f char 0 containing all roots of unity. Let L be the splitting field o f the minimal polynomial of v(A ) over k. Then there is a denested form B of A over k with respect to v such that VC € S (B ), v'(C ) E L fo r some radical valuation v' over k. However, typically we are given a nested radical over Q or m ore generally over an algebraic num ber field. Adding all roots of unity to the ground field is not com putationally feasible. Hence th e theorem above does not lead directly to an j algorithm for denesting. W hat Landau was able to show was th a t by adding a j prim itive root of unity to th e ground field it is possible to find a simplified nested | radical of depth no m ore th an 1 of the m inim um possible depth. To state this result, we will need the following definition. D efin itio n 4.1 Let G be a group and let x ,y E G. The commutator of x and y , denoted [x,y\, is the elem ent x ^ y ^ x y . T he commutator subgroup of G is th e subgroup generated by all the com m utators of G. We shall use G 1 to denote the com m utator subgroup of G and use G4 to denote the com m utator subgroup of G4 " 1 for i > 1. We also let G° = G. W hen G is solvable, the chain of group • G = G° D G 1 D G 2 D ■ ■ ■ is called th e derived series of G. The length of the 1 derived series, denoted 1(G), is the sm allest positive integer n such th a t Gn — 1. T h eo rem (Landau) Let A be a nested radical and v be a radical valuation over k, j where k is a field of char 0. Let L be the splitting field of the minimal polynomial \ of v(A) over k. Let G be the Galois group of L over k and let I be the l.c.m. of the i 3! exponents of the derived series of G. 1 ft is the depth of a denested form of A over k with respect to v. Then there is a nested radical B and a radical valuation v' over k((i) such that v '(B ) = v(A), depthk^ ( B ) < t + 1 and VC € S (B ), v'(C ) G -£(&)• This result guarantees a near optim um solution. However, it rem ains an open question to decide w hether a nested radical can be denested over th e ground field adjoining all roots of unity and to find a denested form when it exists. T he following is the m ain result of this chapter. T h eo re m 4.1 Let k be an algebraic number field and let A be a nested radical over k. Let v be a radical valuation over k and let L be the splitting field of the J minimal polynomial of v(A) over k. Let n be a natural number divisible by [L : & ] and the discriminant of L over Q . Let poo denote the set of all roots of unity. Then the tower of extension fields corresponding to the derived series for G (L((n)/ k((n)) determines a denested form B of A over k(p < * ,) with respect to v. Furthermore, B can be computed in time polynomial in the length of A, the degree of L over k and j n. j I R em ark 4.1 In Theorem 4.1, n can be the discrim inant of any monic, integral 1 I and irreducible polynom ial for L over Q . i 1 T he proof of Theorem 4.1 involves Kronecker-W eber Theorem and a conse- ! 1 quence of th e conductor-discrim inant form ula which will be described in the next j section. T he proof is given in Section 4.3. ! 4.2 C o n d u cto r-D iscrim in a n t F orm ula i i For a brief overview of class field theoretic background, th e readers are referred to i [40] (Section 3 A ppendix), and to [1] for a full account. i i Let K be an algebraic num ber field. A divisor Ad of A T is a finite form al sum j J2?=i PV where e, G Z>o and Vi is either a prim e ideal of K or a real archim edean ' place of K . For a € K , we w rite a = 1 mod*Ad if for all prim es Vi in Ad, a —1 € V^' and for every real archim edean place in Ad, a > 0 at the real em bedding corresponding to the archim edean place. Let Pm denote th e group of principal 32 fractional ideals of K w ith a generator a = 1 mod*Ai. Let Im denote the group of fractional ideals relatively prim e to A i. A fundam ental result in class field theory can be stated as follows (see [40] p. 338). T h eo rem 4.2 L e t L f K be a finite abelian extension. Then there exists a divisor f of K (the minimal such divisor is called the C o n d u cto r of L / K ) such that (i) a prime p ramifies in L / K if and only if p \ f (ii) for all divisors A i with f\A i , there is a subgroup H with Pm C H C I m such that I m / H is isomorphic to G (L /K ), and the isomorphism is induced by the A rtin m ap which associates an unramified prime p with a p £ G (L /K ) where ap is determined by the property that a px = x N^ mod P for all x in the ring of integers of L and for all prime ideals V over p. T he classical Kronecker-W eber Theorem follows from this theorem and can be stated as follows. T h eo re m 4 .3 Let oo denote the real place o fQ . L e tK be a finite abelian extension of Q . Let n £ Z>o so that noo is divisible by the conductor of K / Q . Then K C Q (C »). We also need th e following well-known result [6] p. 160. i I i T h eo re m 4.4 (Conductor-Discriminant formula) Let L / K be a finite abelian ex- | tension. Let D ( L / K ) denote the discriminant of L over K . Then D (L /K ) = j n x fx where x ranges over the group of characters of G (L f K ), and f x is the con- \ ductor of the fixed field of the kernel of x- ! It follows from the conductor-discrim inant form ula th a t for a finite abelian ' I extension K over Q , the conductor of K over Q divides th e discrim inant of K over Q. This will be needed in the proof of Theorem 4.1. 4.3 P r o o f o f T h eo rem 4.1 L em m a 4.1 Let G be a group and let H be a normal subgroup of G. (i) I f G /H is abelian then G 1 C H . (ii) (G / H f “ GiH /H for all i > 0. (iii) l(G /H ) < 1(G). 33 Proof: (i) Let a be the canonical hom om orphism from G to G /H . For x ,y € G, let cr(x) = w, a(y) = v. We have a([x,y\) = [w, n] = 1. Thus every com m utator [x,y\ (E H and therefore G 1 C H. (ii) I t’s easy to see Gz, i > 0, are subgroups of G. Since GZ H = H G \ GZ H is a subgroup of G. Since H C GlH and H (G Z H ) = (GZ H ) H , H is a normal subgroup of GiH . Since [xH, yH ]H = [x, y]H for all x, y € G \ (GZ H )X H = Gi+1H . Let r be th e m ap from GX H to (G /H ) 1 such th a t r([x,y]H ) = [xH,yH]. Then r is a hom om orphism w ith kernel H. Hence ( G /H ) 1 = G 1H /H . Inductively, (G /H ) i+1 = ((G /H ) 1) 1 “ (G 'H /H ) 1 ^ (G>H)X H /H “ Gi+1H /H . (iii) This result follows from (ii). □ i R e m a r k 4.2 Let A : be a field and K be a finite solvable Galois extension of k w ith j solvable Galois group G. Let 1(G) = n and let G = G° D G 1 D ■ ■ ■ D Gn = 1 i is the derived series of G. Then the corresponding tower of fixed fields k = ko C k\ C • • • C kn = K has the property th a t k{ +1 is the m axim um abelian subexten- j sion of K over ki by Lemm a 4.1 (i). In particular, when k contains a prim itive : IV-th root of unity where N is th e l.c.m. of th e exponents of G(ki+i/k i) for all j 1 < i < n, the tower is a pure root tower by Lemm a 3.3. The m axim ality of ; l over ki w ithin K implies th a t any pure root tower from k to K has depth at least n. i The first step in the proof of Theorem 4.1 is a characterization due to Landau [22] th a t when the ground field contains all the roots of unity, th e depth of a denested form of a given nested radical A w ith respect to a radical valuation v over the ground field is th e length of the derived series of the splitting field of the m inim al polynom ial of v(A). The second step in the proof is to find a root of unity sufficient for constructing a denested form. L andau’s result [22] is recast in Theorem 4.5 in order to take radical valuations into account. A slight extension of the result is also needed to deal w ith pure nested radicals. L e m m a 4.2 Let k = k0 C k\ C • • • C kn be a root tower and let Ki be the composite of k f for all a € G (k/k). Then K 0 (fj,oo) C K x(y < * ,) C • • • C K ^ p ^ ) is a root tower and l(G (K n(poo)/ K 0(poo))) £: n - 34 Proof: It is easy to see the com posite of two root tow er is a root tower. Since each ki has a finite num ber of conjugates over kQ , K 0 C K \ C • • • C K n is a root tower. Let KI = Ki(fioo). Since fcf(A*oo)/&f_i(^oo) is abelian for all cr € G (k/k), K'ifK\_x is abelian. Let G = G (K 'JK '0). Let K q = Lo C L\ C • • • C be the subfields corresponding to the 1 derived series of G where m = 1(G). Then Li is th e m axim um abelian subextension j of K n over Inductively, we see th a t Ki C Li-\K { C L{. Therefore, m < n. □ j I T h e o re m 4.5 Let A be a nested radical over a field k. Let a = v(A) for some j radical valuation v over k. Let L be the splitting field of the minimal polynomial of a over k. Then dv(A /k (p oo)) = l(G(L(fx00)f k(/j,oo))). Proof: Let B be a denested form of A over k(p( X > ) w ith respect to a radical valuation j v over k. Let d = depth ^ { ^ ( B ) and let I — /(C?(L(/x00)/A;(//00))). By Rem ark 4.2 J the corresponding tower of fixed fields of the derived series is a root tower. Hence I depthk(fj,00)(B) < l(G(L(fi00)/ k(p 00))). T hat is, d < I. j It is easy to see th a t B is a nested radical over k((n) for some integer n > 0. Let v' be a radical valuation over &(£n) such th a t v'(B) = v(A). Let ko = k ((n) C ki C • • • C kd be th e root tower for B w ith respect to v'. Let Kq = k0 and let Ki, 1 < i < d, be th e composite of k f for all cr € G (k/k). Then K 0 C K \ C • • • C Kj, is a root tower. Since fco(//oo) = &(//<*,), by Lemm a 4.2, l i G i K d i p ^ / k ^ ) ) ) < d. Kd is Galois over k as K f = Kd for all cr € G (k/k). Since v(A) € Kd, we have L C K d. Let G = G (K d(fJ'o 0 )/k(iJ,o0 )) and let H — G (K d(p,< x > )/ L^oo)). Since L(fi00)/k(fUoo) is Galois, H is a norm al subgroup of G. By Lem m a 4.1 (iii), Z(<Z(L(^oo)/*(^oo))) = l(G /H ) < 1(G) < d. T hat is, I < d. Therefore, / = d. □ C o ro lla ry 4.1 Let A be a pure nested radical and u be a radical valuation over a field k. Let L be the splitting field of the minimal polynomial o fu (A ) over k. Then l(G(L(fitx > )/ k(p,< x > ))) = dv(A /k (p oo)) for all radical valuations v over k. Proof: This follows from Theorem 4.5 and Theorem 3.1. □ 35 P r o o f o f T h eo rem 4.1: Let K = i f l k(/j,oo). Then G (L /K ) = Gr(L(/u00)/A;(/u00)). Suppose K C k((n) for some n divisible by the exponent of G (L /k). Since K = L (1 k((n), G (L /K ) = G (L((n) /k ((n)). Hence G {L((n)/I<((n)) “ G (L (^ 00)/fc(^00)). Therefore, a denested form of A over k(fioo) w ith respect to v can be found by constructing the tower of subfields corresponding to the derived series of G (L((n)/ k((n)) by Theorem 4.5. The next step is to construct such an n. Let F = k nQ(/foo). Then G(k(fj,0O) / k ) = G (Q (^ 00) / i r). Let M = /t 'n Q ( //00). T hen K = k M and G (K /k ) “ G (M /F ). Let n = \D(L/Q)\[L : J fc ]. Then n is divisible by D { M f Q). Hence n is divisible by the conductor of M over Q. So by K ronecker-W eber Theorem , M C Q(Cn)- Hence M = K P i Q(Cn) and it follows th at K = k M C k((n). Therefore, G {L(U )} k{(n)) S G ( L / K ) S G(L(/x 00)/A;(A i00)). T he diagram illustrating th e above properties is as follows: (C») Q(C. M F Q I Hence the corresponding tower of subfields of the derived series of G (L((n)/ k((n)) is a pure root tower and determ ines a denested form for the given nested radical A over fc(^oo)- I To com pute a denested form of A, we first com pute th e splitting field L over [ k and a monic, integral and irreducible polynom ial g for a prim itive elem ent of L j over Q by Theorem 2.12. The discrim inant D(g) can be com puted by th e resultant j i 36 ( of g and g'. Observe th a t D(g) is divisible by the discrim inant of L over Q . Let n be the product of \D(g)\ and the degree of L over k. We then com pute the derived series for G(Z(£n )/ k((n)) and th e corresponding tower of subfields by Theorem 2.13. For each extension in the tower, we com pute a set of generating roots. This ! can be done by using Lagrange resolvents described in the next chapter. We then | find th e associated nested radical B for u(A) in the tower. B is a denested form for A over k(g00) w ith respect to v. T he overall tim e com plexity is polynom ial in th e length of A, th e degree of L over k and D(g). □ R e m a r k 4 .3 Let A be a nested radical over a field k and let 5(A ) = { ni^//i,i,..., n2i/? 2,2, •••, nd'^/fd,sd } where f a E f cfai . i, . . . , XiiSl, x3,i, ■ ■ •, ). Let v be a radical valuation over k. Then the m inim al polynom ial f ( x ) of v(A) over k can be com puted by using resultant in tim e polynom ial in n and log |A| where n is th e degree of f ( x ) and \A\ = m a x , - 1 ) . See [24, 22] for the description of the algorithm and tim e com plexity analysis. i I 37 C h a p ter 5 R o o ts o f U n ity 5.1 In tro d u ctio n Let K be a field and let n be a positive integer. A primitive n-th root of unity in K is an elem ent of order n in K*. A cyclotomic field of order n over K is a splitting field over K of th e polynom ial x n — 1 € K[x\. Let Un be th e set consisting of all prim itive n -th roots of unity in C. The n-th cyclotomic polynom ial 4 > n (a:) is defined to be monic polynom ial Tlcei/nC^ ~ ~ 0 - For K = Q , the following results are well-known. T h eo re m Let n be a positive integer and let L be a cyclotomic field of order n over Q . Then (i) $n(x) € z[x\; (ii) $ > n is irreducible over Q[x]; (iii) [L : K] = and G {L /K ) “ ( Z /n Z ) \ T he n -th root of unity (n can be described transcendentally by the form ula £n = cos(27rk/n) + i sm (2 irk/n) where k — 1,2,..., n. T he nested radical \ f \ is not considered as a nested radical for a prim itive n -th root of unity since its value contains 1 which is not a prim itive n -th root of unity for n > 2. 38 Gauss gave a famous m ethod of solving the p-th cyclotomic polynom ial by nested radical [16]. An interested consequence of G auss’s result is the fact th a t the regular 17-gon can be constructed w ith rule and compass. In this chapter, we show how to solve the n -th cyclotomic polynom ial by pure nested radical in m inim um depth. T he results are sum m arized in the following theorem . T h eo re m 5.1 Let n be a positive integer and let $ n(ar) be the n-th cyclotomic polynomial over Q . Then a pure nested radical A n of minimum depth which solves 3 > n(:z) = 0 exactly can be computed in time polynomial in n. Furthermore, let d(n) denote the depth of A n. Then d(n) can be determined recursively as follows ( 1) d (m n ) = m ax{d(m ), d(n)} for m ,n with (m,n) = 1; (2) d(pe) = d(p) + 1 for prime p and e > 1; (3) d(p) = d(p — 1) + 1 for odd prime p; (4 ) d( 1) = d(2 ) = 0 . | R em a rk 5.1 In C hapter 6, we will show th a t the pure nested radical A n is actually j 1 of m inim um possible depth as an exact nested radical. , i i I T he rest of sections of this chapter are devoted to proving this theorem . In sec- j tion 2, we describe Lagrange resolvent which is a basic tool for solving polynomial equations by nested radicals. In section 3, we prove two useful theorem s concern ing the pure depth of roots of unity and th e pure depth of a prim itive elem ent of a cyclic extension. The proof of Theorem 5.1 is given in section 4. Throughout this chapter, let i f be a field. For a1, ..., as € K , K { ^^ai,..., n ^/af) will denote an extension K ( ai , ..., a s) where a "' = a*, 1 < i < s. 5.2 L agrange R eso lv en t Let p > 2 be a prim e and let (p be a prim itive p-th root of unity. Then Q ( C p ) is a cyclic extension of degree p — 1 of Q. Let L / K be an extension field. Recall th a t L is cyclic over K if L is Galois over K and G (L /K ) is cyclic. The following theorem can be found in [26] p. 324. T h eo re m 5.2 Let n be a positive integer and let K be a field such that (n € K , Let L be a cyclic extension of degree n of K . Then there exists an a € K such that 39 (i) x n — a is irreducible; (ii) L is a splitting field of x n — a over K ; (iii) L = K (a ) for every n-th root a of a in L, that is, L / K is a pure root extension with generating root a . To find such elem ent a in K , Lagrange resolvent is used. Let K ( a ) / K be a cyclic extension of degree n and let cr be a generator of the Galois group G (K (a )/K ). Let ( be an n -th root of unity, then th e associated Lagrange resolvent is t(( ,a ) = a 0 + O i H h C n_1On-i where c*, = a la. Let fj(£,a) = for 0 < i < n — 1. It is easy to see th a t to((, a )n = tt ((, «)» = . . . = t n_x(C, « )n = A. Since A is invariant under <r, it m ust be an elem ent of th e field K ((). Therefore, if £ is a prim itive n -th root of unity and ( G K then K (a ) = K ( -\/A). T he elem ent a can be set to A. R e m a r k 5.2 Note th e tim e required to com pute A is polynom ial in n. W hen K = Q and a is a prim itive p-th root of unity. Then Q ( a )/Q is cyclic of degree p — 1. Let g be a prim itive p-th root m od p. Then a , a9, a92,..., a 9*-1 are all th e 1 prim itive p-th root of unity. The Galois group G (Q (a )/Q ) is generated by the elem ent a such th at a (a ) = Let (3 be a prim itive (p — l)-st root of unity and let t(/3, a ) = a + fiat? + • • • + fip~2a gP~2. We have cr(t) = Let t,- = t(fi\ a) then t\~x and tit\~x~l are invariant under a. Therefore, once t \ _1 is known , we have a = (p — l ) _1(ti -I- <2 + • ’ • + tp- 1) is a description of o; in term s of known j quantities if f3 is known. For details, please see [10]. Let K ( a ) / K be cyclic of degree n such th a t K (a ) D K ( ( n) = K . Then a ! consequence of Lagrange resolvent is th a t d^(o;) < d ^ ( ( n) + 1. In th e next section, j we will prove two results concern th e relationship between and djc(Cn)- 5.3 T w o U sefu l T h eorem s Let K (a ) be a cyclic extension of degree n over K . T he following results concern th e relationship betw een ^ ( a ) and dfciCn) which is an im portant consideration of our proof of Theorem 5.1. T h e o re m 5.3 Let K (a ) be a cyclic extension of degree pe of K , where p is a prime and e > 0. Let K ( a ) be contained in a pure root extension over K . Then (p € K . Furthermore, assuming (4 € K when p = 2 and e > 1, if K (a ) fl K ( ( pe) = K then C p° € K . T h e o re m 5.4 Let p be a prime and e > 1 . Assume £4 € K if p = 2 and e > 1 . I f K{ot) is a cyclic extension of K of degree pe and K (a ) P i K ( ( pe) = K then dK{oi) = dfdCpe) + 1. W e need some preparation before proving Theorem 5.3. Let L f K be a finite extension. L is said to have the unique subfield property if, for every d | [L : K \, there is exactly one subfield M of L of degree d over K . Clearly, every cyclic extension has the unique subfield property. T he following | result is proved in [8], L e m m a 5.1 Let M = K(/3), where j3 is a root of an irreducible polynomial x m — b with b € K . Then M J K has the unique subfield property if and only if (i) for every odd prime p, p |m ; £p $ M — K and (ii) if4 \m then (4 ^ M — K . \ I We will use the following special case where m is a prim e power. ; L e m m a 5.2 Let p be a prime. Let K be a field. Assume ( 4 € K if p = 2. Let L = K ( a ) where a is a root of an irreducible polynomial x v > i — a for some a € K . Then L / K has the unique sub field property. Proof: By Lem m a 5.1, we only have to show (p (£ L — K when p is odd. Suppose (p € L — K then K C K ( ( P) C L and 1 < [K((p) : K] < p, a contradiction, since any subfield of L over K has degree divisible by p. Therefore, (p ^ L — K . □ > 1 I 41 I L e m m a 5.3 Let K ( a ) /K be a Galois extension of degree pe where p is a prime and e > 0. I f K (a ) is contained in a pure root extension of K then K(ce) C K ( p' e y/a i, ..., p e Ifal) where ai G K . Furthermore, if p is'odd or (4 £ K when p — 2 then e > e4 - > 0 ,1 < i < s. Proof: Let L = K ( m ^/a) where a G K and (m ,n ) = 1. Then it is easy to see th a t L = K ( \/a, \/a). Therefore, we can assume w ithout loss of general ity th a t K (ct) C K { %/bi,..., V bt) where qi is a power of some prim e pi and 1 bi € K , 1 < * < t. Let 5 = { V&T,..., V h } and K { = K ( S - { V £})- We have [Ki{a) : Kf\ = pe’ for some 0 < e' < e. It is easy to see [i^(S') : K{] = pe f for some ea - > 0. Since Ki C K i(a) C K ( S ), if pi ^ p then e' = 0 and we can delete from S. Therefore, K (a ) C K { ..., Moreover, sup pose p is odd or £4 € K when p = 2. Let T = { p e {/a l, ..., p e (/a7} and let K'i = K ( T — { y^/Oi}). Then by Lemma 5.2, K ( T ) = K-( p\/bi) has the unique subfield property as an extension of K-. Hence, K ^ a ) = K '( p^/ai) for some e'- < e. Therefore, K (a ) C K ( p B y /a i,..., p e £/a^) w ith e' < e. □ Lem m a 5.2 and Lemm a 5.3 will be used to prove the first part of Theorem 5.3. The following result is from [17], see also [20] p. 321. L e m m a 5.4 Let L / K be a finite extension field such that L is a splitting field of the polynomial (arni — a\)(x n2 —af) • • • (x ns —as), where rii are positive integers and ! ai G K . Assume K contains a primitive p-th root of unity for p dividing any ni where p is a prime or p = 4. Let r be the least common multiple of the n* and be a primitive r-th root of unity. Let A G L with A”‘ = a;. Let M be a field such that K C M C L. Then M is generated over K by certain elements of the form Q A*1 A§2 • • ■ \ k ss where 0 < kj < nj, l < j < s , 0 < i < r . That is, M / K is a root extension. j L e m m a 5.5 Let € K where p is a prime and let e be a positive integer. I Let L / K be a finite extension such that L is a splitting field of the polynomial ; (xp S 1 — a i)(xp B 2 — af) • • • (xpSs — as) where 0 < e; < e ,a 8 - G K for 1 < i < s. Let \ M / K be cyclic of degree pe and K C M C L. Then M is a splitting field of the irreducible polynomial xp B — a over K . 42 Proof: Let A , • G L w ith Af * = at - for 1 < i < s. By Lem m a 5.4, M is gener ated by some elem ents /?j,l < j < n, of th e form C^A^A^2 ••• A*s. Since M / K is cyclic of prim e power degree, for all i ,j , 1 < i , j < n, either K((3i) C K(/3j) or K({3j) C K(j3i). Therefore, there is a k such th a t K{(3j) C K (0 k ) for all j , 1 < j < n. Hence, M = K(/3k). Since pk is a root of x p* — a for some a G K and M / K is cyclic of degree pe, the binom ial x p & — a is irreducible and M is its splitting field over K . □ Now we are ready to prove Theorem 5.3. P r o o f o f T h e o re m 5.3 : By Lem m a 5.3, K (a ) is contained in a pure root extension K ( p y /a7 ,. . . , p6 ^ ) , where a* € K and e > ej > 1 for 1 < * < s Let L = K ( pe ^ /a7 ,. . . , pes~^/aflf) and, w ithout loss of generality, assum e a $ L. Then L (a ) is cyclic of degree pd for some d > 0 over L and L(a) C L( p e f/dTs). By ; Lemm a 5.2, we have L (a ) = L( p f/dTs). Therefore, (pd G L. Hence (p G L. Suppose ! C p € L — K then K C K ((p) C L and 1 < [K((,p) : K] < p, a contradiction. Therefore, (p g L — K . Hence, we have (p G K . Since K (a ) is contained in th e splitting field of (x pB1 — ai) • • • (x pe* — as) and C P G K , by Lemm a 5.5, K (a ) is the splitting field of xp B — a for some a G K . Therefore, (pe G K (a ). If K ( a ) D K ( ( pe) = K then (pe G K . □ An im portant consequence Theorem 5.3 is the following C o ro lla ry 5.1 Let K (a ) be a cyclic extension of K of degree pe where p is a prime ! and e > 0. Then dK{Cp) < dxiot) < dx{Cp& ) + 1- 1 Proof: Let K = K 0 C K \ C • • • C Kd be a pure root tower of depth d such th a t j a G Kd and d = Let (3 G K (a ) such th a t K(j3) is cyclic of degree p over K . Consider th e least i such th a t G K{. Then i < d. By Theorem 5.3, £p G 1. Hence dj((£p) < dx{oL) Let L = K (£pe). Then L{o l)/L is cyclic and \L(a) : L]\pe. Hence L {a )/L is a pure root extension by Theorem 5.2. Therefore, £ dK {(pe) + 1. □ | We need two m ore preparatory lemmas to prove Theorem 5.4. ; 43 L e m m a 5.6 L etK (a ) andK (fi ) be finite abelian extensions of K such that K{a)C[ K(fi) = K . Let L / K be any finite extension field and let M = L D If M = K or M = L H K (a ) o r M = LC\ K (p ) then L(a) n L(/3) = L. Proof: Let m = [K{a) : K ],n = [K(fi) : K\. Since K (a) fl K(/3) = K , we have [I<(a,p) : K] = mn. We have [L(a,p) : L] = [L(a) : L] [L{P) : L]/[L{a) D L(p) : L]. (i) If L n K (a ,p ) = K then G (K (a ,p )/K ) “ G (L{a,p)/L). Hence, [L(a,P) : L] = mn. Therefore, [L(a) fi L(/3) : L] = 1, th a t is, L'(o') fl L(ft) — L. T he above argum ent can be illustrated as the following diagrams: L K L(a) L (ii) Suppose M = LC\K(a). Then K C M C K(a). Hence, M ( a ) = K{ot) and M (a,P ) = K (a,P). It follows th a t [M(a, fi) : M] = [K(a,fi) : M] = [K(a,fi) : K]/[M : I<] and [M{a) : M] = [K ( a ) : M\. Since K (a ) fl I<(p) = I< and K C M C K (a), we have M n K(P) = K. It follows th a t G (M (P )/M ) = G (K (p )/K ), hence [M(p) : M] = [K(P) : iT]. Now, [M(a,P) : M] = [I<(a,P) : I<]/[M : K] = [ .K ( a ) : K)[K{p) : K ] /[ M : K \ = [ # ( « ) : M][K(P) : I<] = [M (a ) : M ][M (/3) : M ], Therefore, M ( a ) D M(ft) — M . The above argum ent can be illustrated as the j following diagram: K (a ,P ) = M (a ,p ) K (a ) = M (a) M 44 | I ____l Now we can apply the argum ent in (i) w ith K = M , noting th a t L fl M(a,/3) = L n K (a , /?) = M. (iii) If M = L fl K(/3) then, by sim ilar argum ent, we have L(a) fl L({3) = L. □ L e m m a 5.7 Let M and N be Galois extensions of K with degree m and n respec tively. Assume M fl N = K . Let L / K be a finite extension field of degree I. If ( /: m ) = 1 then L fl M N = L fl N . Proof: Since M f) N = K and M / K is a Galois extension, we have [M N : iV] = m. Suppose on th e contrary th a t L n M N ^ L O N . Let a E L' = L f l M N and a £ N. T hen d = [iV(a) : N] > 1 and d\m. Let K ' = N fl K(a). Since N/I< is Galois, we have [N(a) : K ( a )] = [N : IC] and [N(a) : IC] = [N(a) : N][N : IC] = [N (a) : K(a)][K(a) : IC]. It follows th at [iL(a) : K ’] = [iV(a) : N] = d. Therefore, d\[K(a) : K], Since a £ L, we have d | I. Hence, we have (l ,m ) ^ 1, a contradiction. Therefore, a € N and we have ! V C N. M N N{a) K (a) N IC I< Hence, L fl M N = V = V fl N = (L n M N ) D N = L n N. □ j ! 1 P r o o f o f T h eo rem 5.4 : ; Since d if(a) < o?jc(Cpe) + 1 by Corollary 5.1, we only have to show th a t dx(o;) > j dK((P e) + 1- This is done by induction on d = c?ic(a). For d = 1, we have (pe G K by Theorem 5.3. Therefore, di<(Cpe) = 0- 45 Assume the theorem hold for d = 1 ,2 ,. .. , k, k > 1. Let d jf(a) = d = k 1. If p = 2 then dfc{Cpe) = 1 < d. Therefore, we m ay assum e p is odd. Since dji{oc) = d, there is a pure root tower K = K q C K\ C . . . C Kd such th a t a £ Kd — Kd- 1, K ifK {-\ is a pure root extension. Let M = K x n K (a , Cp«). (i) If [M : K ] > p then there is a cyclic subextension of K\ of degree p over K which is contained in a pure root extension of K. Therefore, (p € K by Theorem \ 5.3. Hence d^(Cpe) = 1 < d. , (ii) If [M : K] = 1 then K i(a) fl ifi(£ pe) = Ki by Lem m a 5.6 applied to K (a), K{(,pe.) and K\. Now d ^ict) < d — 1 = k. By induction hypothesis, dKiict) > d^iCpe) + 1, so g?k(Cp«) < dKxiCp6) + 1 < dKxiot) < d#(a:) — 1 = k. (iii) If 1 < [M : K] < p, apply Lemma 5.7 to K (a), K ( ( pe) and K\, we get M = K \ fl K (a, £P e) = K\ D K(£pe). Therefore , we can apply Lem m a 5.6 to K (a), K(£pe) and K \ to get K\(a) fl A T i^e) = K\. Using th e same argum ent as (ii), we can show dx(CP e) < k. O | 5.4 P r o o f o f T h eo rem 5.1 j i We give one m ore lem m a before proving Theorem 5.1. L e m m a 5.8 Letp be a prime and let g = [Q (0 ?C pe) : Q (0 )]* Then g = (p— l)p e_1 tf{U ve) = 1 and £ = pe~% if (l,pe) = p. Proof: L et’s recall the following facts about cyclotom ic extensions. We have 1 [Q(Cm) : Q ] = <f>(m), Q (C m )H Q (C n) = Q if (m ,n) = 1 and Q (C /)n Q (C P.) = Q(C„) j if (/,p e) = p . ; Since g — [Q(Cpe) : Q ]/[Q (C pe) ^ Q (0 ) : Q L results follow. □. ! Now we give a detailed proof of Theorem 5.1 P r o o f o f T h e o re m 5.1: For each n, we shall construct a pure root tower Tn, Q = Qo< n C Qi,n C • • • C Qd,n where d = d{n) such th a t £n E Qd,n and, for 1 < i < d(n), QitU has th e following two properties: ] I 46 | ____________________________________________________________________________ i PI- Qi,n/Q is th e com posite field of Q ( C ® j ) for some primes p3 and e3 > 1. P j P2. Let c ^ 1 if € Qit7 1 then E Qi— We use induction to prove the second part of Theorem 5.1 It is easy to see d (l) = d(2) = 0 and T\ = T2 = Q. We also have d(3) = 1 and ! the associated tower T3 is Q = Q0j3 C Q 1 ,3 = Qo,3(\/—3)- It is easy to see T \,T2, and T3 satisfy properties P I and P2. Assum e for n — 1,2, ... ,k, Theorem 5.1 holds and there is a Tn satisfies prop erties P I and P2. For n = k -fi 1 > 3, we consider the following cases. C a se 1: n = p > 3, a prime. Let p — 1 = Im where I is odd and m is a power of 2. We first consider the case when / = 1, i.e., p = 2b -f 1 for some 6 > 1. Since d(p — 1) = d(2b) = 1, we only have to show d(p) ^ 1. Suppose on the contrary d(p) = 1 then by Lemm a 5.3, we have (p € Q ( 2y /aL,. . . , 2 e ^/a~s) where e ,- > 0, a; € ; Q . Let M = Q(C4) and let L = M(£p). Then L j Q is abelian of degree 2h+1,L fM is cyclic of degree 2b and L is contained in the splitting field of (x2*1 —a\) • • • (x2*3 —as) over M . Hence, by Lemm a 5.4, L / M is generated by certain elem ents /? i,. .. , j3t of the form ' ^ss where e = max^=1{ej} and A| 3 = a3 0 < kj < 2ej, 1 < j < 5, 0 < i < 2e. Since L /M is cyclic of degree a power of 2, there is a /? = /?j, 1 < j < t such th a t L = M(/3). Since /?2* € Q, there is an a € Q such j th a t j3 is a root of f(x ) = x 2* — a. Therefore, all conjugates of j3 over Q have j the form (% e/3 for some k. Q(/?), being a subfield of an abelian extension L / Q , is also abelian. Let d be the largest integer such th a t (2d0 is a conjugate of /?. Then (2d € Q(/?) and [Q(/3) : Q] < 2d. Since [Q(/?) : Q] > : M] = 2b, we have d > b. However, ^2^ ^ Q(C4:Cp) f°r d > 2. Therefore, d — 2, L = M(f3) = and [L : Q] < 4 < 2b+1, a contradiction. Hence, d(p) = 2 = d(p — 1) + 1. Now we consider the case when / > 1. Since Q ( C p ) / Q is cyclic of degree p — 1, by using Lagrange resolvent, d{p) < d(p — 1) + 1. In the following, we shall prove th a t d{p) > d{p — 1) -f 1. Hence, d(p) = d(p — 1) + 1. Let L and M be subfields of Q ( C p ) such th a t \L : Q] = / and [M : Q] = m. We have M fl Q (0 ) = Q since Q(CP) H Q (6 ) = Q- By induction hypothesis, there is an odd | prim e q such th a t qe\\l for some e > 1 and d(l) =; d(qe). Let Q(o:) be the subfield j of L of degree qe over Q . Applying Theorem 5.4, dq(or) > d(qe) + 1 = d(l) + 1- ' 47 I ___________________________________________________________________________________________ I Therefore, d{p) > c?q(q;) > d(l) + 1. By induction hypothesis, d(l) = d{lm) = d(p — 1). Therefore, we have d[p) = d(p — 1) + 1. T he associated pure root tower Tp is Q = Qo,P C Q\,p C • • • C Qd(p-i),p C <3d(p),p where Qi< p = QitP_i for 0 < i < d(p - 1) and Qd(P),P = Qd(P- i),p( P“ n/A) w ith A £ Qrf(p-i),p. Since inductively Tp_i satisfies P I and P2, it is easy to see Tp satisfies properties P I and P2. C ase 2: n = pe where e > 1 and p is a prime. If p = 2 then it is easy to see d(2e) = 1 and T^e is Q = Qo,2e C < 5 i,2e = Qo,2e( 2‘ y/—1)* If P 7 ^ 2 then since £pe = pe~yfCp for e > 1, w eh av ed (p e) < d(p) + 1. j In th e following, we will use Theorem 5.3 to prove d(pe) > d(p) + 1. Therefore, d(pe) = d(p) + 1. Let Q (a ) be a cyclic subfield of Q(CP e) of degree p over Q. Then it is clear th at d(pe) > dQ(a). We argue th a t dQ(a) > d(p) + 1 and then it will follow th at d(pe) > d(p) + 1. Let d = dQ{a). Then there is a pure root tower Q = K 0 C K\ C • • • C K j-i C Kd such th a t a £ Kd — K j - i ■ Since Kd-i(a) / K j-i is a cyclic of degree p, applying Theorem 5.3 to K = K j - i , it follows th a t (p £ Kd-i- Hence d(p) < d — 1 = , dq ( « ) — 1 - i The associated tower Tp& is Q — QotP e C ^?i,pe C Z * C Qrf(pj} pe C Qd(P)~ j-i,pe where QilP e QitP for 0 ^ z ^ dip) and i,pe — Qd(P )jP si p \JCP)* H is easy to see Tpe satisfies properties P I and P2. C a se 3: n — n*=i pV where s > 1 and pi ^ pj for i ^ j. It is easy to see d(n) > m ax|=1{ d (p f)} since (n = IIL iC p6 '- In the fol lowing, we show d(n) < m ax|=1{ d (p f)}. Thus com plete the proof of d(mn) = m ax{d(m ), d(n)} for (m , n ) = 1 and th e second p art of Theorem 5.1. j We only have to show th e com posite of T «i, 1 < i < s, is a pure root tower and I i satisfies properties P I and P2. T h at is, T n can be chosen to be th e composite of T p et for 1 < i < s. Let (u,v) = 1. The com posite of T u : Q C Q i , u C • • • C Q d ( u ) ,u and T v : Q C Q i , v C ••• C Q d { v ) ,v for (u, v) = 1 is defined to be the tower T U T V : Q c Q i , u Q i , v C Q 2 , u Q 2 ,v C • • • C Q t , u Q t , v where t = m ax{d(u), d(u)}, Q k , u Q d ( u ) ,u for k d(u) and Q k , v — Q d ( v ) ,v for k div). Since (uv - (U (V 1 (uv £ Q t,uQt,v where t = m ax{d(«), <f(v)}, Qk,u = Qd(u),u for k > d(u) and Qk,v = Qd{v),v for k > d(v). Therefore, we have to show TUTV 48 is a pure root tower. T h at is, we have to show Qi> u Qi,v/ Q%-i,uQi-i,v is a pure root extension for 1 < i < max{d(t<), d(v)}. By properties P I and P2, we have Qi— i,uQi— \,v = Q (0 ) some / > 0 and Qi,uQi,v = Q(0>Cp,15 • • • » for some m > 0 where pi are distinct primes such th at pf does not divide I. By property P2, if a > 1 then pi\\l For 1 < j < m , let gj = [Q (0 ,C e i) : Q (0)I- % Lem m a 5.8, we have gj — pj3~ if ej > 1 or gj — pj — 1 if ej = 1. Since [QijU Qi,v • Qi-x,uQi-i,v\ = n £ it Q ( C ^ ) • Q(Cp^) n Q(Cr)] := Q%,uQi,v/Qi-\,uQi-i,v is a pure root extension. Again, it is easy to see Tn = Tp* 1 2 • • • Tp^ satisfies properties P I and P2 from construction. The following procedure is based on th e above and Rem ark 5.2. P r o c ed u re prim itive-root-of-unity (n) In p u t: n > 2, a positive integer. O u tp u t: A n, a pure nested radical of m inim um depth for (n. 1. If n = 2 then retu rn (-l). 2. If n = pe, where p a prim e, e > 1 then let t = prim itive-root-of-uni ty(p) and return( p£ \/t). 3. If n = pem, where p a prim e and (p, m) = 1 then let 11 = prim itive-root-of-unity(pe), let t2 = prim itive-root-of-unity (m ), and j return(<it2)- 4. If n = p an odd prim e then • let t = prim iti ve-root-of-unity (p — 1) • let g — a prim itive root mod p • for * = 1 to p — 2 let U = x -f Vx* + t2ixg2 + • • • + tb -W x 9”’2 • com pute r, = for 1 < i < p — 2 • let si = p-^/rT and let S{ = for 2 < i < p — 2 • let C P = (p - + s 2 + • • • + ( -1 ) ) • return(Cp). 5. End. 49 It is easy to see th e tim e com plexity of the procedure is polynom ial in n. This completes the proof of Theorem 5.1. □ From th e proof, we have the following C orollary 5.2 Let n be a positive integer. Then there is tower of pure extensions of depth d, Q = Qo C Qi C • • • C Qd, such that d = d(n) and Qi are cyclotomic fields for 1 < i < d. R em ark 5.3 Consider the chain of prim es p i,P 2, • • • ,Pk such th a t p* = 2p2 _i + 1 for 2 < i < k. Then d(pi) = d{pi) + (i — 1) where d(pi) = dQ((Pi). This kind of chain is called a Cunningham chain of length k. For example, pi = 2, p 2 = 5, P3 — 11, p4 — 23, ps = 47 is a Cunningham chain of length 5. Nelson has found a chain of length 9 starting at 85864769 (see [30] p: 194). If there are arbitrary long Cunningham chains then the pure depth for a prim itive n-th root of unity can be i 0 (lo g n ). ! 50 C h a p ter 6 S o lv in g P o ly n o m ia l E q u ation s 6.1 In tro d u ctio n N ested radicals represent roots of solvable polynomials. T he problem of solving I polynom ials by nested radicals has been studied for centuries. M ethods for solving linear and quadratic equations go back to th e early Egyptians. In the 16th century, Italian m athem aticians succeeded in solving cubic and quartic equations [21]. In late 18th century, Gauss found a m ethod to solve the cyclotomic polynom ial equa tions [16]. Early in the 19th century, it was shown th a t there are equations of each i degree higher th an four th a t are not solvable by radicals (Ruffini-Abel theorem ). ' In th e m id 19th century, Galois showed th a t a polynom ial equation is solvable by ; i radicals if and only if th e associated group is solvable (Theorem 2.7). i Landau and Miller [25] showed th a t there is a polynom ial tim e algorithm to determ ine th e solvability by radicals of a given monic irreducible polynom ial over the integers and to express the roots in radicals. In [38], van der W areden showed th a t an irreducible polynom ial over a field K solvable by radicals is solvable by exact nested radicals. The exact nested radical he constructed is the associated nested radical of a root of the polynom ial in a pure root tower K = K 0 C K\ C ! • • • C Kd where K{ is generated by a generating root of degree pi over Ki-\ for some prim e p*. The resulting exact nested radical is in general not of m inim um depth. W hen roots of unity are treated as symbol of no nesting depth th en we have the following result for irreducible, solvable polynomials over algebraic num ber fields. 51 T h e o re m 6.1 Let k be an algebraic number field and f be an irreducible, solvable polynomial over k. Let L be the splitting field of f over k. Let n be a natural num ber which is divisible by [L : fc ] and the discriminant of L over Q . Then the tower of pure root extensions corresponding to the derived series for G(L((n) /k ((n)) de termines a nested radical of minimum depth over k(p < * > ) which represents precisely all the roots of f over k when (n is viewed as a symbol which can assume as its value all conjugates of a primitive n-th root of unity over k. The nested radical can be computed in time polynomial in the length of f , the degree of L over k and ; n. I I However, if roots of unity are not treated symbolically then the problem is more com plicated. We will provide some partial results in this chapter. Since pure nested radicals are com putationally favorable, it is im portant to relate the m inim um depth of pure nested radicals for a given irreducible polynom ial and th e m inim um depth of nested radicals for the same polynomial. Let a be solvable over a field K. T h at is, a is a root of an irreducible, solvable polynom ial over K . T he following result shows th a t the pure depth of a is at m ost one more th an its optim al depth. T h e o re m 6.2 Let a be solvable over a field K . If there is a nested radical A of ; depth d over K for a then there is a pure nested radical A! of depth < d + 1 over j K fo r a . Let K be a field. An elem ent of K is said to be abelian (cyclic) if K ( a ) / K is abelian (cyclic). A polynom ial / over K is said to be abelian (cyclic) if its roots are abelian (cyclic) over K . T he following result shows th a t the pure depth is equal to th e optim al depth for abelian elem ents over Q. T h e o re m 6.3 Let a be abelian over Q . Then (i) -D q(a) = dQ(a); and (ii) there is a polynomial time algorithm to find a nested radical of minimum depth for a. j I T he rest of the sections are devoted to proving these three theorem s. I j I I 52 i 6.2 P r o o f o f T h eo rem 6.1 The proof of Theorem 4.1 can be refined to yield a proof for Theorem 6.1. The splitting field of the m inim al polynom ial of v(A ) is now replaced by the splitting field L of the input irreducible polynom ial / . The tower of extensions correspond ing to th e derived series for G(L(£n)/k((n)) is in fact a pure root tower and this can be easily proved using Lemm a 3.3. C om putationally we then have to com pute for each extension in the tower a set of pure generating roots. Let fl be a root of / in L((n). Let B be th e associated nested radical of fl over j k((n). In the following we show th at B represents exactly all the solutions to / I over k when (n is viewed as a symbol which can assume all conjugate values of a ! prim itive n -th root of unity over k. I Let g E &(£„)[£] t»e an irreducible factor of / over &(£n) such th a t g{fl) = 0 ' I and let G = G(k(^n)/k). Then Y[<reG9a £ k[x]. Furtherm ore, any root of / over i A ; is a root of f las.G9a € A r[a r] and vice versa. Let Ap be a pure nested radical j over k(Cn) w ith fl E value(Ap). Then value(Ap) = {A|<7(A) = 0}. Let a E G. Let Ap be th e nested radical obtained by replacing all (n in Ap by (£ ■ Then value(Ap) = {A|^< 7 (A) = 0}. Hence U< T € Gua/ue(A^) is the set of all th e root of / over k. □ 6.3 P r o o f o f T h eo rem 6.2 Let a be solvable over a field K and let A be a nested radical of depth d over K for a. We will prove Theorem 6.2 by showing th a t there is a pure root tower of depth at m ost d + 1 containing or', a conjugate of a over K. Hence the associated nested radical for a ' in this pure root tower is a pure nested radical of depth at m ost d+ 1 over K. This will be done by using the fact th a t A is exact to show th a t j there is a radical valuation v over K such th a t the corresponding root tower for A w ith respect to v is contained in a pure root tower of depth at m ost d + 1 over K . We will first show th a t if (4 E K then there is a radical valuation v over K such th a t K(v(S)) is contained in a pure root extension of K where S is a set of simple nested radicals of depth 1 over K. Then, inductively, there is a radical valuation v over K such th a t the corresponding root tower for A w ith respect to v is contained j I I 53 : in a pure root tower of depth d over K . Since A is exact, v(A) is a conjugate of a. If £4 0 K then we consider A as a nested radical over the field K ' = K (£4)- Now there is a pure root tower of depth d over K ' containing a conjugate of a over K . Since K ' is a pure root extension of K , we have a pure root tower of depth d + 1 over K containing a conjugate of a. We begin w ith some technical preparation. Let K be a field, n > 2 be an integer, p be a prim e, and a € K . The following theorem gives the conditions under which xpn — a is irreducible over K . I T h eo re m 6 .4 Let K be a field, n > 2 be an integer, p be a prime and a be an arbitrary element in K . ! « . . . . . I (i) I f p is odd then xp — a is irreducible over K if and only if a K p. j (ii) Ifp — 2 then x 2" —a is irreducible over K if and only i f a g K 2 and a ^ — 4K 4. j Proof: See, for exam ple, [20] p .224. □ ■ The following theorem gives a necessary condition under which xpn —a is irreducible over a pure root extension of K. T h eo re m 6.5 Let K be a field and let p be a prime. Assume £4 E K if p = 2. Let s € N . Let a{ € K and et - be positive integer for 1 < i < s + 1 . Let a,- be root of irreducible binomial xp B % — at - over K such that M — K(c*i, ... ,c*s) is a pure root extension of degree pe where e = ei- If xpC a+ 1 — a s+i is reducible over M then ' ■ d ^ there exists an element b E K such that a s+i = If aT where 1 < d < ea+i, j b ^ M p and Ui € Z for 1 < i < s. j i We need th e following lemma. L e m m a 6 .1 Let K be a field of characteristic 0 and let p be a prime. If a, (I are roots of irreducible polynomials xp — a, xp — b over K then [K(a,j3) : K] = p2 unless b = cpak for some k £ Z and c G K . Proof: See [20] p .240. □ 54 P r o o f o f T h e o re m 6.5: C la im 1 The assertion holds when e* = 1 for 1 < * < s + 1. T h at is, M = K ( a u . . . , a a), where o f = a4 - G K , is a pure root extension. If xp — as+i is reducible over M then there exists a 6 6 K such th a t a s+1 = IP IT|=i aT where Ui G Z. Proof of claim 1: W ithout loss of generality, assume xv — a s+i is irreducible over K . Otherwise, by Theorem 6.4, a s+1 G K v. We use induction on s to prove this claim. If s = 1 then th e claim follows Lemm a 6.1. Now consider s > 1. Let L = K ( a \ ,..., a s_ i). If xp — aa+1 is reducible over L then, by induction hypothesis, we are done. Otherwise, since xp — as+1 is reducible j over M , [L(as, Q !s+i) : L] ^ p2 where a s+i is a root of xp — a s+i. Therefore, since xp — as, xp — aa+i are irreducible over L, a s+i = cpa“a where c G L and w, G Z by Lem m a 6.1. If c G K then we are done. Otherwise, K(c) C L where c is a root of xp — [ «s+i«7“s € K[x], Hence xp — as+1a~U s is reducible over L. Therefore, by induction hypothesis, a s+1a J U s = bP ni=x aT f°r some 6 G K . Therefore, as+i = IP nf=a aT • C la im 2 T he assertion holds when es+i = 1. Proof of claim 2: W ithout loss of generality, assume xp — a s+i is irreducible over K . Otherwise, by Theorem 6.4, a s+i G K p. Again we use induction on s to prove this claim. Note th a t K { a s) has the unique subfield property by Lem m a 5.2. If s = 1 then, since xp — as+j is reducible over K ( a s), as+j = qP+1 for some a s+i G K ( a s) by Theorem 6.4. Hence K ( a s+1) C K ( a s) and [/^(as+i) : K] = p. By th e unique subfield property of K ( a „), K ( a s+i ) = where a's — (a j)1 ’6 ’-1 . T h at is, a's is a root of the irreducible binom ial xp — as over K . Since xp — as, xp — a s+i are irreducible over K and [K(a's, a s+1) : K ] ^ p2, by Lem m a 6.1, j a s+i = bP a™ * for some b G K and «S GZ. ! Now consider s > 1. Let K s = K ( a i, ..., ors_i). If xp — a s+1 is reducible j over K s then, by induction hypothesis, we are done. Otherwise, since xp — as+1 j is reducible over M , there exists a s+\ £ M — K s such th a t g^+1 = as+1. Hence ; /irs(as+ i) is a subfield of i f s(a s) of degree p. Since M is a pure root extension over j K , xpes — as is irreducible over K s. Hence by Lem m a 5.2 K s(as) has the unique subfield property, therefore, Jfs( a s+i) = K s(a's). where e/s = (o;J ,)pes_1. T h at is, a's is a root of the irreducible binom ial xp — as over K s. It is easy to see th a t K s(a's) = K ( a i, ..., a s- i , oi' s) is a pure root extension over K and xp — a s+1 is reducible over ! K a(a, s). Now we consider K s-x = K { a ^ ..., q;5_2, a ') . If o;s+1 £ K s-x then we are done by induction hypothesis. Otherwise, = /Ts _i(a;'s_1) where a's_x = —1 a s-i 5 th a t is, a'a_x is a root of xp — a s_j. Hence a 5+i € K{ax,..., a ^ , a's). Inductively, < * s+1 £ K ( a[,..., a ') where a\ is a root of xp — < % { . Therefore, by Claim 1, there exists a b £ K such th a t a s+i = IP II;=i aT where u* £ Z for 1 < i < s. Now we can com plete th e proof of Theorem 6.5. If xpSs+ 1 — as+i is reducible over M then as+1 = Xp for some A £ M by Theorem 6.4. Hence xp — as+1 is reducible over M . By Claim 2, as+1 = n i= i where 6i £ K and w4il £ Z. If bi £ M p then xp — b\ is reducible over M . Hence bi = b % n ^ i where b2 £ K and Ui,2 ^ Z. Therefore, a s+i = fein^ior * ’ 1 = $2 Y\i=iaT f°r some Wi £ Z. If b2 $ M p then we are done. Otherwise, repeat this process. Inductively, it is easy to see th a t a s+1 = bPd n i= i aT f°r some b £ K , b $ M p and Ui £ Z for 1 < i < s. □ C o ro lla ry 6 .1 Let K be a field and let p be a prime. Assume £4 £ K if p = 2. ! Let s £ N . For 1 < i < s, let < z e - £ K , e8 - be positive integers. Then there exists a I radical valuation v over K such that K(v( * > e y/ai), • • • ,v( p e ^/a7)) is contained in a pure root extension of K of degree pd for some integer d > 0. Proof: Let e = m ax;=1{ei}. If xpC ' — a t - is reducible over K then, by Theorem 6.4, ai = bp ' for some bi £ K — K p and d{ > 1. Hence we can choose among those radical valuations v such th a t v( p e f/di) is a root of xpe'~d' — bi. Therefore, w ithout ! loss of generality, we m ay assum e th a t xpe‘ — a 8 is irreducible over K for 1 < i < s. i 1 We use induction on s to show th a t there is a pure root extension L over K such th a t ai £ L where Oj = v( py/ai) for some radical valuation v over K . If s = 1 then let fix € K be such th a t fi\ * = ax- It is easy to see Lx = K(fix) is a pure root extension over K of degree pd for some integer d > 0. Let v be a radical valuation over K such th a t ax = v( p e f/dx) — fix *• Then K ( a i) C L\. I I i 56 | Assume th a t th e assertion holds for s = 1 ,- - - ,k where k > 1. T h at is, K(ati, ..., ak) is contained in a pure root extension Lk over K of degree pd for some integer d > 0 where a,- = u( p < ij/a7), 1 < * < k, for some radical valuation v over K . Now consider s = k -f 1. Let 0i, 1 < i < /, be the generating roots for Lk where I < k and 0f = bi E K for I < i < I. If xpC * — a3 is irreducible over Lk then it is ] e - e l easy to see xp — a3 is irreducible over Lk- Let 03 tE K be such th a t 0P = a3 and let a 3 = v( p e y/a^) — 0P & e*. Let L 3 = Lk(/33)• It is easy to see L s is a pure root J extension over K of degree pd for some integer d > 0, K ( a i, ..., a 3) C L s and we I are done. Otherwise, by Theorem 6.5, a3 = bpd f lL i f°r some b € K , b £ Lp k and Ui E Z. Therefore, we can set v( p e f/a^) = a s such th a t a s = {j3pd n!= i 07'Y* where 03 E K and 0P * = b. Then a 3 E Lk{03)- Let L s = Lk(03). It is easy to see L s is a pure root extension over K of degree pd for some integer d > 0 since 03 is a root of th e irreducible binom ial x pe — b over Lk- D L e m m a 6.2 Let K be a field and let s > 1 be an integer. For 1 < i < s, let Ki be pure root extensions over K of degree p f where pi are distinct primes. Then the composite field M of K \ , ..., K s is a pure root extension over K . \ t Proof: W ithout loss of generality, assume p\ > • • • > ps. Let Ki = Kicti i,..., a ; Ui) ’ • p e ‘,j ' ' ’ ’ I where otf- = bij E K. Let d{ — maXj{es j} . We use induction on .s to show th a t j [M : I<] = Y\i=iPV- Hence M = K ( a 1A, ... , a ltUl, a 2,i,... , a StU s) is a pure root extension. T he assertion clearly holds for s = 1. Suppose s = k -j- 1 where k > 1. Assume by induction th a t the composite field M of K i , , Kk is a pure root extension over I( where k > 1. Let m = [M : K], Then P i,.. - ,Pk are the only prim es divide m. Let N = LLS(( ds) and n = [N : K]. Then N is Galois over K since it is the splitting field of i c } th e polynom ial (xP s ’ — 6Sil) • • • (xP a ’ ‘ — b3iU $). It is easy to see only primes < p3 i divide n. Hence M D N = K. Therefore, [M N : K ] = mn. j Since K C M K S C M N , [M N : K) = [M N : M K S][MKS : I<\ = mn. Let i n3 = [Ks : K] then [MKS : K] < m n s. It is easy to see [M N : M I{S] = [N : K s) = n /n 3 from th e following diagram: 57 MN Hence [M KS : K] = m n a. Therefore, M K S is a pure root extension of K. □ T h e o re m 6 .6 Let K be a field containing £4. Let S be a set of simple nested radicals of depth 1 over K . Then there is a radical valuation v over K such that K (v(S)) is contained in a pure root extension of K . Proof: Let m , n , s and t be positive integers such th at (m , n) = 1 and ns — m t — 1. It is easy to see th a t value{ m tfa) = value(( \/a)s/ { an^> f°r a^ radical valuations v, there is a radical valuation v' over K such th a t K(v( m y/a)) = K(v'{ Therefore, we may assume S = { ph/di, ■ ■ ■ p\/a Z } where Pi are primes. Divide S into disjoint subsets S i , ..., Sr such th a t and are in the same subset Sk if and only if pi = pj = qk- For all radical valuation v over K , K ( v ( S )) is th e com posite field of K (v (S i )) for 1 < i < r. By Corollary 6.1, there exists a radical valuation v such th a t K(v(Si)) is contained in a pure root extension Ki of K of degree qf' for some integer d, > 0, for 1 < i < r. By Lemm a 6.2, the com posite field of Ki, 1 < i < r, is a pure root extension of K . Therefore, there is a i radical valuation v over K such th at K(v(S)) is contained in a pure root extension 1 of K . D i P r o o f o f T h eo re m 6 .2 : Let Lq = K 1 = K{Ca) and let A be a nested radical of depth d for a over K. Then A is also a nested radical of depth d over K'. Let S i = { B \B € S ( A ) , depth(B) = i } — { • . . , } where Bj, 1 < j < t, are nested radicals of depth i — 1 58 over K . It is easy to see th a t Si is also a set of simple nested radicals of depth i over K '. By Theorem 6.6, there is a radical valuation v over IC such th a t K'(v(St)) is contained in a pure root extension L\ of IC. Inductively, for i > 1, } is a set of simple nested radicals of depth 1 over L i- 1. Hence, by Theorem 6.6, there is a radical valuation v' over L i- 1 such th at £ j -_1(i/(S , ! -)) is contained in a pure root extension Li of Li- 1. Now choose the radical valuation v such th a t v{ n ^ B j ) — v'{ yjv(Bj)). Hence we have a radical valuation v over k such th a t K(v(Si)) is contained in a pure root extension Li for 1 < i < d. | Hence, th e corresponding root tower for A w ith respect to v is contained in j t a pure root tower of depth d over K '. T h at is, depth d over IC if £4 E IC, or depth d + 1 over K if £4 ^ K . Therefore, the associated nested radical of a (or a conjugate of it) in the pure root tower is a pure nested radical of depth no more th an d + 1 over K . □ T he following corollary follows easily from Theorem 6.6 by induction. C o ro lla ry 6.2 Let K be a field containing £4. Let A be a nested radical of depth ! d over IC. For radical valuation v over K , let K = Kq C C • • • C ILv d be the root tower of depth d for A with respect to v. Then there is a radical valuation u over IC and a pure root tower of depth d over K , K — Lo C L\ C • • • C Lj, such that for 0 < i < d, K f C Li. In th e rest of this section, we show it is possible th at dfc(a) = D k (c c ) T 1 where a is solvable over some field K . More precisely, we have the following I T h e o re m 6 .7 Let IC = Q (a ) where a 2 = 2 . Then £^(£257) = 3 and Dk(C257) = 2. We need th e following theorem by Schinzel from [35]. T h e o re m 6 .8 Let F be a field, let n be a positive integer not divisible by char F and let m be the number ofn-th roots of unity contained in F. Then for any given a E F, the Galois group o fx n — a is abelian if and only if am = A n for some A E F. L e m m a 6.3 Let K = Q (a ) where a 2 = 2. Then dic{(32) = 2. 59 ! Proof: It is easy to see [if (£32) : if] — 8 . The set of all roots of a minim al polynom ial of £32 over i f can be represented exactly by the pure nested radical y 1/a: + yj — 1/a of depth 2 over if . Hence we only have to show d ^(^32) 7^ 1. Assume djc(C32) — 1 then £32 is contained in a pure root extension M = i f (<*1,..., a s) of i f such th a t a 2* ' = a, € i f by Lem m a 5.3. Let K{ = K ( a x, ..., O j-i, a;+ 1,..., a s) for all 1 < i < s. j (a) If £4 Kfiafi — K{ then, by Lemm a 5.1, Kfioti) has the unique subfield J property over Ki. Since [it*(£32) : Ki] < 8, we can restrict e ,- to 3. (b) If (4 € Ki(ai) — Ki then a* = (m -fi n(4)2 for some m , n € ifj by Theorem 6.4 since a;2 6 * — a i is reducible over A,-(£4). Let /f l = K i ( ( 4 ) . Then if/ = i f ^ a 2* * -1) and [if/(£32) : if/] < 4. Let f i i 6 K i such th a t f i 2' 1 1 = m -fi n( 4 . Then K ' ( f i i ) has the unique subfield property over if/ by Lem m a 5.2 since a:2* * 1 — (m -fi n£4) is irreducible over if/ and £4 £ if/. L et’s consider the following three cases, (i) If C 32 € if/ then we can replace by £4. (ii) If [if/(£32) : if/] = 2 then, by the unique subfield property of K ' ( f i i ) over if/, if/(£ 32) = K 'ffi2*' 2) = Ki{ctf' 2). Hence we can restrict e; to 2. (iii) If [if/(£32) : if;] = 4 then, by the same argum ent as (ii), 1 we can restrict e4 - to 3. ! Therefore, (32 is contained in a pure root extension N = K(fix, ..., f3t) such I th a t Pf*1 = bi € i f , d; < 3 for all 1 < i < t. Now, let F = i f (£4). F(C32) is a cyclic extension of degree 4 of F and F ((32) is a subfield of th e splitting field of the polynom ial (x2*1 — bx) ■ • • (x2dt — bt). By Lem m a 5.4, F ((32) /F is generated by elem ents of the form C 2dfii1 "'fit* where d = max{d;} < 3 and 0 < kj < 2di for 1 < j < t. Since F (£ 32) is cyclic of degree 4 over F, there is a fi = ^ f i x 1 * • • fit* such th a t F (( 32) = F(fi). Since F (£32) = if(Cs2) and F (£32) = F(fi), F(fi) = i f (£4,/?) = K(C32). Hence i f (£4,/?) is abelian over K . since [K(fi) : if] > 4, fi has at least 4 conjugates over if . Hence ! £4 € if(/5) since fi is a root of the binom ial x 2d 6 where b = fi2 < 1 6 K and d < 3. ! Therefore, i ?(/ 5) = i f (£4,/?) = K(fi). Since [if(£32) : K] = 8 , we have d = 3 and ar8 — 6 is irreducible over if . By Theorem 6.8 , 62 = A 8 for some A € i f since there are only 2 8-th root of unity in i f . Hence b = ±A4. If 6 = A4 then x 8 — b is reducible over if . If b = — A 4 then 6 = — 4(A /a)4 and x 8 — b is reducible over i f by Theorem 6.4 since A/ct 6 if . Hence, we have a contradiction in either case. Therefore, dj<:(£32) ^ 1. □ 60 P r o o f o f T h e o re m 6.7: It is easy to see [if(£ 257) : K] = 256 since Q(C257) n K = Q arid [Q(C257) : Q] = 256. Assume d^(C257) = 2. Let A be a pure nested radical of depth 2 over K for £257- Let 5(A ) = Si U S 2 where Si = {B \B € S(A),depth(B) = *}. Consider a radical valuation v over K . If (4 K (v(Si)) let M i = i f (£4, u (5 i)). It is easy to see th a t M i is a pure root extension of K . Now 5 2 can be considered as a set of simple nested radicals of depth 1 over M\. Hence, by Corollary 6.1, there is a radical valuation v' over M\ (or over K if VJ5 € S\, v \ B ) = v(B)) such th a t | M i(v'(S 2)) is contained in a pure root extension M 2 of M\. j Therefore, there is a pure root tower K = M q C M i C M 2 such th a t ( 4 € M i ! and £257 € M 2 — M\. From th e following diagram , it is easy to see th a t either [Afi(C257) : M x] > 128 or [if(C257) C M a : K\ > 128 since [if (C257) : if] = 256. A fi (C257) Afi (a) If [iLTj(^257) : M i] > 128 then C 128 G M \ by Theorem 5.3. (b) If [if (£257) H M i : i f ] > 128 then 30 such th a t K (0 ) is a cyclic of degree > 128 over i f where K{0) = K ( ( 257) n M i- Since 0 € M i, dK{0) = 1. By the same argum ent used in the proof of Lemma 6.3, K ( ( 4,0) = if(C 4,A) where A2e = c € if . Since [K((4,0) : if] > 128, e > 6. Hence C e4 € if(A ). T hat is, K(C4,0) = K ((4, \ ) = I<{A). Therefore, C 128 € if(A ) C M i. In either case, d ^ (^ i2s) — 1, a contradiction since d^(C32) — 2 by Lemm a 6.3. Therefore, dj< {(257) = 3. By Theorem 5.1, there is a pure nested radical A257 of depth 2 for £257 over Q. It is easy to see th a t A257 is a nested radical (not a pure nested radical) of depth 2 for <£.57 over if . □ 6 i ; 6.4 P r o o f o f T h eo rem 6.3 To prove Theorem 6.3 (i ), we will use th e facts th a t the ground field is Q , a is abelian and we have the freedom to consider any radical valuation over Q. For a set S of sim ple nested radicals of depth 1 over Q , we will show th a t there is a radical valuation v over Q such th a t Q (v (5 )) has some useful properties as sum m arized in th e following theorem . T h e o re m 6 .9 Let Q (a) be an abelian extension of degree d over Q and let e > 0 be an integer such that 2e|d and 2e+1 fd. Let S be a set of simple nested radicals of depth 1 over Q . Then there is a radical valuation v over Q such that Q (u(S')) C Q (oo, «i> • ■ •, otn) where a ^ 0 = — 1 = a0, e0 > max{2, e}, af*1 = 2 = ai, ex > 2, j ei | and for 2 < i < n, a P i = ai, e ,- > 1, pi, a,- are positive primes, and for 0 < i,j < n \ if i 7^ j then either pi ^ pj or ai ^ aj. Furthermore, let L = Q(a:o, an,.. -, cen), M = Q ( a 0, « 2, • • •, otn), and N = Q (a i,... , a n). Then (i) M and N are pure root extensions of Q ; (ii) N q is the composite field of all quadratic subextensions of N over Q ; (Hi) L% = M q = N q (c xq); (iv) [L(a) : L] = [M ( a ) : M], moreover, if Q (a:)/Q is cyclic of odd prime power degree then [L(a) : L] = [M ( a ) : M] = [Q (a) : Q ]; I (v) if Q (a ) is cyclic of degree r s and Q (« ) fi Q(Crs) = Q where r is a prime, j s > 1 then L{a) fl L(£r*) — L and M{a) fl = M ; < (vi) = dM((u) for all positive integers u. We need some preparation before proving this theorem . Let L , K be fields such th a t K C L. We use L ^ to denote th e m axim um abelian subextension of L over K . L e m m a 6.4 Let L , K be fields such that K C L. Let M be abelian over K . Then {L M )$ = (L $ )M . Proof: Let = M L ff. Then it is easy to see th a t M\ Pi L = L a f) and (L M )^ = (L M i)^. Hence we only need to show th a t (L M = M j. Since L H M i = L $ , [LM1 : L<$\ = [L : L^][MX : L a £\. Since ( L M J g n L = L$, [ .L { L M ^ : L f] = [L : £#][(ZAfi)S£ : L f]. It is easy to see th a t L M 1 = LiLMxfjf. 62 ! I Hence [L : L$][MX : L $ ) = [L : L$][(LM1)$ : L$\. Hence [Mx : L$\ = [(L M i)^ : Z $]. Therefore, (L M )x = M\ since M \ C (L M x) j , □ L e m m a 6.5 Let n > 1 be an integer, px,p2, • • • ,Pk distinct positive primes and v a radical valuation over Q such that v(y/pi) = ai > 0 for 1 < i < k. Then [Q (a i,...,a ;jb ) : Q] = nk. Proof: See [33]. □ L e m m a 6 .6 Let K be a field such that £4 E K . Let a be abelian over K . Then D k (o c ) = dfc(a) = dx{0) where K (0) C K (a ) and 1^(0) is cyclic of prime power ^ degree over K . j Furthermore, let K (a) = K ( a x, a 2) for some 0 1 , 0:2 G K{a). Then c?ic(o:) = max{dK(a1), dK(a2)}. Proof: Let D k (o c ) — d and let A be a nested radical of depth d over K for a. Then, by Corollary 6.2, there is a radical valuation v over K such th a t the root tower for A w ith respect to v is contained in a pure root tower of depth d over K . Therefore, a is contained in a pure root tower of depth d over K . Hence djc(a) < d. Since j D k (< x) < dK(a), DK {oi) = dx{o). j Since 0:1 and 02 are abelian over K , K (« i) can be decomposed into cyclic 1 extensions K (0 X), K {02),..., K (0 S) of prim e power degrees over K and K { a 2) can be decomposed into cyclic extensions K (0 s+i), K (0s+2),..., K (0 n) of prim e power degrees over K. Hence K{a) = K (0 X,02, ..., 0n). Let B t be nested radicals of m inim um depth for 0i over K and let D = max"=1{ % ( ^ ) } . Let A be a nested radical over K such th a t S'(A) = U” _15'(H,). By Corollary 6.2, there is a radical valuation v over K such th a t the root tower of depth D for A w ith I respect to v is contained in a pure root tower of depth D. It is easy to see ! th a t this pure root tower contains 0x, . . . , 0 n and, hence, a. Therefore, djc(a) < j max"=1 {d^(0i)}. It is easy to see dK{0f) < dx(a) for all i, 1 < i < n. Hence j dK {a) = m ax” =1{ d ^ (^ )} . Therefore, there is a j, 1 < j < n , such th a t dx(ce) = j dicifij)- Since dK{0f) < m a x fd ^ C ^ i)? ^ (a ^ )} and dxfak) < ^ ( a ) for k = 1,2, ] we have cfo(a:) = max{d/c(o:i), efofa^)}. n 63 P r o o f o f T h e o re m 6.9: Let to, n, a, b be positive integer such th a t to > 1, n > 1, (m, n) = 1 and (a, 6) = 1. Then th e following relations are easy to verify: value{y/—a) = value( \ / — 1 \/a); value{yfab) = value{tfa\/b)\ value(^jafb) = value(y/a/\/b); value( m y/a) = value((^/a)s/( y/a f) where sm — tn = 1; value( p s/aqT) = value(( p ^/a)qt) where p,q are prim es, p ^ q, and s,t > 1; value{ Vo**7) 5 value( pS~\/a) where p is a prim e and s > t > 1. Therefore, w ithout loss of generality, we m ay assum e th a t S = { 2\ / — 1, ( ? 1 \ / —13 • • • p^ /a i, ..., p^ /a ^ } where e0 > m ax{2,e}, pi, a,- are positive primes for 1 < i < n and qj are odd prim es for 1 < j < m. Since p ^ Q 9 for all prim es p, q, ei , t xp> — a{, 1 < i < n, are irreducible binomials. Let v be th e radical valuation over dj Q such th a t ai = u( p^/aJ) > 0 for 1 < i < n and —1 = u( % /— 1) for 1 < j < to. Then Q (n (5 )) C L = Q («o, • • •, atn) where ct£° = —1 = a0, = 2 = ax, ei and for 2 < i < n, = a 4 such th a t e4 - > 1, p j,a 4 - are positive prim es, and for 0 < i ,j < n if i 7^ j then either pi ^ pj or a, 7^ aj. (i) The fact th a t M, N are pure root extensions of Q is a consequence of Lemma 6.5 as we argue below. Let 1 < ix, * 2,..., iu < n such th a t «4 l = a l2 = • • • = a!u = b. e* l e* u Then it is easy to see th a t Q(o4l,..., o !u) = Q(/?) where f3 > 0 and "~ Pi* = b. Hence, N = Q(/?i,... ,0k) where k > 1, / ? , • > 0, / ? , " * ’ = bi, rii > 1 and bi are j distinct prim es for 1 < i < k. Let I = Jl£=i n i and m i = l / ni f°r I < i < k. Let ! N ' = Q(/S',..., (3 'k) where > 0 and (f3')‘ = Then [N1 : Q] = lk by Lemma 6.5. Since Q C N C N', [N' : Q] = [T V ' : N][N : Q]. Since [N : Q] < n L i a»d [iV 7 : iV] < n L i m ii we ^ ave [-W : Q] = n L i ni- T hat is, [N : Q] = n "= i pV where i Pi = 2. Therefore, N is a pure root extension of Q. Let M ' = Q (a 2, ..., a n). J Then N = M'{ai), [M' : Q] = YVl^vV and [Q («i) : Q] = 2ei. Hence y/2 g M ' I since y/2 € Q (ax). Since \/2 is contained in every nontrivial real subextension of ; Q (a 0)/Q and M ' is real, Q(a;o) H M ' = Q . Since Q (a 0) is abelian over Q and j Q («o) n M ' = Q , [M : Q] = 2e° n r =2 Pi' • Therefore, M is a pure root extension of ; Q- I 64 (ii) Let Qi = Q(o:i,..., « i- i, a 4 +1, ..., a n). Assume 3fi such th a t Q(/3) is cyclic subextension of N of degree q over Q where q is an odd prim e or q = 4. Since Qi n Q2 n • • • n Qn = Q , there is an i, 1 < i < n, such" th a t fi 0 Qi. Hence | fi E N — Qi. Since N is a real extension, (g N. Hence (g ^ Qi. By Lemma 5.1, Qi(aii) has the unique subfield property of over Qi. Hence Qi(fi) = Q i(fii) where fii E N and fif = a 4 . Since Qi(fi) is abelian over Qi, (g E Qi a contradiction. Therefore, N q is the com posite field of all quadratic subextensions of N over Q. (in) By Lemm a 6.4 and the fact th a t -v/2 € Q(ao)> it is easy to see th at L g = M q = N $ ( a 0). (iv ) By Theorem 2.5, we have [L (a ) : L] = [Q (a) : Q ]/[L fl Q(o;) : Q] and [M(a) : M] = [Q (a) : Q ]/[M fl Q (a ) : Q]. It is easy to see th a t L fl Q (a) = L q fl Q(c*) and M fl Q (a ) = M q H Q (a ). Since L q = M q , we have [L (a ) : L] = [M(a) : Afj. Moreover, if Q ( a )/Q is cyclic of odd prim e power degree then I f l Q ( a ) = Q and MC\Q(a) = Q since L q = M q is of degree a power of 2 over Q and Q (o;)/Q is abelian. Therefore, we have [L(a) : L] = [M ( a ) : M] = [Q (a) : Q]. (u) If r = 2 then £2s E L and C 23 € M by the construction of L and M . Hence L(a) n L(C2.) = L(a) n L = L and M (a) D M(£2s) = M (a) n M - M . If r is odd then L q fl Q K < -r.) = £ ^ n q ( c . . ) by Lemma 5.7 since Q ( a ) n Q ( ( r.) = I Q. Hence LOQ(a, £ra) = L q fl Q (a , (V 3) = LQflQ(^r») = LnQ (£r»). By Lemma 5.6, L (a ) fl L((r*) — L. Similarly, M (a ) D M ((rs) = M. (vi) We use induction on u to show th a t d^^u) = d]tf((u). Let (m,n) = 1. Then L(Cmn) = L(Cm,Cn) and M ((mn) — M(C,m,Cn)- Hence di(Cmn) = max{c?L(Cm),di,(Cn)} and dM (Cmn) = max{dM(Cm),dM(Cn)} by Lemma j 6.6 since £4 E L and £4 E M . Hence we only need to show th a t d^Cpf) — djwiCpf) where p is a prim e and / > 0 . lip = 2 and / < e0 then di,((2f) = ^m(C2-0 = 0 since (2 f E L and (2 f E M . lfp = 2 and / > e0 then [L((2j) : L] = [M((2f) : M] = 2/_e°. Since (£f S ° = a o, C 21 is a pure generating root for L((2/) and M(Ct2f). Hence ^i(C 2/) = dftf(^2f ) = 1 since L((2f)/L and M(C,2j ) / M are pure root extensions. Therefore, we m ay assume p is odd. Assum e / > 1. Since [Q(CP/ ) : Q] = (p — 1 )pf ~1, Q(CPf) = Q(Cp>fi) where P € Q (C p/) and [Q (0) : Q] = Since [ L q : Q] is a power of 2 by (in), | Q(fi) n L = Q (fi) fl L q = Q. Since Q(CP/ ) H L = Q (fi, (p) fl L q , applying Lemm a 5.7, we have Q(CP>/?) H L = Q(CP) H L. By Lemm a 5.6, L(fi) fl L((p) = L. Hence 65 [L((pf) '• -£(Cp)] = [L(fi) '• L] = [Q (^ ) : Q] as illustrated in the following diagram: Therefore, there is a 8 € L(CPf ) such th a t L(S) is cyclic of degree p over L. Since L((pf) is a pure root extension of L((p), di,(^pf) < + 1- It is easy to see th a t di,(6) < dL((pf). Since L(S) fl L((p) = L , d ^ S ) = dL(Cp) + 1 by Theorem 5.4. T h at is, dL((pf) > dL((p) + 1. Hence dL((pf) = dL((p) + 1. Similarly, dM(CP f) = ^m(Cp) + 1- Therefore, we only need to show th a t dj^iCp) = j By Lem m a 6.6 , there is a 7 ' € M ((p) such th a t M(-y')/M is cyclic of prim e j I power degree and dM((p) = Y)- D ecom pose M ( 7 ,)nQ(Cp) into cyclic exten- ! sions Q (7 i), ..., Q(7<) of prim e power degree over Q . T hen M ( 7 ') = M (71,..., 7*). j Since M ( ^ ’) f M is cyclic of prim e power degree, there ex ists a j such that for all j i, 1 < i < i ikf(7i) C M(7_? -). T hat is, M ( j j ) = M(-yi, ... ,7<) = M ( 7 '). Let ! 7 = 7j and let [ ^ ( 7 ) : M ] = ra w here r is a prim e, s is an integer. T hen L{pf) is also cyclic of degree ra over L by (iv). Since Q(Cp) H Q(Cr*) = Q and 7 G Q(Cp)> Q ( t ) n Q(Cr.) = Q . B y (v ), L (7 ) n L ((r.) = L and M (7 ) H M(Qrs) = M . B y T he orem 5 .4 , d^i 7 ) = dx,(Cr») + 1 and 7 ) = + 1- B y in d u ction hypothesis, <*!.(CrO = dM ((r•)- H ence <^(7) = dM (l)- Therefore, dM ((P) = dM(7 ) = dL(7 ) < dL(Cp)- Sim ilarly, there is a 6 € Q(Cp) such th at dz,(£p) = d ^ S ) = c?m(^) < ^m(Cp)- Therefore, c?m(Cp) = ^l(Cp)- n Theorem 6.9 is very useful. For example, in C hapter 5, we show how to find a pure nested radical A n of m inim um possible depth for a prim itive ro-th root of unity (n over Q . By using Theorem 6.9, we can show th a t in fact depth,Q(An) is th e optim al depth of (n over Q. T h at is, we have the following 6 6 C o ro lla ry 6.3 Dq(Cn) = dq(Cn) = d(n). Proof: Let -Dq(Cn) = D and let A be a nested radical of m inim um depth D for £n over Q . Let S j = {B € S(A)\depth(B ) = j }. A pply Theorem 6.9 to Si and Q and let radical valuation v and fields L, M be as defined in th e theorem . Then Q (u(S'i)) C L and £4 6 L. Let A ' be th e nested radical obtained by replacing all B € Si of A w ith v(B). Then A ' is a nested radical of depth D — 1 over L. By Corollary 6.2, there is a radical valuation vr over L such th a t the root tower for A' w ith respect to v' is contained in a pure root tower of depth D — 1 over L. Therefore, there is a pure root tower of depth D — 1 over L th a t contains £n If D = 1 then by Theorem 6.9 (Hi) £n € M . Since M is a pure root extension of Q , dq(Cn) = 1. If D > 1 then o?i,(Cn) < D — 1. By Theorem 6.9 (ui), c?m(C0 = dz,(£n) < D — 1. Since M is a pure root extension of Q , dq(C„) < 1 + (D — 1) = D. Hence dg(£n) = D. □ C o ro lla ry 6.4 Let Q (o ) be cyclic of degree p over Q where p is an odd prime. Then D g(a) = d q (o ) = d(p) + 1. Proof: Let Z>q(o) — D. Let A be a nested radical of m inim um depth for a over Q and let Sj = {B 6 S(A)\depth(B) = j} . As in th e proof of Corollary 6.3, apply Theorem 6.9 to Si and Q (q ) and then apply Corollary 6.2. We see th a t a is contained in a pure root tower of depth D — 1 over L where L is as defined in the Theorem 6.9. Therefore, D > d ^(a) + 1. We m ay assume th a t a f: L d- i- Sincep is a prim e, it is easy to see th a t L d- i (o c ) is cyclic degree p over Ld_i. Therefore, by Theorem 5.3, £p € L b - i - By Theorem 6.9 (vi), dM(£P) = dL(£p) < D - 2. T h at is, dg(£p) < dM(£p) + 1 < D - 1. By Theorem 6.9 (u ), Q (a ) C\L = Q (« ) C\M = Q . Hence L(a) is cyclic of degree p over L and M ( a ) is cyclic of degree p over M . By Theorem 5.4, d q (a ) = dq(Cp) + 1 < D — 1 + 1 = D. Since D g(a) < dg(a), D q(o;) = dq(o:) = dq(CP) + 1 = d(p) + 1. □ C o ro lla ry 6.5 LetQ (a) be cyclic of degree pe overQ such that Q («)nQ (£pe) = Q where p is an odd prime and e > 1. Then /} q (a ) = d q (a ) = d(pe) + 1. Proof: Let D q (a ) = D and d q (a ) = d. It is easy to see th a t D < d. In the following, we show th a t D > d. \ i 67 | Let A be a nested radical of m inim um depth for a over Q and let Sj = {B € S(A)\depth(B ) = j). As in the proof of Corollary 6.3, apply Theorem 6.9 to Sx and Q (a ) and then apply Corollary 6.2. We see th a t a is contained in a pure root tower of depth D — 1 over L where L is as defined in th e Theorem 6.9. Therefore, D > dL(a ) + 1. By Theorem 6.9 (iv) and (u), we have [£(<*) : L\ = [M ( a ) : M] = pe, L (a ) fl T(Cp«) = L and M (a) fl M ((pe) = M. Therefore, d ^ a ) = d ^ p e ) + 1 and dM{oi) = dM((pe) + 1 by Theorem 5.4. By Theorem 6.9 (m), d^fip*) = dM(Cpe)- T h at is, d ^ a ) = djif(a). Since M is a pure root extension of Q , dM(&) + 1 > d. Therefore, D > di,(oc) + 1 = dM(ot) + 1 > d. □ To prove Theorem 6.3, the first step is to extend Corollary 6.5 to the following T h e o re m 6.1 0 Let Q(/3) be cyclic of degree pe over Q where p is a prime and e > 0. Then D q (/3) = dQ(/?) □ We need two preparatory lemmas to prove this theorem . L e m m a 6.7 Let K be a field and let K i , K 2 be abelian over K such that K\ fl K 2 = K . Let : K] = pe where p is prime, e > 0 and ft € K \ K 2. Let M = C K\ and N = M(fi) 0 K 2- Then [M : K] = pa, [N : K] = ph for some a ,b > 0 and M N = M (fi). • 1 Proof: Since K\ D K 2 = K and 6 K \ K 2, by Theorem 2.4, we have [K\K2 : K] = ! [Kx : K)[K2 : K) = [K2(P) : K][KX : K}/[K2{fi) D I<x : K] = [Kfifi) : K 2][K2 : | K][Ki : K ]/[M : K). Therefore, [M : K] = [K2(P) : K 2) = pa for some a > 0. Since K\ fl K 2 = K and M C K x, we have M fl K 2 = K . Hence [M K2 : K] = [M : K][K2 : K] = pa[K2 : K]. Let [M fl K(P) : K] = pd for some d > 0. Then [M{P) : K] = [M : K][K{P) : K]/[M D K(P) : K] = pa+e~d. Since M C K 2(p), we have M(f3)I<2 = I<2(p). Therefore, [K2(P) : K] = [M (p)K2 : K] = [M{P) : K][K2 : K]/[M(P) D K 2 : I<] = pa+e~d[K2 : K ]/[N : K] since M(P) n I<2 = N. Since [I<2(P) : K] = pa[I<2 : K\, [A T : K] = pe~d. Since K x n K 2 = K and M C K X, N C I<2, we have M D N = K . Hence [M N : K] = pa+e~d. Since ! N C M(P) and [M N : K] = [M(fi) : K], we have M N = M(ft). The diagram illustrating our properties is as follows: 6 8 KiKi K □ L e m m a 6.8 Let K be a field and let (2p €: K where p is a prime. Let M and N be pure root extensions of K of degree pa and ph respectively. Then there is a pure root extension L of K such that M N C L and L / K is Galois of degree pe for some integer e > 0. i Proof: Let M = K{ot\ , ..., a s), N = K ( a s+i, ..., a t) where a f ‘ = ae - G K and x v*' — a, is irreducible over K for 1 < i < t. Let L 0 = K ( a 0), where a 0 = C pe and e = max^=1{ej}. Since (p € K , L0 is a pure root extension of degree p° of K for some c > 0. In the following, we show th a t Li = L,_i (a,) is a pure root extension I of K of degree some power of p. Therefore, inductively, L = L t is a pure root | extension of K , L / K is Galois of degree pe for some e > 0 and M N C L. J c e ! Let Li-i = K(ao, fix,. . . , fii-i) where Oq = b0 £ K and /5J = 6 * G K for 1 < i < i — 1 be a pure root extension of K such th a t IL(ao, ..., a{-i) C L{-\. If Li-1 fl K{ai) = K then Li — Li-i(fii) is a pure root extension of K where fii = bi = ai and a* € Li. Otherwise, by Theorem 6.5, there is a bi € K such th at bi £ LV i_x and a,- = b\ n}=o where 1 < d < e,- and Ui € Z for 1 < j < i — 1. Let Li — Li-\(fii) where j3f — bi. It is easy to see a; G Li, Li is a pure root extension of K since x p & — bi is irreducible over L,-_x, and L i/K is Galois of degree some power of p. □ 69 j I ____t P r o o f o f T h eo rem 6 .1 0 : I Since Q {0)/Q is cyclic of degree pe where p is prim e a n d 'e > 0, there exists a 7 € Q(/3) such th a t Q (7 ) is cyclic of degree p over Q . By Corollary 6.4, £>(3 (7 ) = d(p) + 1. Since D q (7 ) < D q (0), d(p) + 1 < D q (0). It is trivial th a t D q (0 ) < <7q(/?). By Corollary 5.1 and Theorem 5.1, dQ(0) < d(p) + 2. Hence d(p) + 1 < D q (0) < d{p) + 2. If D q (/3) = d(p) + 2 then it is easy to see D q (0) = c?q(/?). If D q ({3) = d(p) + 1 then it is sufficient to show < ^Q (cp)(/3) < 1 since dQ^p)(0) < 1 implies d§{0) < d(p) + 1 = D q (0). Hence D q (0) = dq(0). If p — 2 then 1 < D q (0) < 2. Assume D q (0) = 1 and let A be a nested radical of depth 1 for 0 over Q. Let S = •S'(A). T hen, since A is exact and 0 is cyclic over Q , 0 E Q (v(S)) for all radical valuation v over Q. Let L and M be as in Theorem 6.9. Then Q(/?) C L q — M q C M . Since 0 € M and M is a pure root extension of Q , dQ(0) = 1. T h at is, dQ(£a)(/?) = 1 since Q (^2) = Q- If e = 1 then the assertion follows from Corollary 6.4. In th e following, we assum e th a t p is odd and e > 1. By Kronecker-W eber Theorem , 0 € Q(Cm) for some positive integer m. Let m = pdn where (p ,n ) = 1. (i) If n = 1 then 0 E Q(£pd). It is easy to see d > 1. Let 0' E Q (0) such th a t Q(0') is cyclic of degree p over Q . It is easy to see D q (0) > D q (0'). By Corollary 6.5, D q {0') = d(p) + 1. By Theorem 5.1 and Corollary 6.3, d(p) + 1 = dQ(CP“) = Dq(Cpd). Hence D q (0) > D q (0') = D Q((pd). Since 0 € Q(CP<0, we have D q (0) < Z>Q((pd). Therefore, D q (0 ) = -Dq(Cpd). Since d > 1 and 0 € Q(Cp« 0 - Q(Cp)» ^Q(Cr)(P) = dQ«p)((pd) = L (ii) If d = 0 then Q (0) H Q(CP e) = Q- Hence, by Corollary 6.5, D q (0) = d(pe) + 1 = d(p) + 2 since e > 1, and as observed before, D q (0) = c?q(/?). (iii) Assume p is odd, e > 1, d ^ 0 and n / 1. Let M ' = Q ((n, 0) n Q(Cp< i) and N = M '(0) fl Q(C»i)- Then, by Lemm a 6.7, M 'N = M '(0 ) and [M ' : Q] = pa ,[AT: Q] = pb for some a, b > 0. If a = 0 or b = 0 then we have th e same case as (ii) or (i). Therefore, we may assum e 1 < a, b < e. Let M 1 — Q ( x) and N = Q (y). T hen dQ^p^(x) = 1 and | N fl Q(CP0 = Q- Hence by Corollary 6.5, JDq(t/) = dq(j/) — d(pb) + 1. Since ' M ' n N = Q and M 'N = M '{0), [M 'N : M'] = [M'{0) : M '\ = [N : Q]. Hence j 70 [Q(/J) n W : Q] = [Q(/J) : Q]/[Q(/J) : Q(/J) n M'] = p7 [M'(/J) : M'} = y"/[JV : Q] = p - ‘. If 6 = e then [Q (/3)nM ' : Q] = 1. Since Q(£pe) has the unique subfield property, M ' is a cyclic subextension of degree pa (a > 0) of Q(Cp«) over Q , and Q(/?) is cyclic of degree pe, over Q , Q(/?) and Q (Cpe) have nontrivial intersection over Q if and j only if Q(/3) and M' have nontrivial intersection over Q. Hence Q (/?)n Q (£ Pe) = Q. ) By Corollary 6.5, we have Dq(@) = d(pe) + 1 = d(p) + 2 = dQ(/3). ! If b — 1 then d Q ^ ( y ) — 1 by Theorem 5.4. By Lemm a 6.8, there is a pure root J extension L / Q(Cp) such th a t x ,y 6 L. Since /? € Q (x,y), 0 £ L. Therefore, we J have c/q(c p)(0) < I- If 1 < b < e then we have D q (? /) = dQ(y) = d(ph) + 1 = d(p) + 2 by Corollary 6.5. In the following we prove by contradiction th a t D q (/3) ^ d(p) + 1. Therefore, D a m = d(p) + 2, and as observed before, D q (/3) = dQ(0). Assume Dq(/3) = d(p) + 1. Let B be a nested radical of depth Dq(/3) = d(p) + l for j3, and X be a nested radical of depth D q (x ) = d(p) + l for x. For nested radical A, we shall use Si (A) to denote the set of all simple nested radicals of depth 1 of A. Let Z be a nested radical such th a t S (Z ) = S ( B ) U S (X ). Now apply Theorem 6.9 to y and S \(Z ), and let v, L, M be as in th e theorem . Then Q (u (S i(Z ))) C L, (4 G L, [L(y) : L] = [M(y) : M ] = [Q („ ) : Q J, [ ! ( / J) : L] = [M (0 ) : M] = [Q (/3) : Q ], and [L(x) : L] = [M{x) : M ] = [Q(a:) : Q]. Furtherm ore, L(y) fl = L and M (y) fl M(Cpfc) = M by noting th a t Q (y) fl Q(CP ») = Q. Let Z ' be the nested radical obtained by replacing all C G S \(Z ) of Z w ith v(C ) and let B ' and X ' be defined similarly. Then B ' is a nested radical of depth d(p) ! for /3 over L and X ' is a nested radical of depth d(p) for x over L. By Corollary 6.2, j I there is a radical valuation v' over L such th at the root tower for Z ' is contained j in a pure root tower T of depth d{p) over L. Therefore, v'(B') and v'(X ') are j contained in T. Since f3 and x are abelian over L, (3 and x are contained in T. Since y £ Q (/3,x), y is contained in T. T h at is, d ^ y ) < d(p). Since L(y) fl L ((pb) = L and M {y) fl M ((pb) = M , we have d ^ y ) — dL((pb) + 1 and djw(y) = dM{Cpb) + 1 by Theorem 5.4. By Theorem 6.9 (vz), di,(£pb) = dM(Cpb)- Hence djvf(y) = d ^ y ) < d(p). Since M is a pure root extension over 71 | Q> dQ(y) < i + dM (y) < d(p) -f 1, a contradiction since dq(y) = d(p) -f- 2. There- j fore, D q (0 )=£ d(p) + 1. □ T he following corollary follows directly from th e above proof. C o ro lla ry 6 .6 Let Q (/3)/Q be cyclic of degree pe where p is a prime and e > 1. Then D Q(p) = dQ(fi) = d(p) + dQ{Cp)(/3). Let Q ( <*)/Q be abelian. Then Q(o:) can be decomposed into cyclic extensions j Q (ft) i < i < s, of prim e power degree over Q such th a t Q (a ) = Q (/?i,. . . , /3S) j and Q ( # ) fl Q (/?i,. . . , /?;-i, fii+i, ■ ■ ■ , /?*) = Q for 1 < i < s. T he second step of | proving Theorem 6.3 is to show the following j T h e o re m 6.11 dq (o ) = maxf_1{dQ(/ dl -)}. T he following two lemmas are used to prove this theorem . L e m m a 6.9 Let Q (/?)/Q be cyclic of degree pe where p is a prime and e > 1. Let n be a positive integer such that p\n and d(n) = d(p). I f dQ^p)(/3) = 1 then Q(Cn) H Q(/?) = Q and dQ(<n)(/3) = 1. Proof: If p = 2 then d{2) = 0. Hence n = 2, otherwise d(n) > 0 by Theorem I 5.1, a contradiction. Hence Q(Cn) = Q- Therefore, th e assertion is trivial. In the , following we assume p is odd. ! Let d(p) — d. By Corollary 5.2, there is a tower of cyclotomic extensions ! Q = Qo c Qi C • • • C Qd such th a t Qd D Q(£„). Since d(p) = d, (p Qd-x■ If ; Qd H Q(/5) 7^ Q then Qd has a cyclic subextension of degree p over Q. Consider j th e least i such th a t Qi Pi Q(/?) = Q . By Theorem 5.4, € E Qi-t- T hat is, j (p € Qd- i, a contradiction. Hence Q d fl Q(/3) = Q. Therefore, Q(Cn) D Q(/3) = Q- ' Let K = Q (£*> )• Since = 1, by Lemma 5.3, /? is contained in a pure root j extension L = K ( a i , ... , a s) such th a t cdf' = a2 - € K for some e2 - > 0. By j Lem m a 5.5, K(/3) = K { a ) where a p & = a £ K . Since K = Q(£p) Q Q(Cn) Qd, x p B — a € Q d[x] and fl € Qd(a). Since QdL\Q(/3) = Q , [Qd(fl) ■ Qd\ — pe• Therefore, x p* — a is irreducible over Q d. Hence Qd(a ) is a pure root extension of Qd. Since / 3 e Q d(a), <*Q(c„)(0) = 1 . □ 72 L e m m a 6 .1 0 Let p i, ... ,ps be primes (not necessary distinct). Let K be a field such that 2(P i < E K for all i, 1 < i < s. Let Li be pure root extensions of degree pf* for 1 < i < s. Then there is a pure root extension L / K such that L\ • • • L s C L. Proof: Divide S = { L i, ..., L s} into subset S i , ... ,St such th a t S = Si U • • • U St, SiDSj — 0 for i ^ j and L,-, Lj € Sk if and only if p, = pj = qk• By Lem m a 6.8, there j is a pure root extension Mi such th a t Lj C Mj for all Lj € Si and M {/K is Galois of degree qf' for some di > 0. It is easy to see th a t Mj fl M i • • • M i-iM i+i • • • Mt = K since qi, 1 < i < t, are distinct. Hence the composite field of M i, ..., M t is a pure root extension of if . □ P r o o f o f T h e o re m 6.11: Let D = max®=1{dQ (/?,■)}. It is easy to see th a t dQ(a) > D. In th e following, we show th a t there is a pure root tower of depth D containing a. T h at is, dQ(ct) < D. Therefore, <7q(o:) = D. If D = 1 then the assertion follows from Theorem 6.9. Hence we assume D > 1 in th e following. Let [Q(/?i) : Q] = pf‘ where p,- is a prim e and e; > 0. Let m = nf= i where qi — Pi if d{pi) < D — 1 and d (p f) > D — 1; otherwise qi — p f . By Theorem 5.3, d(pi) < D — 1 for all 1 < * < s. By Corollary 5.2, there is a pure root tower of cyclotom ic extensions Q = Qo C Qi C • • • C Q d - i such th a t C m € Q d - i- Let Qd - i = Q(Cn)- Then m\n. W ithout loss of generality, assum e C 4 G Qi- ! I It is easy to see th a t QD-i(fii) is cyclic over Q d- 1 and [QD-i(fii) ■ Qd - i ]\pV■ B C p« * € Qd- 1 then Li = Q d -i(A ’ ) is a pure root extension of Qd - 1 by Theorem 5.2. If C p £ Q d- 1 and C pe* ^ Qd- i then <iQ(cPt)(A ) = 1 by Corollary 6.6. By Lemm a j 6.9, QD-i(fii) is contained in a pure root extension Li of Qd - i- By Lem m a 6.10, | there is a pure root extension L of Qd- 1 such th a t L{ C L. j Therefore, Q = Qo C Qi C • • • C Qd - 1 C l is a tower of pure root extensions , of depth D and (3 G L. □ , j P r o o f o f T h e o re m 6.3 (i): By Theorem 2.5, Q(ck) can be decomposed into cyclic extensions Q(fii) 1 < | i < s, of prim e power degree p f over Q such th a t Q (a ) = Q(/?i,..., fis) and { Q (A )n Q(Pi, •••, Pi-i, Pi+i, Ps) = Q for 1 < i < s. It is easy to see th a t j 73 D q (a ) = max*= = 1 {I>Q(^i)}. By Theorem 6.11, dQ(a) = max^=1{^Q(/?,)}. There fore, D q (q !) = c?q(q:) since, by Theorem 6.10, D q (/%) = dQ(/?*) for allz, 1 < i < s. □ 6.5 A lg o rith m s An irreducible polynom ial is said to be abelian (cyclic) over a field if its Galois group is abelian (cyclic). In this section, we give algorithm s to find nested radicals of m inim um possible depth for roots of abelian irreducible polynomials over Q, thus com plete the proof of Theorem 6.3. We first consider the case when f ( x ) is cyclic of degree pe over Q where p is a prim e and e > 1. Let j3 be a root of f(x ). Corollary 6.6 implies th a t to solve f ( x ) exactly by nested radical of m inim um depth is equivalent to solve it exactly by pure nested radical of m inim um possible depth over Q ((p). Assum e p ^ 2. Let K = Q (C ?> ) then dxi/S) < 2. By Lem m a 5.5, if d^(/5) = 1 then (pe e K(/3) and K(j3) = K ( z ) for some 2 € K (0 ) such th a t zp* € K . Let G be th e Galois group of K (/3 )/K and a be a generator of G. To find such 2 , we first express cr(/3) and in term s of some linear com binations of 1, /3, /32, ..., /3pe_1 over K . Then, for 1 < j < pe — 1 and (j,p) = 1, checking if there is an 2 G K(/3) such th a t < 7(2 ) = QeZ is reduced to solving systems of linear equalities. (Note th a t for 2 e K (ft) such th a t < 7(2 ) = (pez, K ( z ) = K(/3) if and only if (j , p ) = 1.) The ! following algorithm is based on this idea. ( ! A lg o rith m cyclic-1: Input: f(x ): an irreducible cyclic polynomial over Qof degree pe where p > 2 is a prim e and e > 1; Output: A: a nested radical of of m inim um possible depth which solves f( x ) exactly. d: the depth of A. 1. Let K = Q(CP). (Note f ( x ) is irreducible and cyclic over K .) 74 2. Let P be a root of f(x ). 3. If C pe & K (fi) then do th e following (Note in this case D q (/3) = d(p) + 2.) (a) Let K \ = Q(Cpe)- (b) Find th e m inim al polynom ial f ' ( x ) of ft over K \. Let [KtiP) : I<i] = pe > , e' < e. (c) Find the Galois group of f'(x ) over K \. (d) Use Lagrange resolvent to find 7 such th a t 7 pe — b E K i. ( N o te ir 1(/?) = /G ( 7 ) = Q (C ^ ,7 )-) (e) Find g(x) E Q (#) such th a t b = g((p*). (f) Find h ( x ,y ) E Q (a:,y) such th a t P = h((pe, 7). (g) Let be th e corresponding nested radical for C ,pe and let B = g(Ape). (h) O utput A = h(A pe, v\fB ) and d = d(p) + 2. 4. If E K{P) then do th e following (a) Find G, the Galois group of f( x ) over K . (b) Let a be a generator of G. (c) Find bi such th a t cr(fi) = E flo 1 bi/3% . (d) F ind c* such th a t = E f lo 1 ciPl- (e) Let 2 = E f lo 1 ZiP- (f) For 1 < j < pe — 1 such th a t (j,p) = 1, do the following (i) Solve for Zi such th a t a (z ) = Qez. T h at is, *(*) = E & 1 Zi(a(P)Y = ( E & 1 Z i ^ i Y t l 1 = & z . (ii) If such z exists then do th e following (N ote K (P ) = K ( z ) = Q(Cp? z)-) • Let zpe = b. • Find g(x) E Q(a:) such th at b = g((p). • Find h ( x ,y ) E Q(x,y) such th a t p = h((P, 7 ). • Let A p be th e corresponding nested radical for £p and let B = g(A p). • O utput A = h(A pey p \fB ) and d — d(p) -f 1. (g) If step (f) fails then do step 3 (a) to (h). 5. End. R e m a r k 6.1 1. We can check if £Pe € K{P) by factoring $ pe(x) over Q ((p, P) using Theorem 2 .11. If $ pe(a:) has a linear factor over Q ((p,P) then C pe € K{P)- 75 2. The polynom ial h (x ,y ) can be found by factoring f ( x ) over Q ((pe,fl) using Theorem 2.11. 3. A p, A pe and d(p) can be found by using procedure primitive-root-of-unity in C hapter 5. 4. T he Galois groups can be found by Theorem 2.12. 5. T he tim e com plexity is polynom ial in pe and log |/( x )|. I I Now we consider th e case when p — 2. T h e o re m 6.12 Let (3 be a root of an irreducible, cyclic polynomial of degree 2d j over Q such that dQ(fl) = 1. (i) I f d = 1 then 35 G Q(fl) such that S2 G Q and Q(/3) = Q (5). (ii) I f d > 1 then 35 such that 52 G Q and fl G Q(£2d+2,5). I Proof: (i) Trivial. (ii) Let L = Q(o>i,... ,ces) where a f = a{ G Q be a pure root extension such th a t fl G L. By Lem m a 5.3, we m ay assume qi = 2e* where et - > 0 for all 1 < i < s. Let M — Q(C4) and let N = M (fl). Then N /Q is abelian. It is easy to see N is cyclic over M and N is contained in the splitting field of (x2* 1 — ax) • • • (a:2* 5 — as) over M . Hence, by Lem m a 5.4, N /M is generated by certain elem ents fli, ... ,flt of the form C 2eail a 22 ’'' a ss where e — max*_1{ej} and 0 < kj < 26j , 1 < j < s, 0 < i < 2e. Since N / M is cyclic, there is a f l j , 1 < j < t such th a t N = M ( f l j ) . Since flY € Q , there is an a G Q such th at f l j is a root of f( x ) — x 2* — a. Let |a| = b2" for some b G Q — Q 2, b > 0 ,s > 0. Then f l j G Q (^2e+15Q :) where a > 0 and a 2 e 3 = b. Note th a t Q (a) has th e unique subfield property by Lem m a 5.1. Let L be the m axim um abelian subextension of Q (£2e+i, a ) over Q and let F be the m axim um abelian subextension of Q(o;) over Q. By Lemm a 6.4, L = Q(C2e+i)ir . It is easy to see F = Q(5) where 5 = a 2* -3-1, th a t is, 52 = b. j Since Q (/?,5) is cyclic of degree no greater than 2d over Q (5), fl G Q(C2^+2,5). □ i J I To find such 5 in the second case is easy. If fl G Q(C2d+2) then there is no need j to find 5. Otherwise, 5 G Q(£2d+2,/?) and Q (5) can be any quadratic subfield of | I Q(£2d +2 ,fl) other th an Q ( \ / — 1)- Since Q ((2d+ 2,fl) is abelian over Q , it is easy to ! I find Q (5). We om it the algorithm . ! 76 | Now we consider th e case when f( x ) is abelian over Q. Let a be a root of f(x ). Then Q (o )/Q is abelian and Q (a ) can be decomposed into cyclic extensions Q (/?») I < i < s, of prim e power degree over Q such th a t Q(ck) = Q (& ,...,& ) and Q(A) 0 Q(/?i,..., A -i, ..., /?*) = Q for 1 < i < s. By Theorem 6.11, dQ(a) = m a x ^ J d q (/?;)}. A lg o r ith m abelian-1: Input: f(x ): an irreducible abelian polynom ial over Q (assum e the degree of f(x ) is not a power of 2). Let a be a root of f(x ). Output: A: a nested radical of m inim um possible depth which solve f ( x ) exactly. 1. D eterm ine D — dQ(o). • Decompose Q(c*) into cyclic extension Q (/%), i = 1,..., s. • Use A lgorithm cyclic-1 to find c ? q (/5 s). • D = m ax{dq(/?,)}. 2. Let p? = [Q(A') : Q] and let Qd - i be a cyclotomic extension such th a t for 1 < i < s (P j € Qd - i and Q* £ Q d- i if ) < D — 1. 3. Decompose Q d- i (c x ) into Qd - i(A 0 > 1 — ® — s - 4. Use A lgorithm cyclic-1 to find ^ 5. Express a in term s of ..., /?'. 6. O utput th e corresponding nested radical of a. 7. End. R e m a r k 6.2 1. It is easy to see th a t the com putation of above algorithm is dom inated by step 1 and step 4. Since Algorithm cyclic-1 runs in polynom ial tim e, the above algorithm runs in polynom ial tim e as well. 2. W hen the degree of f( x ) is a power of 2 and dq(a) = 1 then, by Theorem 6.12, it is easy to see th a t a £ Q(C2d> ■ • ■ , fis) where d > 1 and (If £ Q for , 1 < i < s. We om it the algorithm for this case. 77 ! 1 C h a p ter 7 C o n stru ctib le N u m b ers 7.1 In tro d u ctio n Constructible numbers over Q axe num bers th a t can be constructed by ruler and compass. It is well-known th a t a complex num ber is constructible if and only if it is contained in a tower of quadratic extensions over Q. More generally, we say j th a t an elem ent a is constructible over a field K if and only if a is contained in ; a tower of quadratic extensions over K . It is clear th a t the corresponding nested radicals of constructible num bers in a tower of quadratic extensions are nested radicals involving only square roots called constructible nested radicals. Let a be a constructible num ber. T he length of a is defined to be the m inim um num ber of different square root extractions needed to construct a. The depth of a is defined to be the m inim um num ber of parallel square root extractions needed to construct a. Let A be a constructible nested radical. The length of A is defined ! i to be the num ber of different square roots in A and th e depth of A is the nesting i i depth of A. An irreducible polynom ial is said to be constructible if its roots are ( constructible. j In this chapter, we study the following problem: Given an irreducible polynomial f ( x ) over a field K , decide if f( x ) is constructible and, if it is the case, find the length and depth of the roots and a constructible nested radical o f minimum length and minimum depth respectively for representing exactly all the roots of f( x ) . I 78 T he classical study of Euclidean construction w ith ruler and compass led to the theory of solvable Galois extensions. Landau and M iller [25] gave a determ in istic polynom ial tim e algorithm for deciding if a polynom ial over Q is solvable by radicals, and for constructing a nested radical for th e roots of the polynomial. However, constructible num bers have m inim al polynomials solvable by radicals in volving only square roots. In fact, one can show th a t even an irreducible solvable polynom ial of degree 2n m ay not have constructible roots. Hence a polynomial tim e decision algorithm for solvability does not im ply a polynom ial tim e decision algorithm for const ructibility. It is not difficult to show th a t th e splitting field of a constructible num ber a over a field K decomposes into a tower of quadratic extensions [19]. This observation leads to a decision algorithm for constructible num bers. B ut its tim e com plexity is not polynom ial in th e input size. To get a more efficient algorithm , we will derive a stronger characterization of constructible numbers. A tower of quadratic extensions Kq C K \ C • • • C Kd is said to have length d. T he length of a constructible num ber or over a field K is therefore the m inim um possible length of a quadratic tower over K th a t contains a . O ur first result shows I th a t in fact K (a) itself m ust decompose into a tower of quadratic extensions over I K and such a tower has m inim um length among all quadratic towers over K \ containing or. i T h e o re m 7.1 Let a be constructible over a field K . Then there is a tower of quadratic extensions from K to K (a ). Moreover, the length d of the tower deter mined by K (a ) is the length of a. The theorem above gives rise to a determ inistic polynom ial tim e algorithm for constructing a constructible num ber w ith m inim um num ber of square roots. More precisely, T h e o re m 7.2 There is a deterministic polynomial time algorithm which on input an irreducible polynomial f over an algebraic number field, determines if f is constructible, and if it is the case, returns a constructible nested radical of minimum j length for representing all the roots of f . ; I R e m a r k 7.1 In fact given a root a of an irreducible polynom ial, th e algorithm of ! Landau and M iller [25] constructs a tower of fields from K to K {a ) which can not j 7 9 ; _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ i be refined. Theorem 7.1 implies th a t a is constructible if and only if such a tower is a tower of quadratic extensions. Hence this algorithm can be easily modified into a polynom ial tim e algorithm for Theorem 7.2. However, we will present a sim pler algorithm which is based directly on Theorem 7.1. Let K be a field. An extension L over K is called an abelian extension of exponent 2 or an abelian 2-extension over K if L = K {o t\,..., a s) such th a t a f £ K . A tower of abelian 2-extension of depth d over K is a tower of field extensions K = K q C K i C • ■ • C Kd such th at Ki is an abelian 2-extension over K i-\ for all i, 1 < i < d. The maximum abelian 2-subextension of L over K is the subextension M of L over K such th a t M is the composite field of all quadratic subextensions of L over K . One sees th a t the depth of a constructible num ber a over K , denoted by j depth‘ s (a) or depth2 ( a ) , is the m inim um depth of a tower of abelian 2-extension j over K th a t contains a. In fact, such a m inim um tower exists w ithin the splitting | field of th e m inim al polynom ial of a. More precisely, T h e o re m 7.3 Let K be a field and let a be constructible over K . Let f ( x ) be the minimal polynomial of a over K and let L be the splitting field of f( x ) over K . Let L q = K and let Li be the maximum abelian 2-subextension of L over L { -\. Then depth2(a ) = d if and only if L = L& and L Ls,-1- From th e theorem we derive: T h e o re m 7.4 There is a deterministic algorithm which on input an irreducible, constructible polynomial f over an algebraic number field, returns a constructible I nested radical of minimum depth for representing all the roots of f . The time complexity depends polynomially on the length of f and the degree of the splitting field o ff. 1 I i T he algorithm in Theorem 7.4 is not a polynom ial tim e algorithm as it depends j on th e splitting field of / . i I A problem related to th e depth of constructible num bers is th a t of simplifying I constructible nested radicals. This was studied by Borodin, Fagin, Hopcroft and I 80 i _ _ _ _ _ _ _ i Tom pa in [3] where simplification w ith and w ithout the restriction of using only square roots was investigated. P artial solutions were obtained in both cases. The m ore general problem of simplifying nested radicals was studied in C hapter 4. G ati [15] showed th a t for infinitely m any n, there exist irreducible polynomials of degree n over Q th a t require n — 1 square-root extractions to obtain all the roots. T he issue of solving polynomials with nested radicals of m inim um nesting depth was investigated in C hapter 6 . The general case where roots of unity are not given for free rem ains open. Theorem 7.4 tackles the special case where only square roots are the only radicals used in solving the polynomials. T he rest of this chapter is devoted to proving these results. 7.2 P r o o f o f T h eo rem 7.1 and 7.2 L e m m a 7.1 Let K be a field and let K = K 0 C K \ C • • • C K d be a tower of i quadratic extensions of length d > 2. Let M be a field such that K C M C Ka and [ [Kd : M \ = 2. Then M has a quadratic subextension over K . j I Proof: If M D K 2 7 ^ K then it is easy to see th a t M has a quadratic subextension , of K . In the following, we assume M fl K \ = M fl K 2 = K . Then it follows th at ' Kd = M K t = M K 2. ■ W ithout loss of generality, let K x = K 0(J3) such th a t /?2 = b E K 0 and K 2 = 1^0(7 ) such th a t 7 2 = u + v/3 where u,v E K q. Let 7 ' be such th a t 7 1 2 = u — v/3 and let K 2 be the splitting field of the m inim al polynom ial of a prim itive element of K 2 over K 0. Then it is easy to see th a t K 2 = K (/3,7 , 7 '). Hence [K2 : K 0] < 8 . Let F = K 2 n M . If [ K 2 : K 0 ] = 4 then f<2 = K 2 and [T1 : K 0] = [ K 2 : K 0 ] [ M : Kq]/[K2M : K q] = 2, a contradiction. Hence, [ K 2 : Kq] = 8 . There are two cases: Case 1: K 2 C K d - Let 6 = 7 7 ' 6 K . Then 8 2 = (u + vf3){u — v/3) = u 2 — v2b € K 0 . Since [ K 2 : K 0 ] = 8 and K 2 C IQ , 8 I<2. Let L = K t ( 8 ) . Then [ L : K ] = 4. By Theorem 2.4, [M D L : I<0] = [ M : K 0 ) [ L : K 0] / [ M L : K 0 ] = 2rf_1 x 4 /2 d = 2. Therefore, M has a quadratic subextension over K 0 . Case 2: I<2 % K d. Then [K2K d : K d] = 2. T h at is, [K2K d : K 0] = [ .K*M : | K 0] = 2d+1. Hence [F : I<0] = [K2 : K 0][M : K 0]/[K2M : K 0) = 8 x 2d" 1 /2 d+1 = 2. | Therefore, M has a quadratic subextension over K 0. □ 81 L e m m a 7.2 Let K be a field and let K = K q C K i C • • • C Kd be a tower of quadratic extensions of length d > 2. Let M be a field such that K C M C Kd and [Kd : M] — 2. Then there exists a tower of quadratic extensions K = Mo C M \ C • • • C M d-i such that M = M d -i- Proof: By Lem m a 7.1, M has a quadratic subextension over Kq. Let it be M \. It is easy to see [M\Ki : M iA j_i] < 2 for all i, 1 < i < d. Hence M l = M 1I\0 C M \K \ C . . . C M \K d = 7 0 is a tower of extensions where each step is an extension of degree 1 or 2 . B ut, since [Kd ' ■ Mi] = 2d~1, we get from the tower a tower of quadratic extensions of length d —1 from M i to Kd- Since M i C M C M \K d = Kd and [Kd ■ M] = 2, by Lem m a 7.1, M has a quadratic subextension over M \. Inductively, there is a tower of quadratic extensions K = M q C M i C • • • C M d-i such th a t M = Md-i- d i T h e o re m 7.5 Let K be a field. Let K = No C N i C • • • C N a be a tower of quadratic extensions. Let M = K (a ) C N a. Then there exists a tower of quadratic extensions K = M q C M\ C • • • C Md such that M = Md = M d-i(a). j i Proof: Let K = K 0 C K \ C • • • C K e be a tower of quadratic extensions of j m inim um length e such th a t a € K e. We shall use induction to show th a t [K(a) : K] = 2e. Hence K e = K (a ). ■ Consider /< "e_2(o:) over J le_i. If [A'e_2(o;) : K e- 2] < 2 then e is not m inim um , j a contradiction. Hence [A” e_2(a) : K e- 2] = 4 and A T e- i C K e~2(a) = K e. j Assume inductively, for i = l , . . . , n > 2 , K e-(i- X ) C A "e_j(Q!) and [Ae_i(o:) : J A "e_j] = 2*. In the following, we show th a t [Ke-.(i+i){a) : A e-^+i)] = 2,+1 and JCe_i C A T e_( m ) (a ). Therefore, [AT(a) : K] = 2e. i It is easy to see th a t [A'e_(J+i)(o;) : Are_(,+i)] = 2® +1 or 2* since / f e_(4 +i) C ' K e-i C K e-i(a ) and A'e-(«+1) ^ A T e-(i+i)(Q:) C /Ce_ i(a) and, by induction hy- J pothesis, [K e- i( a ) : A T e-*] = 28 . If [A'e_(j+i)(o!) : K e-(i+1)] = 2 * then [/^ -{ (o ) : j Ae_(i+i)(a;) = 2. So by Lem m a 7.2, there is a tower of quadratic extensions I of length i from /Ce-(j+1) to A T e_(j+i)(o!). Hence there is a tower of quadratic j extension of length e — (i + 1) + * = e — 1 < e, a contradiction. Therefore, ! [A T e_(i+i)(a) : -Ke-(*+1)] = 28+1. It follows th a t I<e-i C = K e^ iJri){a). □ 82 It is easy to see Theorem 7.1 follows from Theorem 7.5. Let K be a field and let a be constructible over K such th a t [K(a) : K] = 2d where d is an integer and d > 1. Let f ( x ) be th e m inim al polynom ial of a over K . In th e following, we outline a procedure to find a tower of quadratic extensions from K to K (a ). By Theorem 7.5, there is a tower of quadratic extensions K = Kq C K \ C • • • C K i = K (a ). Since [Kd : K d - 1] = 2, a can be expressed as u + Vy/w where u ,v,w € K d - 1. Hence a' = u — vy/w € Kd is a conjugate of a over K . By factoring f( x ) over K (a ), we can find a' expressed in term s of a. Therefore, u = (a + a !)/2 and Vy/w = ( a — o l')/2 can be expressed in term s of a too. (1) If u ^ K then K (u ) is a nontrivial subextension of K (a ) over K since y/w $ K (u). (2) If [K(vy/w) \ K] — 2 then K (vy/w ) is a nontrivial subextension of K ( a ) over K since d > 1. (3) If [K (v y /w ) : K] > 2 then K ( v 2w ) is a nontrivial subextension of K (a ) over K since v2w K (otherwise [K (v y/w ) : K] < 2) and y/w K ( v 2w). At least one of (1) — (3) m ust be true. Let M = K(/3) be a nontrivial subextension of K (a ) over K . Let g(x) be the m inim al polynom ial of a over M and let h(x) be th e m inim al polynom ial of /? over K . Then we can repeat the same procedure to find nontrivial subextensions of K d /M w ith g(x) and M f K w ith h(x). Eventually, we shall have a tower of quadratic extensions from K to K (a ). T he above procedure depends on factoring polynom ial and finding the m inim al polynom ials of prim itive elem ents. B oth of them have polynom ial tim e algorithm when th e ground field K is an algebraic num ber field by Theorem 2.11 and Theorem 2.14. More formally, we have the following algorithm . A lg o rith m le n g th input: An algebraic num ber field K and an irreducible polynom ial f(x) of degree 2l over K . output: A tower of quadratic extensions K — K q G K x C .- -- C .K i ~ K {a) and a constructible nested radical of m inim um length for representing exactly all the roots of f(x) if f ( x ) is constructible. 83 1. Let a be a root of f(x ). 2. Call Find-Tower(A’ , a, f(x )). 3. Find the corresponding nested radical for a in Ki. 4. End. Procedure Find-Tower(L, fl,g(x)) (where g (x ) is th e m inim al polynomial of fl over the field L) 1. If degree(g(x)) = 2 then retu rn else do the following: 2. Call Find-Subextension(X, fl,g(x)). 3. Assume the procedure retu rn M ,'y,h i(x). If [M : K) = 2 * then let I<i = M . 4. Find th e m inim al polynom ial h2(x) of fl over M . 5. Call Find-Tower(X, 7 , Ai(x)). 6 . Call Find-Tower(M , fl, h2(x)). 7. End. Procedure Find-Subextension(L, fl, g(x)) (where g(x) is the m inim al polynom ial of fl over the field L) 1. Factor ^(x) over L[y]/(g(y)). 2. Find a linear factor x — fl1 other th an x — fl. (N ote fl1 is a conjugate of fl over L .) 3. If fail then outp u t “ the input polynomial is not constructible” and stop else do the following. 4. Let u = fl + fl' and v = fl — fl'. 5. Find th e m inim al polynom ial h (x ) of u over L. 6 . If degree(h(x)) > 1 then retu rn M — L {u),u and h(x) else find the m inim al polynom ial h(x) of v over L and if degree(h(x)) = 2 then retu rn M = L (v ),v and h(x) else find the m inim al polynom ial h(x) of v2 over L if deg(g(x)) > deg(h(x)) > 1 then retu rn M = L (v 2),v 2 and h(x) else ou tp u t “ the input polynomial is not constructible” and stop. 1 t 7. End. | i 84 | P r o o f o f T h eo re m 7.2: T he correctness of A lg o rith m len g th follows from the discussion before it. In the following, we analyze th e tim e complexity. Assume th e input polynom ial f ( x ) is constructible. Let F0(x) be the m inim al polynom ial of a prim itive elem ent 70 of the given field K , th a t is, K = Q ( : r ) / ( F o ( x ) ) . Let deg(f(x)) = n = 2l and let deg(Fo(x)) = m. Let 7i be a prim itive elem ent of K { found by the algorithm and let Fi(x) be the m inim al polynom ial of 7 , ■ over Q . Let f i j ( x ) be the m inim al polynom ial of 7 j over K i . Initially, we are given fo,i = f and F0. To simplify the analysis let A = m ax{|Fo(a:)|, |j/(aO[|} and let d = mn. j It is easy to find th e m inim al polynom ial Fj, of a = 7 ; over Q as follows: ; Firstly, the resultant F (x) = rest(F o (t),f(x ,t)) € Q[z] is com puted where f (x ,t ) is f(x) w ith Fs substituted in for 70’s. Then Fi(x) can be com puted by factoring F (x). T he size of F(x) is bounded by (2m)!A2m < (dA)d. By Theorem 2.8, |F ( x )| is bounded by 2rf|F(a:)| < (2dA)d. Since u = + (3', v = fd — {¥ and v2 = {32 — 2j3P' + /3'2, inductively, it is easy to see th a t the roots of F{(x) are sums of elem entary sym m etric functions of a*, the roots of f(x). By Theorem 2.9, |[o r * 1 1 < ||/(:r)|| < A. Hence the root of F{(x) is bounded by 2,+1||/( x ) |j2i+1 < j dA d. Hence |Fj(a:)| is bounded by 2d(dAd)d < (d A )^, th a t is, Fi(x) have succinct ; descriptions. Note the bounds presented are not best possible. : The polynom ial /,-j can be com puted by factoring Q (y)/(F{(y)). Hence, by j Theorem 2.9, the size of is bounded by dlF/KdlFil)^ < {dA)di. Hence /,-j have I succinct descriptions too. T he com putation of the algorithm is dom inated by th e calculations of the m inim al polynomials F; and /,-j. Each is com puted by factoring some polyno m ial over some algebraic num ber field which takes 0 {d ie log2(dA ) rf2 log2(dA)(i3) = 0 (d 29 log4(dA)) tim e by applying Theorem 2.11. We need to apply such factor ing algorithm 0(1) tim es. Therefore, the algorithm runs in 0 (d 30 log2(dA)) tim e, j Theorem 7.2 follows. □ 85 | ____ 1 7.3 P r o o f o f T h eo rem 7.3 an d 7.4 L e m m a 7.3 Let K be a field. Let a be constructible over K and depth2 (a) = d. Let f ( x ) be the minimal polynomial of o c over K and let L be the splitting field of f ( x ) over K . Then there is a tower of abelian 2-extensions of depth d, K = M q C M i C • • • C Md such that L C M d. Proof: Since depth2(a ) = d, there is a tower of abelian 2-extensions of depth d, K — Ko C K i C ••• C Kd such th a t a € K d • Let K f = {AC T |A £ K f) where < r £ G(K/K). Then K f is an abelian 2-extension over Kf_x. Let Mi be the com posite field of K f for all a £ G. Then it is easy to see th a t Mi is an abelian 2-extension of M ;_i and L C Md = M . □ P r o o f o f T h e o re m 7.3: If Ld = L then it is easy to see depth2 K (ct) < d. Therefore, we only need to show if depth2 K(a ) = d then L = Ld- Let G = Go be the Galois group of L over K and let G% be the subgroup of G such th a t Li = L Gi. By induction on d, we show th a t Ld = L G d = L. Consider d = 1. It is easy to see th a t L = K (a ) and L\ = L. Assume th e assertion holds for d = 1,..., k. T h at is, let F be a field, ft be a constructible num ber over F of depth d < k and N be th e splitting field of the m inim al polynom ial of (3 over F. Then there is a tower of extensions F — No Q Ni C • • • C Nd = N such th a t Ni is the m axim um abelian 2-subextension of N over N i- 1. : Now consider d — k + 1. By Lemm a 7.3, there is a tower of extensions K = j M q C M \ C • • • C Md such th a t L C Md and Mi is an abelian 2-extension of Af,-_i- W ithout loss of generality, assume Mi is a m axim um abelian 2-subextension of Md over M i- 1. It is easy to see depth2 Mi (a) = d — 1 = k, L M i is a Galois extension of M i and L fl M i = L i . Let ax = a and a 2, ..., a; be the conjugates of a over K . Then depth\fi (a,) = d — 1 = k for i = 1,.../. Let Fi be the splitting field of th e m inim al polynomial of a ; over M i. Then, by induction hypothesis, there are towers of extensions M i = Fh0 C Fiti C ••• C Fitk = Fi such th a t F ij is the m axim um abelian 2- subextension of Fi over Ft j_i. \ I 86 Let M ij be the com posite field of Fij for all i , l < i < I. Then — M \L and M i = M i> 0 C M iti C • • • C M \y k is a tower of abelian 2-extensions. By Theorem 2.4, G\ = G {L /L \) = G (L M i/M \). Therefore, th e tower of m ax im um abelian subextension from L\ to L is of depth k. Hence Ld = L. □ T he following corollary shows th a t there are constructible num bers whose length and depth are equal. C o ro lla ry 7.1 Let n be a positive integer. Let a = C 2"+2 + C ^+2 where (\ is a primitive l-th root of unity over Q . Then the length of a = depth2(a ) = n. Proof: Let L — Q (o ). Then L is cyclic of degree 2n over Q. Therefore, depth2[a) = n. □ P r o o f o f T h e o re m 7.4: i Theorem 7.3 yields the following algorithm: A lg o r ith m d e p th input: An algebraic num ber field K and the m inim al polynom ial f(x) of a constructible num ber a over K . output: T he depth of a and a constructible nested radical of m inim um depth for representing all the roots of f(x) over K. 1. Find the splitting field L of f(x) and the Galois group G of L over K. 2. Find a tower of abelian 2-extensions from K to L, K = K q C Ki C • • • C Kd = L such th a t Ki is th e m axim um abelian 2-subextension of L over Ki- 1. 3. Find A, the corresponding nested radical expression of a in f . I 4. O utput d and A. i 5. End. | The correctness of th e above algorithm follows from Theorem 7.3. In the fol lowing, we analyze its tim e complexity. In step 1, th e splitting field and the Galois group can be found by applying Theorem 2.12 in tim e polynom ial in [L : K] and log \f(x)\. In step 2, Ki can 87 i _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ i be found by com puting th e m axim um abelian subextension Li of L over K i-i by applying Theorem 2.13. Then decompose Li into cyclic subextensions M j over K i-1 and for each M j find a quadratic subextension Nk over K i- 1. It is easy to see Ki is th e com posite field of all Nk- We can apply Theorem 2.14 to find Ki. All these can be done in tim e polynom ial in [L : K] and log |/( x )|. Therefore, th e tim e com plexity of th e algorithm is polynom ial in [ L : K ] and log !/(rc)| (the length of f(x)). Theorem 7.4 follows. □ i I 88 C h a p ter 8 C o n clu d in g R em ark s 8.1 S u m m ary o f th e D isse rta tio n We have developed a framework for studying the simplification of nested radicals and solving polynomials by nested radicals in m inim um depth. This framework provides form al definitions of nested radicals, their depth and value; the notion of pure nested radicals and their field theoretical properties; and a field theoret- 1 1 ical translation of the problems m entioned above. Based on this framework, the ! following results are obtained. I i 1. Given a nested radical A and a radical valuation v over an algebraic num ber field K, there is an algorithm to find a denested form of A over K([i00) w ith respect to v where ^ is the set of all roots of unity. 2. Let f ( x ) be an irreducible, solvable polynom ial over an algebraic num ber field K. Let L be th e splitting field of f(x) over K . Let n be a positive J integer divisible by [L : K] and th e discrim inant of L over Q and let ( n be a j prim itive n -th root of unity. Then th e tower of root extensions corresponding to the derived series for G (L((n)/ K (Cn)) determ ines a nested radical B of m inim um depth over K{ji oo) which represents exactly all the roots of f ( x ) over K when (n is viewed as a symbol which can assume as its value all conjugates of a prim itive n-th root of unity over K. Furtherm ore, this nested radical B can be com puted in tim e polynom ial in the length of f(x), [L : /<], and n. 89 3. Given a positive integer n, there is an algorithm to find a nested radical of m inim um depth over Q which represents exactly all th e roots of the n- th cyclotomic polynom ial < frn(x). The tim e com plexity of th e algorithm is polynom ial in n. 4. Given an irreducible, abelian polynomial f ( x ) over Q , there is a polynomial tim e algorithm to find a nested radical of m inim um depth which represents exactly all the roots of the polynom ial f(x). 5. An elem ent a is constructible over a field K if and only if there is a tower of extension fields K = K q C K \ C • • • C K\ = K(ct) from K to K(a) such th a t [K{ : Ki-i] = 2 for 1 < i < I. 6 . Let f(x) be an irreducible polynomial over an algebraic num ber field K. Then there is a polynom ial tim e algorithm to determ ine if f ( x ) is constructible, and, if it is the case, return a constructible nested radical of m inim um length for representing exactly all the roots of f(x). 7. Let a be constructible over an algebraic num ber field K. Let f(x) € K[x] be th e m inim al polynomial of a over K and let L be a splitting field of f(x) j over K. Let L0 = K and let Li be the m axim um abelian 2-subextension of j L over Li- 1. Then th e depth of a is d if and only if L — Ld and L ^ Ld- 1. Furtherm ore, there is an algorithm , which runs in tim e polynom ial in the degree of the splitting field of f(x) and l'og|/(a:)|, to find a constructible nested radical of m inim um depth for representing exactly all th e roots of /(*)• t 8.2 D irectio n s for F urther R esea rch i t I We propose th e following questions for further research. O p e n Q u e s tio n 8 .1 ; Given a nested radical A and a radical valuation v over an algebraic number field K . \ Find a denested form of A over K with respect to v. j 90 Q uestion 8.1 is th e denesting problem over an algebraic num ber field in its general form. One can show th a t dv(A) = d(L ) where L is the splitting field of v (A ) and d{L) is the m inim um possible depth of a root tower containing L. T he m ain diffi culty of this problem is th a t in general d(L) / l(G(L/K )) where L is the splitting field of the m inim al polynom ial of v{A). Therefore, we can not solve this problem by standard Galois theoretical m ethods. We are m ainly interested in the denesting problem over algebraic num ber fields. However, it is also im portant to denest a nested radical which involves polynomials or rational functions. O p en Q u estio n 8.2 : Given a nested radical A over K ( x ) and a radical valuation v where K is a j field. I Find a denested form of A with respect to v. Cases of particular interest are when K = Q or K = F p. O p en Q u estio n 8.3 ; Given a constructible number a over a field K . Find a polynomial time algorithm to decide the depth of a over K . N ote we can not directly apply Theorem 7.3 since the degree of the splitting field may be exponentially large. I O p en Q u estio n 8.4 ; Let p > 2 be a prime and let K be a field containing £p, a primitive p-th root o f unity. Let a be contained in a tower of pure root extensions K = Kq C K \ C • • • C Kd such that, for 1 < i < d, Ki is generated by a generating root of degree p over K {-i. Prove or disprove that there is a tower of pure root extensions K = Lq C Li C j • • • C L e = K(a) such that, for 1 < i < e, Li is generated by a generating root of | degree p over L i- \. j N ote it is not easy to generalize the m ethod used in the proof of Theorem 7.5 to this case. 91 I O pen Q u e s tio n 8.5 ; Let a be solvable over Q . Prove or disprove that dQ{a) = D q(o;). Let a be solvable over a field K . We have shown th a t the pure depth of a is at m ost one more th an its optim al depth. However, we do not have an algorithm to find such a pure nested radical for a. O p e n Q u e s tio n 8 .6 ; Given an irreducible, solvable polynomial / ( x ) over an algebraic number field K . Find a pure nested radical of minimum possible depth over K which represents exactly all the roots of f(x). T he m ost challenging problem is perhaps the min-depth exact solution problem in its general form. O p e n Q u e s tio n 8.7 ; Given an irreducible, solvable polynomial f(x) over a field K . 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