Lecture 9

Primary decomposition continued : Fix a meth . ring R .

Recall,N is a primary submodule of M if Assr Mk, is a

Singleton ,ie .

, if MIN is co - primary .

Also,M is co-primary iff H zero din . ron M

,It me M - to } , In>o

sit . rum = O .

Example : If I is an ideal of Rtt . VI is maximal,then

I is primary .Indeed

, suppp.pe/I=lY(AnnRRlI)= HCI) = HUI) = HE} .

Since Assr RII E Supp ,zRII and Assr Rt to, get

Assr RII = LII} .

I o

.Let N

, , Nz be p - primary submodule of M .

-

Then N,n Nz is also p - primary .

Pf : I a canonical injection MIN ,nµ, ↳ MIN,Ot MIN

,.

Also, MIN

,to ⇒ MIN ,nNz FO i

% 4 F Asse MIN,nm ,

E Assr ( MIN ,MIN,) E

,

Assr MIN ;11

hp}.

I

Defn : Let N be a submodule of M .

A primarydecompositionof N is an equation

N = Q,n Qun . . . now

,

where Qi are primary submodules of M .

Such a primary decomposition is irredundant if none of the Qi 'scan be omitted and the associated primes of MIQ ; are

all distinct.

Remark : Using'Lemma I

, any primary decomposition can be simplifiedto an irredundant one .

Lem : If N = Qin . - rn Qu is an irredundaut primary-

decomposition and Qi is pi - primary , then

Assr MIN = Lp , , ooo , Pn} .

Pf : The natural injection May ↳ 0¥,

Mla;gives

Assia Ma, I ¥,

Assa MIQ;= LP , , ooo, Pn} -

Conversely , for IE i En . Qin . . - n Qian Qian . . - Qn/µ ¥0

( by irredundance) .

Exercise : Check the canonical map

Qin . . - n Qi,

n Qin . . -nQnk, → MIQi

is injective .

Ooo 4 F Assr Qin - - - n Qi - in Qin h . . - nQn/µ E Assa Mio; =L P;} .

I. P ; E Assr MIN

II

Upshot : If O E M has an irredundant primary decompositionthen Assam is finite

Proposition : Let N be a p -primary submodule of M and SER-

be multiplicative . Let

g : M → s- ' M

be the canonical map .

C) If p n s F 4 ,then S - ' N = s - ' M

.

c) If pn s =p ,then N is a p -primary submodule of M iff

N = g-' ( N ') , where N

'

is a S- ' p - primary submodule

of 5'M . ( g- ' ( N 't = N ⇒ N' = s - ' N)

Pf : C ) Asssi,

S- '

M/s - in = Asss.ir ( s - ' ( ""ki ) )** Asset'4N= Lp}← LIE Assa Mlm : f n s =p} = of .

ooo s- '

M/s - in = 0.

(2) N is p - primary ⇒ Assr MIN = Lp } ⇒ Ma, F O .

Choose I E MIN - Loy .Then H SES

,Sim t O as otherwise s

would be a zero divisor on MIN ⇒ s e p , contradicting pns = ¢ .

ooo mi et O in S- '

( Mlm) ⇒ s- '

Mls - in ¥0

By HH , Asss.ir s- '

Mls - in, = { S- ' p} .ooo s

- 'N is S- '

p- primary .

Clearly , g-' ( s - IN ) Z N . Suppose me g-

' ( s- ' N) ⇒ my E S- 'N

⇒ F NEN,se s s - t . my = ng ⇒ Ites it . Cts ) m = En EN

.

If my N ,then ts is a zerodivisor on Ma, ⇒ ts E p ,

a

contradiction by TSE S.I . m EN

,ii., g- ' ( S

- 'N) IN .

Thus,

g-' ( s- IN) = N ,

proving the forward implication .

Conversely , suppose N = g-' ( NY , where N

'is S

-'p - primary in S - ' M .

Then

N'= S

- ' N ( exercise)and by 88

,

{ I e Assa MIN : In 5=94 ←> Asss. . , S-'

MIN , - IS - '

p} .

so IF C- Assr Min : Ins - oh = Lp} . ( lol)Let IE Assn Min , and I = Anne (mt N ) .

Then met N .

If I n S F f ,then F SE S it . Sm E N = g-

' (NY

⇒ my E t ( SF ) E N' ⇒ me g-

' (NY = N,a contradiction .

~ ~Thus, p t Nese MIN ⇒ p n s = of .

Hence,

Asar'

MIN = IF C- Assr Min : In 5=4 } Lp} . a