50 Andre Gthmnn 6. LOCALIZATION Locliztion i very powerful technique in commuttive lgebr tht often llow to reduce quetion on ring nd module to union of mller locl problem. It cn eily be motivted both from n lgebric nd geometric point of view, o let u trt by explining the ide behind it in thee two etting. Remrk 6.1 (Motivtion for locliztion). () Algebric motivtion: Let R be ring which i not field, i. e. in which not ll non-zero element re unit. The lgebric ide of locliztion i then to mke more (or even ll) non-zero element invertible by introducing frction, in the me wy one pe from the integer Z to the rtionl number Q. Let u hve more precie look t thi prticulr exmple: in order to contruct the rtionl number from the integer we trt with R = Z, nd let S = Z\{0} be the ubet of the element of R tht we would like to become invertible. On the et R S we then conider the equivlence reltion (,) (, ) = 0 nd denote the equivlence cl of pir (,) by. The et of thee frction i then obviouly Q, nd we cn define ddition nd multipliction on it in the expected wy by + := + nd :=. (b) Geometric motivtion: Now let R = A(X) be the ring of polynomil function on vriety X. In the me wy in () we cn k if it mke ene to conider frction of uch polynomil, i. e. rtionl function X K, x f (x) g(x) for f,g R. Of coure, for globl function on ll of X thi doe not work, ince g will in generl hve zeroe omewhere, o tht the vlue of the rtionl function i ill-defined t thee point. But if we conider function on ubet of X we cn llow uch frction. The mot importnt exmple from the point of view of locliztion i the following: for fixed point X let S = { f A(X) : f () 0} be the et of ll polynomil function tht do not vnih t. Then the frction f g for f R nd g S cn be thought of rtionl function tht re well-defined t. In fct, the bet wy to interpret the frction f g with f R nd g S i function on n rbitrrily mll neighborhood of, uming tht we hve uitble topology on X: lthough the only point t which thee function re gurnteed to be well-defined i, for every uch function there i neighborhood on which it i defined well, nd to certin extent the polynomil f nd g cn be recovered from the vlue of the function on uch neighborhood. In thi ene we will refer to function of the form f g with f R nd g S locl function t. In fct, thi geometric interprettion i the origin of the nme locliztion for the proce of contructing frction decribed in thi chpter. Let u now introduce uch frction in rigorou wy. A bove, R will lwy denote the originl ring, nd S R the et of element tht we would like to become invertible. Note tht thi ubet S of denomintor for our frction h to be cloed under multipliction, for otherwie the formul + := + nd := for ddition nd multipliction of frction would not mke ene. Moreover, the equivlence reltion on R S to obtin frction will be lightly different from the one in Remrk 6.1 () we will explin the reon for thi lter in Remrk 6.4. Lemm nd Definition 6.2 (Locliztion of ring). Let R be ring. () A ubet S R i clled multiplictively cloed if 1 S, nd b S for ll,b S.
(b) Let S R be multiplictively cloed et. Then 6. Locliztion 51 (,) (, ) : there i n element u S uch tht u( ) = 0 i n equivlence reltion on R S. We denote the equivlence cl of pir (,) R S by. The et of ll equivlence cle { } S 1 R := : R, S i clled the locliztion of R t the multiplictively cloed et S. It i ring together with the ddition nd multipliction + := + nd :=. Proof. The reltion i clerly reflexive nd ymmetric. It i lo trnitive: if (,) (, ) nd (, ) (, ) there re u,v S with u( ) = v( ) = 0, nd thu uv( ) = v u( ) + u v( ) = 0. But thi men tht (,) (, ), ince S i multiplictively cloed nd therefore uv S. In imilr wy we cn check tht the ddition nd multipliction in S 1 R re well-defined: if (,) (, ), i. e. u( ) = 0 for ome u S, then u ( ( + )( ) ( + )( ) ) = ( ) 2 u( ) = 0 nd u ( ( )( ) ( )( ) ) = u( ) = 0, nd o + = + nd =. It i now verified immeditely tht thee two opertion tify the ring xiom in fct, the required computtion to how thi re exctly the me for ordinry frction in Q. Remrk 6.3 (Explntion for the equivlence reltion in Definition 6.2 (b)). Compred to Remrk 6.1 () it might be urpriing tht the equivlence reltion (,) (, ) for S 1 R i not jut given by = 0, but by u( ) = 0 for ome u S. Agin, there i both n lgebric nd geometric reon for thi: () Algebric reon: Without the dditionl element u S the reltion would not be trnitive. In fct, if we et u = v = 1 in the proof of trnitivity in Lemm 6.2, we could only how tht ( ) = 0 in ( ). Of coure, thi doe not imply = 0 if S contin zero-divior. (On the other hnd, if S doe not contin ny zero-divior, the condition u( ) = 0 for ome u S cn obviouly be implified to = 0.) (b) Geometric reon, continuing Remrk 6.1 (b): Let X = V (xy) in A 2 R be the union of the two coordinte xe in the picture on the right, o tht A(X) = R[x,y]/(xy). Then the function y nd 0 on X gree in neighborhood of the point = (1,0), nd o y 1 nd 0 1 hould be the me locl function t, i. e. the me element in S 1 R with S = { f A(X) : f () 0}. But without the dditionl element u S in Definition 6.2 (b) thi would be fle, ince y 1 0 1 0 A(X). However, if we cn et u = x S, then x(y 1 0 1) = xy = 0 A(X), nd hence y 1 = 0 1 S 1 R deired. Remrk 6.4. Let S be multiplictively cloed ubet of ring R. () There i n obviou ring homomorphim ϕ : R S 1 R, 1 tht mke S 1 R into n R-lgebr. However, ϕ i only injective if S doe not contin zero-divior, 1 = 0 1 by definition only implie the exitence of n element u S with u( 1 0 1) = u = 0. We X ( )
52 Andre Gthmnn hve lredy een concrete exmple of thi in Remrk 6.3 (b): in tht ce we hd y 0 R, but y 1 = 0 1 S 1 R. (b) In Definition 6.2 () of multiplictively cloed ubet we hve not excluded the ce 0 S, i. e. tht we wnt to llow diviion by 0. However, in thi ce we trivilly get S 1 R = 0, ince we cn then lwy tke u = 0 in Definition 6.2 (b). Exmple 6.5 (Stndrd exmple of locliztion). Let u now give ome exmple of multiplictively cloed et in ring R, nd thu of locliztion. In fct, the following re certinly the mot importnt exmple probbly ny locliztion tht you will meet i of one of thee form. () Of coure, S = {1} i multiplictively cloed ubet, nd led to the locliztion S 1 R = R. (b) Let S be the et of non-zero-divior in R. Then S i multiplictively cloed: if,b S then b i lo not zero-divior, ince bc = 0 for ome c R firt implie bc = 0 ince S, nd then c = 0 ince b S. Of prticulr importnce i the ce when R i n integrl domin. Then S = R\{0}, nd every non-zero element i unit in S 1 R, with invere. Hence S 1 R i field, the o-clled quotient field QuotR of R. Moreover, in thi ce every locliztion T 1 R t multiplictively cloed ubet T with 0 / T i nturlly ubring of QuotR, ince T 1 R QuotR, i n injective ring homomorphim. In prticulr, by () the ring R itelf cn be thought of ubring of it quotient field QuotR well. (c) For fixed element R let S = { n : n N}. Then S i obviouly multiplictively cloed. The correponding locliztion S 1 R i often written R ; we cll it the locliztion of R t the element. (d) Let P R be prime idel. Then S = R\P i multiplictively cloed, ince / P nd b / P implie b / P by Definition 2.1 (). The reulting locliztion S 1 R i uully denoted by R P nd clled the locliztion of R t the prime idel P. In fct, thi contruction of loclizing ring t prime idel i probbly the mot importnt ce of locliztion. In prticulr, if R = A(X) i the ring of function on vriety X nd P = I() = { f A(X) : f () = 0} the idel of point X (which i mximl nd hence prime, ee Remrk 2.7), the locliztion R P i exctly the ring of locl function on X t in Remrk 6.1 (b). So loclizing coordinte ring t the mximl idel correponding to point cn be thought of tudying the vriety loclly round thi point. Exmple 6.6 (Locliztion of Z). A concrete exmple, let u pply the contruction of Exmple 6.5 to the ring R = Z of integer. Of coure, loclizing t Z\{0} in (b) give the quotient field QuotZ = Q in Remrk 6.1 (). If p Z i prime number then locliztion t the element p in (c) give the ring { } Z p = p n : Z,n N Q, where loclizing t the prime idel (p) in (d) yield { } Z (p) = :,b Z, p b Q b (thee re both ubring of Q by Exmple 6.5 (b)). So note tht we hve to be creful bout the nottion, in prticulr ince Z p i uully lo ued to denote the quotient Z/pZ. After hving een mny exmple of locliztion, let u now turn to the tudy of their propertie. We will trt by relting idel in locliztion S 1 R to idel in R.
6. Locliztion 53 Propoition 6.7 (Idel in locliztion). Let S be multiplictively cloed ubet of ring R. In the following, we will conider contrction nd extenion by the ring homomorphim ϕ : R S 1 R. Proof. () For ny idel I R we hve I e = { : I, S}. (b) For ny idel I S 1 R we hve (I c ) e = I. (c) Contrction nd extenion by ϕ give one-to-one correpondence {prime idel in S 1 R} 1:1 {prime idel I in R with I S = /0} I I c I e I. () A 1 S 1 R for S, the idel I e generted by ϕ(i) mut contin the element 1 ϕ() = for I. But it i ey to check tht { : I, S} i lredy n idel in S 1 R, nd o it h to be equl to I e. (b) By Exercie 1.18 (b) it uffice to how the incluion. So let I. Then ϕ 1 (I) = I c ince ϕ() = 1 = I. By () pplied to Ic it follow tht (Ic ) e. (c) We hve to check couple of ertion. The mp i well-defined: If I i prime idel in S 1 R then I c i prime idel in R by Exercie 2.9 (b). Moreover, we mut hve I c S = /0 ll element of S re mpped to unit by ϕ, nd I cnnot contin ny unit. The mp i well-defined: Let I R be prime idel with I S = /0. We hve to check tht I e i prime idel. So let, b t S 1 R with bt I e. By () thi men tht b t = u c for ome c I nd u S, i. e. there i n element v S with v(bu tc) = 0. Thi implie tht uvb I. Since I i prime, one of thee four fctor mut lie in I. But u nd v re in S nd thu not in I. Hence we conclude tht I or b I, nd therefore Ie or b Ie by (). (I c ) e = I for ny prime idel I in S 1 R: thi follow from (b). (I e ) c = I for ny prime idel I in R with I S = /0: By Exercie 1.18 () we only hve to check the incluion. So let (I e ) c, i. e. ϕ() = 1 Ie. By () thi men 1 = b for ome b I nd S, nd thu there i n element u S with u( b) = 0. Therefore u I, which men tht one of thee fctor i in I ince I i prime. But nd u lie in S nd hence not in I. So we conclude tht I. Exmple 6.8 (Prime idel in R P ). Applying Propoition 6.7 (c) to the ce of locliztion of ring R t prime idel P, i. e. t the multiplictively cloed ubet R\P in Exmple 6.5 (d), give one-to-one correpondence by contrction nd extenion {prime idel in R P } 1:1 {prime idel I in R with I P}. One hould compre thi to the ce of prime idel in the quotient ring R/P: by Lemm 1.20 nd Corollry 2.4 we hve one-to-one correpondence {prime idel in R/P} 1:1 {prime idel I in R with I P}, thi time by contrction nd extenion by the quotient mp. Unfortuntely, generl idel do not behve nicely: where idel in the quotient R/P till correpond to idel in R contining P, the nlogou ttement for generl idel in the locliztion R P would be fle. A we will ee now, nother conequence of the one-to-one correpondence ( ) i tht the locliztion R P h exctly one mximl idel (of coure it lwy h t let one by Corollry 2.17). Since thi i very importnt property of ring, it h pecil nme. Definition 6.9 (Locl ring). A ring i clled locl if it h exctly one mximl idel. ( )
54 Andre Gthmnn Corollry 6.10 (R P i locl). Let P be prime idel in ring R. Then R P i locl with mximl idel P e = { : P, / P}. Proof. Obviouly, the et of ll prime idel of R contined in P h the upper bound P. So the correpondence ( ) of Exmple 6.8 preerve incluion, we ee tht every prime idel in R P i contined in P e. In prticulr, every mximl idel in R P mut be contined in P e. But thi men tht P e i the only mximl idel. The formul for P e i jut Propoition 6.7 (). Note however tht rbitrry locliztion in Definition 6.2 re in generl not locl ring. Remrk 6.11 (Ring of locl function on vriety re locl). If R = A(X) i the coordinte ring of vriety X nd P = I() the mximl idel of point X, the locliztion R P i the ring of locl function on X t in Exmple 6.5 (d). By Corollry 6.10, thi i locl ring with unique mximl idel P e, i. e. with only the mximl idel correponding to the point. Thi fit nicely with the interprettion of Remrk 6.1 (b) tht R P decribe n rbitrrily mll neighborhood of : lthough every function in R P i defined on X in neighborhood of, there i no point except t which every function in R P i well-defined. Exercie 6.12 (Univerl property of locliztion). Let S be multiplictively cloed ubet of ring R. Prove tht the locliztion S 1 R tifie the following univerl property: for ny homomorphim α : R R to nother ring R tht mp S to unit in R there i unique ring homomorphim ϕ : S 1 R R uch tht ϕ( 1 r ) = α(r) for ll r R. Exercie 6.13. () Let R = R[x,y]/(xy) nd P = (x 1) R. Show tht P i mximl idel of R, nd tht R P = R[x](x 1). Wht doe thi men geometriclly? (b) Let R be ring nd R. Show tht R = R[x]/(x 1), where R denote the locliztion of R t in Exmple 6.5 (c). Cn you give geometric interprettion of thi ttement? Exercie 6.14. Let S be multiplictively cloed ubet in ring R, nd let I R be n idel with I S = /0. Prove the following loclized verion of Corollry 2.17: () The idel I i contined in prime idel P of R uch tht P S = /0 nd S 1 P i mximl idel in S 1 R. (b) The idel I i not necerily contined in mximl idel P of R with P S = /0. Exercie 6.15 (Sturtion). For multiplictively cloed ubet S of ring R we cll S := { R : S for ome R} the turtion of S. Show tht S i multiplictively cloed ubet well, nd tht S 1 R = S 1 R. Exercie 6.16 (Alterntive form of Nkym Lemm for locl ring). Let M be finitely generted module over locl ring R. Prove the following two verion of Nkym Lemm (ee Corollry 3.27): () If there i proper idel I R with IM = M, then M = 0. (b) If I R i the unique mximl idel nd m 1,...,m k M re uch tht m 1,...,m k generte M/IM n R/I-vector pce, then m 1,...,m k generte M n R-module. So fr we hve only pplied the technique of locliztion to ring. But in fct thi work eqully well for module, o let u now give the correponding definition. However, the computtion to check tht thee definition mke ene re literlly the me in Lemm 6.2, we will not repet them here.
6. Locliztion 55 Definition 6.17 (Locliztion of module). Let S be multiplictively cloed ubet of ring R, nd let M be n R-module. Then (m,) (m, ) : there i n element u S uch tht u(m m ) = 0 i n equivlence reltion on M S. We denote the equivlence cl of (m,) M S by m. The et of ll equivlence cle { m } S 1 M := : m M, S i then clled the locliztion of M t S. It i n S 1 R-module together with the ddition nd clr multipliction m + m := m + m nd m := m for ll R, m,m M, nd, S. In the ce of exmple 6.5 (c) nd (d), when S i equl to { n : n N} for n element R or R\P for prime idel P R, we will write S 1 M lo M or M P, repectively. Exmple 6.18. Let S be multiplictively cloed ubet of ring R, nd let I be n idel in R. Then by Propoition 6.7 () the locliztion S 1 I in the ene of Definition 6.17 i jut the extenion I e of I with repect to the cnonicl mp R S 1 R, 1. 10 In fct, it turn out tht the locliztion S 1 M of module M i jut pecil ce of tenor product. A one might lredy expect from Definition 6.17, it cn be obtined from M by n extenion of clr from R to the locliztion S 1 R in Contruction 5.16. Let u prove thi firt, o tht we cn then crry over ome of the reult of Chpter 5 to our preent itution. Lemm 6.19 (Locliztion tenor product). Let S be multiplictively cloed ubet in ring R, nd let M be n R-module. Then S 1 M = M R S 1 R. Proof. There i well-defined R-module homomorphim ϕ : S 1 M M R S 1 R, m m 1, ince for m M nd S with m = m, i. e. uch tht u(m m ) = 0 for ome u S, we hve m 1 m 1 = u(m m ) 1 u = 0. Similrly, by the univerl property of the tenor product we get well-defined homomorphim ψ : M R S 1 R M with ψ ( m ) = m, for R nd S with u( ) = 0 for ome u S we lo get u(m m) = 0, nd thu m = m. By contruction, ϕ nd ψ re invere to ech other, nd thu ϕ i n iomorphim. Remrk 6.20 (Locliztion of homomorphim). Let ϕ : M N be homomorphim of R-module, nd let S R be multiplictively cloed ubet. Then by Contruction 5.16 there i well-defined homomorphim of S 1 R-module ( ϕ id : M R S 1 R N R S 1 R uch tht (ϕ id) m ) = ϕ(m). By Lemm 6.19 thi cn now be conidered n S 1 R-module homomorphim ( m ) S 1 ϕ : S 1 M S 1 N with S 1 ϕ = ϕ(m). We will cll thi the locliztion S 1 ϕ of the homomorphim ϕ. A before, we will denote it by ϕ or ϕ P if S = { n : n N} for n element R or S = R\P for prime idel P R, repectively.
56 Andre Gthmnn The locliztion of ring nd module i very well-behved concept in the ene tht it i comptible with lmot ny other contruction tht we hve een o fr. One of the mot importnt propertie i tht it preerve exct equence, which then men tht it i lo comptible with everything tht cn be expreed in term of uch equence. Propoition 6.21 (Locliztion i exct). For every hort exct equence ϕ ψ 0 M 1 M 2 M 3 0 of R-module, nd ny multiplictively cloed ubet S R, the loclized equence i lo exct. 0 S 1 M 1 S 1 ϕ S 1 M 2 S 1 ψ S 1 M 3 0 Proof. By Lemm 6.19, locliztion t S i jut the me tenoring with S 1 R. But tenor product re right exct by Propoition 5.22 (b), hence it only remin to prove the injectivity of S 1 ϕ. So let m S 1 M 1 with S 1 ϕ( m ) = ϕ(m) = 0. Thi men tht there i n element u S with uϕ(m) = ϕ(um) = 0. But ϕ i injective, nd therefore um = 0. Hence m = u 1 um = 0, i. e. S 1 ϕ i injective. Corollry 6.22. Let S be multiplictively cloed ubet of ring R. () For ny homomorphim ϕ : M N of R-module we hve ker(s 1 ϕ) = S 1 kerϕ nd im(s 1 ϕ) = S 1 imϕ. In prticulr, if ϕ i injective / urjective then S 1 ϕ i injective / urjective, nd if M i ubmodule of N then S 1 M i ubmodule of S 1 N in nturl wy. (b) If M i ubmodule of n R-module N then S 1 (M/N) = S 1 M/S 1 N. (c) For ny two R-module M nd N we hve S 1 (M N) = S 1 M S 1 N. Proof. () Loclizing the exct equence 0 kerϕ M ϕ imϕ 0 of Exmple 4.3 () t S, we get by Propoition 6.21 n exct equence 0 S 1 kerϕ S 1 M S 1 ϕ S 1 imϕ 0. But gin by Exmple 4.3 thi men tht the module S 1 kerϕ nd S 1 imϕ in thi equence re the kernel nd imge of the mp S 1 ϕ, repectively. (b) Thi time we loclize the exct equence 0 M N N/M 0 of Exmple 4.3 (b) to obtin by Propoition 6.21 0 S 1 M S 1 N S 1 (M/N) 0, which by Exmple 4.3 men tht the lt module S 1 (M/N) in thi equence i iomorphic to the quotient S 1 N/S 1 M of the firt two. (c) Thi follow directly from the definition, or lterntively from Lemm 5.8 (c) locliztion i the me tking the tenor product with S 1 R. Remrk 6.23. If n opertion uch locliztion preerve hort exct equence in Propoition 6.21, the me then lo hold for longer exct equence: in fct, we cn plit long exct equence of R-module into hort one in Remrk 4.5, loclize it to obtin correponding hort exct equence of S 1 R-module, nd then glue them bck to loclized long exct equence by Lemm 4.4 (b). Exercie 6.24. Let S be multiplictively cloed ubet of ring R, nd let M be n R-module. Prove:
() For M 1,M 2 M we hve S 1 (M 1 + M 2 ) = S 1 M 1 + S 1 M 2. 6. Locliztion 57 (b) For M 1,M 2 M we hve S 1 (M 1 M 2 ) = S 1 M 1 S 1 M 2. However, if M i M for ll i in n rbitrry index et J, it i in generl not true tht S 1( ) i J M i = i J S 1 M i. Doe t let one incluion hold? (c) For n idel I R we hve S 1 I = S 1 I. Exmple 6.25. Let P nd Q be mximl idel in ring R; we wnt to compute the ring R Q /P Q. Note tht thi i lo n R Q -module, nd uch iomorphic to (R/P) Q by Corollry 6.22 (b). So when viewed uch module, it doe not mtter whether we firt loclize t Q nd then tke the quotient by P or vice ver. () If P Q we mut lo hve P Q ince P i mximl nd Q R. So there i n element P\Q, which led to 1 1 = P Q. Hence in thi ce the idel P Q in R Q i the whole ring, nd we obtin R Q /P Q = (R/P) Q = 0. (b) On the other hnd, if P = Q there i well-defined ring homomorphim ( ) ϕ : R/P R P /P P,. 1 We cn eily contruct n invere of thi mp follow: the morphim R R/P, end R\P to unit ince R/P i field by Lemm 2.3 (b), hence it pe to morphim R P R/P, 1 by Exercie 6.12. But P i in the kernel of thi mp, nd thu we get well-defined ring homomorphim ( ) R P /P P R/P, 1, which clerly i n invere of ϕ. So the ring R P /P P i iomorphic to R/P. Note tht thi reult i lo ey to undertnd from geometric point of view: let R be the ring of function on vriety X over n lgebriclly cloed field, nd let p,q X be the point correponding to the mximl idel P nd Q, repectively (ee Remrk 2.7 ()). Then tking the quotient by P = I(p) men retricting the function to the point p by Lemm 0.9 (d), nd loclizing t Q correpond to retricting them to n rbitrrily mll neighborhood of q by Exmple 6.5 (d). So firt of ll we ee tht the order of thee two opertion hould not mtter, ince retriction of function commute. Moreover, if p q then our mll neighborhood of q doe not meet p, nd we get the zero ring the pce of function on the empty et. In contrt, if p = q then fter retricting the function to the point p nother retriction to neighborhood of p doe not chnge nything, nd thu we obtin the iomorphim of (b) bove. A n ppliction of thi reult, we get the following refinement of the notion of length of module: not only i the length of ll compoition erie the me, but lo the mximl idel occurring in the quotient of ucceive ubmodule in the compoition erie in Definition 3.18 (). Corollry 6.26 (Length of loclized module). Let M be n R-module of finite length, nd let 0 = M 0 M 1 M n = M be compoition erie for M, o tht M i /M i 1 = R/Ii by Definition 3.18 () with ome mximl idel I i R for i = 1,...,n. Then for ny mximl idel P R the locliztion M P i of finite length n R P -module, nd it length l RP (M P ) i equl to the number of indice i uch tht I i = P. In prticulr, up to permuttion the idel I i (nd their multiplicitie) bove re the me for ll choice of compoition erie.
58 Andre Gthmnn Proof. Let P R be mximl idel. From the given compoition erie we obtin chin of loclized ubmodule 0 = (M 0 ) P (M 1 ) P (M n ) P = M P. ( ) For it ucceive quotient (M i ) P /(M i 1 ) P for i = 1,...,n we get by Corollry 6.22 (b) (M i ) P /(M i 1 ) P = (Mi /M i 1 ) P = (R/Ii ) P = RP /(I i ) P. But by Exmple 6.25 thi i 0 if I i P, where for I i = P it i equl to R P modulo it unique mximl idel P by Corollry 6.10. So by deleting ll equl ubmodule from the chin ( ) we obtin compoition erie for M P over R P whoe length equl the number of indice i {1,...,n} with I i = P. We cn rephre Corollry 6.26 by ying tht the length of module i locl concept : if n R-module M i of finite length nd we know the length of the loclized module M P for ll mximl idel P R, we lo know the length of M: it i jut the um of ll the length of the loclized module. In imilr wy, there re mny other propertie of n R-module M tht crry over from the loclized module M P, i. e. in geometric term from conidering M loclly round mll neighborhood of every point. Thi hould be compred to propertie in geometric nlyi uch continuity nd differentibility: if we hve e. g. function f : U R on n open ubet U of R we cn define wht it men tht it i continuou t point p U nd to check thi we only need to know f in n rbitrrily mll neighborhood of p. In totl, the function f i then continuou if nd only if it i continuou t ll point p U: we cn y tht continuity i locl property. Here re ome lgebric nlogue of uch locl propertie: Propoition 6.27 (Locl propertie). Let R be ring. () If M i n R-module with M P = 0 for ll mximl idel P R, then M = 0. (b) Conider equence of R-module homomorphim If the loclized equence of R P -module ϕ 0 M 1 ϕ 1 2 M2 M3 0. ( ) (ϕ 1 ) P (ϕ 2 ) P 0 (M 1 ) P (M 2 ) P (M 3 ) P 0 re exct for ll mximl idel P R, then the originl equence ( ) i exct well. We y tht being zero nd being exct re locl propertie of module nd equence, repectively. Note tht the convere of both ttement hold well by Propoition 6.21. Proof. () Aume tht M 0. If we then chooe n element m 0 in M, it nnihiltor nn(m) doe not contin 1 R, o it i proper idel of R nd thu by Corollry 2.17 contined in mximl idel P. Hence for ll u R\P we hve u / nn(m), i. e. um 0 in M. Thi men tht m 1 i non-zero in M P, nd thu M P 0. (b) It uffice to how tht three-term equence L ϕ M ϕ P L P M P poible poition in the five-term equence of the ttement. ψ N i exct if ll locliztion ψ P N P re exct for mximl P R, ince we cn then pply thi to ll three To prove thi, note firt tht { } ψ(ϕ(m)) ( ( m )) } (ψ(ϕ(l))) P = : m L, S = {ψ P ϕ P : m L, S = ψ P (ϕ P (L P )) = 0,
6. Locliztion 59 ince the loclized equence i exct. Thi men by () tht ψ(ϕ(l)) = 0, i. e. imϕ kerψ. We cn therefore form the quotient kerψ/imϕ, nd get by the exctne of the loclized equence (kerψ/imϕ) P = (kerψ)p /(imϕ) P = kerψp /imϕ P = 0 uing Corollry 6.22. But thi men kerψ/imϕ = 0 by (), nd o kerψ = imϕ, i. e. the equence L ϕ M ψ N i exct. Remrk 6.28. A uul, Propoition 6.27 (b) men tht ll propertie tht cn be expreed in term of exct equence re locl well: e. g. n R-module homomorphim ϕ : M N i injective (rep. urjective) if nd only if it locliztion ϕ P : M P N P for ll mximl idel P R re injective (rep. urjective). Moreover, in Remrk 6.23 it follow tht Propoition 6.27 (b) hold for longer exct equence well. Exercie 6.29. Let R be ring. Show: () Idel continment i locl property: for two idel I,J R we hve I J if nd only if I P J P in R P for ll mximl idel P R. (b) Being reduced i locl property, i. e. R i reduced if nd only if R P i reduced for ll mximl idel P R. (c) Being n integrl domin i not locl property, i. e. R might not be n integrl domin lthough the locliztion R P re integrl domin for ll mximl idel P R. Cn you give geometric interprettion of thi ttement?