Lecture 8

Recall, PE Speck is an associated prime of an R- mod M ifI me M s - t - p = Anne (m ) ⇐ 7 injection Rip ↳ M .

Lemma 1 : Let M be an R -module .

=L ) If N is a submodule of M ,then

Assr N E Assr M I Asse N u Assa MIN .

E) If O = Mo EM ,E . . . E Mn = Nl is a chain of R - submodule,

then n

Assia M E U Asse Militia .

i -- i

Pf : E) follows by C ) using induction on n.

C) Clear that Assr N E Assam .

Let PE Assam and me M

sit .

p = Annp

m.

If Rm n N f- O ,choose RER lit . rm # O and rm E N .

Exercise : Show Anna nm =p and to p E Assr N .

If Rm n N = O ,then consider in := mt N E MIN . It is

clear that

p E Ann pent .

Let rt Ann ,zm- ⇒ rm EN .

But rm e Rm n N =0

⇒ re Ann em =p . 8 . Anne in =p ⇒ PE Assr MIN .

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Theorem 2: Let R be a meth . ring and M be a fin . gen .R-mod

.

=

Consider a filtration0 = Mo E M E . . . E Mn = M e. t . H l E ie n

,

Mihai,

I RIP;,for Pi C- Speck .

Then Assa M E Lp , > ooo , Pn} E Suppa M .

Exactness

Pf : Mi 1mi. .

I RIP ; ⇒ (Mi )Pi/(Mie ,)p;

= (" ilMi. . )p;of localization

= Frae ( Rip ;) to ⇒ O # (Milp;

E Mp ;

⇒ pi E Suppa M ⇒ Lp , ,. . . ,Pn } I Supp

,M

.

Exercise : If Pe Speck ,then Assr Rip = Lp} .

Then Lemma I ⇒ Assr M Ei!U,

Asse Mihai. .

= Assr Rtp ;

= Lp , ,. . .

, Pn} .

II

Coro : If M is a fin - gen .module over a meth . ring R ,

-

then Assam is finite .

Pf : Apply Theorem 2 above t Lecture 7, Corollary 6 .

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Proposition 4 : Let R be meth. and M be an R- mod .Then

#

C ) Assr M E Supp ,zM .

(2) If PE Supp ,zM ,I q E ASSRM sit. 9 Ep .

③ Assr M and Supp, M have the same minimal elements .

Pf : C ) follows by Lee 't,Lemma 7 -

El PE Supp, M ⇒ Mp to ⇒ Asspep Mp ¥ 0 .

SinceLee7

,Lem 8

Assr,Mp = Lg E Assam : g E p} ,

we win tot

(3) Follows from C ) t (2) ( convince yourself ! ! ) .

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Corollary 5 : If R is meth.and M is a fin . gun . R -module

,=

then the minimal elements of Assam are preciselythe minimal primes of Ann ,zM .

Pf : Apply Proposition 4 t fact that M f.g .

"

Supp,zM= ly ( Annam) .

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Example : Assway,

Klay) 4×2 ,×y,has unique minimal

element (x ) .

Last time we saw 4. y ) E Asskcx.pkk.DK#xgg .

Upshot : Assam may have primes that are not minimal in

Supp ,aM .

Def " : The non- minimal elements of Assam are called

embedd@primes.Defni.rERisazcroisorowMif7mfOe.t . rm=O.

Proposition 6 : Let R be a meth . ring and M to be an R-mod .

⑧

#

Then

{ zerodivisor on M } = U p .

PEASSPNM

Pf : you will show this in HW 2.

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Primarydecompsitinoo Fix a meth . ring R .

Def " : An R-mod M is coprimary if it has only one associated

prime .

A submodule N is a p - primary submodule of M-

if Asse Ma, =hp} .

Dropout TFAE :

① M is co -primary .

② M FO and if re R is a zero divisor on M ,then

r is lpotmt, ie , H MEM ,

In > O et . Mm -

-O

.

Pf : ① ⇒ ② : duppose Assen = Lp 's.

Then M¥0 .

Let RER sit . rm -

- O for me M - Lot . By Proposition 6,re p .

Take any x E M - Lo} .

Then f f Assr Rx E Assr M -

- hp} .

⇒ Asse Rx = Lpl p is the unique min. prime of Ann ,zRx

⇒ t¥R =p bio VI = n min - primes of I

⇒ rep = iA ⇒ F n> 01 rn×=0 ⇒ r is locallynilpotent .

② ⇒ ① o

.Let p :=L re R : r is locally nilpotent of M} .

Exercise : p is an ideal of R .

By Prop . 6 t hypothesis of ② , q E Assen ⇒ 4 I p .

Let g = Anup m.Then rep ⇒ F n > o at . r

"m -_ o

O⇒ rn Eg ⇒ req - oo p I 9 -

T. It ge Assam , g =p ⇒ Assam = Lp } .

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Remark : locally nilpotent = nilpotent if M=R.

Corollary8 : Let R be a meth . ring and I be an ideal.

-

TFAE :

① RII is co - primary ( ie . I is primary to R).

② H x.y ER , xy E I ⇒ x C- I or y E ITI .

Remark : A primary ideal I is usually defined using ② in the

literature . See Atiyah - MacDonald .

Pf : ① ⇒ ② : xyt I and x¢ I ⇒ y is a zerodiv . on RIIPEPE y is locally nilpotent on Rye ⇒ F n> o e - t .

yn ( ITI) = O

⇒ yn c- I ⇒ YETI .

⑦ ⇒ ① : Show every zero din . on RII is locally nilpotent using② and apply Prop 7 .

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