Lecture to

1 Integralclosurecont'd2 LocalizationBONUS ProofofThm I in Section1.3Refs AM Section5.3 Chapter3 V3 Section 9.3

1 A is comm'veunital ring B is an A algebra RecallSection2of Lecture 9 the integral closureAB beBI 6 is integral over A A subalgebra in B

PO Fraction fieldsSuppose A is adomain We can form its fractionfield FraeABy definition it consists of fractions w a beA 6 eIn Free A we have f f ad be Thefractions areadded multiplied in the usual way A directcheck showsthat Free A is a fieldNote that A has a ring embedding into Free A at G

Moreover every embedding q Acs L factors asA Free A L where the latter is 849476Examples Free T2 Oh

For a field I we have Frac FCx xn Alg xnthe field of rationalfunctions in n variablesLet deTL not a completesquare Then TL52 is adomain

IpFree I 52 Q 52

I11 NormaldomainsLet A be a domain

Definitioni The normalizationofA is A Freed integralclosureofA in Free Aill A is normal if A coincides w its normalization

Special cases1 L is a field ACL is a subring Claim It is normalIndeed It is integrclosed in L Free At CLIt closed in Frac AT2 UFD normal let A be UFD GEfree A w

coprime abeA Need to show f isintegral over AGEA i e 6 is invertible Let f x Xtc Y't t GXtcCiEA best f f so 0 61719 a'sEEGai6k

i

The sum isdivisibleby 6 So a b Since a 26 are coprime thisimplies that 6 is invertible

Exercise Let L be a field AgAzCL benormal subringsProve

that AAAz isnormal Prove the analogous claimforinfiniteintersections

12 Examples of computation of integral closure

I

Example1 compute theintegral closure of A TLin Q 52LeTL is a squarefree number Weneed to understandwhen

fed a p at652 la be0h is integral ever ILemmat TFAEi f is integral ever Iii La AZ622ETL

Proof We set 5 a 652 Note that ftp 2a ppa 6d rational numbers So xp xp x 2axt a b'dISo ii i

On the otherhand notethat peg is a homomorphismTL a TL 25 So for fateTax wehave Ftp ftp.TSoif Fcp o thenftp.t e Inparticular if p is integral over TLthen p is integral By Proposition 1 ofSection2 ofLectures

ftp.pp ed are integral over TL But TL is UFD hencenormalSo elements of A integral ever I are integers ii fellows I

Exercise Numberthy If 2 2 or 3 medo then in

Gbe TLif 2 1modsthen in either a b et er s be It

Corollary i TL 52 isnormal 2 2 er 3 mod tIf 2 1mad4 then thenormalin of TL 52 is

at652la be E or a be It

ii 7155 is normal but not UFD

Example2 Consider thering As Elxy y fall fweek

Exercise This ring is adomain fx is not acompletesquare

Factgoingbeyondwhat we cover inthisclass A isnormalf x has no repeatedroots the curve xy ly fix is

nonsingular computethe Jacobianvector of xy tsy fx

Weconsiderthespecial case f x x anddescribethenormalizationinthis case Let'swrite I j for the imagesof xy in A

Exercise the elements Ik Iky for k e form a basis in A forany f in fact

Lemma2 Theintegral closure of A in K Frac A coincides w

Lyle and so is isomorphic to thepolynomialalgebra InparticularA isn't normal

Proof Wehave g I in A So for t gleek wehave t Id

t is transcendental ever e Wehave I t J t so Act Butfor anydomain A CK haveFree A CK Since Acthave KFree A Since A isgeneratedby an element integraloverA

pits integral over A Aswe've seen in Section 1.1 A is

normal So A is thenormalization of A andA is notnormalo

13 Finiteness of integralclosuresLet A be adomain K Free A Kc L finitefieldeatinh Is A finite ever AA It's complicated

Theorem Assume thatone ofthefollowingholdsI A is Neethin normal char K OII A is afingen'dalgebra over a field or ever TL

Then At is finite ever APreet under I will appear as a bonus

Example let A TL L is afiniteextensionofOh TheringAt is called the ring of algebraic integers in L cruciallyimportantfor AlgNumberthy Both I II applyso Thm A is finite ever I i.e is a fingendabeliangroup

Fact Premium exercise As an abeliangroup A's TL where

n dimalHint observe that since At is a domain na o for let LeAlimplies n e or 2 0 so A's TL forsomen Thenobserve thatBEL I KE The s t Kpe A

2 LocalizationLocalizations of ringsgeneralize thefractionfields of

domains Toconstruct a localization of A we needasuitablesubset in A

Definition Asubset SCA is multiplicative ifYESs te S Ste S

Examplese S all invertibleelements in A ismultiplicative1 If A is a domain then S A a is multiplicative1 Forgeneral A S all nonzerodivisorseA is

multiplicative2 For fed thesubset S f Inna is multiplicative2 For f f eA S f fi ni a is multiplicative3 LetpCA beprime ideal S Alp is multiplicativeres rep s te S step step thelast implication uses thatp isprime stes

I

BONUS Proofof Theorem in Section 1.3 underassumption IProof Let Lime En Every element Le LgivestheK linearoperator say Ma on L via multiplication So for Le L itmakes sense to speak about tr a tr Ma Ek

Steph Weclaimthat for Le A we have trine ALet fateAG be a monicpolynomial w f x so Choose analgebraic extension I of L where fat decomposesinto linearfactors All eigenvalues ofMa are roots offix hence are

integral over A Therefore tr k the sum ofeigenvalues isintegral over A But tr a Ek andsince A is normal we seetrine AStep2 For aje l define sp trap This is a symmetric

K bilinearform x L K Weclaim that since char ksothis bilinearform is nondegenerate Moreprecisely for nellaF mo s t Cuun trcumi te Let u u u un betheeigenvaluesofmu counted w multiplicities Then tr um Ehaim If therh s s are l for all m then thx to char Eschak d we

get up Un o which is impossible since U e

Step3 Thx to the exercise in Section P3 we can finda Kbasis l en of L w lie A Let et e bethedualbasis w rt i.e tr lies Sig it exists 6k is

nondegenerate Let M Span et en Note that forLEA we have 2E.la lie Wehave gli trCali EAb c LEGA So LEM ATCM But A is NoetherianA

andM ismanifestlyfinitelygenerated A module HenceAt is afinitelygenerated Amodule and we are done A

Corollary Theorem is also trueunder assumption II ifA is a finitelygenerated F algebra charEseSketchofproof Thanks to the Noethernormalizationlemmewe can replace A w FIX Xm for some M thisdoesn'tchange At exercise Now we are inthe situation of Iof the theorem 5

For a proofof II w a finitelygeneratedalgebra overan arbitrary field see E Section 13.3