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Local Fields 2011-2012 Lecture Notes & Questions

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Local Fields

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Local Fields

Lectured by T. A. Fisher

Michaelmas Term 2011

1 Introduction top-adic numbers 1

2 Valuations 7

3 Dedekind domains 13

4 Extensions of complete fields 19

5 Inverse limits 27

6 Ramification 30

7 Norm index computations 40

8 Quadratic forms 50

Examples Sheets

Last updated: Mon 6thJan, 2014 In progress! So please let me know of any corrections: glt1000@cam.ac

Course description

The theory of local fields was introduced by Hensel in the early 1900’s as an alternative approach to algebraic number theory. The basic idea is to consider the completions of a number fieldKat all absolute values, not just the ones arising from the embeddings ofK into the reals or complexes. One can then borrow techniques from analysis to studyKand its finite extensions in a way that focuses on their behaviour at justone prime. For instance the analogue of the Newton-Raphson method for root finding goesby the name of Hensel’s lemma.

Nowadays, local fields have established themselves as a natural tool in many areas of number theory and also in subjects like representation theory, algebraic topology and arithmetic geometry (e. elliptic curves). So this course is likely to be useful for those taking the “Class Field Theory” and “Elliptic Curves” courses.

The course will begin by introducing the field ofp-adic numbersQp(wherepis a prime). This is the completion of the field of rational numbersQwith respect to thep-adic absolute value defined for non-zerox∈Qby|x|p= 1/pnwherex=pna/bwithpnot dividinga orb. Topics to be covered will then include: absolute values on fields, valuations, complete fields and their extensions, the different and discriminant, decomposition groups, inverse limits, Hensel’s lemma and ramification theory. If time permits, then possible further topics include: Skolem’s method, local class field theory (statements only), the Hilbert norm residue symbol, and the Hasse-Minkowski theorem.

Pre-requisite Mathematics

Basic algebra up to and including Galois theory is essential. It will be assumed students have been to a first course in algebraic number fields.

Literature

J.W. Cassels,Local fields, CUP, 1986.
G. Janusz,Algebraic number fields, AMS, 1996.
N. Koblitz,p-adic numbers,p-adic analysis and zeta-functions, Springer, 1977.
J. Neukirch,Algebraic number theory, Springer, 1999.
J.-P. Serre,A course in arithmetic, Springer, 1973.
J.-P. Serre,Local fields, Springer, 1979.

Corollary. If char(K)&gt;0, then all absolute values onKare non-archimedean (asZ→R has finite and thus bounded image).

Example 1. K=Q,p= 5,|·|=|·| 5.

Define the sequence:a 1 = 3,a 2 = 33,a 3 = 333,a 4 = 3333,...

We haveam≡an(mod 5n) for allm&gt;n, so|am−an| 65 −nfor allm&gt;n. So this is a Cauchy sequence.

Butan= 13 (10n−1), and so|an+ 13 |= 5−n→0 asn→∞. I.,an→− 13 w.r.|·| 5.

Slogan number isp-adically small if it is divisible by a large power ofp.

Example 2. We construct a sequence of integersansuch that for alln&gt;1,

a 2 n+ 1 ≡ 0 (mod 5n) an+1 ≡ an (mod 5n) (∗) Takea 1 = 2. Suppose thatanis already chosen, and writea 2 n+ 1 = 5nc, somec∈Z. Then (an+b 5 n) 2 + 1 = a 2 n+ 1 + 2· 5 nanb+ 5 2 nb 2 ≡ 5 n(c+ 2anb) (mod 5n+1). We pickb∈Zso thatc+ 2anb≡0 (mod 5). This is possible, since (2an,5) = 1.

Then takean+1=an+ 5nb.

(∗) implies that (an) is Cauchy. Supposean→ℓ, someℓ∈Q.

Then|ℓ 2 + 1| 6 |a 2 n+ 1|+|a 2 n−ℓ 2 |→0 asn→∞. Soℓ 2 =−1.//\

So this shows that (Q,|·| 5 ) isnotcomplete.

Definition 1. Thefield ofp-adic numbers,Qp, is the completion ofQw.r.|·|p.

Note.+,×,|·|ponQextend by continuity to +,×,|·|ponQp. Easy to check that (Qp,|·|p) is a non-archimedean valued field.

Definition 1. Thering ofp-adic integersisZp ={x∈Qp :|x|p 61 }. (This is closed under addition by the ultrametric inequality. Being a closed subset ofQp, it is complete.)

Lemma 1. Zis dense inZp. (In particular,Zpis the completion ofZw.r.|·|p.)

Proof dense inQp. AndZpis open inQp(by the ultrametric inequality: x∈Zp, y∈Qp,|x−y|p&lt; 1 ⇒ |y|p 61 ⇒y∈Zp).

SoQ∩Zpis dense inZp.

ButQ∩Zp={x∈Q:|x|p 61 }={ab∈Q:a, b∈Z, p∤b}=Z(p). (Localisation in the sense of commutative algebra.)

Letab∈Z(p), i. a, b∈Z,p∤b. For eachn&gt;1, pickyn∈Zsuch thatbyn≡ 1 (modpn). Thenbyn→1 asn→∞, soayn→abasn→∞.

SoZis dense inZ(p), which is dense inZp. 2

Global situation Local situation [K:Q]&lt;∞ [K:Qp]&lt;∞ OK= integral closure ofZinK OK= integral closure ofZpinK OKneed not be a UFD OKis always a UFD (in fact a DVR, (e.g=Q(

√

−5),OK=Z[

√

−5]) i. a PID with just one prime)

LetKbe a number field,p⊂ OKa prime ideal, and 0&lt; α &lt;1. Forx∈K∗, letvp(x) = power ofpin the prime factorisation of (x) as fractional ideals.

Define|x|p=

{

αvp(x) ifx 6 = 0 0 ifx= 0 . This gives an absolute value onK.

For suitableα, this extends|·|ponQp, wherep∩Z=pZ.

Kpis the completion ofKw.r|·|p.

Remarks/Facts. (i) [Kp:Qp]&lt;∞. (Proof later.)

(ii) Every finite extension ofQparises as the completion of some number field. (Proof later.) (iii) In Example 2 we showedi=

√

− 1 ∈Q 5. (See Hensel’s Lemma later.) This is related to the fact thatp= 5 splits inK=Q(i), i. (p) =p 1 p 2 inOK=Z[i].

Lemma 1. Let|·| 1 and|·| 2 be non-trivial absolute values on a fieldK. The following are equivalent: (i)|·| 1 and|·| 2 induce the same topology (ii)|x| 1 &lt; 1 ⇐⇒ |x| 2 &lt; 1 (iii)|x| 2 =|x|c 1 for some fixedc &gt;0. If these conditions hold then|·| 1 and|·| 2 areequivalent.

Proof.

(i)⇒(ii).|x| 1 &lt; 1 ⇐⇒ xn→0 w.r.|·| 1 ⇐⇒ xn→0 w.r.|·| 2 ⇐⇒ |x| 2 &lt;1. (ii)⇒(iii). Picka∈K∗with|a| 1 &lt;1 (possible since|·| 1 is non-trivial). Letx∈K∗, letm, n∈Zwithn &gt;0.

log|x| 1 log|a| 1

&gt;

m n

⇐⇒ nlog|x| 1 &lt; mlog|a| 1

⇐⇒

∣

xn am

∣

&lt; 1

⇐⇒

∣

xn am

∣

&lt; 1

⇐⇒

log|x| 2 log|a| 2

&gt;

m n (note|a| 2 &lt;1)

Butmn∈Qwas arbitrary, so

log|x| 1 log|a| 1

=

log|x| 2 log|a| 2

.

Hence log|x| 2 =clog|x| 1 for some fixedc &gt;0, and so|x| 2 =|x|c 1. (iii)⇒(i). Clear. 2

WriteK=Q(α) (theorem of the primitive element). Then insideK̂, we haveKdense inR(α), andR(α) =RorC, since these are complete, and using thatCis algebraically closed.

ThusK̂=RorC. 2

Remarks. (i) We should check that the only extension of|·|∞onRtoCis|·|∞itself.

(ii) In fact,RandCare the only complete archimedean fields. See, e., Chapter 3 of Cassels, “Local Fields”. (iii) The non-archimedean places of a number fieldKare the|·|pforp⊂Oka prime ideal. (Proof later.)

Theorem 1 (Weak approximation). Let|·| 1 ,.. .,|·|nbe pairwise inequivalent non- trivial absolute values on a fieldK. Givenβ 1 ,.. ., βn∈Kandǫ &gt;0, there existsα∈K such that|α−βi|i&lt; ǫfor alli.

Proof show by induction onnthat there existy 1 ,.. ., yn∈Ksuch that|yi|i&gt;1 and |yi|j&lt;1 for alli 6 =j.

Casen= 2.

Swapping|·| 1 and|·| 2 if necessary, Lemma 1 shows there existsw∈Ksuch that |w| 1 &lt;1 and|w| 2 |&gt;1. Since|·| 2 is non-trivial, there existsz∈Kwith|z| 2 &gt;1.

Takingy=wrzforrsufficiently large gives|y| 1 &lt;1 and|y| 2 &gt;1. Then puty 1 = 1/y andy 2 =y.

Induction step.

Suppose we havey∈Kwith|y| 1 &gt;1 and|y|j&lt;1 forj= 2,.. ., n−1. Casen= 2 implies there ist∈Ksuch that|t| 1 and|t|n&lt;1. Then, for sufficiently larger,

y 1 =







y if|y|n&lt; 1 ( yrt if|y|n= 1 yr 1+yr

)

t if|y|n&gt; 1

satisfies|y 1 | 1 &gt;1 and|y 1 |j&lt;1 for allj= 2,.. ., n.

By symmetry, gety 1 ,.. ., yn. To finish the proof, put

α=

∑n

(

yri 1 +yir

)

βi

and takersufficiently large. 2

Exercise an alternative proof in the caseK=Qusing Theorem1 the Chinese Remainder Theorem. (Swinnerton-Dyer.)

Lemma 1. Let (K,|·|) be non-archimedean. Then

(i)|x|&lt;|y|=⇒ |x±y|=|y| (“all triangles are isosceles”) (ii)|x 1 +···+xn| 6 max|xi|, with equality if|xi|&lt;|x 1 |for alli &gt; 1 (iii) if (K,|·|) is complete, then

∑∞

i=1anconverges iffan→0 asn→∞.

Proof. (i) We’re given|x|&lt;|y|. Then|x+y|= max(|x|,|y|) =|y|. But also,|y| 6 max(|x+y|,|x|) since|y|&gt;|x|, so|y| 6 |x+y|. Therefore|x+y|=|y|.

(ii) Ultrametric law and induction. To prove the condition for equality,takex= x 2 +···+xnandy=x 1 in (i).

(iii) Letsn=

∑n i=1ai. Ifsn→ℓasn→ ∞, thenan=sn−sn− 1 →ℓ−ℓ= 0 as n→∞.

Forx∈K,r &gt;0 (r∈R), defineB(x, r) ={y∈K:|x−y|&lt; r}andB(x, r) ={y∈K: |x−y| 6 r}.

Lemma 1. Let (K,|·|) be non-archimedean. Then

(i) ify∈B(x, r) thenB(y, r) =B(x, r) (ii) ify∈B(x, r) thenB(y, r) =B(x, r) (iii)B(x, r) is both open and closed (iv)B(x, r) is both open and closed.

Proof. (i) Use ultrametric law.

(ii) Use ultrametric law.

(iii)B(x, r) is open by definition of topology. And it is closed since ify /∈B(x, r) then B(x, r)∩B(y, r) =∅.

(iv)B(x, r) is closed since|·|is continuous. And it is open since ify∈B(x, r) then B(y, 12 r)⊂B(y, r) =B(x, r), by (ii).

(Note we needr &gt;0 here, and we could have usedrinstead of 12 r.)

Remark is totally disconnected, i. the only connected subsets are the singletons. Indeed, ifx, yare distinct, putr= 12 |x−y|. ThenB(x, r) and its complement partition Kinto open sets, one containingxand the othery.

(iii)⇒(iv). Writem=x 1 Ov+.. .+xnOv, with wlog|x 1 |&gt;.. .&gt;|xn|. Then m=x 1 Ov. (See Remark (ii) above.) (iv)⇒(i). Writem=πOv, and letv=v(π)&gt;0. Ifx∈K∗thenv(x)&gt;0, so ifx∈mthenv(x)&gt;c. Thereforev(K∗) is a discrete subgroup. (“Nothing between 0 andc.”) 2

Definition 2. Adiscrete valuation ring(DVR) is a PID with exactly one non-zero prime ideal (= with one maximal ideal).

Lemma 2. (i) Ifvis discrete thenOvis a DVR.

(ii) IfRis a DVR then there exists a discrete valuationvonK= Frac(R) such that R=Ov. (Andvis unique if we normalise it.)

Proof. (i) Recall:Ovlocal and not a field Lemma 2⇒ Ovis a PID

}

=⇒ Ovis a DVR.

(ii) LetRbe a DVR with prime elementπ. Everyx∈R{ 0 }can be written uniquely asx=uπr, for someu∈R∗,r&gt;0.

Everyx∈K∗can be written uniquely asx=uπr, for someu∈R∗,r∈Z.

Definev:K∗→Rbyuπr7→r. ThenOv={x∈K:v(x)&gt; 0 }=R. 2

Examples.

1(p)={x∈Q:|x|p 61 }is a DVR with field of fractionsQ,π= (p), and residue fieldZ/pZ. 2 a DVR with field of fractionsQp,π= (p), and residue field is stillZ/pZ. 3 field,K=k(t) – “rational functions”. Definev 0

(

tnfg((tt))

)

=n∈Z, wheref, g∈k[t], withf(0), g(0) 6 = 0. O={f∈k(t) :f(0) defined} O∗={f∈k(t) :f(0) defined and non-zero} m={f∈k(t) :f(0) defined andf(0) = 0} ViaO →k,f7→f(0), we seeO/m∼=k.

Likewise, fora∈k, defineva

(

(t−a)nfg((tt))

)

=n∈Z, wheref, g∈k[t], with f(a), g(a) 6 = 0. (“Order att=a”.) Andv∞

(

f(t) g(t)

)

=v 0

(

f(1/t) g(1/t)

)

= deg(g)−deg(f). (“Order att=∞.)

Remarks.

(i) Ifk=kthen these are the only discrete valuationsvonk(t) withv(k∗) = 0. (ii)K=k(t) is the function field ofP 1. Similar examples exist for any smooth point on an algebraic curve / Riemann surface (takek=C).

Examples (ctd).

4=k((t)) = field of Laurent series =

{∑

n&gt;n 0 ant

n:an∈k

}

, (n 0 any integer). v(

∑

nant

n) = min{n:an 6 = 0} O=k[[t]] = power series ring m={f∈k[[t]] :f(0) = 0} O/m=kviaf7→f(0)

Lemma 2.

(i)k[[t]]∗=

{∑

n&gt; 0 ant n:a 06 = 0

}

(ii)k((t)) is a field, containingk(t), andvextendsv 0 (iii)k((t)) is the completion ofk(t), andk[[t]] is the completion ofk[t], w.r.t 0.

Proof.

(i) Let

∑∞

n=0ant n∈k[[t]] witha 06 = 0.

Solve forbnsuch that (

∑∞

n=0ant n) (∑∞ n=0bnt n) = 1.

I.e 0 b 0 = 1⇒solve forb 0. Thena 0 b 1 +a 1 b 0 = 0⇒solve forb 1. Etc.

(ii) By (i),k((t)) is a field. It containsk[t] and so it contains its field of fractions.

Iff(t) =tnqp((tt)) (withp, q∈k[t],p(0), q(0) 6 = 0), then by (i),p, q∈k[[t]]∗, so v(f) =n=v 0 (f).

(iii) Sincek[t] is dense ink[[t]] (can truncate a series whenever we want), it suffices to show thatk[[t]] is complete w.r.t.

Letf 1 , f 2 ,.. .be a Cauchy sequence ink[[t]]. Then givenr, there existsNsuch that for allm, n&gt;N,v(fm−fn)&gt; r– i.e≡fn(modtr+1).

Letcr= coefficient oftrinfN. Thenfn→g, whereg=

∑∞

r=0crt

Thereforek[[t]] is complete.

Similarly,k((t)) is the completion ofk(t). 2

Letv:K∗→Zbe a normalised discrete valuation,O=Ov. Pickπ∈Kwithv(π) = 1. Thenm=πO. (πis called anormaliser.)

Hensel’s Lemma (version 1).AssumeKis complete w.r. v. Letf(X)∈ O[X]. Sup- pose that the reductionf(X)∈k[X] has a simple root, i. there existsa∈ Osuch thatf(a)≡0 (modπ) andf′(a)6≡0 (modπ) – i.,f(a) = 0 andf ′ (a) 6 = 0.

Then there exists a uniquex∈Osuch thatf(x) = 0 andx≡a(modπ).

Hensel’s Lemma (version 2).AssumeKis complete w.r. discrete valuation. Suppose f(X)∈O[X] anda∈Osatisfies|f(a)|&lt;|f′(a)| 2.

Then there exists a uniquex∈Osuch thatf(x) = 0 and|x−a|&lt;|f′(a)|.

Proof=v(f′(a)). (Note, version 1↔r= 0.)

We construct a sequence (xn) inOsuch that: (i) f(xn)≡0 (modπn+2r) (ii) xn+1≡xn(modπn+r)

.

Putx 1 =a. Suppose thatxnhas been constructed, satisfying (i), i.e(xn)≡cπn+2r somec∈O. We’ll takexn+1=xn+bπn+rfor someb∈O.

Notef(X+Y) =f 0 (X) +f 1 (X)Y+f 2 (X)Y 2 +.. ., withfi∈O[X],f 0 =f,f 1 =f′.

Sof(xn+1) =f(xn+bπn+r)≡f(xn) +f′(xn)bπn+r(modπn+2r+1).

Proposition 2.6∗ 3 /(Q 3 ) 3 ∼=(Z/ 3 Z) 2.

Proof∈Z∗ 3 withb≡1 (mod 9). Writeb= 1 + 9cfor somec∈Z 3.

Thenb≡(1 + 3c) 2 (mod 27). Apply Hensel’s Lemma withf(X) =X 3 −b.

|f(1 + 3c)| 63 − 3 &lt; 3 − 2 =|f′(1 + 3c)| 2 , so we getb∈(Z∗ 3 ) 3.

ThereforeZ∗ 3 /(Z 3 ) 3 ∼=(Z/ 9 Z)∗/{± 1 }∼=Z/ 3 Z, and soQ∗ 3 /(Q 3 ) 3 ∼=(Z/ 3 Z) 2. 2

K a field,v:K∗ →Za normalised discrete valuation,πa uniformiser,k=O/πOthe residue field.

Lemma 2. SupposeA⊂Ois a set of coset representatives fork=O/πO. Then

(i) Everyx∈Ocan be written uniquely asx=

∑∞

r=0arπ

rwithar∈A

(ii)Kis complete ⇐⇒ every

∑∞

r=0arπ

r(withar∈A) converges.

Proof. (i) Exercise – sheet 1, question 6.

(ii) (⇒). Lemma 1(iii) – terms tend to 0.

(⇐). Suppose (xn) is a Cauchy sequence inO. Writexn=

∑∞

r=0br,nπ

rfor some br,n∈A.

(xn) Cauchy⇒there isN 0 such that form, n&gt;N 0 , we havexm≡xn(modπ). Sob 0 ,m=b 0 ,n, say =a 0.

(xn) Cauchy⇒there isN 1 such that form, n&gt;N 1 , we havexm≡xn(modπ 2 ). Sob 1 ,m=b 1 ,n, say =a 1.

Continue in this way, gettinga 0 , a 1 , a 2 ,.. .inA.

Thenx=

∑∞

r=0arπ

rconverges andx n→x. SoOis complete. Let (xn) be a Cauchy sequence inK. Then there existsN such that for all m, n&gt;Nwe have|xn−xm| 6 1, and soxn∈xN+Ofor alln&gt;N.

ButOis complete, soxN+Ois complete, so (xn) converges. ThereforeKis complete. 2

The Teichm ̈uller map

Kcomplete w.r. discrete valuationv. Suppose the residue field is finite, say|k|=q.

Putf(X) =Xq−X∈O[X]. Eachα∈kis a simple root off(X) =Xq−X∈k[X].

Hensel (version 1)⇒there is a uniquea∈Osuch that

{

aq=a a≡α(modπ)

.

Definition∈Ois theTeichm ̈uller representativeforα∈k. Writea= [α].

Lemma 2. [·] :k→Ois multiplicative.

Proofα, β∈k. Then

(

[α][β]

)q = [α]q[β]q= [α][β], so [αβ] = [α][β]. 2

Corollary. k∗←֓O∗. E.,μp− 1 ⊂Qp.

Theorem 2.9 w.r. discrete valuationv. If char(K)&gt;0 andkis finite then K∼=k((t)).

Proof(K) = char(k) =p. And|k|=qis a power ofp.

Letα, β∈k. Sincepdivides

(q i

)

for all 0&lt; i &lt; q, we have ( [α] + [β]

)q = [α]q+ [β]q= [α] + [β]

and so [α+β] = [α] + [β].

Therefore [·] :k←K ֓is a field embedding, as [·] repsects addition and multiplication.

By Lemma 2,K=

{∑

n&gt;n 0 arπ r:ar∈k

} ∼

−→= k((t)), viaπ7→t. 2

Proof of Theorem 3(⇐).LetRbe a local Dedekind domain with maximal idealm 6 = 0.

Note: forI ⊂Ran ideal, eitherI⊂morI=R(as every ideal is contained in a maximal ideal).

Step 1 principal.

Pick 0 6 =x∈m. Lemma 3 implies (x)⊃mnfor somen&gt;1. Pick the least suchn, somn⊂(x),mn− 1 6⊂(x).

Picky∈mn− 1 (x). Thenym⊂mn⊂(x), soyxm⊂R, soπ− 1 m⊂R, whereπ=x/y.

Ifπ− 1 m⊂mthen Lemma 3⇒π− 1 ∈R⇒y∈(x), contradicting the choice ofy.

Thereforeπ− 1 m=R, and som= (π) is principal.

Step 2 a PID.

LetI⊂Rbe a non-zero ideal. ConsiderR-modulesI⊂π− 1 I⊂π− 2 I⊂...

Lemma 3 impliesπ−tI 6 =π−(t+1)Ifor allt Noetherian, so there is a largestn such thatπ−nI⊂R. Ifπ−nI⊂m= (π) thenπ−(n+1)I⊂R, contradicting the choice ofn.

Thereforeπ−nI=R, and soI= (πn). SoRis a PID, and hence a DVR. 2

Theorem 3=⇒Theorem 3 (sketch). LetS=R\p, withpprime, and letI, J⊂R be ideals.

We quote the following properties of localisation. - S− 1 Ris a local with, with unique maximal idealS− 1 p. - IfI(pthenS− 1 I=S− 1 R. - I=J ⇐⇒ S− 1 I=S− 1 Jfor all primesp(recallSdepends onp). - RDedekind =⇒ S− 1 RDedekind =⇒ S− 1 Ris a DVR by Theorem 3.

LetI ⊂Rbe a non-zero ideal. Lemma 3 impliespβ 11.. .pβrr ⊂Ifor some distinct primesp 1 ,.. .,pr.

S− 1 I=

{

R ifp∈{/ p 1 ,.. .,pr} (S− 1 pi)αi ifp=pi,some 0 6 αi 6 βi

.

ThenI=pα 11.. .pαrr.

For uniqueness, supposepα 11.. .prαr=pβ 11.. .pβrrwithp 1 ,.. .,prdistinct primes. Taking S=R\piwe get (S− 1 pi)αi= (S− 1 pi)βi, soαi=βiby uniqueness of factorisation in a DVR. 2

LetKbe a field, andL/Ka finite extension of degreen. Forx∈L, writeT rL/K(x) and NL/K(x) for the trace and determinant of theK-linear mapL→L,y7→xy.

Letσ 1 ,.. ., σmbe the distinctK-embeddings֒ L→K. Recallm 6 n, with equality iffL/K separable. In this case,T rL/K(x) =

∑n i=1σi(x) andNL/K(x) =

∏n i=1σi(x).

Lemma 3. IfL/Kis separable then the trace formL×L→K, (x, y)7→T rL/K(xy) is non-degenerate.

Proof=K(x) (primitive element theorem). ThenLhasK-basis 1, α, α 2 ,.. ., αn− 1 , andσ 1 (α),.. .σn(α) are distinct.

Then det

(

T rL/K(αi+j)

)

= det

(

σi(α)j

) 2

6 = 0 – Vandermonde determinant. 2

Remarks.

(i) More generally, it can be shown that for any finite-dimensional commutativeK- algebraR, the trace formR×R→Kis non-degenerate ⇐⇒ R∼=L 1 ×.. .×Lr with eachLi/Ka separable field extension. (ii) (⇐) follows easily from Lemma 3. (iii) Suppose 0 6 =x∈Rwithxm= 0, somem. Then (xy)m= 0 for ally∈R, so T rL/K(xy) = 0 for ally∈R, so the trace form is degenerate.

Theorem 3.7 Dedekind domain,K= Frac(OK),L/Ka finite extension. LetOLbe the integral closure ofOKinL. ThenOLis a Dedekind domain.

Remark caseOK=Zshows that the ring of integers in a number field is a Dedekind domain.

Proof will prove only the case whenL/Kis separable. ForL/Kinseparable, see e. Janusz, Chapter I, section 6.

We must check four things: (i)OLis a domain – done sinceOL⊂L. (ii)OLis Noetherian. (iii)OLis integrally closed – done by properties of integral closure. (iv) Every non-zero prime ideal inOLis maximal.

Proof of(ii). Note: trace form is non-degenerate. Letx 1 ,.. ., xnbe aK-basis forL. Multiplying through by suitablec∈K∗, we may assumex 1 ,.. ., xn∈ OL. Let y 1 ,.. ., ynbe the dual basis w.r. the frace form, i.e rL/K(xiyi) =δij, alli, j.

Givenz∈OL, writez=

∑n i=1λiyi, someλi∈K. Thenλj=T rL/K(zxj)∈OK. HenceOL⊂OLy 1 +···OLyn. AndOKis Noetherian, soOLis a finitely-generated OK-module – (∗). HenceOLis Noetherian.

Proof of(iv).LetP ⊂ OLbe a non-zero prime ideal. Thenp=P ∩OKis a prime ideal inOK.

Pick non-zerox∈P, then 0 6 =NL/K(x)∈P ∩OK=p, sopis non-zero.

OKis Dedekind, sopis maximal, i.e/pis a field.

k=OK/p֒→ OL/Pis injective. (∗)⇒ OL/Pis a finite-dimensionalk-algebra (vector space).

Pprime⇒OL/Pis a finite integral domain⇒OL/Pis a field (apply rank-nullity toy7→xy)⇒ Pis maximal, as required. 2

Let’s assume thatOKis a PID. By the structure theorem of modules over a PID,OL is a freeOK-module, say of rankn.

ThenOL∼=OnK, which implies [L:K] =n, andOL/pOL∼=

(

OK/p

)n ∼=kn.

Then dim

(

OL/pOL

)

=n= [L:K].

In general, letS=OK\p, then replaceOLandOKbyS− 1 OLandS− 1 OK(a DVR). Note: - S− 1 OLis the integral closure ofS− 1 OK. - S− 1 pandS− 1 Piare primes inS− 1 OKandS− 1 OL, respectively. - eiandfidon’t change when we localise. 2

Lemma 3. The diagram S− 1 OL −→ OL/pOL=R

T rL/K↓ ↓T rK/k S− 1 OK −→ OK/p=k

commutes.

Proof’s assume that OKis a PID. ThenOL is a freeOK-module, say with basis x 1 ,.. ., xn. Thenx 1 ,.. ., xnare aK-basis forL, andx 1 ,.. .,xn(reductions modp) are ak-basis forR.

Forz∈OL, havezxi=

∑n i=1aijxjfor someaij∈OK. LetA= (aij).

ThenT rL/K(z) =T r(A) =T r(A) =T rR/k(z).

In general, letS =OK\p, thenS− 1 OL is a freeS− 1 OK-module, say with basis x 1 ,.. ., xn− 1 OKis a DVR – follow proof as before. 2

Proof of Theorem 3 continued.

(i)pramifies inL =⇒ OL/pOLcontains nilpotents =⇒ trace form is degenerate =⇒ ∆(x 1 ,.. .,xn) = 0 for anyx 1 ,.. ., xn∈OL =⇒ ∆(x 1 ,.. ., xn)≡0 (modp) for anyx 1 ,.. ., xn∈OL.

(ii)punramified inL =⇒ OL/pOLis a product of field extensions ofk(separable sincekis finite) =⇒ trace form is non-degenerate =⇒ ∆(x 1 ,.. .,xn) 6 = 0 for somex 1 ,.. ., xn∈OL (3) =⇒ ∆(x 1 ,.. ., xn)6≡0 (modp) for somex 1 ,.. ., xn∈OL. 2

Definition the the idealdL/K⊂OKgenerated by ∆(x 1 ,.. ., xn) for all choices ofx 1 ,.. ., xn∈OL.

Corollary. pramifies inL ⇐⇒ p|dL/K.

In particular, only finitely many primes ramify inL.

Remark a PID, we havedL/K=

(

∆(x 1 ,.. ., xn)

)

, wherex 1 ,.. ., xnis a basis for OLas anOK-module.

Proposition 3.11/Kis Galois, thenG=Gal(L/K) acts transitively on the prime ideals abovep.

Proof not, i. there is noσ∈Gsuch thatσ(Pi) =Pj.

By CRT, we can pickx∈ OLsuch thatx≡0 (modPj) andx≡1 (modσ(Pj)) for allσ∈G.

NL/K(x) =

∏

σ∈Gσ(x)∈Pj∩K=p⊂Pi =⇒ τ(x)∈Pifor someτ∈G

=⇒ x∈τ− 1 (Pi)

=⇒ x≡0 (modτ− 1 (Pi)), contradicting the choice ofx. 2

Corollary. IfL/KGalois, thene 1 =.. .=er(=e, say), andf 1 =.. .=fr(=f, say). And [L:K] =ef r.

Definition groupisGP={σ∈G:σ(P) =P}.

Note.|GP|=ef.

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