Lecture 3

Lecture 2: ① If I, I are ideals of R, then H CI) EYED

Grotta ⇒ IT E II.

② The bijection of Lecture 2 , Proposition 5 induces a bijection between theirreducible closed subsets of Spec CR) and the prime ideals ofR.

Pf : ① Y CI) EYES) ⇐ anew,Q Ep?µ⇐,P ⇐ IT ETI .

② If PE Speck ,let HCP) = HCI) U H = HCI 's)

,

for ideals I,T of R .

Then P 2 IT.

Exercise : If I , , . . . . In , Q are ideals of R , Q is prime andn

Q 2 IT Iiit

then I i s . t . Q Z Ii .

o? P 2 IT ⇒ P ? I or P z T ⇒ H (p) E X (I)or HCP) E YG) ⇒ HCP) = Y CI) or HCP) = YET⇒ HCP) is irreducible fits to be PE HCM] .

Conversely , if Z is irreducible, lappose Xy Ephezp .

Lec 2

⇒ 2- = Vip?zP) E HCxy ) ⇒ Cx) u My)Props

"""bZ E Hex ) or Z E HCy) ¥ X EIN Ep? P

or y Etty E phezp .Moreover

,2- to ⇒ pezh P FR .

o! N P is prime .

PEZ

a

Units and Jacobson radical 8-

X ER is a unit if 7 getsit . xy =L .

Equivalently , x is a unit ⇐ (x ) = R .

Lemma 1 : x ER is a non- unit ⇐ I a maximal ideal m of R-

s-t. X EM -

tf : Apply Lecture l, Corollary 5 .

I

Recall,nil CR) = if ÷÷n!

-

n P.

Ptspec R

DIE : The TawtonofI , denoted Tac CRI , is the h

of all maximal ideals of R .

Remark : For finitely generated algebras over a field , nil CR )= Tac CR)

'

I : X E Tac CR) ⇐ H YER , ltxy is a unit in R .

Lemma I

1¥ It xy is not a unit ⇒ 7 max . ideal

m sit . Itxy em .

But x Em ⇒ I = fltxy ) - xy Em,a contradiction

.

Let m be a max ideal of R and assume x¢m .Then

Mt Cx) = R ⇒ a y ER , z em s - t . Z = It yx⇒ z is a unit

,which is impossible by Lemma I .

II

Notation : R× ' = multiplicative subgroup of units of R .

Mobiles : I will assume you know basics of module theory ,including tensor products . If not , read Atiyah - MacDonald

chap 2 or take grad algebra .

Def An R - module M is free if I a set I and anR- linear isomorphism ROTI I M .

M is generated by {mili⇐ EM if the unique R- linear map

q : ROTI →Mei1-7 mi,it I

is surjective .

In other words, every element of M can be expressed as an R -

linear combination of some finite subset of Lmi 's ice .

Nakayama 's lemma 8 Let x be an ideal contained in Tac CR) .

-

If M is a finitely generated R-mod it .

M = AM,then M = 0

.

1¥ Assume M t O and let {mi , . . . , mn} be a minimal generatingSet of M as an R - mod .

Since M = x M,I a

, , ooo , an Ex s - t .

m,= a

, m,t o . . t an Mn

⇒ ( I - a , ) m ,= azmz t ooo

t an Mn- I

⇒ m,= C - a , ) ( azmzt . . . tan Mn )

Lemma 2

⇒ m, E R - linear span of Lma , . . . , mn } contradicting minimality .

ooo ML = 0 .

A

Most commonly used in the setting of local rings .

Def A local ring (R , m) is a ring with a uniquemaximal ideal m .

A Semite ring is a ring with

finitely many maximal ideals.

GmmonformofNakayamaislem.mu# If M is a finitely generatedmodule over a local ring

42 , m ) sit . M = MM,then M = O .

Pf : Exercise.Hint : what is the Jacobson radical of a local

ring ?

Another form of Nakayama 's lemma : Let a E TacCR) and M be-

a finitely generated R- mod.

If N is anR- submod oof M sit . M = N t x M,then M = N .

Pf : Exercise.

Hint : Look at MIN .

Recall that if 11,W are vector spaces over a field k s- t .

V petal =0then 11=0 or 1×1=0 because V

,W are free K- modules .

We get an analogous result for finitely generated modules over

a local ring , even when such modules are not a priori free .

Corollary 3 : If 42 , m) is a local ring and M,N are finitely

-

generated R- modules such that

M ④RN = 0

then M = O or N = O .

Pf : M*N = o ⇒ R1m④r( Mox, N ) =O

See

⇒ (Rim ④←M ) ④p,m(RlmQrN) = O

exercise below

Rtm⇒ Mlm µ ④ Rim Nim N

= O ⇒ Mlmpq = O or N/mµ=0is a field

Nakayama⇒ M = O or N = 0 .

I

Exercise (Properties of tensor products) : Let M,N be R - mods and-

R→ S be a ring map -

① If I is an ideal of R ,then RII Qr M = MIM .

② Sa (MQRN) E (Sox,zM) ④s (SQRN) .