primary - university of illinois at chicago
TRANSCRIPT
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