Indices of Principal Orders in Algebraic Number Fields Citation Knight, Melvin John, II (1975) Indices of Principal Orders in Algebraic Number Fields. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/hcvt-yx83. https://resolver.caltech.edu/CaltechTHESIS:08262021-155312274 Abstract Let K be an extension of Q of degree n and D K the ring of integers of K. If θ is an algebraic integer of K and K = Q(θ), then Z[θ] is a suborder of D K of finite index. This index is called the index of θ. If k is a rational integer, the numbers θ and θ + k have equal indices. Define two numbers to be equivalent if their difference is a rational integer. Using Schmidt's extension of Thue's Theorem it is shown that in any field of degree less than or equal to four there exist only a finite number of inequivalent numbers with index bounded by any given number. This is true for every finite extension of Q and a proof is given using a slight generalization of Schmidt's Theorem. An application of Schmidt's Theorem to a problem on the units in a cyclic field of prime degree is given. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Mathematics) Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Kisilevsky, Hershy Harry Thesis Committee: Unknown, Unknown Defense Date: 16 May 1975 Funders: Funding Agency Grant Number NDEA UNSPECIFIED ARCS Foundation UNSPECIFIED Caltech UNSPECIFIED Record Number: CaltechTHESIS:08262021-155312274 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:08262021-155312274 DOI: 10.7907/hcvt-yx83 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 14342 Collection: CaltechTHESIS Deposited By: Benjamin Perez Deposited On: 30 Aug 2021 16:44 Last Modified: 14 Jul 2025 15:51 Thesis Files PDF - Final Version See Usage Policy. 15MB Repository Staff Only: item control page

INDICES OF PRINCIPAL ORDERS IN ALGEBRAIC NUMBER FIELDS Thesis by Melvin John Knight In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1975 (Submitted May 16, 1975) ii ACKNOWLEDGMENTS I want to thank my wife Beth for her love and encouragement. I would also like to thank Dr. Olga Taussky Todd for the many talks we had on this problem and the help that she gave me. This topic was suggested to me by my adviser, Dr. Kisilevsky, and he also gave me the most important references. I appreciate the help he gave me and the opportunity I had to work with him. I received financial support from the U.S. Department of Health, Education, and Welfare in the form of an NDEA Fell owship , from the ARCS Foundation through their Fellowship, and from the California Institute of Technology through research and teaching assistantships. iii ABSTRACT Let K be an extension of Q of degree n and OK the ring of integers of K. If 9 is an algebraic integer of K and K = Q(e ), then Z[8 J is a suborder of OK of finite index. index of e. This index is called the If k is a rational integer, the numbers have equal indices. e and e + k Define two numbers to be equivalent if their difference is a rational integer. Using Schmidt's extension of Thue' s Theorem it is shown that in any field of degree less than or equal to four there exist only a finite number of inequivalent numbers with index bounded by any given number. This is true for every finite extension of Q and a proof is given using a slight generalization of Schmidt's Theorem. An application of Schmidt's Theorem to a problem on the · units in a cyclic field of prime degree is given. iv TABLE OF CONTEN T S ACKNOWLEDGMENTS ii .. ..... . . ABSTRACT . . . . INTRODUCTION iii 1 CHAPTER 1 A SYSTEM OF DIOPHANTINE EQUATIONS . 2 NORM FORM EQUA TIONS . 3 CERTAIN FIELDS WITH THE BOUNDED INDEX PROPER TY . . . . 4 . . . . .... . . .. . .. .. 14 30 ... 41 ....... ..... ............ 53 THE PROOF OF THEOREM 2 REFERENCES 4 1 INTRODUCTION Let K be a finite extension of Q with [K:Q] = n. An order () of K is a finitely generated Z-module which contains a basis for K over Q and is also a ring with 1. if it is of the form Z[ 8 J for some An order is called principal e E K. It is well-known that K contains a unique maximal order and that it is the ring of integers of K, denoted ()K. It is easy to see that Z[e J is a principal order of K if and only if K = Q(9) and 9 E ()K. From the theory of finitely generated modules over principal ideal domains it follows that every order of K is a free Z-module and contains a free integral basis. then there exist numbers ill]_, ••• , wn Thus, if () is an order of K, in () such that and each number in () is uniquely represented in ( 1 ). The vector (Uii., w2 , ••• , wn ) is called an int e g r al b a sis for (). The principal order . 1 b as1s . ( 1 , e, 0 2 , ••• , 0 n -l ) • Z [ 9 J h as t h e 1ntegra Such a basis is called a power basis. Let 0 1 C () be two orders of K with bases (A1 , 11. 2 , ••• , An) and (0i, e2 , ••• , en) respectively, then there exists unique integers ai, j satisfying (i = 1,2, ... ,n), and the integral matrix A = (ai,j) satisfies 2 A 2) = Since o' and O each have rank n, the determinant of A is not zero. The index of O' as a subgroup of O is (0 :O') = l det Al. If o' = Z[A] the special notation, Index0 (>..), will be used for (0 :Z[;>..] ). If O = OK, then ( .D: Z[ 'A]) will be denoted by Index ( 'A ). The number 8 is a generator of the principal order .() if O = Z[e ]. It is easy to see that e + k is also a generator of () if k is any rational integer, so there are always an infinite number of generators of any principal order. Call two numbers equivalent if they differ by a rational integer. The following theorem was proved by Hall [ 9 J. The proof is given in Chapter 1. Theorem 1: Let [K:QJ ~ 3. For every c > 0 there exist only a finite number of inequivalent ;>. E OK satisfying Ind ex ( A) = c . It is not difficult to show that for every order O in such a field and c > 0 there exist only a finite number of inequivalent A E () such that Index 0.) = c. 0 lemma 1 in Chapter 1. This is the corollary following An order O of an arbitrary field K with this property is said to have the bounded index property since 3 there exist only a finite number of inequivalent 11. E O with a given non-zero index. The major tool required for the proof of Theorem 1 is Thue' s Theorem on the integer solutions of certain diophantine equations, but this cannot be used to prove the corresponding result when [K: Q] > 3. The general result which is proved here uses the generalization of Thue' s Theorem proved by Schmidt [ 13 ]. This result, the proof of which is given in Chapter 4, is given by the following theorem. Theorem 2: Let K be a finite extension of Q, 0 an order of K , and c > 0 constant, then there exist only a finite number of inequivalent A E O satisfying Indexo ( A ) = C • In other words, every order of every finite extension of Q has the bounded index property. The end of each proof is indicated by the symbol § near the right mar gin. 4 CHAPTER 1 A SYSTEM OF DIOPHANTINE EQUATIONS The following lemma is useful for the proof of Theorem 1 and also shows that the theorem generalizes to every order of the field. Lemma 1: Let K be any finite extension of Q and 0 1 , 0 2 any two orders of K, then 0 1 has the bounded index property if and only if () 2 has it. Proof: It suffices to show that if '°1_ does not have it then 0 2 also will not have it. Suppose that [\ 1 , A 2 , ••• } is an infinite set of inequivalent numbers in zero constant c. ~ each with index in '°1_ equal to the non- The result will follow from showing that a cer- tain multiple of the AI s will all lie in the order 0 2 and have a common non-zero index in it also, so 0 2 will not have the bounded index property. If O' c; () are two orders of a field K with (0: O') = m, then mO c; O', since the factor group 0/0' has order m. Let (OK:01 ) = m 1 and (OK:02 ) = m 2 , then for each "-i' i = 1, 2, ... , m 2 11.i E 0 2 • The set [m 2 11.1 , m 2 11.2 , ••• } is an infinite set of inequiva - lent numbers in 0 2 which satisfy Cn-2)Cn+1) 3) Index (m 2 \i) = m 1 m 2 02 2 C ' ( i = 1, 2, ... ) . It is clear that they are inequivalent because if m 2 11.i - m 2 >,_j is a rational integer, then tv -11.• 1nust be rational, but it is also an l J 5 algebraic integer, so a rational integer and ")... is equivalent to "). .. 1 J To calculate the index notice that m 2 \ is also in Z [Ai J and calculate Index (m 2 >..i) two ways. First by using the fact that Z[m 2 11.i] is contained in Z[>..iJ The last index follows easily since the matrix of the transformation is diagonal. Similarly, Z[m 2 >..i] is a suborder of 0 2 , so its index also satisfies Combining these two equations gives (3 ). Therefore, 0 2 does not have the bounded index property and § the lemma follows. Corollary: If [K :Q] ~ 3, then every order has the bounded index property. Proof: Theorem 1 shows that OK has it, so every order must by lemma 1. The proof of Theorem 1 requires Thue' s Theorem, which is stated here for convenience. A proof is given in Mord ell [ 12]. 6 Theorem 2 (Thue ): The equation +ay n n =m-# 0, where n ~ 3 and f(x, y) is irreducible in the rational field, has only a finite number of integer solutions. Proof of Theorem 1: If n = 1, so K = Q, then every integer is equivalent to 1 and the result is trivial. _Suppose now that n = 2. principal order O = Z[e]. orders. By lemma 1 it suffices to conside r a Every finite extension of Q has such If >-.. E O with index c > 0, then 11. = a 0 + a 1 8 for unique inte - gers a 0 , a 1 and the matrix transforming the basis (1, e) of O into the This has determinant a 1 , so I a 1 I = c and A is equivalent to either c0 1 or to -c8 1 • If n = 3 it again suffices to prove the principal order O = Z[ e J has the bounded index property, where K = Q(e ). Let 8 be a root of the irreducible cubic equation 4) where r 1 , r 2 , and r 3 are integers. Every element in O can be written uniquely as an integral combination of the power basis 7 (1, e, 8 2 ). In particular, 8 3 = - r3 - r 28 5) because of (4). If A E O, then 11. = a 0 + a 1 8 + a 2 8 2 for some integers The determinant of the matrix A which transforms the basis (1, 8, e2 ) is easily calculated from the expressions of 1, 11., and 11. 2 in terms of the basis (1, 8, 8 2 ), giving Notice that a 0 does not appear in (6) since equivalent numbers have equal index. where f(t) = g(t - r 1 ). The polynomial in (6) may be written aff(a 1 /a 2 ), Thus f is irreducible over Q since it is rationally equivalent to g. The index of 11. in O is c if and only if the integers (a 1 , a 2 ) determined by 11. = a 0 + a 1 e + a 2 e2 are solutions to the diophantine equation (6 ), where det A is either c or -c. By Thue' s Theorem there are only a finite number of such integer pairs and thus only a finite number of inequivalent such "-• index property and OK must also. Thus O has the bounded § 8 Let K ~ C be a finite extension of Q, .0 = Z [ e J an order of K an_ 1 en-i any element of .0, where [ K: Q J = n and each ai is a rational integer. There are precisely n isomor - phisms of K into the complex numbers. Let e = 0i, e2, ... , en denote the images of e under these isomorphisms, so that A = Ai, A2 , • •• , An are the corresponding conjugates of A• Therefore, (i = 1, 2, ... , n) . 7) . Let K~:~ = K(81, e 2' • • • ' en) and Xi, X2' ... 'Xn-1' pendent variables over K~:~. Since K = 0(0 ), Define the polynomial 01, 02 be indep_ by the n conjugates of 8 are distinct. If ei, 8 j are distinct, then A• - A• 1 8) J e.1 , e.) . J e-1 - e.J Suppose A transforms ( 1, 8, .•. , en-i) into ( 1, A, ... , A.n-i ), then the determinant of A depends only on a 11 ••• , an-i. A useful expres - sion for this determinant was given by Dade and Taus sky [ 5 J. Lemma 2: 9) With notation as above, let z(X1, X2, • • • 'Xn-1) = n i<j £(x1, • • • 'Xn-1 9 then z(a1 , a 2 , • • • , an_ 1 ) is the determinant of A. Proof: By (7) the matrix A transforms the basis (1, e., ... , 9;i- 1 ) to 1 1 (l,11.., ... ,11..n-1) in the field Q(e. ), so A satisfies the matrix equation ' 1 l l 1 1 1 1 1 1 61 82 en A1 A2 ~ At-1 11. n-1 Ann A = 9 n-1 1 9 n-1 2 S n-1 n 2 Taking determinants in this expression gives (det A) n i<j (8. - 9.) l J = n i<j § The result (9) follows from this by using (8 ). Let G be the Galois group of K':~ over Q and for each er E G define the action of er on the variables x 1 , x 2 , ••• , Xn-i t o be trivial, then Since the polynomial i. is symmetric in O 1 , 0 2 and each factor in (9) is different, each er E G permutes these factors. er z(x1 , . . . , Xn-1) = n Therefore, er £(Xi, ... , Xn-l e. , e.) = n i_ (xl , • • • ' Xn -1 e. , e.) . i<j J 1 i<j 1 J 10 So z is fixed by the group G, has algebraic integer coefficients and thus has rational integer coefficients. For each choice of 1 ~ i < j ~ n let Gi,j be the subgroup of G consisting of all <YE G which stabilize the set [ei, ej}. <YE Gi J. if and only if <Y fixes both , them. Thus e.l and e.J or else interchanges This is the subgroup of G which fixes the polynomial 10) e.l , e.) . J Let Li,j be the fixed field of Gi,j• The group G induces exactly j GI / I Gi,j j isomorphisms of Li,j into C given by the cosets of Gi,j in G. This is exactly the number of factors in (9) which are con- jugate to the factor given in (10 ). Let where <Y .runs over a set of coset representatives of Gi,j" There are exactly !GI/ JGi,jl factors in (11) and this polynomial is fixed by G, so it has rational integer coefficients. Not every pair of these polynomials need be conjugate. Let S be a maxim.al set of pairwise non-conjugate factors £(x1 , ••• , ~ - i ; 8 ., e.) of (9 ), l J element of S. Lemma 3: so each factor in (9) is conjugate to exactly one 11 Proof: Each factor of (9) is conjugate to exactly one element of S and appears only once in the product for the norm in ( 11 ). There- § fore, this product is the same as (9 ). To prove that every order in a given field K over Q has the bounded index property, it is sufficient to prove it for a principal order () = Z[e J because of lemma 1. Since each element ). _ E () may be written ). _ = a 0 + a 1 8 + · • • an_ 1 en-i with uniquely determined integers a 0 , • • • , an-i, this element has index in O equal to c if and only if e.,e.)J 1 J 12) = c .. 1,J for each polynomial in S and with the constants c . . such that their 1,J product is ± c. gers. It is clear that all of these constants must be inte- Since the integers ± c have only a finite number of factori- zations, the bounded index property is equivalent to the system of diophantine equations (12) having only a finite number of simultaneous integer solutions (ai, ... , an_ 1 ) for every choice of non-zero constants c ... 1,J As an example of these lemmas, let n = 3 as in the proof of Theorem 1. Now equation (6) gives with 9 a root of (4 ). 12 As a second example, let K = Q(I; ), g a primitive fifth root of one. This is a normal fourth degree field and OK = Z [ g J. The Galois group G is cyclic and generated by er, where er (g) = ~ 2 • Write 81 = g, e~ = g2 , 03 = 1;4, and 8 4 = 1; 3 - For the set S take The group G 1 , 2 is just the identity, while G 1 , 3 = ( 1, er 2 ). Therefore, L 1 , 2 = K and L 1 , 3 = Q({s ), the unique quadratic subfield of K. By lemma 3, Both factors have integer coefficients, the first is of fourth degree and the second is quadratic. Notice that for each pair 8i, 8 j the set is a Z-module. Thus, the system of norm form equations (12) is the same as a system of norms from certain modules. Thue 1 s Theorem may be restated in terms of modules. K = Q(a) be an extension of degree greater than 2. Thue I s Theorem says that there exist only a finite number of elements in the module [x + y~ I x, y integers} Let 13 with a given non-zero norm from K to Q. Chapter 2 gives the details of Schmidt's generalization of Thue' s Theorem to more general modules. 14 CHAPTER 2 NORM FORM EQUATIONS This chapter covers the material on Schmidt's Theorem [ 13 J needed to prove Theorem 2. here are given in [3]. The proofs of some statements made Every module is assumed to be a finitely generated Z-module. Let K be an extension of Q of degree n = s + 2t, where s is the number of real imbeddings and t the number of complex imbeddings of K into C, and let er 1, er 2 , ••• , er n be the corre spending isomorphisms. Let N(xi, x 2 , • • • , xm) be a norm form from K to Q; that is, for some fixed numbers a 1 ,a 2 , • • . , am in K n a ) n er 1. (x1a 1 + · · • x mm i= 1 where each variable is fixed by all of the automorphisms. Of interest here are the integer solutions (x1 , x 2 , .•• , xm) of the norm form equation ( 13) for some constant c. The set of numbers of the form x 1a 1 + • • • ~Q'm where x 1 , x 2 , • • • , xm take on all possible integer values is denoted by 15 and is a finitely generated Z-module. Z-basis ({\, f3 2 , • • • , f3k). It follows that M has a free The integer k is the rank of M and is less than or equal to n. The solutions to (13) are easily found from the solutions µ of the equation µEM. ( 14) The integers (x1 , x 2 , . . . , xm) are found by solving a system of equations which depends on the relationship between the set (cx 1 , . • . , cxm) and the basis (f31 , • . • , f3k) of M. solutions to ( 14 ). Therefore, it suffices to find all Sine e only µ = 0 is a solution if c = 0, assume c -/= 0 for all that follows. Suppose now that M is full in K, i.e. that rank M = n. It is possible to solve (14) for all solutions in a finite number of steps. First, for a full module define the coefficient ring OM = [ \) E K l v µ E M for all µ E M} . It is proved in Lemma 5 that this is an order in K. Let uKIQ be M the group of units in DM with norm from K to Q equal to +l, a subgroup of index 2 or 1 in the full group of units of OM depending on whether there exists a unit with norm -1 or not. By the Dirichlet Unit Theorem for orders, there exists a set of r = s +t - 1 units 'Th, •.. , T1r such that every unit in ~ can be written uniquely as 16 where ~ is a root of 1 and a 1 , . . . , ar are integers. Since the units 'fh, ... , llr can be calculated in a finite number of steps, so can U KIQ M • For every µ E M the set 15) is the set of all associates of µ over ~ with the same norm and is called an (M, K)-family. Tinally, there exists an effectively computable constant y such that every solution to (14) has an associate over a solution and with height bounded by y. ~ which is also Each of these solutions can be calculated in a finite number of steps, so all solutions to (14) are contained in the union of a finite number of families of the form (15 ), where µ is one of these solutions with height bounded by Y· For full details on this method see [3, p. 122]. As an example of a norm form equation for a nonfull module take the equation µEM 16) where M = {Xi ff + x 2 V5 + ¾ffi} and K = Q(V3, 'VS). All solutions to (16) with x 1 x2¾ = 0 can be given. If x 1 = 0, then µ must lie in the submodule This is proportional to the full module {x 2 + x3 ✓3} of the field 17 L 1 = Q( "✓3) and therefore Using this with (16) gives and all solutions are given by for every integer t. If x 2 = 0, then µ will lie in the submodule which is proportional to a full module of the field L 2 = Q('VS ). Again, (16) reduces to an equation from a subfield: The solutions to this equation are all given by for every integer t. When ~ = 0, µ will lie in the submodule 18 and for such a µ = where L 3 = Q(fil). Thus, equation (16) gives .Since both 3 and 5 extend to unique nonprincipal ideals in this field, there are no solutions to this last equation. Therefore, every solution to (16) with x 1 x03 = 0 is given by ± VS (2 + 1 7) for all integer values of t. {s/ That only a finite number of other solutions exist will follow from Sclunidt' s Theorem. A similar example was given by Schmidt. Let M be a module in K and L any subfield. Define the submodule to be the set of µ in M such that for every "A in L there is a nonzero rational integer z such that z Aµ E M. If M denotes the vector space over Q generated by M, then ML is the set of µ in M such that 19 The last form of the definition shows that ML is a vector space over L, since for µ in ML and A in L and thus µ"A is in ML. Because of this, [ L: Q] divides rank ML. It is easy to calculate ML in terms of the bases for M and L. Extend the basis 0\, ... , (3k) of M to a basis of K, say ((31 , ••• , (3k, Let (Ai, ... , "'l) be any basis for L, then µ in M is in f3k+l' ... , f3n). ML if ·and only if µA. E M 1 for each 1 s;; i ~ 1. This will be so if the representation of µA· l with respect to the basis ((3i, ... , f3n) terminates with n-k zeroes. Thus, µ is in ML if and only if the k coefficients of µ with respect to the basis ((31 , ••. , (3k) satisfy the set of l(n-k) linear conditions required above. A basis for ML can be found using these conditions . - For example, MQ = M for every module is easily seen and if the module M is full in K, then M = K, so MK = M. However, for any nonfull module M, M is a proper subspace of K and must have dimension over Q smaller than n. If µ were a non-zero element of M\ then dim µ K = n ~ dim MK < dim K = n . So MK = {O} if M is not full in K. Let AB be the composite of the two fields A and B. 20 Lemma 4: Proof: For any two subfields A, B of K Assurrie first that µ E (MA) 8 , then µ is an element of MA such that Every element of AB is of the form a 1 (31 + · · · + a r(3r with ai E A and (3i E B, so µ is in MAB if and only if µa (3 E for every a E A and (3 E B. µa (3 E Therefore, (MA) 8 (: M But MA is a vector space over A, so a MA C MA C M MA 8 • Conversely, suppose µ E MA 8 , then µA ~ µAB C M. Therefore, µ E MA and it is only necessary to show But B is a subfield of AB, so µ B ~ µ AB ~ MAB ~ MA 21 since being in MAB is more restrictive than being in MA. Thus MAB C (MA)B and the lemma follows. With the module M = [x1 Y3 + x 2 VS + X:3ffi} from the previous example (16) and the fields Li, L 2 ., L 3 , the submodules corr es ponding to these fields are M 1 , M 2 , and ~ respectively. These are the same submodules which appeared in the example. To calculate ML3 , for example, extend the basis for M to the basis [ l, 'V3, VS, {Ts} of K and use the basis [ l, ffi} for L 3 • Since 1 • µ is in M for every µ in M, the condition for µ to be in ML3 reduces to the one relation µ{Ts E M. If µ is written as x 1 Y3 + x 2 VS + X:3 vfs, then µ ffi E M requires rationals a, b, c such that L Therefore, X:3 must equal zero and M 3 = ~ of the example. Let D~ denote the ring of coefficients of ML in L, i.e. the set 1 of all A in L such that Aµ E ML for all µ in ML. Lemma 5: Proof: If ML -1- [ O}, then 0~ is an order of L. It is clear that D~ is a ring with 1, so it only remains to prove that it is a full, finitely generated module in L. Let µ be any non-zero element of ML, then µO~ C ML and 22 Since ML is a finitely generated module so is µ -l ML and thu s also the submodule .O~. To show that .O~ is full in K, it suffices to show that there exists an integer z f: 0 such that zOL ~ 0~, where OL is the maximal order of L. For each element Ai of a basis for .O L and each µ- of a basis for ML there is a non-zero integer z . . such that J 1,J 1,J. A·µ1 J E ML • Z· The integer z = flzi,j satisfies the above conditions, so O~ is full in L and is thus an order of L. The procedure used for full modules can be followed in general to some extent by using the modules ML for each subfield. For each of the finite number of subfields L of K construct the module ML. If ML is not just zero, O~ is an order of L and has a group of units as given by Dirichlet's Unit Theorem. Denote by U LIQ the subgroup of the group of all units in O~ which consists of M those units with norm from K to Q equal to +l. This group actually depends on K also, but for simplicity this dependency is assumed. As when M is a full module, the index of U L/q in the M full group of units of L is finite. The set µ U L/Q is the set of M associates of µ over .O~ which have the same norm and is called an (M, L/Q)-family. A family is maximal if it is not properly contained in any (M, L'/Q)-family for any L' contained in K. 23 Every element µ of M is contained in a maximal family, since for each field L either µ {:_ ML or else it is contained in the (M, L)family µ UL/q and no others. M One of the finite number of families containing µ must be maximal. Theorem 3 (Schmidt): Of course, every µ is in MQ = M. There exist a finite number of maximal families which give the solutions to (14). Applying this theorem to the equation (16 ), since MK = [ 0} for the nonfull module M, it is necessary to look at It has already been calculated that field. ML3 ML1 ML = M1 , for each subML2 = M2 , = 1vfs, with M 1 , M 2 , and ~ the submodules' given in the example. Also, it is always true that MQ = M. lies of solutions for the fields ML1, Equation (17) gives the famiML2, and ML3 and by Schmidt's Theorem there are only a finite number of maximal families altogether. It follows that there are a finite number of other families of solutions which, if any exist, will be of the form However, ~IQ is a subgroup of the group [ l, -1 }, so all but a finite number of solutions to (16) are given by (17). In a general norm form equation it is always possible to calculate ML, O~, and u~IQ for each subfield L in a finite number of steps, but it is not known how to find all representatives for the maximal families. Still, for some modules it is possible to prove that only a finite number of solutions exist. called non-degenerate if ML The module M is = [ 0} for every subfield L except 24 possibly for quadratic imaginary fields or Q. By Dirichlet's Unit Theorem, these are the only extensions of Q for which the group U ~/Q will be finite. It follows from Schmidt's Theorem that equa - tion (14) has only a finite number of solutions when M is nondegenera te. It is easy to show that Schmidt's Theorem generalizes Thue' s. It is necessary to prove that the equation has only a finite number of integer solutions, where K = Q(e) and [K: Q] ~ 3. This :will follow by showing that M = [x +ye} is non- degenerate. Clearly MK= [O}, so let L be any field strictly between Q and K. 2 ~ If ML -f. [ 0 }, then [ L: Q J ~ rank ML ~ rank M = 2 . The ineq uali~ between [ L: Q J and rank ML is because ML is a vec tor space over L. It follows that ML = M. Let L = QO,..), then A e E M and Ae = x for some rational numbers x, y. + ye Since "A is not rational, this implies e E L, contradicting the choice of L. So ML = [ 0} and M is non- degenerate. Schmidt's Theorem has a nice extension to the more general problem 25 µEM, 18) when MF = M. For each field L satisfying F C L C K define ML L and .OM as before, but now let U L/F M be the group of units in .O~ whose norm form K to F is +l. For each µ in ML the set µ U ~IF is an (M, L/F )-family and a maximal family is defined as before. Lemma 6: There exist a finite number of maximal families which give the solutions to (18). Proof: Taking the norm of both sides of (18) from F to Q gives µEM . So µ lies in one of a finite number of maximal families of solutions to this equation which are of the form 11' UM L/Q I , for fields L below K. The fields which appear in these families need not be above F, but since Mf = M, and the field FL is a field above F. It follows that the solutions to (18) are contained in a finite number of families of the form 26 with La field above F. It is clea r that if µ'U ~/Q has any solutions, then µ' can be taken to be one. The only units of UL/Q which will M give other solutions are precisely those with norm from K to F equal to +l, i.e. those in the group u~IQ. § The following lemma, a form of which was proved a different way by Gyory [ 7], will be used in the proof of Theorem 2. Lemma 7: Let Yi, Y~, and y be any integers in K with y -i 0. There exist only a finite number of units u 1 , u 2 in K satisfying Proof: Every unit of K can be wr itten uniquely as I; a root of one in K and Th, ... , 11r a fixed fundamental system of units. Let p be any prime. Those units fo r which O $;a . $; p-1 1 form a finite set and will be called p-power free (with respect to 'fl 1 , ••• , 11r ). Every unit in K is the product of a p-power free unit and a unit which a pth power. Choose p > 3 so that the field Q(sp) is disjoint from K, where Sp is a pri1nitive p th ro ot of 1. Then the representation i s unique for each unit of K. If the equation has an infinite number of solutions in units of K, there must be an infinite subset of solutions u 1 , u;j with p-power free parts equal to Us, u 4 respectively, for some pair (\.¼, u 4 ). This is clear since there are only a finite number of such pairs. There- fore, the lemma will follow from showing that for every y 1 , y ;j 27 and y f. 0 in K, there exist only a finite number of solutions x, y in units of K to the equation This will now be proved. If either Y1 or Y2 is zero, the result is obviouso If Y1 Ya f. 0, then and without loss of generality the equation can be assumed to be xP + y2 yP = y for some integers Yr,;, y in K and Ya y f. 0. Also, since pth powers in y2 can be absorbed in y, either Ya is not a p th power or can be taken to be L If y2 is not a p th power, then zP + y2 is irreducible over K, see for example [ 11 J. If Y;a = 1, then (zP + 1)/{z + 1) is irreducible over Ko follow since K and Q{sp) are disjoint. p-1 zP + 1 = TT Write ~ This w ill for Sp, then s (z + i) i=O Let S be any nonempty subset of [O, l, 2,. o .,p-1} such that g (z) = n {z + ~ i) i ES is in K[z]. The coefficients of g are in K factorization is exactly the same as over Q. n Q(~p) = Q, so the 28 So suppose first that y 2 is not a p th power. y2 and call it /3. Fix a p th root of If E = K(/3), then it is sufficient to prove that there exist only a finite number of solutions x, y in integers of K to the equation Since [E:K] = p > 3, ME= [O} for M = fx + y/3} and there are no subfields between E and K. of K. Also MK = M since x, y are integers Thus, there exist a finite number of maximal families of solutions each of which is of the form But UK/K is a finite group, in fact just [ l} in this case, so this M gives the required result. Now suppose y 2 = 1, then E = K(s ) and the equation is p 19) (x + y) NEIK (x + S y) p = y . Take norms from K to Q to get and each of the norms is an integer, so where c' is one of a finite set of integer factors of c. The rank of 29 M = [x + I; y} is 2[K: Q] and MK = M. Since p was chosen greater than 3, [E:KJ = p-1 > 2, so ME = [ O}. The same argument as the p one used to establish Thue' s Theorem from Schmidt's shows that ML = [ O} for every field L strictly above K. Every solution is thus contained in a finite number of families of the form The integers x, y of K are determined by these numbers since 1, sp are independent over K. If µ leads to a solution to (19) and µ = Xo + spYo, then for u E u~IQ, u EK so and Thus, the only solutions to ( 19) in the family are those for which · uP = 1 and this is only u = 1. So in this case there again are only a finite number of solutions. 30 CHAPTER 3 CERTAIN FIELDS WITH THE BOUNDED INDEX PROPERTY Because of lemma 1 it is reasonable to say K has the bounded index property if any order of K has it. D= Z[8] and 8 =8i,•••,8n,the conjugates of Let K = Q(e ), e, then K has the bounded index property if and only if there only exist a finite number of integer solutions (x,_, ... , ~-l) to the system of norm form equations 20) N L . . / Q [i(x1, ... '¾-1 1,J e.,e.)J 1 J = ( l s i < j s n) c .. 1,J for every set of non-zero constants c. .. l,J The field L . . is the 1,J field generated by the coefficients of £(x1 , ••• , xn-l ; 8. , 8. ). 1 J For each pair e.1 , 8.J define the module M 1,J . . to be so each norm form in (20) is the norm of an element from some M. .. 1,J However, equation (20) requires a dependency between the elements of the various modules with these norms in order for the solution (x1 , ••• , xn-l) to be integral. The values of x 1 , x 2 , . . . , ~-l are determined uniquely by the set of individual solutions µ .. to the equations 1,J NL . . /Q (µ 1,J .. ) = c .. 1,J l,J µ .. E M .. l,J 1,J 31 This is because µ . . is equal to l,J (>,. - A.)/(e. - e.) J 1 J 1 for some A E D and if A,w are two numbers in D such that A.. = J A_. 1 w. - w. J 1 for each i < j, then A - w is fixed by G and A, w are equivalent. Thus, x 11 x 2 , ••• , xn-l are uniquely determined. It will be useful to know the rank of M. .. 1,J Let F .. be the field fixed 1,J by the group generated by all automorphisms in G which take 8. to e .. 1 J Lemma 8: Proof: F·. 1 The rank of M . . is n-[F . . : Q] and M . . ,J = M ... 1,J 1,J 1,J 1,J The rank of M. . is equal to the rank of (8. - e. )M . . and 1, J 1 J 1, J equals the dimension of the vector space they generate over Q. Consider the linear transformation T from Qn-i to (e. - e. )M . . 1 J l,J given by xi(e. - e.) + ... 1 J If (x1 , • . • , Xn-i) is in the null space of T, then x__ --u-1 8 n-1 i = and this element is in F . . since it is fixed by all automorphisms 1,J taking 0. to 1 eJ.. 32 Since every element of F . . can be written uniquely in terms 1,J of the basis 1, e ., ... , e~- 1 of Q(8. ), only those elements of F . . 1 1 1, J 1 whose representations have zero for the first component with respect to this basis result from an element of the null space. Therefore, the dimension of the null space is [F . . : Q] - 1 and 1,J this gives the rank of M . . as n-[F . . : QJ. 1,J 1,J The last statement is easy since an element µ E M . . is in 1,J F·. M l,J if i,j µF . . ~ M . . . 1,J 1, J Therefore, let µ be in M . . and a an element of F. . . l,J It follows l,J that there exists A E .0 such that µ = o..l - A-)/(e. - e.} , J l J and since a is in F . . it is fixed by every automorphism of G taking 1,J 9. to l eJ.. w- =>...a. J J So there is a unique w in K such that w. =}...a and l 1 Thus = ( A. - "A..) a/ (e. - 8 . ) 1 J 1 . J = (w. - w.) / (8. - 8 . ) E 1 J 1 J M. . 1, J F·. and µ is in M. _i,J l,J Although the proof of Theorem 2 covers every finite extension of Q, it is interesting to apply Schmidt's Theorem in a different manner for fourth degree fields and certain other special fields. The simplest fields are covered by the following theorem. 33 Theorem 5: The field K with [K: QJ = n has the bounded index property if n is a prime \and G(K>:~/Q) = en, or any n if G(K :;,~/ Q) is either An or Sn. ,,, Proof: If n is a prime and G(K'''/Q) contains only Q as a proper subfield. en, then K is normal and = Each of the modules M . . l,J is nonfull since their ranks are n-1 or less, so they are all nondegenerate. It follows that (20) can have only a finite number of integer solutions. Suppose that G(K~!</Q) is Au or Sn. For n ~ 3 Theorem 1 is equivalent to this theorem, so assume n ~ 4. The system (20) reduces to one norm form equation since both An and Sn are 2-ply transitive for n ~ 3. The result will follow provided that the module M 1 , 2 is non-degenerate. It will be shown that for n > 4 the field L 1 , 2 contains no proper subfields except Q, so M 1 , 2 is non-degenerate for these fields. When n = 4 a simple argument will be given. The fact that L 1 , 2 contains no proper subfields except Q when n > 4 follows from the following lemma. Lemma 9: either Proof: G 1, 2• The group G 1 , 2 is maximal in G(K>:~/Q) when G(K>:~/Q) is An or Sn provided n is larger than four. Let G = G(K* /Q) and rr be any element of G which is not in Let H be the group generated by G 1 , 2 and rr, so it is neces- sary to prove H = G. These are permutation groups on the set [ 1, 2, ... , n} realized by their actions on the conjugates 8 1 ,9 2 , ... , an. 34 Suppose er is an arbitrary element of G and that there is a er' in H such that er(l) = er'(l) er(2) = er'(2), then er-1 er' fixes both 1 and 2 and is thus in G 1 , 2 . this shows that er must also be in H. Since er' is in H, Therefore, to prove the lemma it is only necessary to show that for every pair of distinct a, b ip. [ 1, 2, . . . , n } there is a er in H with er (2) = b . er(l) = a, Let r and s be the numbers which satisfy = rr(l) rr (2) r , = s , then sine e rr is not in G 1 ,2 [r,s} :/: [1,2}. A number of cases arise depending on which of the numbers a, b, r, s are in the set [ 1, 2} and which ones are equal to each other. In each case, since n ~ 5, an element er as required above can be constructed using only rr and automorphisms from G 1 ,2 • A case where this lemma fails for n = 4 will be worked out in detail. Let r, s l [ l, 2} and a = 1, b > 2. er' is in H with er'(l) E [ 1, 2}, Suppose a permutation 35 If o-'(l) -1- 1, then by following a-' with the even permutation (1, 2)(r, s) will give another permutation in H which now fixes 1. So suppose o-'(1) is 1 and o-'(2) is not b, but is still larger than 2, then a- is easily constructed from a-'. Since there are at least five elements in [ I, 2, ... , n}, there is a c > 2 with m = o- 1 (2), b, and c distinct. So (m, b, c) = T is even and fixes both 1 and 2, and a- = -ro-' is the desired automorphism. An automorphism o- 1 can be easily constructed. Since TT rf:. G 1 , 2 there is a number c > 0 with r, s, c distinct and such that either one or two of rr(r), rr(s) or rr(c) is in [ I, 2}. T Using some power of the even permutation (r, s, c) will give 0- 1 = TT T TT • It follows that H = G and G 1 ,2 is maximal in G. Since the order of the group G 1 , 2 is (n-2)! if G(K~:'/Q) is An and 2 (n-2 )! for Sn, the field L 1 , 2 is of degree n(n-1) over Q. 2 It follows that M 1 , 2 is not full in L 1 , 2 for n :2: 3 and thus M 1 , 2 is nondegenerate, except possibly for n = 4. If n = 4, then L 1 , 2 is a sixth degree extension of Q and contains only one cubic subfield L as a proper subfield. shown before, M 1 , 2 of rank 3 implies As has been 36 which is a contradiction since M 1 , 2 contains a generator for L 1 , 2 . So again M 1 ,2 is non-degenerate. The following lemma is interesting by itself and is also needed to prove Theorem 6. Lemma IO: If L 1 = Q(a 1 ) and L 2 = Q(a 2 ) are two distinct quadratic fields, then the system of norm form equations 21) has only a finite number of integer solutions (x1 , x 2 , ¾, x 4 ) satisfying the equation when neither side of the equation is identically zero. Proof: Let a 1 , a 2 denote the conjugates of a, 1 , a, 2 in the appropriate fields. Since either a or b is not zero, Similarly, N(d - ca 2 ) I- 0. NL 1 ;q(b - aa, 1 ) I- 0. Without loss of generality, assume bd f. 0, then if Y1 = Y2 = ax1 + bx 2 NL1/Q {b - a 0'1) Xi Y3 = NL2IQ (d - CQ'2)X3 Y4 = Y2 equation {21) becomes 37 since for instance y 1 + y 2 Q'(b-aci1 ) = = (b-aa 1 )(b-aa 1 )x1 + ai(b -aci 1 )(ax1 + bx 2 ) b(b - aa 1 )(x1 + X 2 Q'1) • Therefore, the lemma will follow from proving the case when So assume a = c = 0 and b = d = 1 and without loss of generality that x 2 > 0 for all solutions. Since (x1 + x 2 Q' 1 ) times (x1 + x 2 a 1 ) is a constant, any infinite set of solutions must contain a subset of solutions for which x 1 + x 2 Q' 1 , say, goes to zero. Otherwise there would exist an infinite number of quadratic algebraic numbers with bounded height, which is ilnpossible. Assume that there are an infinite number of integer solutions to (21) with x 1 + x 2 Ql 1 and x 3 + x 2 az approaching zero, then since x 1 /x 2 approaches -a 1 , x 1 /x 2 +ci 1 is bounded away from zero and similarly for a 2 • There- fore, I C1 I X2- 2 = IX1 /x2 +e¥1 I 1 < E > 0 X 2+E 2 and the corresponding inequality for a 2 with the same E will have an infinite numb er of simulta~eous integer solutions. However, by another theorem of Schrnidt 1 s [14], it is impossible to silnultaneously approximate two quadratic irrationals cr 1 , cr 2 such that 1, a 1 , a 2 are independent over Q, by an infinite number of 38 fractions x 1 /x 2 and ~/x 2 to an exponent larger than 3/2. So (21) has only a finite number of solutions. Another proof may be given which uses only Thue' s Theorem. The solutions to (21) are related to rational points on the conic All rational points on such a curve are given by a two parameter formula in the parameters p and q. The solutions to (21) come from these parameters if and only if the norm from the fourth degree field Q(Ql 11 Ql 2 ) of the element p + qa is a given constant, where QI is a fixed number such that Q(a) = Q(a 1 , a 2 ). Thue' s Theorem proves only a finite number of such p and q exist. Theorem 6: Proof: § If [K: Q] = 4, then K has the bounded index property. If G(K':</Q) is A 4 or S4 , Theorem 5 gives the result . of the remaining Galois groups is treated separately. lemma 3 gives e., e.)J . 1 J Each 39 Each of the fields L . . is quadratic, each module has rank 2, 1,J and the hypotheses of lemma 10 hold, so only a finite number of solutions (x1 , x 2 , x 3 ) exist. (er 2 ). Take 9 to be one of the integers of K such that K = Q(e) and L = Q(9 2 ). Let 9 = 9 1 , 9 2 = ere, 9 3 = o- 2 8, 8 4 = er 3 9, then with L 1 , 2 = K, L 1 , 3 = L. Since F 1 , 2 is Q and F 1 , 3 is L, the modules M 1 , 2 and M 1 , 3 have ranks 3 and 2 respectively. Since M 1 ,2 has rank 3, x 1 , ~ , ¾ are uniquely determined by the element in this module, so if an infinite number of solutions exist they must belong to the module ~ , 2 as L is the only proper subfield of K. Since the element 8 12 + 8 1 8 2 + e; is fixed by a- must be a generator for L over Q. form a basis for L and L ~ M1,2 • so 1 and et+ 9 1 8 2 + 8 22 are in M~ ,2 • Thus, 1 and 2 , but not by er it e? + 91 8 2 + et It follows that Because [ L: Q] divides rank Mi, 2 , these must both be 2 and The system of norm form equations (20) has been reduce d to 40 22) since 83 = -9 1 gives An infinite number of solutions can exist to the equation (22) only if et+ 8 8 + e; and 8 1 2 2 1 are conjugates, which they are not. Again K has the bounded index property. Case 3: Let G(K~:</Q) = (a-, T) where Dihedral Galois Group. a- 4 = -r 2 = 1 and -ra- = a- 3 -r. The field K can be taken to be the fixed taken to be and F 1 ,2 = Q, F 1 , 3 the quadratic subfield of K, so the rank of M 1 , 2 is 3 and the rank of M1 , 3 is 2. Again the only possibility for an infinite number of solutions is in Mi 2 , but L 1 2 is a fourth degree field which is disjoint from K. ' ' Thus ~, 2 will be full in a quad- ratic subfield which is distinct from F 1 , 3 and only a finite number of solutions can exist again by lemma 10. These are the only Galois groups for n = 4 and thus the theorem is proved. 41 CHAPTER 4 THE PROOF OF THEOREM 2 It turns out that lemma 7 is exactly the required tool for proving that the system of equations (20) has only a finite number of solutions. Gyory [ 7] proved a stronger result than lemma 7 by using the results of Baker [ 1] and Baker and Coates [2 J on the existence of a fundamental system of units of a special type. He used his result to prove that there exist only a finite number of inequivalent algebraic integers of a given bounded discriminant [ 6 ], which is equivalent to Theorem 2. The use of Baker's results gives explicit bounds on the heights of the representatives of the equivalency classes and thus a theoretical method of finding them all. This is almost always impossible in practice, but an example of a type of field for which it may be done will be given. Schmidt's Theorem has already been used in the proof of lemma 7. The rest of the proof uses lemma 7 to get the result, following Gyory' s original proof. Proof of Theorem 2: By considering the module (8. - 8. )M . . 1 J which is similar to M . . , equation (20) becomes 1,J 23) NK/ Q o..1 - AJ• ) = C .• 1,J for some ;\ E O and all non-zero constants c . . has only a finite 1,J number of s olutions. It has been shown that the set of values ;\. - A· determine 11. up to equivalence. 1 J 1, J 42 The set of solutions to any one of the equations in (2 3) is described by Schmidt's Theorem, but the weaker result that every number with a given norm is associated with one of a finite set of numbers will be sufficient here. Thus, there exists a finite set of numbers µ 1 , ••• ,µk such that for every pair of distinct i,j and some µ in the set [ µ 1 , ••. , µk}. The group of units UK is the group of units with norm t l in the maximal order of K. Because of the finiteness of the set [~, ... , µk}, the only way there can exist an infinite number of "A. satisfying (2 3) i s for some infinite set of such A to exist satisfying A. - "A_. E µ . . u K 24) 1 J 1, J for each i, j and a fixed number µ . . in the set [ ~, ... , µ.k} inde- 1,J pendent of A• It suffices to prove that only a finite number of A exist satisfying (24). since it follows that 43 By lemma 7, the units e1 e-1 , e 2 e-1 can have only a finite number of values and so for every A satisfying (24) is one of a finite number of possible triples for some unit E. This will be extended to show that for some unit e 2 5) is one of a finite set of numbers for each choice of i < j. Repeating the same technique with "-i, >.. 2 , ''-i for each i f:. 1, 2 gives for some unit 'Tl and with 1 = 3 as before I I I for some finite set of numbers Q' 1 , 2 , cr 1 , 3 , Q' 3 , 1 , a 1 , 2 , cr 2 ,i, ai,i. and I crl,2 It follows that Thus, 44 I I Tl Q' i, 1 = Cl' i, 1 so that e- 1 (\. - 11.1 ) is also contained in a finite set of values with 1 the same unit E. This same trick will work for each \. - A. by 1 J using the triple ()..1 - A.,), · ()... - >i_.), (A. - )..1 ) and gives (25). 1 1 J J It only remains to show that E is restricted to some finite set. However, for any integer \ in K the discriminant of >.., D(>._) satisfies DO,) = Index2 ( A) D(K) with D(K) the discriminant of K, and thus D(,.) = n = En(n-1) a i<j with ex one of a finite set of numbers depending on the numbers given by (25) for each i,j. The unit E is therefore restricted to a certain finite set of values and only a finite number of 11. exist satisfying (24). This proves Theorem 2. ff K is a field such that s + t - 1 = 1, i.e. with unit group generated by one fundamental unit, then every number with a given index can be calc ulated. As an example let g be a primitive thirteenth root of one and Ki 3 = Q(~ ). Then if K is the unique fourth degree subfield of Ki 3 , it is generated by trK 13 ;K (!;) = s + g + !; 3 and this number and its conjugates form an integral basis for the integers of K. 9 45 Since the prime ideal {3) splits completely in K, for every A E £\ the index of >.. in OK is a multiple of 3 [ 10] and thus the re is no power basis for this order. There are integers A with index 3 and all of these will be found. Let 81 = s + s3 + !;9' 8 4 = I; 7 + !; 8 + s11 . gral basis for DK. 82 = !;2 + !;6 + !;6, 83 = s4 + slO + g12, and These are the conjugates of 8 1 and form an inteUsing this basis gives D(K) = 13 3 • This can also be shown from ramification theory sinc e 13 is tamely ramified. Every integer A can be written uniquely in terms of the basis {l, 0i, 8 2 , 8 3 ) as for integers x 0 ,x1 ,x 2 ,X:3 and satisfies D {;U = Index2 { A) D(K) . 26) Suppose Index OJ = 3 and number the conjugates of >... such that 2 7) then {26) becomes It is easy to find every integer in "A with norm equal to 3 or to 13, since {3) splits completely into the four prime ideals generated by the conjugates of 0 1 and (13) ramifies. Thus, every 46 number with norm 3 is an associate of 8 1 , 8~:p 8 3 , or 0 4 and eve r y number with norm 13 is an associate of 81 - 0 3 , a generator of the unique ideal above (13 ). It is also easy to check that the no r m of each 0. - 0. is a multiple of 13, so A· - A· is a multiple of 8 1 - 93 1 J 1 J for every pair i,j and each A. Thus, the norms in (28) must have either or since there is no unit with norm -1. The fact that (13) ramifies from Q(J13) to K can be used t o prove that every unit of K is actually in the field Q(JTI ). the fundamental unit for K can be chosen to be Tl = Thus , Tl where 3 + ,/"IT 2 Pick a particular complex thirteenth root of l for the val ue of ~ which makes -1 + ,/"IT 2 Then an easy calculation shows 2 9) In solving (28) the calculations are simplified by notin g that ±A and all conjugates of A have the same index. Thus, a conjugate 47 might be used in the calculation in place of >.. without any specific mention of the fact. The final set of solutions must be checked to insure that the conjugates of >.. and - A are included. First take the cas e where 11.1 - >.. 2 has norm 13. above, take 11. Then as such that Applying cr 2 to this equality gives which with (2 7) implies Therefore, x 1 - x 2 + ~ = 0 and the two no rm forms are These are both proportional to full modules in Q(jTI). In the second form, dividing out the factor 0 1 - 83 and using (29) gives Thus, this reduces to The sign is required since this norm is only from Q(JTI) to Q. 48 The other norm is easier and reduces to The only common solutions are (x11 ¾) = (± 1, 0), (0, ± 1 ), and the triples (x1 , x 2 , ¼) are given by ±(1, 1, 0) and ±(0, 1, 1 ). The full set of solutions will be determined from these by taking conjugates. Now suppose 11.1 - 11. 2 has norm 3 · 13, then it is eq ual to (8 1 - 83 ) times an associate of 8 1 , 8 2 , 83 , or 8 4 • Therefore, = for i = 1, 2, 3, or 4 and some integer a. Changing A to -11., if necessary, will eliminate the minus sign. It is alw ays possible to take a conjugate to get the form 30) or But 8 2 - 8 4 is an as s oc ia t e of 8 1 - 8 3 , s o ( 3 0 ) is always reachable by these transformations. Assuming 11.1 - 11. 2 satisfies (30) and applying cr 2 to it, it follows that With (2 7) this implies 49 and thus Xi = ¾ = 0. Index A = 6 X2 Index 8 2 • It just happens that 8 2 has index 3, so x 2 = ± 1. Finally, by checking all possible signs and conjugations, it follows that every A with index 3 in this field is equivalent to, or a conjugate of a number equivalent to, either ± 8 1 or ± (8 1 + 8 2 ) . Furthermore, it can be checked that the four conjugates to each of these two sets of numbers give different orders. So there are exactly eight distinct principal orders of index 3 in this field . A similar problem which can be partially answered using Schmidt's Theorem involves the fundamental system of units for certain fi.e lds. Brumer [ 4] considered cyclic extensions of prime degree over Q. Let K be such a field and let E denote the units of K with norm +l. conjugates. Sometimes E is generated by a unit E and its Such a unit is called a Minkowski unit. The cyclo- tomic units of K form a subgroup of E of index h, where h is the class number of the field K [8]. There is always a unit T1 which together with its conjugates generate H. If the Galois group of K over Q is G and is generated by er, then E is a module over the ring Z [G] / ( 1 + er + · • • + erP- 1 ) 50 The multiplication of an element in t he module by an element of the ring is written exponentially, so that for . t es JUga E = E1 , E2 = <TE, ••• , EP = CT p-l E an d th e = Xo t X 1 t <T • •• t in E with con- E . ring e 1ement X p- 2 CT p-.2 the product /:i is given by = (El ) X 0 ( E2 ) X 1 . . . Since this ring is isomorphic to the ring of integers in KP , denoted here by £), E is a module over O also. The multiplication in this case is exactly as described abov e, except with CT replaced by s· The ring O is also isomorphic with H by the isomorphism sending 1 to T1 and this extends uniquely to an isomorphism between E and il-1 , for some integral ideal fil in (). Since the index of H in E is equal to the norm of this ideal, There exists a Minkowski unit if and only if E is a free ()module, which is exactly when the ideal fil is principal. Brumer used this setup to give sufficient conditions both for a field to have such a unit and for a field not to have such a unit. The problem considered here is to find every Minkowski unit which is generated by some fixed finite set of units in E . 51 Since the group H is of index h in E, every unit E of E satis fies Eh E H. Thus, there exists a unique number °' in .O such that 17°' is the multiplication described above. Therefore, every unit E in E can be written uniquely as 17 °' with hQI in o. where Let E1 , . • • , Es be arbitrary units in E with for each 1 s; is; s. If E is in the group generated by these units and is a Minkowski unit, then there exist integers a 1 , • . . , as such that E = 11 a and °' = a1 al + . . . + a s a s is a generator of the ideal ~.i- 1 • NK /Q(Ot) p = This requires 1/h, with the module M = [x1 cx 1 + • • • +xsas} . °' E M The structure of possible solutions can be given using Schmidt's Theorem . For example, let p-1 be twice an odd number, so the quadratic subfield of KP is imaginary, and let m be the smallest non-trivial divisor of (p-1 )/2. There can only exist a finite number of Minkowski units in any group generated by fewer than m E. units of This is because the module M has rank less than m and thus less than the degree of all of the subfields of KP, except the 52 imaginary quadratic subfield. Therefore, M is non-degenerate and Schmidt's Theorem shows the number of solutions must be finite. The most interesting case is when p = 2q + 1 with q a prime, since as many as q-1 units can be allowed and still only a finite number of Minkowski units will be possible. 53 REFERENCES 1. Baker, A. "Contributions to the Theory of Diophantine Equations." Philos. Trans. Roy. Soc. London, ser. A, 263 (1968): 173-208. 2. Baker, A., and Coates, J. "Integer Points on Curves of Genus l." Proc. Cambridge Philos. Soc . 67 (1970): 595-602. 3. Borevich, Z. I., and Shafarevich, I. R. Number Theory . Translated from the Russian. Academic Press, 1966. 4. Brumer, Armand. "On the Group of Units of an Absolutely Cyclic Number Field of Prime Degree " J. Math. Soc. Japan 21 (1969): 357-358. 5. Dade, E. C., and Taussky, 0. "On the Different in Orders in an Algebraic Number Field and Special Units Connected With It. 11 Acta Arithmetica IX ( 1964) : 4 7 -51. a 6. Gyory, K. "Sur les Polynomes Coefficients Entie~s et de Discriminant Donne'." Acta Arithmetica XXIII (1973): 419-426. 7. Gyory, K. 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