Lecture 5

1 Noetherian rings modules

2 Hilbert's Basis theorem

3 Noetherianmodules overNoetherianringsReferences AMI ChapterG introto Chapter 7 E Section1.4BONUS NonNoetherianrings in Complex Analysis

Why Hilbert cared

1 A is commivering1 1 Main definitionsKexamplesDefinition

4AnA module M is Noetherian if Asubmodule ofMincludingM is finitely generatedin A is a Noetherian ring if it's Noetherian as amoduleover itself i e every ideal isfinitelygeneratedExamples

e Every field F is Noetherianring ideals in F are o F a

1 As TL is Noetherian ble t ideal isgeneratedby oneelement

Nonexample FLxi ie for infinite I

12 Equivalentcharacterizationsof NoetherianmodulesDefinition M is A module By an ascendingchain AC of

qubmodulesofM we mean collection Nilien ofsubmodules of

M St Ni sNit ti o N SNasN SWesay that the AC Nilien terminates if I k est

Nj Nk Aj k

Proposition For an Amodule M TFAE1 M is Noetherian

2 AAC ofsubmodulesofM terminates3 A nonempty set of submodules ofM has amaximal

element w v t inclusion

Proof 2 3 iseasy generalstatementaboutpositsexercise

a 2 AC Ni eN SN S N sikNi is a

submodule compare toproofof F ofmax ideal 2.2ofLec2This N is fingenid so I ma MeeN W N Spang Ma MeNow MieNic forsomeKci Me MeeNk for K Max Klis bcNuncN thxto AC condition N SpanCm me NKso AC Ni terminates atNe2 A Know A AC ofsubmodules terminates Let Nbe a submodule that is nt fingenerated constructNi's by induction pick M EN N SpangmisAme Now

suppose we've constructedMa Mi EN Ni Spangme MiN isnot fingen N Ni F mi E NI Ni setNitSpang me Min Ni So Ni is is AC doesn'tterminate Contradiction II

2 1 Hilbert basistheoremIt turns out that there are a lot ofNoetherianrings infactmost rings we are dealingwith are Noetherian Thefollowingis a basic result in thisdirectionThm Hilbert If A is NoetherianthenACNis NoetherianProof Let I CAN be an ideal Assume it's notfinitelygenerated We construct a sequence of elementsto fie EI asfollows f is an elementof I withminimalpossibledegree Oncef fie are constructed we cheese face I f fi this set is

nonempty bc I fanfic again ofminimalpossible degreeFor Kee define areA niceTL from fi ax lowerLegterms

By the construction hshes shas New let I g an CA kidThis is an ascending chain of ideals in A Since A is Noetherianit must terminate So am c g am am Ebiai forsome mSetgm sfm Eybix minify EI ble fanfmEI Amini aCameInbiai x tlowerLeg

termsLeggmacdegfm

Sobythechoiceof fmiofminimaldegreegm elffm Sincegm EI we

have fm elf fm contradictionD

22 Finitely generatedalgebrasNowweproceed to ageneralizationofthe basistheoremDefinition Let B be an Aalgebra Then B isfinitelygeneratedas an A algebra if I 6 GEB s t beB FEAlg xdt b Fly be

Hence P A x N B F to fly bid issurjective So Bis finitelygenerated over A it's thequotientofAlg xfor some K

Corollary Let A beNoetherian Bbe a finitelygeneratedAalgebra Then B is a NoetherianringProof Use Hilbert's Thm k times to see that AEx xD isNoetherian Let I B be idealneedto show it'sfingendJ P I CAlte this is ideal so J Fa Fe ButthenI 90 J ECE 4 Fe is finitelygenerated 5

Since fields F 272 are Noetherian rings anyfinitelygenerated F or I algebra isalsoNoetherianIn fact as we will see later many constructions e.glocalizationand completion produce Noetherian ringsfromNoetherian rings This is whyNoetherianrings are so wide

spread

3 Noetherian modules over NoetherianringsThefollowing result to beprovednexttime comparesthe

property ofbeingNoetherianfor M its subs quotientsProposition let M be Amodule NCMbe a submoduleTFAE 4 M is Noetherian

2 Both N MINare NoetherianI

We now proceed to characterizing Noetherianmodules overNoetherian

rings Ingeneral Noetherian fingenidBut when AisNoetherian we alsohave

Corollary Let Abe Noetherian Then A fin genidAmoduleMis Noetherian

Proof By 2.1 ofLeck M is a quotientof A By 1 2

of Proposition it's enoughtoshow A is Noetherian Since Ais Noetherian it's enough to check that thedirect sum of2 Noetherianmodules say MaMa is Noetherian

thenwe'llbedoneby induction Note that wehave inclusion M GM MeM HMO projection M MissMe mama tome whosekernel isthe

image ofM so MAM M Me We use 2 a ofPropositionto conclude MOM is Noetherian 5

BONUS I NonNoetherianrings in Complex analysisMost ofthe rings we deal with in Commutativealgebra

are Noetherian Here is however a very natural example ofanon Noetherian ring that appears in Complex analysis

Complex analysis studies holomorphic a.k.a complexanalyticor complex differentiablefunctions let Hel E denote

the set of holomorphicfunctions on G These canbe thoughtaspowerseries that absolutely converge everywhereHel E has a natural ringstructure vie addition multi

I

plication offunctionsPreposition Hd E is not Noetherian

Preet Weillproduce an AC of ideals Ig foe 76calf 204Fi k so t integerKaj jets It's easy to checkthat all ofthese are indeedideals It is alsoclearthattheyform an AC when we increase j we relax the conditionon zeroes We claim that Ij 4 Ij hencethisACdoesn'tterminate Hd E is not Noetherian Equivalently weneedto show that foreachj there f6 e7ld Q suchthat f204Fi k so t k j while tj120Fig eConsiderthefunction f z é t Thisfunction isperiodic withperiod 205T Alsofee E itZSo z so is an order f zero offeel Since 20ft is aperiodevery 20 FP K KER is an orderAzero We set

f z et 1 E 204Fij Thisfunction is stillholomorphicon theentire 0 we have Fj 2045.1j to E f 21TFi kso for Kaj A

BONUS It WhydidHilbert care about theBasistheoremHilbert was interested in Invarianttheory oneofthecentralbranchesof Mathematicsofthe Nth century Let Gbe agroupacting onfindimAvectorspace V by linear transformations g begWe want to understand when two vectors v v lie inthesameorbitI

Definition Afunction f V G is invariant if f is constanton orbits fig fol fgeG vetExercise Y keV lie in the same orbit fcu fly finvariantfunction f we say Ginvariants separate GorbitsUnfortunately all invariant functions are completely out ofcontrol However we can hope to controlpolynomialfunctionsThese are functions that are written aspolynomials in coordinatesof u in a basis if we change a basis then coordinateschangeviaa linear transformation so if a function is apolynomialin one basis then it's apolynomial in every basis The Galgebraofpolynomial functions will bedenotedby ACV itdim knthen a choice ofbasis identifies KEV withCfx tnBy GIV we denote thesubsetof Ginvariantfunctions in NvExercise It's a subring of Ecu

Exampled Let V G G Su thesymmetricgroupactingon Vbypermuting coordinates Then GUT consistsprecisely ofsymmetricpolynomials

Example2 Let Us0 G E f Ake w.at multiplicationLet G act on Vby rescalingthecoordinates t g k s

Ha txt We have fly xie duh feta ta fix xH ted I the Q This isonlypossible when f isconstantAs Example 2 showspolynomial invariantsmay fail toseparate

orbits However to answer our originalquestion it's still worth

to studypolynomial invariantsPremium exercise When G isfinite thepolynomialinvariantsstill separate CorbitsNow suppose wewant to understand whenfor V.V EV

we have for flu f fe Gcv It's enoughtocheckthisforgenerators fof the Q algebra cut So a naturalquestionis whether this algebra is finitelygeneratedHilbertproved this for reductivealgebraicgroups Ghedidn't know theterm butthis iswhathisproofusesFinitegroups are reductivealgebraic andso are Gln E

thegroup of all nondegeneratematrices Sh O matricesof determinant 1 Once orthogonalmatrices andsomeothers for these infinitegroups one needs to assume thattheir actions are reasonable in someprecisesense Latermathematiciansfoundexamples where thealgebra ofinvariantsare not finitelygenerated counterexamples to Hilbert's16thproblemBasis theorem is an essential ingredient in Hilbert'sproofoffinitegeneration For more details on this seeE TGP PS P3 containssome more background onInvariant theory

A