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Number Fields 2011-2012 Course Notes & Questions
Number Fields
University of Cambridge
Related Studylists
Abstract Algebra IPreview text
Number Fields
Lectured by T. Fisher
Lent Term 2012
0 Some Diophantine Equations 1
1 Algebraic numbers and algebraic integers 3
2 Ideals 11
3 Class group and units 18
4 Cyclotomic fields 28
Examples Sheets
Last updated: Sun 12thFeb, 2017 Please let me know of any corrections: glt1000@cam.ac
Course schedule
NUMBER FIELDS (D) 16 lectures, Lent term Part IB Groups, Rings and Modules is essential and Part II Galois Theory is desirable. Definition of algebraic number fields, their integers and units. Norms, bases and discrimi- nants. [3] Ideals, principal and prime ideals, unique factorisation. Norms of ideals. [3] Minkowski’s theorem on convex bodies. Statement of Dirichlet’s unit theorem. Determina- tion of units in quadratic fields. [2] Ideal classes, finiteness of the class group. Calculation of class numbers using statement of the Minkowski bound. [3] Dedekind’s theorem on the factorisation of primes. Application to quadratic fields. [2] Discussion of the cyclotomic field and the Fermat equation or some other topic chosen by the lecturer. [3] Appropriate books Z. Borevich and I. ShafarevichNumber Theory. Elsevier 1986 (out of print). J. Esmonde and M. MurtyProblems in Algebraic Number Theory. Springer 1999 (£38. hardback). E. HeckeLectures on the Theory of Algebraic Numbers. Springer 1981 (out of print) D. MarcusNumber Fields. Springer 1977 (£30 paperback). I. Stewart and D. TallAlgebraic Number Theory and Fermat’s Last Theorem. A K Peters 2002 (£25 paperback)
Note thaty−√−2 =y+√−2, so that y+√−2 =±(√−2)απβ 11.. .πrβr y−√−2 =±(√−2)απ 1 β 1.. .πrβr where√−2,π 1 ,.. ., πr,π 1 ,.. .,πrare irreducible and coprime. Thenx 3 =±(√−2) 2 απ 1 β 1.. .πβrrπ 1 β 1.. .πrβrand henceα≡β 1 ≡.. .≡βr≡0 (mod 3). Soy+√−2 is a cube, sayy+√−2 = (u+v√−2) 3 for someu, v∈Z. I.,y+√−2 = (u 3 − 6 uv 2 ) + (3u 2 v− 2 v 3 )√−2. Equating the√−2 parts, we find 1 = (3u 2 − 2 v 2 )v, sovis a unit inZ, i. v=±1. Then±1 = 3u 2 −2, sou=±1. Then equating rational parts givesy=u(u 2 − 6 v 2 ) =±5. So the only solutions are (x, y) = (3,±5). Later in the course we will solve problems of this type in situations where we do not necessarily have unique factorisation.
Lecture 2 1. Algebraic numbers and algebraic integers
Definition.α∈Cis analgebraic numberiff(α) = 0 for some non-zero polynomial f∈Q[X]. It is analgebraic integerif furthermoref∈Z[X] andfis monic. Lemma 1. Ifα∈Qthenαis an algebraic integer if and only ifα∈Z. Proof haveαn+an− 1 αn− 1 +.. .+a 1 α+a 0 = 0 for some integersai. Ifα∈Q, writeα=rswithr, s∈Z, coprime. Then rn+an− 1 rn− 1 s+.. .+a 1 rsn− 1 +a 0 sn= 0. Sos|rn. Ifpis a prime withp|sthenp|rn, sop|r, contradiction. Sos=±1 and henceα∈Z. The converse is obvious – takef(X) =X−α. 2 Remark the phrase ‘algebraic integer’ is abbreviated to just ‘integer’. To avoid confusion, elements ofZwill be called ‘rational integers’. Notation⊆Sbe rings, andα 1 ,.. ., αm∈S. ThenR[α 1 ,.. ., αm] denotes the subring ofSgenerated byRandα 1 ,.. ., αm. Note R[α 1 ,.. ., αm] = im(R[X 1 ,.. ., Xm]→S, Xi7→αi) Theorem 1.2 algebraic numbers form a field. Proof 0 6 =α∈Cis a root off(X) andd= degfthen 1 αis a root ofXdf(X 1 ). So it will suffice to show that ifα, βare algebraic numbers then so areα±βandαβ. Ifα, βsatisfy polynomials (with coefficients inQ) of degreesmandn, thenQ[α, β] is aQ-vector space of dimension 6 mn. Indeed, it is spanned by {αiβj: 0 6 i 6 m− 1 , 06 j 6 n− 1 }. Ifx∈Q[α, β] then themn+ 1 elements 1, x, x 2 ,.. ., xmnmust be dependent, soxis algebraic. In particular,α±βandαβall lie inQ[α, β], so are algebraic. 2 Theorem 1.3 algebraic integers form a ring. Proof must check closure – i., ifα, βare algebraic integers then so areα±βandαβ. Ifα, βsatisfy monic polynomials (with coefficients inZ) of degreesmandn, then Z[α, β] is generated as aZ-module by{αiβj}as before. To complete the proof, we takeR=ZandS=Z[α, β], andx=α±β,αβin the following lemma. 2 Lemma 1. LetR⊆Sbe rings. Suppose thatSis finitely generated as anR-module, so that for some finite subset ofS, each element ofSis someR-linear combination
It is clear that ifKis a number field thenK=Q(α 1 ,.. ., αm) for someα 1 ,.. ., αm∈K. (E., we could takeα 1 ,.. ., αmas a basis forKoverQ.) By the primitive element theorem (see theGalois Theorycourse), we can takem= 1 so that K=Q(α) for someα. Example(√ 2 ,√3) =Q(√2 +√3). Remarkαis highly non-unique – e.,Q(√2) =Q(√2 + 37) =Q(√ 12 −5) =... LetK=Q(α) be a number field and letg∈Q[X] be the minimal polynomial ofα, which exists sinceKis finite-dimensional as aQ-vector space. The ring homomorphismQ[X]→K,f7→f(α) has kernel〈g〉and imageQ[α]. So by the isomorphism theorem for rings, we haveQ[X]/〈g〉−→∼= Q[α]. Remark[X] is a principal ideal domain, sogirreducible =⇒ 〈g〉is maximal =⇒ Q[X]/〈g〉is a field. Remark inverse of non-zerof(α)∈Q[α] may be computed by running Euclid’s algo- rithm onfandg. SinceQ(α) = FracQ[α] andQ[α] is a field, we haveQ[α] =Q(α) =K. Note that this relies on the fact thatαis algebraic – ifαis transcendental thenQ(α) 6 =Q[α]. If degg=nthenK=Q(α) has basis 1, α,.. ., αn− 1 as aQ-vector space, so [K:Q] =n= degg. (It follows that ifQ(α) =Q(β) then the minimal polynomials ofαandβhave the same degree.) Lemma 1. LetKbe a number field of degreen. Then there are exactlyndistinct field embeddings ofKinC(written֒ K→C). Proof=Q(α), and letgbe the minimal polynomial ofαoverQ, with degg= n= [K:Q]. Sinceg∈Q[X] is irreducible,gandg′are coprime, and henceghasndistinct roots α 1 ,.. ., αn∈C. The embeddings֒ K→Care given byK=Q[X]/〈g〉 →C,X7→αi. Each element ofQis fixed. 2 Sinceghas real coefficients, it hasrreal roots andspairs of complex conjugate roots, with r+ 2s=n. HenceKhasrreal embeddings andspairs of complex conjugate embeddings. Remarks. (i)n, r, sdepend only onK, not on the choice ofα.
Indeed,n= [K:Q] andr= #{embeddings֒ K→R}, ands= (n−r)/2. (ii) In the definition of a number field, we could demand thatK⊆C, then one of the embeddings֒ K→Cis the identity map. Tower Law/LandL/Kbe field extensions. Then [M:K] = [M:L][L:K]. Sketch proof. Ifx 1 ,.. ., xmis a basis forLoverK, andy 1 ,.. ., ynis a basis forMoverL, then{xiyj: 1 6 i 6 m, 16 j 6 n}is a basis forMoverK. 2 Lemma 1. LetKbe a number field andβ∈Kwith minimal polynomialgoverQ. Let β 1 ,.. ., βm∈Cbe the roots ofg(called theconjugatesofβ). Thendi= #{σ:֒ K→C : σ(β) =βi}is independent ofi. Proof have field extensionsK/Q(β) andQ(β)/Q, with [Q(β) :Q] = degg=m. Lethbe the minimal polynomial ofαoverQ(β). Thendi 6 degh, since if we fixσ(β) then we fix the images of the coefficients ofh. I.,di 6 [K:Q(β)]. Theni∑=1mdi 6 [K:Q(β)]m= [K:Q(β)][Q(β) :Q] = [K:Q]. Lemma 1 tells us thati∑=1mdi= [K:Q], and sodi= [K:Q(β)] for alli. 2 Definition/Kbe a finite extension. Ifx∈Lthen there is aK-linear mapφx:L→L given byy7→xy. We define thetraceTrL/K(x) =Tr(φx)∈Kand thenormNL/K(x) = det(φx)∈K. Remarks.
- NL/K(x 1 x 2 ) =NL/K(x 1 )NL/K(x 2 ) forx 1 , x 2 ∈L.
- TrL/K:L→Kis aK-linear map.
- Iff∈K[X] thenf(φx) =φf(x). Lecture 4 Theorem 1.8 be a number field of degreen, andσ 1 ,.. ., σndistinct embedding ֒ K→C. Forβ∈K, the characteristic polynomial of theQ-linear mapφβ:K→K, y7→yβ, is f(X) = ∏n i=
(X−σi(β))
In particular,TrK/Q(β) =i∑=1n σi(β) andNK/Q(β) =i∏=1n σi(β). Proof the minimal polynomial ofβoverQ. Bothfand the characteristic poly- nomial ofφβare monic polynomials of the same degree. We show that they are equal by showing each is a power ofg. Letm= degg, and letβ 1 ,.. ., βmbe the roots ofg. Sog(X) =∏mi=1(X−βi), and
∆(x 1 ,.. ., xn) = det(σi(xj)) 2 = det
σ 1 (.. 1 )... σn(.. 1 ) σ 1 (xn)... σn(xn)
det
σ 1 (.. 1 )... σ 1 (..) σn(x 1 )... σn(xn)
= det(TrK/Q(xixj))∈Q Note thatx′i=j∑n=1aijxjforaij∈Q. WriteA= (aij). Then ∆(x′ 1 ,.. ., x′n) = (detA) 2 ∆(x 1 ,.. ., xn) (∗) Recall discriminant off(X) =i∏=1n(X−αi) is disc(f) =i<j∏(αi−αj) 2. Lemma 1. LetK=Q(α) be a number field of degreen, andfbe the minimal polynomial ofαoverQ. Then (i) ∆(1, α, α 2 ,.. ., αn− 1 ) = disc(f). (ii)x 1 ,.. ., xnis a basis forKoverQ ⇐⇒ ∆(x 1 ,.. ., xn) 6 = 0. Proof. (i) Writeαi=σi(α). Then
∆(1, α,.. ., αn− 1 ) =
∣∣∣
∣∣∣
∣∣∣
∣∣
α 1... 1 α 12... αn . 1... α 2 n .. ... αn 1 − 1... αnn− 1
∣∣∣
∣∣∣
∣∣∣
∣∣
2 =∏i<j(αi−αj) 2 = disc(f)
(ii) This follows by (∗) in the note above. 2 Definition basis{x 1 ,.. ., xn}forKoverQis called anintegral basisif OK={ ∑n i=1λixi:λi∈Z
}.
Theorem 1. An integral basis exists for anyK. (In particular,OK∼=Znas a group under addition.) Proof Lemma 1, there is a basis{x 1 ,.. ., xn} ⊆ OKforKoverQ. Lecture 5 Then ∆(x 1 ,.. ., xn)∈Z\ { 0 }by 1. Choose a basisx 1 ,.. ., xn∈ OKforK as a Q-vector space such that|∆(x 1 ,.. ., xn)|is minimal. Ifθ∈ OKthenθ=∑ni=1λixifor someλi∈Q. If, say,λ 1 =λ+μwithλ∈Zand 0 < μ <1, then setx′ 1 =θ−λx 1 =μx 1 +λ 2 x 2 +.. .+λnxn∈ OK.
Thenx′ 1 , x 2 ,.. ., xnis a basis forKoverQand ∆(x′ 1 , x 2 ,.. ., xn) =μ 2 ∆(x 1 ,.. ., xn). But since 0< μ <1, this contradicts the minimality of|∆(x 1 ,.. ., xn)|, so we must haveλ∈Zand hence we have an integral basis. 2 Definition a number fieldKis ∆(x 1 ,.. ., xn) wherex 1 ,.. ., xnis an integral basis. This is well-defined: the determinant of the transformation from one integral basis to another is±1, so by (∗) the discriminant is independent of the choice of integral basis. Definition fieldis a number field of the formK=Q(√d) withda square-free integer,d 6 = 0,1. Note that:
- K={x+y√d:x, y∈Q}.
- TrK/Q(x+y√d) = (x+y√d) + (x−y√d) =Tr
(x dy y x
)
= 2x.
- NK/Q(x+y√d) = (x+y√d)(x−y√d) = det
(x dy y x
)
=x 2 −dy 2.
Proposition 1.14 a quadratic fieldK=Q(√d), we have
OK=
{Z[√d] ifd≡ 2 ,3 (mod 4) Z[ 12 (1 +√d)] ifd≡1 (mod 4) Proof+y√d∈ OKthen 2x, x 2 −dy 2 ∈Z, so 4dy 2 ∈Z. Sincedis square-free, we must havex=u/2,y=v/2 for someu, v∈Z. We haveu 2 −dv 2 ≡0 (mod 4). (i) Ifd≡ 2 ,3 (mod 4) then since squares modulo 4 are either 0 or 1, we know that u,vare even, sox,yare rational integers. So{ 1 ,√d}is an integral basis, and OK=Z[√d]. (ii) Ifd≡1 (mod 4) thenu,vhave the same parity. Then{ 1 , 12 (1 +√d)}is an integral basis, andOK=Z[ 12 (1 +√d)]. Note that 12 (1 +√d) has minimal polynomialX 2 −X+ 14 (1−d)∈Z[X]. 2 Remark discriminantDKofKis given by
Dk= det
(√ 1 1
d −√d
) 2
= 4d ifd≡ 2 ,3 (mod 4) Dk= det
( 1 1
12 (1 +√d) 12 (1−√d)
) 2
=d ifd≡1 (mod 4)
Lemma 1. LetM 6 Znbe a subgroup. ThenM∼=Zrfor somer 6 n. Ifr=nandA is ann×nmatrix whose rows are aZ-basis forM, then|Zn:M|=|detA|.
2. Ideals
Example=Q(√−5). ThenOK=Z[√−5]. OKis not a UFD – e., 3×7 = (1 + 2√−5)(1− 2 √−5), but 3, 7 and 1± 2 √−5 are irreducible and not associates. To show that 3 is irreducible, write 3 =αβ, withα, β∈ OK. ThenNK/Q(α)NK/Q(β) = 9, butx 2 + 5y 2 =±3 has no solutions forx, y∈Z. Ideals were used by Kummer, Dedekind, etc to restore the property of unique factorisation. RecallOKis an ideal if (i)ais a subgroup under + (ii) Ifr∈ OKands∈a, thenrs∈a. Theorem 1 tells us thatOK∼=Znis an abelian group. So ifaOKis an ideal then by Lemma 1,ais finitely generated as aZ-module. Soais finitely generated as an OK-module. We have shown: Lemma 2. OKis a Noetherian ring. Notation=〈α 1 ,.. ., αr〉meansa={i∑=1r λiαi:λi∈ OK}.
Lecture 6 Definition idealsaandbinOKis
ab=
{∑n i=1aibi:ai∈a, bi∈b, n∈N
}
Ifa=〈α 1 ,.. ., αr〉andb=〈β 1 ,.. ., βs〉, thenab=〈{αiβj: 1 6 i 6 r, 16 j 6 s}〉. We say thatbdividesaif there is an idealcsuch thata=bc. Definition idealais an ideal such thata=〈α〉for someα∈ OK. Note that if〈α〉=〈β〉thenαβ∈ O∗Kand we say thatαandβareassociates. Our aim is the following: Theorem 2.2 any ideala∈ OKthere exists a non-zero idealbOKsuch thatabis principal. Example [K:Q] = 2 anda=〈a, β〉for somea∈Z,β∈ OK. WriteK=Q(√−d) ford∈Z, andβ=u+v√−dforu, v∈Q, soβ=u−v√−d. Then
〈a, β〉〈a,β〉 = 〈a 2 , aβ, aβ, ββ〉 =〈a 2 , aβ, a(β+β), ββ〉 = 〈a 2 , aβ, aTr(β), N(β)〉 = 〈c, aβ〉 wherec= hcf(a 2 , aTr(β), N(β)). Letx=acβ. ThenTr(x) =acTr(β)∈ZandN(x) =ac 22 N(β)∈Z. Butxis a root of the polynomialX 2 −Tr(x)X+N(x), sox∈ OK, so〈a, β〉〈a,β〉=〈c〉. Lemma 2. For a non-zero idealaOK, we havea∩Z 6 = 0 andOK/ais finite. Proof any non-zeroα∈a. Sayαhas minimal polynomialg(X) =Xm+cm− 1 Xm− 1 + .. .+c 1 X+c 0 ∈Z[X]. Thenc 06 = 0, since otherwisegis reducible, andc 0 =−(αm− 1 + cm− 1 αm− 2 +.. .+c 1 )α∈a∩Z. Now, if 0 6 =d∈a∩Z, then there is a surjective homomorphismOK/〈d〉։OK/a. Theorem 1 implies thatOK∼=Znas a group under addition, soOK/〈d〉∼=(Z/dZ)n, which is finite. HenceOK/ais finite. 2 Recall idealp∈ OKisprimeifOK/pis an integral domain – i.,p 6 =OKand if a, b∈ OKsatisfyab∈pthena∈porb∈p. Lemma 2. Letpbe a prime ideal. Ifa,bare ideals andab⊆pthen eithera⊆porb⊆p. Proof not, picka∈a\pandb∈b\p. Thenab∈ab⊆p, so by primality ofpeither a∈porb∈p, contradiction. 2 Lemma 2. Every non-zero prime idealpOKis a maximal ideal. Proof a finite integral domain and 0 6 =x∈R, then the mapR→R,y7→xy, is injective, hence is surjective sinceRis finite. In particular, there existsy∈Rsuch thatxy= 1. This shows thatRis a field. Ifpis prime thenOK/pis an integral domain, soOK/pis a field by Lemma 2, and sopis maximal. 2 Convention this course, ‘prime ideal’ means ‘non-zero prime ideal’ unless stated oth- erwise. Lemma 2. Every non-zero ideala∈ OKcontains a product of prime ideals. Proof not. SinceOKis Noetherian, we can pickato be maximal among ideals not having this property. Sinceacannot be prime (otherwise the lemma holds trivially), there existx, y∈ OKsuch thatx /∈a,y /∈aandxy∈a. Thena(a+〈x〉, by choice ofa, and sop 1.. .pr⊆a+〈x〉for some prime idealspi. Similarly,q 1.. .qs⊆a+〈y〉for some prime idealsqj.
Theorem 2′.Every non-zero fractional ideal is invertible. Remark an invertible fractional ideal thena− 1 ={x∈K:xa⊆ OK}. Proof of 2′.Suppose false. SinceOKis Noetherian, we can takeaOKmaximal subject to not being invertible. Letb={x∈K:xa⊆ OK}. To see thatbis a fractional ideal, pickα∈a\ { 0 }and note thatαb⊆ OK. We haveOK⊆b, soa⊆ab. Ifa=abthen Lemma 2 tells us thatb⊆ OK. But Lemma 2 implies there is somex∈b\ OK, which contradictsb⊆ OK. Soa(ab⊆ OK. The choice ofatells us thatabis invertible, soais invertible. Contradiction. 2 Corollary 2.10,b,cbe integral ideals, withcnon-zero. Then (i)b⊆a⇐⇒ bc⊆ac (ii)a|b⇐⇒ ac|bc (iii)a|b⇐⇒ b⊆a. (‘To divide is to contain.’) Proof. (i),(ii) (⇒) is clear. For (⇐), multiply byc− 1. (iii) (⇒) is clear. (⇐). By (i), (ii) and Theorem 2, we may reduce to the case whereais principal. Ifa=〈α〉 ⊃b=〈β 1 ,.. ., βs〉, thend=〈βα 1 ,.. .,βαs〉OKandad=b, soa|b. 2 Theorem 2. Every non-zero idealaOKcan be written uniquely as a product of prime ideals. Proof. Existence claim that ifa(OKthen eitherais prime ora=bcwitha(band a(c. Then apply the same argument tobandc, and so on. This process must stop sinceOKis Noetherian. To prove this claim, note thata⊆pfor somepmaximal and hence prime. By Corollary 2(iii), we havep|a, soa=pbfor someb. Ifp=athenais prime. Ifb=athen a=pa, so multiplying bya− 1 givesp=OK, contradicting the primality ofp. So the claim is proved. Uniqueness 2 now says thatp|abimpliesp|aorp|b. So ifp 1.. .pr=q 1.. .qs thenp 1 |q 1.. .qs, sop 1 |qifor somei, wlogi= 1. Sop 1 ⊃q 1 , and sop 1 =q 1 by maximality. Multiplying byp− 11 givesp 2.. .pr=q 2.. .qs, and repeating givesr=sandpi=qifor alli. 2
Theorem 2. The non-zero fractional ideals ofKform a groupIKunder multiplication. It is a free abelian group generated by the prime ideals – i., anya∈IKcan be written uniquely as a=pα 11.. .pαrr withp 1 ,.. .,prdistinct prime ideals, andαi∈Z. (Note: notN!) NoteOK ⇐⇒ allαi>0. Proof only difficult part in checking thatIKis an abelian group is checking inverses, but this is Theorem 2. Ifa∈IK, writea=b− 1 cfor some integral idealsb,c, then use Theorem 2. 2 There is a group homomorphismK∗→IK,x7→ 〈x〉, with kernel the group of unitsO∗Kand image the subgroupPKofIKconsisting of principal fractional ideals. Definition class group ofKisClK=IK/PK. Remark can also defineClKas the set of equivalence class of integral ideals, where a∼b⇐⇒ γa=δbfor some non-zeroγ, δ∈ OK. Write [a] for the equivalence class ofaand define [a][b] = [ab]. Lecture 8 Proposition 2.13 following are equivalent: (i)OKis a PID. (ii)OKis a UFD. (iii)ClKis trivial. Proof. (i)⇒(ii) See Part IBGroups, Rings & Modules. (ii)⇒(i) It is enough to show that prime ideals are principal. LetpOKbe a prime ideal. Pickx∈p\ { 0 }and writex=π 1.. .πkwithπi irreducible. Thenπi∈pfor somei. Sinceπiis prime,〈πi〉 ⊆pis a prime ideal. But by Lemma 2,〈πi〉is maximal, so〈πi〉=p. By Theorem 2,OKis a PID. (i)⇔(iii) By definition ofClK. (Think: ‘ClKmeasures the failure of ideals inOKto be principal.) 2 Remark,bOK, definea+b={x+y:x∈a, y∈b}. It is the smallest ideal containingaandb. So Corollary 2, ‘to divide is to contain’ and hencea+bis the hcf ofaandb. We may think ofa=〈α 1 ,.. ., αr〉as the hcf ofα 1 ,.. ., αr.
Proposition 2.16α∈ OK\ { 0 }thenN(〈α〉) =|NK/Q(α)|. Proof{x 1 ,.. ., xn}be an integral basis forK. Takeγi=αxi. Then ∆(γ 1 ,.. .γn) = det(σi(αxj)) 2 =(∏σi(α)) 2 ∆(x 1 ,.. ., xn) =NK/Q(α) 2 DK SoNK/Q(α) 2 =N(〈α〉) 2 , and henceN(〈α〉) =|NK/Q(α)|, sinceN(〈α〉) is necessarily positive. 2 Corollary 2.17OKis a prime ideal. Then there is a unique rational prime p∈Zsuch thatp|p. Moreover,N(p) is a power ofp. Proof Lemma 2,pcontains a positive integera∈Z(⇐⇒ p|a). The least suchais prime, sincep|ab=⇒ p|aorp|b. Ifp|pandp|qforp, qdistinct, then there exist r, ssuch thatpr+qs= 1 sop|1 and hencep=OK, contradiction. Sopis unique. Ifp|pthen〈p〉=pafor someaOK. Thenpn=N(p)N(a), soN(p) =prfor some r. 2 Remark havep∩Z=pZ. In particular, ifp|afora∈Z, thenp|a. Letpbe a rational prime. Then〈p〉=pOK=pe 11.. .perr, with thepidistinct prime ideals. SayN(pi) =pfi. Thene 1 ,.. ., erare theramification indices, andf 1 ,.. ., frare theresidue class degrees. Lecture 9 Take norms: Corollary 2.18∑=1r eifi= [K:Q]. 2 Terminology.
- pramifiesinKif someei>1.
- pisinertinKifpOKis prime.
- psplits completelyinKifei=fi= 1 for alli.
3. Class group and units
Definition subsetX⊆Rnisdiscreteif for allx∈X there existsε >0 such that X∩B(x, ε) ={x}. (This is equivalent to saying that the subspace topology onXis the discrete topology.) Lemma 3. Let Λ⊆Rn. Then the following are equivalent: (i) Λ ={i∑m=1aixi:ai∈Z}for someR-linearly independentx 1 ,.. ., xm∈Rn. (ii) Λ is a discrete subgroup of (Rn,+).
If these conditions are satisfied then we call Λ alattice. Proof conditions are invariant under all linear automorphisms ofRn. ReplacingRn by a subspace, we may assume that Λ spansRn(i.,m=n). (i)⇒(ii) Wlog, take{x 1 ,.. ., xn}to be the standard basis. Then Λ =Znandε= 12 shows that Λ is discrete. (ii)⇒(i) Wlog, Λ⊃Zn. LetX={(a 1 ,.. ., an) : 0 6 ai 61 }. ThenXis closed and bounded, so is a compact subset ofRn. So Λ∩Xis discrete and compact, and hence finite. So Λ/Znis finite, say of orderd∈N. ThenZn⊆Λ⊆ 1 dZn. Then Lemma 1 implies that Λ∼=Zn, so Λ ={i∑=1naixi:ai∈Z}for some x 1 ,.. ., xn∈Rn. But Λ spansRn, so thexiare linearly independent. 2 Recall(from Part IBGroups, Rings & Modules). IfAis a finitely-generated abelian group, thenA∼=T×Zr, whereTis a finite abelian group andris therankofA. LetKbe a number field with [K:Q] =n=r+ 2swith embeddingsσ 1 ,.. ., σr:֒ K→Rand σr+1,.. ., σr+s,σr+1,.. .,σr+s:֒ K→C. There is a group homomorphismL:O∗K→Rr+sgiven byu7→(log|σ 1 (u)|,.. .,log|σr+s(u)|). Lemma 3. IfB⊆Rr+sis bounded thenL− 1 Bis finite. Proof.∏nIfu ∈L− 1 Bthen |σi(u)|is bounded for alli. So the coefficients off(X) =
and hence ofi=1(X−σi(uu.))∈Z[X] are bounded, so there are only finitely many choices off 2 , By Lemma 3, kerLis finite, and hence consists of roots of unity because it is a subgroup, and imL⊆Rr+sis discrete. By Lemma 3,O∗Kis a finitely-generated abelian group of rank 6 r+s. By Lemma 1, foru∈ O∗Kwe haveNK/Q(u) =∏ni=1σi(u) =±1.
Number Fields 2011-2012 Course Notes & Questions
Module: Number Fields
University: University of Cambridge
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