Lecture 8

Continuation ofprooffrom last lecture

See refsfor lect

1 Reminder A is PID M is a finitelygeneratedA module

Thm fromLec F 1 F KER primespainPeeA da dethes t

Me e Apti2 K pt Pete are uniquely determinedby M

In Lect we'veproved 14 18 e A fi fort.fmeAlle

11 Finishingpart 1 of Thin

Proposition Let f eAke f ep pitwhere E isinvertiblepal pr are pairwisedistinctprime maximal ideals

Then A f I E A pti as Amodules

Lemma Let A be any comm've unital ring I ICAbeideals If I IgA thenhave an Isomin

A I I A I A I of Amodules andofringsH

ProofofPrepn moduleLemma induction on r w Fo nothingtoprove For r P I pi Ia pi pin Elements

pipi pt are coprime theirGCD P A is PID

It Is A Apply the lemma A f A I I A I A IsBy induction on r have A I A pi lo A pi whichfinishes theproof I

Proofof lemma I IEA Face I latestSteph We construct a homomorphism A I Ia ALI A IzHi A I I A Ii Ci P2 at I 124 at Ii makessense ble I I Ii Then we defineIT A I A I x A Iz R H X homomorphismofboth rings and of A modules

Step2 JT is injective kero kero A Ker Iz HaveKerto I I I is1,2 So keroAkerD I AIa I IzSowhat we need toprove is I II I A IzAEI n Ians e at g I a grey ha t aez Since he IsG E I have ha EI Iz Similarly age I I use a e ISo we have a e I I whichproves I I I AIsStep3 prove it is surjective ie FGGEA F GeAst6 I Iz 6 I beIz 6 bi e Ii for it 2

Wecheck that 6 abt 96 works E.g for Et wehave6 6 g 1 6 tabs qt qt hit a qbtake IGk GE I DI

Rem Thelemma is a special K 2 case of a meregenieclaimIn I W Iit Ij A for it A I I ItA IiChinese remainderthm

12 Proofofpart2 of Thm uniquenessFix a prime ideal p CA se ThConsider psM p M an A submoduleof M It hasa submodule pls'M so can consider thequotient module

PsMpsM Note that p M p p M so p'MpsMis annihilated by p and hence can be viewed

as a module over A p As we've remarked in Sec 1.2 ofLect p is amaximal ideal in A Hence A p is a fieldSince M is finitelygenerated so are pSM pMlp mThereforethe latter is a finitedimensionalvectorspace overA pNotation LpsM dim

ploppsmpsM

Recall that Me At e A ptiProposition Lps M Kt it pi p di s

Once we know the numbers on the right 2 isprovedthe number of occurrences of A ps is dps m dps M

gland K Lps M for all s sufficiently large

Proofof PrepnStepp explain howdps behaves on direct sums

Claim Lps M Mz LpsMi t Lps MaPreetat the claim

PsMOM p'MAp'M as submodules in MOM w

PSYM Me s painpain p MicpsMi

PsMeme ps M MeI psmp M ep'MIpsMe

and the claimfollows the dimension ofthedirectsum ofvector spaces is the sum of dimensions ofsummands

Step2 Need to compute Lpsofpossible summandsofMA A pt A qt g p

i A

I EPIA is a module isomorphism

p PsApsalpsaA A p as vectorspaces

over thefield A p Idps A 1it A pt M if s t psm a dpsMt so

if set ps a pt so

p'MIps M'spsap A as Akp modulesso dps M p

A

iii M A qt but gp are coprime so gt p A

psM s

p m M psmps M so

Summing the contributionsfromthesummand together wearrive at the claimof thetheorem 5

Example A F x F is dg closedfield M finitedime Fk o p x 7 he F X is theoperatorgivenbyx

PSM Im X XI Lp M re X AI rk X 71

Corollaryofpart 2 Twomatrices XDeMatn A are

similar r X XI rk XXI H XE f se 740similar matrices a isomorphic FEA modules

2 Upnext We've seen a bunchof constrins ofringsdirectproducts

ringsofpolynomialsquotientringscompletions AWA

In the next couple ofweeks or so we'll covertwo moreconstructions

Integral extensions closures of rings Thisgeneralizesalgebraic extensions closuresfromfield theory and

is motivatedby Algebraic Number theory theringsofalgebraic integers are obtained as integralclosuresofTL in finite fieldextensions ofQLocalizations of rings andmodules

Toprepare for our discussion of integralextensionsofringsrecall that if Kc L are two fields thenone can talk about

n L beingfinitelygenerated as afield ever K2 L being algebraic ever k3 L beingfinite over K dim L o

New let A be a commutative ring and B be a commutativeAalgebra note that we donot require that the correspondinghomomorphism A B is injective In Lecture 5 wehavedefined what it means that B is afinitelygeneratedas an A algebra This is an analog of a An analogof3 is as follows

Definition Wesay that B is finite over A if it is afinitelygenerated Amodule

Ananalog of beingalgebraic B is integral ever A will bedefined in the next lecture

I