Modules with Integral Discriminant Matrix

Citation

Maurer, Donald Eugene (1969) Modules with Integral Discriminant Matrix. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/BQG6-4P65. https://resolver.caltech.edu/CaltechTHESIS:03282017-155524180

Abstract

Let F be a field which admits a Dedekind set of spots (see O'Meara, Introduction to Quadratic Forms) and such that the integers Z _F of F form a principal ideal domain. Let K|F be a separable algebraic extension of F of degree n. If M is a Z _F -module contained in K, and σ ₁ , σ ₂ , ..., σ _n is a Z _F -basis for M, the matrix D(σ) = (trace _K|F (σ _i σ _j )) is called a discriminant matrix. We study modules which have an integral discriminant matrix. When F is the rational field, we are able to obtain necessary and sufficient conditions on det D(σ) in order that M be properly contained in a larger module having an integral discriminant matrix. This is equivalent to determining when the corresponding quadratic form

f = Σ _ij a _ij x _i x _j (a _ij = aa _ji ),

with integral matrix (a _ij ) can be obtained from another such form, with larger determinant, by an integral transformation.

These two main results are then applied to characterize normal algebraic extensions K of the rationals in which Z _K is maximal with respect to having an integral discriminant matrix.

Item Type:

Thesis (Dissertation (Ph.D.))

Subject Keywords:

(Mathematics)

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California Institute of Technology

Division:

Physics, Mathematics and Astronomy

Major Option:

Mathematics

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Public (worldwide access)

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Taussky-Todd, Olga

Thesis Committee:

Unknown, Unknown

Defense Date:

7 April 1969

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Funding Agency	Grant Number
NSF	UNSPECIFIED

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CaltechTHESIS:03282017-155524180

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https://resolver.caltech.edu/CaltechTHESIS:03282017-155524180

DOI:

10.7907/BQG6-4P65

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10114

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CaltechTHESIS

Deposited By:

Benjamin Perez

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29 Mar 2017 14:32

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03 May 2024 20:40

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Modules With Integral Discriminant Matrix Thesis by Donald Eugene Maurer In Partial Fulfillment of the Requirements For the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 91109 1969 (Submitted April 7, 1969) 11 Acknowledgements I wish to thank my advisor, 0. Taussky Todd, for suggesting the problem studied in this thesis, and for her patience and encouragement in its preparation. I am also grateful to Dr. D. Estes and Dr. H . Kisilevsky for helpful conversations. Finally, I would like to thank Professor A . Garsia and Professor R. Dean for their encouragement . The research for this thesis was carried out, in part, while the author was a National Science Foundation Graduate Fellow. 111 Abstract Let F be a field which admits a Dedekind set of spots (see 0' Meara, Introduction to Quadratic Forms) and such that the integers ZF of F form a principal ideal domain. algebraic extension of F in K, and o- 1 , o- 2 , ••• , of degree n. Let KIF If M is a ZF-module contained o-n isa Z'F-basisfor M, (traceK I F(o-io-j)) is called a discriminant matrix. which have an integral discriminant matrix. be a separable the matrix D(Q:)= We study modules When F is the rational field, we are able to obtain necessary and sufficient conditions on det D(a) in order that M be properly contained in a larger module having an integral discriminant matrix. This is equivalent to determining when the corresponding quadratic form f = ~ a lJ. . x.x. l J (a .. =a .. ), lJ Jl i,j with integral matrix (a .. ) can be obtained from another such form, with lJ larger determinant, by an integral transformation. These two main results are then applied to characterize normal algebraic extensions K of the rationals in which respect to having an integral discriminant matrix. ZK is maximal with iv Title Part Page Acknowledgements 11 Abstract 111 I In tr oduc ti on 1 II A Characterization of RA.- matrices 4 III · Almost-Fundamental Modules 11 IV The Discriminant of Almost-Fundamental Modules 30 v Normal Almost-Fundamental Fields 39 References 45 1 I. Let K j F degree n. Introduction be a separable algebraic extension of a field F For each aE K, If al> a 2 , ••• , Dx_(a) will denote the an a the trace of is a basis for over F KjF, will be denoted by and X.EK, nXn F-matrix (SKjF(X.aiaj)). of then Matrices of this [5], Taus sky [11] and type have appeared in the work of Faddeev Bender (2] in connection with representations. In the special case X. = 1, the matrix D(a) = (SK jF(aiaj)) is called a discriminant matrix. Taus sky [13] has studied the characteristic roots of discriminant matrices when K j F is an algebraic number field. The main purpose of this thesis is to study integral discriminant matrices. In Chapter II we generalize, slightly, a theorem of 0. Taussky [11], but the main result is a characterization of the representation matrices Rx. (a) defined by the e qua ti on R (a)· a = X.a A.- (for A.EK}, where g denotes the column vector a= an Specifically, we prove that, if RA.(a) =(SKI F(X.C\aj<)). RA.- matrices. II • • is the dual basis, then This clarifies the connection between DX. and This result appears in [2], independent of [2]. , but the proof given here is 2 In Chapters Ill through V we make additional assumptions about K jF. We assume that F ZF that the integers any admits a Dedekind set of spots [10, p; 42] and of F If M form a principal ideal domain. ZF-module of K with an n-element CT we define the discriminant dK jF(M) to be the ideal in F det D(~). Since K jF tions under which modules M is separable, D(~) contain is integral. ZK is l dK F(M) :f. ( 0). n generated by We study condi- In particular, we assume that our (the integers of K). fundamental if D(CT) is integral and M We call M almost- is not contained in a larger ZF-module which has an integral discriminant matrix (it is easy to see that this is a property which does not depend upon the basis chosen). Let A KjF be the set of all ZF-modules which contain an integral discriminant matrix. ZK and have We construct an ideal which is maxi- mal with respect to belonging to AK j F contains every other ideal in AK jF' and we show that this ideal · We then discuss some special fields in which this ideal is almost-fundam.ental; e.g. quadratic fields, cubic fields and cyclotomic fields. In Chapter IV we obtain necessary and sufficient conditions on dK jF(M) in order that M be almost-fundamental. We adopt a proof similar to the method used by G. Pall (unpublished) to study the dis criminant of a fundamental quadratic form. here, the ground field F In the theorem. we prove is assumed to be either the rationals or a p-adic field. In Chapter V the results of the previous sections are applied to characterize normal extensions mental. K jF for which Z. K is almost-funda- The ground field here is assumed to be either the rationals or 3 a field complete with respect to a rational field, p-adic valuation. When F is the K must be either quadratic or biquadratic. The material prerequisite to reading this thesis can be found in LeVeque [9], Artin [l] and Zariski, Samuel [14]. The treatment of the local theory is based on the concepts and notation of 0 1 Meara [10]; and the properties of the Hilbert symbol and Hasse symbol used in Chapter IV can be found in the appendix of [2], in 0 1 Meara [10] or B. W. Jones [8]. 4 II. Let KjF A Characterization of RA. - matrices be a separable algebraic extension of F Then there is some element aU·}, •.. , e(n} are the conjugates of such that e, (i) we let Ci. (i = 1, 2, If K .= F( 0). the mapping "'(i). i th conjugate ..... each element Ci. into its belong to K, 0 in K of degree e - e( i) N ow, i"f Ci. , 1 1 0=fl( a. 2 , n. ), sends e • • j Ci. n ....' n) denote the colum.n vector (i) Ci.1 ( i) Ci. ( i) Ci. 2 = and we let Ci.(l) = a. The be denoted by From now on we assum.e that Ci. K ~:c: a over F. M( Ci.). Then n Xn matrix whose det M(Ci.):F 0 since is the dual basis (i.e., SKj F(a..a*:" } = . 1 J KIF i th column is Ci.(i) will is a basis for is separable. Also, if olJ.. ), a simpl~ calculation shows that M(a*) M'( ~) = I where A1 is the transpose of the matrix A, and I is the identity matrix. In this chapter we are interested in matrices of the type defined below. 2.1 Definitio n . Let ~ and ~ ~set be two bases for TA.(Q.~) = (SKjF( ACi.ii3j)}, of these matrices, -------- and~ let KjF. For each A.EK !(~'.~_} be the set of all 5 These matrices are F-matrices, and it is clear that can be regarded as a vector space of dimension n TA.(~,~) over F since each can be uniquely expressed as an F-linear combination of the matrices TQ'. , Ta , ... , TQ'. . In the special case where I z n obtain the DA. (g) matrices. We note that where T (Q'.,.§.) J(A.) = diag [A. (i), A. (2), ••• , A. (n)]. ~ =Q'., we This result appears in a paper of 0. Taus sky [11]. The purpose of this chapter is to develop the connection between the DA. -matrices and the RA-matrices, and we begin with the following lemma. 2.2 Lemma. For each = (s lJ.. ) (s lJ.. E F) be the matrix, relative to Q'. *, of the linear transformation of K deter-mined by Ci..* - A.a. (i = 1, 2, ..• , n). Then 1 1 Proof: Let S A. ai = ~ s ik a: ' k and so A. a. a. 1 J By taking traces in the last expression, we obtain sij =SK I F(A.aiaj). Therefore we have When A. = 1, this becomes 6 D(~) Ci. * =Ci. hence, in general, we obtain and the proof is complete. As a consequence of 2.2 we have the next corollary. -1 2.3 Corollary. R>-..'(Q) = Dµ (Q'.) RA.(Q)Dµ(q)for each pair >-..,µEK. Proof: Let A.,µE K, and set p = >-..µ. Then D (Ci.) = R, (ex.) D (Ci.) = R, (Ci.) D (Ci.). p /\.µ µ/\. µ Now take the transpose of both sides to obtain Multiply both sides of the last expression by D~\Q) to complete the proof. 0. Taus sky [11] proved that if F is the rational field every matrix S which satisfies the hypotheses of theorem 2.4 below must be of the form S =DA. (Q), for some A.EK. In 2.5 we prove a converse for the more general setting. 2.4 Theorem. Let · A be an integral matrix with characteristic poly- nomial f(x) zero of f(x). which is irreducible ~the rationals Q. Then~ Q-matrix S satisfies if and only if S = DA.(~) for some. A.EK; for some ideal contained in Z 0 [e], rational integral coefficients. Let 8 be a where ~ is an integral basis the ring of polynomials in e with 7 Proof: In (11] it was shown that Ci. could be chosen to be a characteristic vector of A, and so Aa = 8 · .a. Hence A= R 8 (a). The proof follows from the remark preceding the statement of 2.4, and from 2.3. The converse to 2.4 can be proved in a more general setting: 2.5 Lemma. Let A= (a .. ) be an F-matrix such that lJ -- ---- 1 A' = D- (a) AD (a) µ holds for all µEK . - µ- Then there exists a A.EK such that A= RA. (a). Proof: The hypothesis is equivalent to the condition that AD fJ. (Q'.) be symmetric for each µEK. Now let A. 1 , A. 2 , ••• , A. n be determined by the equation Ci.I Ci.2 A· ~ Ci. n We must show that A. 1 (µ) (C .. ), lJ = A. 2 = = A. n . A I Ci.I = ) Az C'i.z . A. a n n To do. this we set ADµ(a) = where (µ) c lJ.. ~ aiksK \ F(µ~aj) = 8 K \ F(µCi.j ~ aik~) k Now , we are assuming that therefore we obtain C~~-d = C.(_µ) for all i,j and all µEK and lJ Jl 8 SK I F (µa. a . (A.. - A. . ) ) = o. 1 J 1 J Since the trace is non-degenerate, A. . = A.. and the lemma is proved. 1 J We now consider the set T(g_,f}_) . 2.6 Lemma. T(g_,fi) is a field containing the F-scalar matrices if and * only if there is~ yEK Proof: is a such that M(J2) = M(Q' )J(y). (I) Suppose T(a,fl) pEK such that T, T /\. µ Choose any A., µEK, is a field. = T . then there But for any A.EK we have p and so TATµ= M(a) J (A.) M' (J2) M(a) J (µ) M 1 (fl) = M(a) J (p) M' (fl). It follows that M 1 (fl) M(a) = J(y), where y = p/A.µ. Since >l< M(a ) M(a) = I ' we obtain >l: M(J2) = M(a') J(y). (II) Suppose that the last equation holds for some yE K; M 1 (fl) M(~) = J(y). Hence T(~, fl) Let A., µEK , then is closed under multiplication, and T, T l\. over , each TA has an inverse in T(~,fl). then then µ = T T,. fJ· /\. For we let µ = l/yA., Moreand 9 1 :::: M(Ct')J(y- )M'(§J -1 1 = M(Q'.)M- (a)[M'(.\2)] M1 (m -- I n· In view of the remark following 2.1, the proof is complete The next theorem is the principal result of this chapter . 2. 7 Theorem . basis. Let Q'. be ~ basis for KIF J1 is any basis of KIF , If then and let T(a,J1) coincides with the set of all RA. (Ct') matrices if and only if there is ~ yE K Moreover, Proof: a* be the dual such that fi =ya*. Rf...(Ct') = (SK IF(f...Ct'iCt'/)). In view of 2.6 it is sufficient to show that if f(g,.J1) is a field then it coincides with the set of RA. (Ct') matrices; and because of 2.5 it is sufficient to show that for any µEK, for then T(Q'.,f}_) is contained in the set of RA. (Q'.) matrices, and since T(Q'., f}_) is a vector space of dimension n over F, coincide . We therefore assume that the two sets must T(a,f}_) is a field; and then * M(f}_) = M(Ct' ) J(y) for some y EK. We have D~ 1 (q_) TA. (a,f}_) Dµ (Q'.) = [M' (a)J-l [J(µ) J- 1 [M(Q')J-1 M(Q'.)J(A.)M (£2)M(Q'.)J(µ)M' (Q'.)= 1 [M'(a)JOn the other hand, 1 J(yf...)M'(Q'.). 10 T~(Q,.£2.) = M(Q)J(A)M'(9') = [M' (Q)J- l J(yA) M' (Q). This establishes the first part of the theorem. * To prove the last assertion of 2. 7, we take {i.=Q. * TA (Q,Q) D(a) Then = M(a) J(A) M' (a) * M{Q) M' (a) = M(q_) J(A) M' (Q) >'< By 2.2 we obtain TA (Q,Q') = RA (a). According to this theorem both the DA -matrices and the RA matrices can be obtained from the TA -matrices by taking .§. = Ci*, respectively. fl = a and We can prove the following rather interestiJ?-g corollary. 2.8 Corollary l . Let dual basis. l Then a be a field basis for and let Ci * be the * = 1. Ci.Ci. l Proof: For each AE K, l we have Hence we see that SK IF(A) = ~ SK IF(aid~A) = SK IF(A laia). i Therefore for all AE K. 1 This '\"' * SK[F(/.-(1- Li aiai)) = 0 The corollary follows since the trace is non- degenerate, ::l< follows directly from the equation M(a ) M' (Ci} = I. ll III. 1. Almost-Fundamental Modules Introduction and Definitions. the separability of K \ F, p. 42], K. and we let ~ We shall now assume, in addition to that F admits a Dedekind set of spots [10, denote the set of all extensions of spots on F We also assume that the ring of integers ZF of F to is a principal ideal domain, in which case ZK has an n-element ZF-basis [14,p.265]. The discriminant of the field K \ F is the discriminant of ZK; ~\F = dK\F(Li{). write, for short, and we The following lemma will be used repeatedly: 3.1 Lemma. If~ is any ideal in K \ F, then Proof: [l, p. 133]. In the remaining chapters we will study ZF-modules which have an integral discriminant matrix; in particular we will be interested in modules which are maximal with respect to this property. a module with basis ![_, then with M Let M we cari associate a quadratic form, which may be written as follows: f(x) = ~ a lJ.. x.x., 1 J a .. = a .. , lJ Jl i,j where aij = SK\ F(O""io}· The discriminant matrix is integral if and only if each a.. is integral. lJ be G. Pall has called a quadratic form 12 fundamental if its coefficients are integral and it cannot be obtained from another integral form of smaller determinant by an integral trans formation. Our condition that the requiring f(x) a.. lJ be integral is stronger than to have integral coefficients; however, in any field for which 2 is a unit the two conditions are identical. We are thus led to make the following definition, 3.2 Definition. Jr. Let M be~ Z F-module with~ n-element ZF-basis almost fundamental if D(cr) is integral, and M We call M properly contained in any Z F-module with this property, We will generally assume that M define the set contains ZK and so we A K [ F below. 3. 3 Definition. Let AK IF be the set of all z F-modules M which satisfy the two conditions: (a) M =:i (b) If is not ZK a, f3 € M, 3.4 Lemma . Let then SK [ F(a,B) € Z. ~K[F be the different of K\F. - l Proof: Since ~KIF is defined by the first part of the lemma is obvious. Then 13 >~ Now let w be an integral basis for is a ,ZK-basis for integral. then the dual basis w ,ZK, -1 ~KjF· Now, * D(~ ) * :i:~ = M(w ) M'(~ ) = [M'(~)] -1 [M(~)] -1 = D Therefore ~~IFEAKIF - 1 (w). if and only if dKIF = zF. Since ,ZF is a principal ideal domai~, and ZK has an n-element basis, it follows that every module in AK jF has an n-element ,ZF-basis (and so condition (b) in 3. 3 is equivalent to requiring that integral discriminant matrix). M have an Suppose now that M l_ c M 2. C · · · is a chain of elements from AKjF' M* =l l and let mi j m.EM. i d l finite >'< 0 for (a) is obvio_us, and if .Q'., f3 E M for which a, f3E Mk ; hence that flKjF (b} is satisfied. there is some k By Zorn's lemma, we see contains almost-fundamental modules. In the remainder of Chapter III we study modules belonging to AKIF' 2. The Lar g est Ideal in AKlF" W e would like to say something about the almost-fundamental modules in AK IF . This seems to be difficult, but we can say something about the ideal in AKI F . In this section, therefore, we study the ideal which are maximal with respect to the 14 property of belong ing to AKjF 3. 5 Lemma2, M Let MEAK j F . If M is an ideal then M 2 C ~~ 1 F. 1 If 1 is almost-fundamental it i!. ~ideal if and only Ji M 2 C ~~ jF· Proof: (I) Suppose is an i deal, and let a, f3 EM. M Then f3M S M, -1 ~KjF · and so (II) Suppose Let aE M M is almost-fundamental, and suppose and consider the ZF-module M 1 ~ ZK. and ay+ m 2 Let ax+ m 1 two elements of M 11 aZK + M = M 1 • 2. -1 M C~KjF· Clearly (x,yELK; m 1,m 2 EM) be any then (ax+ m 1 ) (ay+ m 2 ) = a 2 xy + am 1y+ m 1 m 2 + am 2 x and So M 1 EAK IF . implies that 2a. is almost-fundamental and hence M 1 = M, which But M aZK C M. The Local Case . Therefore M is an ideal. We now consider the case where F is complete with respect to a single spot. 3. 6 Theorem. Let KjF ~~extension of the complete field F and 13 denote the prime ideal in K ; also, let ~KIF =J3°. Then the 2 ideal M~ ='fJ -[o/ ] belongs to AK IF. Here [x] denotes the g reates t let 2 The main theorems of this chapter can be proved directly from lemma 3. 5. However, we wish als·o to develop the connection between the local and global theory; therefore no attempt has been made here to give the shortest possible proofs. 15 inte ge r in x. Proof: /3, and every ideal of K is Since there is only one prime ideal a power of the prime ideal, it is sufficient to show that the largest ideal in Let rI .AKIF . be a prime element and let n = ef, ramification index and f Wp w2 , • •• j wf in 1j3 =p -[o/2.] is is the inertial index. where e is the Then there are units K such that ZK =TT ZF willj, l~i~f l~j~e [l, p. 84]. Now clearly Mp ~ ZK. So suppose and write so that for e MfSEJiKjF. Suppose J3s from 3. 5, 2s ~ o, is the largest ideal in and so s = [o/2]. The proof is complete, AK jF then s ~ [ o/2]. But 16 3.7 Corollary. The discriminant of the ideal fM ) = (N d KIF'?> M}3 of 3.6 is (~))(6-2[6/2]) KIFj-J Proof: The discriminant of KIF is 6 (NK IF~)) . . The corollary follows by applying 2.1. Now suppose that T contained in K, Proof: so that f = [T:F] If M 3.8 Lemma. is the largest unramified extension of F is a LT-module contained in AK IT' Clearly M is a LT-module containing ZK, sufficient to verify that (b) of 3.3 is satisfied. Proof: We have M :=:> ZK :=:> 7 T . a and e = [K:T]. 7 T-module, and so it is If Ci, !3E M Now suppose Ci, [3E M. [3LT CM, and so SK I F(Ci.[3,ZT) C SKIF(Ci.[3,l'T) = STIF(SKIT(Ci[3LT)) then ME~ IF" ZF. then Since M But = ST I F(SK I T(Ci [3) LT) Therefore, since TI F ZT. Hence M is unramified, satisfies (a) and (b) of 3 .3. We now need the concept of the complementary module 3.10 Definition. If M is a ZF-module, the complementary module ((M) is defined to be the following set: is 17 are true has ari n-element basis (a) * is a basis for then ..Q:., <J ((M1). then ((M 1 ) ~ ((M 2 ) (b) If M 1 C M 2 ( c) ( ( ( M1 ) = M1 (d) If M 1 is an ideal, then~ is ((M 1 ) (e) If M 1 is~ ideal, then ((M1 )M1 = ~~ \ F Proof: See [14] • 3.12. Lemma. Let M if and only if ((M) ~ M. Proof: If ME.AK\F' mE ((M). be a 7 F-module containing ZK. then for each mEM, SK\F(mM) C Then MEi{K\F ZF and so The converse is obvious. Note that 3.10 through 3.12. hold for the general extension K \ F. We can now prove the following theorem. 3 .13 'J3 Theorem. Let K \ F be the prime ideal. contained in K; exponent of "(3 the largest If and let be an extension of the complete field F, Let T be the largest unramified extension of F !J 6 , (6 is called the differential ~K \ F = [14, p. 2.98] ), with M'fi ZT-module contained in 2. and the ideal of 2..6. !IK \F . di vi des the differential exponent of almost-fundamental. Also if K \ F Then M/j '/!>, then Mjj is is unramified or fully ramified, is 18 is almost-fundamental. Proof: M'f3 Suppose that is properly contained in a We show that this leads to a contradiction. belong to i4K IT. ZT-moduie N. By 3.9 both M,B and N Also KIT is fully ramified, and so NK I T('{J) =~ , the prime ideal in T. Now, because of (e) in 2.11, we see that M,E ( If N ~ Mf-i , (M)3) = // * If 2 di vi des theorem is proved. and by 3.12 we obtain * * !...J Since this contradiction can also ZF-module, the second assertion of the Now suppose 2 does not divide must differ by at least a fourth power of.4 d otherwise. (N) =i N =:i Mn 6 this is not possible. be obtained if N is only a Io p - [6/2] -1 then ((N) ~ ((M,!3 ), ((Ml)) =i ( if 2 6. Since We use 3.1 to obtain (M)= KIT /3 ~ -2(6/2].A/o=b ~~ ~~ ' and d (f(M·))-=,.L/-2(6/2] .- 2 o_ -1 KIT \..., --;,~ · ~ -=~ . 73 Comparing the two discriminants, we obtain a contradiction. Therefore Mf-? = N. To complete the proof we notice that if KIF is unramified, then 19 77 MJ-3 = LK , -1 ~KIF = and so fully ramified, then T = F M/3 is almost-fundamental. If K I F is and again MJ3 is almost-fundamental. The proof is complete. 2b, The General Case. "/3 E ~ let KJ-3 We return to the general case. be the completion of K completion of F K,E in at For each fJ , and let %t' denote the so that ~ I~ . K F ~ F Let .7:r3 and Z~ We will also let /3 denote the integers of K/J _and ~ denote the prime ideal in the prime ideal in K determined by ~ . respectively. KJ3 , and set '(3 =J3n ZK' Now if_,/()Z,, is any ideal in K, then it can be factored uniquely into a product of prime ideals = where n or13 /(Jl is the exponent of the highest power of fj ~ divide~. Finally, if U is any ideal in which ~ (ord~U) KTJ, we let U ="[3 f-l . The method of this section is to consider KIF at each of its completions K,R I~ and apply the results of 2a to K,13 I~ Hence we must now establish the connection between the general case and the local case. 20 If N is a 3. 14 Definition. (Then N~ is a Nf3 = N ~ . -Z 'F-module rn K, we let Z -module in K}3 . ) We will need the following lemma. 3. 15 Lemma. Proof: Let Z13 = ZK z~ . z be the quotient .ring !J7/3 = [a /b a , b E ZK ; b ~ '(3 } , J and let be the quotient ring 7-_-7 = [a/bja,bEF;b~~}. The inclusion relations are diagrammed below: Now, it is known ([3], P· 13) that Z13 = 7 K z~ and we see that 7 -73 Since that 7 -- -7 7 K 7 c -;Jt -- 7 - K -_~/ .r Zn is the closure in Kn of ,7._ , it will be sufficient to show !-' K £~ is closed. Let w be an integral basis for K ; then -73 e ZK 7 -;:L Consider Z"f3 Z = ~i w. ;z-;Y- . 1 Then as a topological g;roup under.addition. closed, compact subgroup of 7 -p_ , and so is w. Z. {./ 1 ~ . 7 -:-:;::: is a Therefore, 21 £.;fa'.- is closed ([7], p. 48). ?K 3. 16 Lemma. LetA be an ideal in K. '(3 E dSJ , ~"(3 Then for each 1<;3 ; and is an ideal in __,,{)-{_ = rr ---07_/3 J3 Ec1) Proof: It is clear that~ is an ideal since ---0-Cf3 L 'f3 =/()7.. !!_j£ L K Z-1L £ __A>L/3 In order to complete the proof, we must show that__,A?Z'/3 fJ ( o rdJ3 ,,,{)[_} __Arl."(3 . Now c ~ ( ord;Q .4 7.) /.J = [x E K{J I orc;a x ~ ord'(5 --0-C} ; but__AJ[~ contains an element a. with ordE a. = ord%> /.J ( orc;3 ,,oz) /:J an ideal, so we must have_,A)z',8 = '(3 • . All, andA12 Lemma. Let~ is e We now study the connection between AK\ F and 3. 17 = /IK "{J IF ~ • be an ideal in K , and suppose that/O'ZEAKjF• Then ,,(J? -;/~~ Proof: First, Now let a,, f3 E/0?-;J ; we wish to show that (write SKE IFcY- ) sfJ lj( (a., (3) E z.A/ . sum of terms of the form uv cient to assume Then a. = a 1 'Y , Since any element (u EA, Sf3 j.y for o:yir0 is a finite v E .7 -jt. ) , it is clearly suffi- j3 = 13 1 6 , where a 1 , 13 1 SAY[ and 'Y• o E £_# . 22 We will be finished if we can show that be the principal ideal generated by a /3 sf; I~ (a1/31) E z--;1£ . 1 1 Let U in K, and let rrD'z,,. U= J3' Ed\! p It is known ([6], p. 430) that SK\F(U) is integral if and only if SfJ '\~ 1 (Up ) 1 gral, and so is integral for all "(3' EJ9. But SK\F(a. 1 13 1 ) is inte- S{J \j'.' (a. 1 13 1 ) must be integral also. This completes the proof of the lemma. We can now prove one of the main theorems of this chapter. 3. 18 Theorem. in 3. 6 . J3 E d9 , let ME For each _T_h_e_s_e_t_o_f_i_d_e_a_l_s_i_nII KIF has a unique maximal element rr ~ = M f3E J9 M(3 From 3. 4 and the fact that F has only finitely many ramified Proof: primes, we see that ~ = 7?-i for almost all . Moreover, since M that M ~ L K. Now :;;. 2 /i E J9 . Hence, Mis f3 :::)- 73 for all f3 E J9 , we see an ideal in K. and M be the ideal determined -1 ~ :;;..'/3\~ K\F (by3.5). 77 = So M 2c -1 :;;.K \ F , and, again by 3. 5, M E.A'K\F • Suppose thay07 is an ideal in AK IF. is an ideal, and by 3. 1 ~7.2 ~ M/j complete. e because of 3. 6. Then, by 3. 16 , at each EA K fJ IF ; hence, -'Y- Therefore,pz c M , and the proof is We have, in fact, proved that M contains every ideal be- longing to II K \ F • 23 The previous theorem generalizes 3. 6 , and the next theorem is a generalization of 3. 7. But before stating the theorem, we '/3 E J9 let f( '(3 l-:1- ) be the inertial introduce some notation. For each index of ~ IF~ , let e ( f3 I~) be the ramification index of K73 IF-;$( , and let 6((3 1--;;t) be the differential e~ponent of J3 (see 3. 13). Let µ(,B I~) be defined by µ(/9 3. 19 Theorem. 21 0 ( 73 1-;t ) 0 if 1 otherwise. I j() = Let M be the ideal determined in 3. 18. Then where Proof: By 3. 1 we have and NKIF(M) = rr NKIF( M7)) = ~~TTEJ9 NK~ IF_.,,_,/- (MJQ) f-J 73 E J9 _ rr ~ IdK I F /J TT N K F ( M"/3 ) . E 1-;t- 73 1 ~ The last expression is obtained since Also, ([6], p. 429). Hence, we obtain M/3 = J3 if ·jL f dKIF • 24 dKIF(M) = 1T N~ I Fjt (~ ) • dK IF 13 -;£ ~idKIF 7311L 13 rr rr = rr = ~ Tf 11- f (13 ·~d µ (,73 I ~ ) 1 ·7£ jdKI F ,731 ;1 d-;z idKIF by applying 3. 1 and 3. 7 • We will use this theorem in the next section. 3. Special Fields. We shall consider the following question: when is the ideal of 3. 18 almost-fundamental? As we show in chapter IV, it is not always almost-fundamental; but it is in the special fields we investigate in this section. In theorem 3. 13. , the ideal M;B was seen to be almost-fundamental if the differential exponent is divisible by 2 , or if K"/3 IF~ is either unramified or totally ramified. question of whether or not ing case is still open. ME The is almost-fundamental in the remain- In this section, we shall investigate quadratic, cubic, and cyclotomic number :fields over the rationals, and show that in each case the ideal of 3. 18 is almost-fundamental. The problem of determining for what other number fields the ideal is almost-fundamental is still open. 3a. 3. 20 Quadratic Fields Theorem . rational field Q . Let K = Q(JD) be a quadratic extension of the Then there is exactly one almost-fundamental module, call it M , containing .ZK ; and (a) If D = 1 (mod 4) , then M = Z. K • 25 (b) If D = 2 (mod 4) , then M = 7 QEB ~ ZQ . (c) If D = 3 (mod 4) , then M -- z Q EB 1 +JD 2 In each case M is an ideal. If D Proof: = 1 (mod 4), then dKIQ = D is square-free, and so ZK is the only almost-fundamental module in /iKIQ. We assume, there- fore, that D ¥ 1 (mod 4). /31 p. Let p be an odd prime and suppose p II dK IQ or p ~ dK J Q , it is clear that K)3 J Then, since Qp is unramified and so MJ3 = ~ • Now we consider the cas.e p = 2 , and suppose 2 2r II dKIO. The prime 2 ramifies, and (2) = f3 ; the local extension is fully ramified, and so the differential exponent 2 or 3 (also note that where J Ip o = r ). o is either Thus, is a valuation determined by J3 . Now, in view of 3. 18, it is clear that M = [ x ~ K I Ix Ip ~ 1 if p i- 2 ; Ix I2 ~ Ji. } . In particular, M =f. Z'K , and it follows from this that M is almost- fundamental, since dK IQ (M) is obtained from dK J Q by dividing out the square factor. We now determine an integral basis for M. polynomial of JD is f(x) = x 4 -D , and The defining 26 ~ ~lJ Q has 1 I 2 , JD/ 2D as an integral basis. Therefore , 3. 4, Mc Now, by -1 ~ K I Q ; s 0 if a. E M > b' = 2a' + 2D JD ; a'' b ' E l' Q • a. Now M is an ideal and M ::::> ZK, so 2a.-a' = ~ Jf5 E M. Suppose that p (p-:/. 2) is a prime divisor of D; then '~JD' p = Jb'lpJr)' and this is larger than 1 unless p divides b. Therefore, every odd prime divisor of D must divide b. =2 (mod 4). Then, in view of the last remark, Case I: Suppose D we can write a. in the form o:. Now, = ~ + ~ JD , with a, b E Z. Q • !~I = 2 Ia j 2 and 2 2 j b and 21 a • j ! JD 2 j = 2,j2 Jb j 2. Thus, we must have So but every element of the module on the right side of this expression belongs to M, and so we have equality. Case II; Suppose D o:. Now, j ~I otherwise 2 =3 (mod 4). Then we can write a. in the form = a + bJ.5 ; a, b E 2 = 2jaj 2 and I ~JD j = 2jbl 2 Io:. l 2 > J2 . Therefore, M c- 77 (1'\ l+./D 2 /L. Q w 1!. Q • 2 ; hence, a= b (mod 2), for z Q . 27 In order to establish equality, it is sufficient to show that (1 +Jfj)/2 ...1 2, then If p -t belongs to M. l+Jfj 2 11 +Jf51 "". . . 1. 2 p Suppose p = 2 ·; then = ., 2 =3 (mod 4). Therefore, since D -M - 77 ffi l+Jfj £.Qw 2 The analysis used above can be applied to any AK\ Q to show that it must be contained in M. 3b. Cubic Fields 3. 21 Theorem. z 0-module in The proof is complete. Let K be a cubic extension of the rationals Q. Then the ideal, M, determined by 3. 18 is almost-fundamental. Proof: By 3. 19 we have 1 r.-:'/ I d fJ \P. p where ~ f (13 IP )µ (,8 IP ) d = pJ3\p p We will show that dK\Q(M) is square-free. Let p \ dK\Q, and suppose p = is the factorization of p in terms in the product. rr P.~ 'f3\p p 7 K, and suppose r is the number of Now, at least one /3 > 1 and so r ~ 2. Hence, we have the following possibilities: 3 These results, in the case of a quadratic field, can also be obtained without p-adic analysis. 28 or 3 ~ = ~ 1 p Suppose p = ~ ~ f(/3 2 -Q 3 Suppose p = p =p p Jp) • e(J3 ponent of d er(3 2 Jp) = 1; hence, the exponent of \p) = l, for otherwise d 1 =p p dp is at most 1. But f(',B 2 1J 1 '(3 2 . Since e(/j 1 Jp) = l, µ(,8 1 \p) = 0, and f ( 73 2 Ip )µ (73 2 Ip ) d Now, p • 1 f( . 1 Then /J 1 IP )µ V3 1 IP ) Jp) = 3, and so f( '(3 \p) = 1; therefore, the ex1 is again at most 1. The proof is complete, since this is true for every prime divisor of dK 3c. Cyclotomic Fields 3. 22 Theorem. IQ . Let K be a c y clotomic extension of Q of de g ree p-1 Then the ideal, M , determined by 3. 18 (w here p is an odd prime). is principal and almost-fundamental . Proof: The field discriminant is given by p-1 d I = (-1)_2_. p p-2 • K Q Hence, if q is a prime and q =f. p, then for anyfl"j E ~ such that divides q, M-"'! = Z.-4"/ • p Now suppose = rr ~ ,73 Ip is me rnc-i;onzat.wn u.i. p in K. It follows ([ 14 ] , P· 302) that e(fj \p) Then e(J.3 IP) S: p-1, sop o(/3 Jp) = e('(J IP) - 1. f e{ '(J \p). Hence, we obtain 29 p-2 = = = 6 f(/3 Jp)(e(,8 Jp)-1) 6 f(/3 Jp)e(J3 Jp)- /51P 731P (u-1)- ~ 6 'j3 IP f(J3 jp) . 6f('i.Jlp) , J3Jp f-> since p-1 = 6 f('/J Jp)e(,73 IP). 731P ,() .y)p-l So f(p Jp)= 1, and p= p • Therefore, We then see that d dKJQ(M) is square-free. Moreover, C is a primitive p th root of unity and h = (p-3 )/ 2. since "(J = ( 1-C ), and p =p and d M = (1-C) -h q , where This follows o('13 Ip) = p ~ 2 , s o [ oI 2 J = (p - 3 ) I 2 . = 1. 30 IV. The Discriminant of Almost-Fundamental Modules Throughout this chapter we shall assume that F is either the If a, b E F, then (a, b) rational field Q or a p-adic field. p will de- note the Hilbert symbol of a and b with respect to p; and if A is an C (A) will denote the Hasse-symbol of A. p F-matrix, The Hilbert symbol, the Hasse symbol, and their properties are presented in [8]; we assume these properties . We shall also make use of the following technical lemma. .!£A= diag[a , a , ••• , an], then 1 2 Lemma. = T( (a., a.) C (A) p Proof: "5:" l J l J p . We will need this only for odd primes p, and so we assume that p is odd. We have rr C (A)= (-1,A) (A.,-A .+l), n p i<n i i p P where A . is the determinant of the principal i-rowed minor. l is diagonal, A. l = so (A., -A .+l) l l p T( a • ai+l)p rr ak' k5:i k = k 5: i = ( n -n ak, k 5: i = Also, k5: l" a ) ( r r ak' ai+l)p k -p k 5:.l rr (ak, ai+ 1 )p . (k5:i rr ak' ai + 1 )p = k5:i When A 31 rr a k )P (-1, - = k~n n' (ak' -1)P = k~n Hence, C (A) p rr (ak' a.)p rr (a.,J a.)J p rr = l l<i~n rr = rr k~i i~n j~n k<i (ak' a.) l p = rr j~i (a., a.) J l p . In order to prove the main theorem of this chapter, we apply a method used by G. Pall to prove a similar result in case F = Q. Although Pall 1s proof is as yet unpublished, the result appears in [ 4], and the technique was sketched for me by D. Estes. 4. 1 Theorem. Suppose that M is a Z F-module with an n-element ? F-basis a , and suppose that D(~:) is integral. Then M is almost- fundamental if and only if the following conditions are satisfied: (a) li_ p is an odd prime, p 3 ~ dK j F(M} ; and if p 2 j dK j F(M), then C (D(cr)) =· - 1 • -- p (b) If 2.t jj dKjF(M), then Proof: O~t~ 1. (I) First define an 11 inner product" (· , • ) on K in the follow- ing way. For each x, y E K , let We see that M is not almost-fundamental if and only if the following condition holds: there is an m EK, m $ M such that (m, M) S and (m, m) E :,Z F. ,., For then the module M' ZF = M + m Z F properly contains M and the trace of the product of any two elements of M,:, is integral. Since M * ::i M, the module M,,~ has an n-element Z. F-basis and so must have an integral discriminant matrix. We wish to ex- 32 press this condition in terms of the matrix D(a) and therefore we will express the inner product (x, y) as follows. Then let For each x E K write x denote the column vector = x" x n Since D(a) = (SKJ F(C\O"j)) , an easy calculation shows that (x,y) = x 1 D(~)y. ( 4. 2) Now suppose that there is an element m EK , m ~ M and such that (m, M) c Z F and (m, m) E Z. F. m = Write + x n O" n d where the x. and d belong to 1 and d does not divide every x . • 1 Let p be a prime divisor of d and let d = pd 1 • Then m 1 $ M but x = pm 1 ~ M. Now set m 1 = d 1m. The following conditions are then fulfilled: x =/. 0 (mod p) (that is, not every entry is divisible by p) (ii) (x, M) = 0 (mod p) 2 (iii) (x, x) = 0 (mod p ) . (i) Conversely, suppose we can find an x E M which satisfies (i), (ii), and (iii) for some prime p. (m, M) c Then let m = x/p. .Z F and (m, m) E 7 F. Clearly, m ~ M but Therefore, we see that M is al- most fundamental if and only if conditions (i), (ii), and (iii) cannot be 33 satisfied for any prime p and any x E M. By (4. 2) conditions (i), (ii), and (iii) can be written as x ~ 0 (mod p) ( 4. 3) x'D(a) = 0 x'D(cr)x = 0 (mod p 2 ) . (mod p) Hence, we have shown that M is almost-fundamental if and only if (4.3) does not hold for any prime p and any integral vector x . Note that (4. 3) has a rational integral solution x if and only if it has a p-adic integral solution y, for we can write y = x + pa. z where x is a ra- tional integral vector and z is a p-adic integral vector. If we choose a.~ 2, then x must satisfy ( 4. 3 ). Therefore, M is almost-funda- mental if and only if there is no prime p such that ( 4. 3) has a p-adic integral solution x. It is this last formulation that we will use throughout the remainder of the proof. (II) Suppose M is almost-fundamental. and (b) are necessary. We now show that (a) F irst, suppose p is an odd prime . Then there is an integral (p-adic integral) unimodular matrix U such that U'D(cr )U is diagonal ([8], p. 84). Now (4. 3) has an integral solution if and only if it does with D(cr) replaced by U' D(2:)U. And the Hasse symbol remains invariant under this transformation. Hence, we can assume D(cr) is diagonal. If p 3 divides detD(2:) , then D(cr) is one of the following: 0 al a2 a2 0 01 al 3 p a or n 0 pan-1 2 p a n 34 0 pa 0 where, in the last case, a n- 2 , a n , and a n- 1 n are not divisible by p. In the first two cases it is evident that 0 = x" 0 1 will satisfy 4. 3 • In the third case, there will be an x satisfying 4. 3 if the congruence a n- 2 x 2 + a n- 1y 2 + a n z 2 :: has a non-trivial solution. and since (-a , a n trivial solution. n- 2 , -a a ) n n- 1 p ~I 2 (mod p) But this congruence can be rewritten as Therefore, p Assume that p 0 = 1 , the congruence does have a non- 3 { det D(cr) . II <let D(cr). Then D(cr) is one of the following: 0 2 p a , or n pa where no a. is divisible by p. l must be of the second form. By the argument used above, D(cr) In the second case, fundamental if the congruence a n- 1 x 2 n + a ny 2 - 0 (mod p) M is not almost- 35 has a non-trivial solution. There are non-trivial solutions if and only -a a if ( n;l n) = 1 , where (p) is the Legendre symbol. We can express this condition in terms of C (D(a)) as follows. p - Since D(a) is diagonal, we apply the lemma at the beginning of the chapter to obtain C (D(a)) P - = 11 (a.,pa 1 i:5: n - 2 ) · n- 1 P 11 ) (pa ,pa) (pa (a .,pa) (pa ,pa ,pa) i:5:n- 2 . 1 n p n- 1 n- 1 p n n p n- 1 n p So it is necessary that C (D(a)) = -1. p Now suppose that p = 2. We cannot assume that D(a) is diagonal, but we may take ( [ 8], pp. 84, 85) 0 A = D(a) 0 where A, B , B , . . . 1 2 are integral block matrices. The matrix A is diagonal, and the B. are 2X 2 blocks of one of the following types: 1 t. 2 1 t. -1 2 1 0 t. -1 2 1 t. -1 2 1 0 or t. -1 2 1 If 2 2 t. 2 1 divides det A, then A can be written in one of the following forms: 0 0 al a2 . 2ak-l or ak-1 0 2 2 ak 0 2~ 36 where in the second case a and a are not divisible by 2. In the . t h e kth position . . ) is . a so1ution . (w h ere th e 1 occurs in of 4 • 3 . In the n-1 n first case it is clear that 0 0 x" ::: 1 0 second case, we can find a solution ak_ 1x 2 + aky 2 = 0 x if the congruence (mod 2) has a non-trivial solution. Since ak- l and ak are both odd, it does have a non-trivial solution. We therefore reach a contradiction un- less at most one entry in A is divisible by 2. If, for any block B. , t. > 1 , then t. 2 t. t. 2 2 ix + 2 ixy + 2 iy = 0 (mod 4) i i or t. i 2 xy - 0 (mod 4) has a non-trivial solution, and hence 4. 3 does also. each t. l = 1, and j det B . j = 1 . i Hence, if zt jJ det D(a) , then (III) Now assume that (a) and (b) are satisfied. that there is no prime p and no integral Again, if p = 2 we may take A Therefore, We must show x which satisfies 4. 3. 37 Then each of the blocks B. must be such that t. = 1 ; hence, if 1 1 t > 0, one of the entries in A must be divisible by 2, and A must have the form 0 • t 2 a 0 n = 1 (modZ), and O:S:t:S: 1. For A, B 1 , B 2 , ... satisfying these conditions, it is not possible to find an x which will satisfy with each ai 4. 3 . .If p is odd, and p r det D(a) , then D(cr) has the form a where each a . =/. 0 (mod p). 1 n Clearly, there is no x which satisfies II <let D(~) a similar argument suffices to show that there 4.3 . If p is no x satisfying 4. 3 • Finally, suppose p 2 II <let D(~). Then C (D(a)) = -1 , and D(CY) can be put into one of the fallowing forms: . p al az pan-1 l al az or 2 p a n panJ where each a . =/. 0 (mod p). 1 By computing the Hasse-symbol for each of these forms, we see that D(a) must be of the first type. there is an x satisfying 4. 3 only if 2 2 a x + a y = 0 (mod p ) n- 1 n Now 38 has a non-trivial solution; but in view of the condition C (D(cr )) = -1, p this is not possible. We have proved 4. 1 • 39 V. Normal Almost-Fundamental Fi e lds Suppose that K J F is an extension of F which satisfies the conditions imposed in Chapter Ill. 5. 1 Definition. The field K J F is said to be almost-fundamental if 7 K is almost-fundamental. 5. 2 Theorem. then K is an extension of the complete field F is almost-fundamental if and only if the differential exponent satisfies Proof; Immediate from 3. 6. In the remainder of this chapter we assume that F = Q, the rational field, and we study normal almost-fundamental extensions. The results of section 3a are restated in the next theorem. 5. 3 Theorem. and only if K If KI Q is quadratic, then K is almost-fundamental if. = Q(JD), where D = 1 (mod 4). We now prove the following theorem. 5. 4 Theorem. If Kj Q is normal and almost-fundamental, then Kj Q is quadratic or non-cyclic of degree four. Proof: For any almost-fundamental extension K j Q we have, in view of 3. 19 , for each prime divisor p of the discriminant dKj Q. the definition (p. 33) of µ(J3 jp), it is clear that Now in view of 40 µ('(3 jp) s: o(J3 jp) Hence, o = I: f('j3 jp)[o(,E jp) - µ(}3 jp)J ,B Ir implies that µ((3 Ir) = o([3 jp) . Now ( [ 14], p. 302) 6(/J jp) ~ e(/J jp) - 1 ~ 0 with equality on the left if and only if p does not divide e ( see that if o(p j p) is even If 0((3 jp) is odd then -;J j p). We it must be zero, and then e ("/3 j p) = 1 • o(/!> jp) = 1 ' and e(p jp) = 2. In this case p :/- 2. Now assume that Kj Q is a normal extension. pair Then for every p , Arj of divisions of p, e C'/3 jp) = e (,ffi-I jp) and f ( IB jp) =. f(-.-Oj'jp). It follows that e(;E jp) = 2 for every ,E jp. But if e and f are the common values of e ('(!; j p) and f("/3 j p), respectively, and g is the number of divisors of p, then n = = efg , [K:Q] and so 2 must divide [K:Q] . Now by 3. 19 and 4. 1 we must have I: f(/3 jp)6(/3 jp) 'fJ Ir = I: f(/3 jp) s: 2. f3 Ir So the possible factorizations of p in Z. K are the following: (a') p f(-rJ jp) = 2 or 1 (b I) p f(,23ijp) = 1. Since n is even, we see that n = 2 or n = 4. The case n = 2 has been dealt with in 5. 3 ; we therefor~ suppose that n = 4. 41 Let L be a quadratic subfield of K. ~ K jQ = fundamental, for ~K j L · Then L must be almost- ~L j Q , and by taking norms we obtain ( 5. 5) Since 2 does not divide dK j Q, and dKj Q has at most square factors, it follows that 2 does not divide dL j Q and so dL j Q is square-free. We now show that K must contain more than one quadratic subfield. Suppose it is not true, and let L be the unique quadratic subfield. For each ?> Ip we have shown that e (/3 j p) = 2, and so since e('j3 IP) is the degree of K over the inertial field of~ ([14], p. 292), and L is the unique quadratic subfield, it follows that L must be the inertial field of p factors in J3 . In view of (a 1 ) and (b'), we see that 7 L in one of the following ways: p = 13 or p 731 132 . = In either case, p is unramified in L. But in view of 5. 5 , we see that it is possible to choose p so that p divides both dKIQ and dL IQ, in which case p must ramify in L. contradiction. fields: Therefore, L. = Q ( $ ) , l l Hence, we are led to a K contains two distinct quadratic sub- D. = 1 (mod 4), l (for i = 1, 2). If (D , D ) = 1, we can prove the following: 2 1 5. 6 Theorem • . If D 1 and D 2 are relatively prime, then the field K = Q(5i, ~) is almost-fundamental if and only if the following conditions are satisfied: 42 (a) D 1 and DZ = 1 (mod 4) (b) If p and q are primes such that p divides DZ and · q divides D 1 , then -D (-1 p Proof: -D = ( -qz ) = -1 . We have already shown that condition (a) is necessary . In order to show that (b) is necessary, we now determine an integral basis for KI Q . Since D basis for L 1 = 1 (mod 4), then ( 1, 1 l+~ Z ) is an integral !a . Further, we shall show that it is also an integral basis for KI Lz. Let l+~ z The discriminant matrix of A)-[_ over Lz is z 1 1 it is integral, and dKI LZ {,(Jl) = (D 1 ) is square-free in Q. (D ) has no square factors in 1 D 1 ?'Q = 7L Further, , for we can write 2 (p 1)(p2) .. . (pr ) ' where the p . are distinct primes; hence 1 Now pi does not divide dL j Q = DZ and so is unramified in 2 therefore the prime ideal factors of D ZL are distinct. 1 Novy{JZ ~ 7 Lz ; 2 ?'K, and we have just shown that dK j Lz (AJZ) is 43 Z L -module square-free, so/{. cannot be properly contained in a with integral discriminant. Therefore~= Z. K. 4 2 From this, it follows that 1 +J"D"i" l+~ 2 2 1' 1 +J"D"i" l+~ 2 2 is an integral basis for KI Q. Using the results of the preceding paragraph, we are able to compute the discriminant of the field: dK I Q 2 = NL 1 Q ( dK L 1 )d L 1 Q I I 2 = NLl jQ(D2)Dl = I (DlD2) 2 We now apply 4. 1 to the matrix 2 2 D(~) 2 1 1 l+Dl -2- = 2 1 l+D2 -2- 1 l+Dl --2- l+Dl l+D2 -2- -2- where w is the integral basis for KI Q which was determined above. Let o.1 be the principal i-rowed minor determinant; then o1 =4, o2 = 4D 1 , o3 = 4D 1 D 2 , o4 = (D 1 D 2 ) 2 , and [8] 4 Although 7 K may not have a 2-element ZL -basis, one can define its discriminant as in [3 J, page 11. follows from [3], proposition 4, page 12. Then the c'6nclusion above 44 3 C (D(w)) = p If qjD - 1 rr (-1, -o ) (o., -o.+l) 4 p i= 1 1 1 , then C (D(w)) = q - and if p j D 2 -D (-2 q , then -D C (D(w)) = ( - 1 ) p p - . In view of 4. 1 , we must have -D _l) p -D = (-2) q = -1 . The sufficiency of conditions (a) and (b) follows from 4. 1 and the fact that we can use the argument above to show that dK (DlD2) 2 • 5. 5 Example. is not. IQ= Q(,/5, ,/I3) is almost-fundamental, but Q(,/IT, ,/Pf) 45 References [l] E. Artin, Theory of Algebraic Numbers, Gottingen notes, Gottingen, 19 59. [2] E. Bender, Symmetric Representations of an Integral Domain Over a Subdomain, Ph.D. thesis, California Institute of Technology, Pasadena, 1966. [3] J. W. S. Cassels andA. Frohlich, Algebraic Number Theory, Thompson Book Co., Inc., 1967. [4] D. Estes and G. Pall, Modules and Rings in the Cayley Algebra, to appear in the Journal of Number Theory. [5] D. K. Faddeev, On the Characteristic E uations of Rational Symmetric Matrices, Dokl. Akad. Nauk USSR N. S. 58 (1947), 753-754. [6] H. Hasse, Zahlentheorie, Akademie- Verlag, Berlin, 1963. [7] T. Husain, Introduction to Topological Groups, W. B. Saunders Co., 1966. [8] B. W. Jones, The Arithmetic Theory of Quadratic Forms, Carus Monographs, Math. Assoc. Amer., 1950. [9] W. J. LeVeque, Topics in Number Theory, Vol. II, Addison Wesley , 1956. [10] 0. T. 0 1 Meara, Introduction to Quadratic Forms, Academic Press, Inc. , 1963. [ 11 J 0. Taus sky, On the Similarity Transformation between an Inte ral Matrix with Irreducible Characteristic Polynomial and Its Transpose, Math. Annalen _l_ 19 0- 3. [12] 0. Taussky, Ideal Matrices I, Archiv der Mathematiks 13 (1962), 275-282. [13] 0. Taus sky, The Discriminant Matrices of an Algebraic Number Field, J. London Math. Soc . . 43 (1968), 152-154. [ 14] 0. Zariski and P. Samuel, Commutative Algebra, Vol. I, University Series in Higher Mathematics, Van Nostrand Co., Inc., 19 58.