On injectivity of maps between Grothendieck groups induced by completion

ON INJECTIVITY OF MAPS BETWEEN GR OTHENDIECK GR OUPS INDUCED BY COMPLETION HAILONG DA O Abstract. W e giv e an example of a lo cal normal domain R such t hat the map of Grothendiec k groups G ( R ) → G( ˆ R ) is not injectiv e. W e also raise some questions about the kernel of tha t map. 1. Introduction Let ( R, m, k ) b e a lo cal ring and ˆ R the m -adic completion of R . Let M ( R ) be the ca tegory o f finitely genera ted R -mo dules. The Gr othendiec k g roup of finitely generated modules ov er R is defined a s: G( R ) = M M ∈M ( R ) Z [ M ] h [ M 2 ] − [ M 1 ] − [ M 3 ] | 0 → M 1 → M 2 → M 3 → 0 is exa c t i In [K K], Kamoi and Kurano studied injectivity of the ma p G( R ) → G( ˆ R ) in- duced b y flat base-change. They sho w ed that such ma p is the injectiv e in the following cases : 1 ) R is Hensenlian, 2) R is the lo calizatio n at the irrelev ant ideal of a po sitiv e ly grade d r ing ov e r a field, or, 3) R has only isola ted singularity . The ir results raise the question: Is the map b et ween Grothendieck group induced by completion always injective? In [Ho1], Hochster announced a counterexample to the a bove question: Theorem 1 .1. L et k b e a field. L et R = k [ x 1 , x 2 , y 1 , y 2 ] ( x 1 ,x 2 ,y 1 ,y 2 ) / ( x 1 x 2 − y 1 x 2 1 − y 2 x 2 2 ) . L et P = ( x 1 , x 2 ) and M = R/P . Then [ M ] is in the kernel of the ma p G( R ) → G( ˆ R ) . However [ M ] 6 = 0 in G( R ) . Ho c hster ’s example comes fr om the “direct s ummand hyper s urface” in dimension 2 and is not normal. He pr edicted that ther e is also an ex ample which is no rmal. The main pur pose of this note is to provide such an ex a mple. W e hav e: Prop osition 1.2. L et R = R [ x, y, z , w ] ( x,y ,z ,w ) / ( x 2 + y 2 − ( w + 1) z 2 ) . R is a normal domain. L et P = ( x, y , z ) and M = R /P . Then [ M ] is in the kernel o f the map G( R ) → G( ˆ R ) . However [ M ] 6 = 0 in G( R ) . This will b e proved in Section 2. W e note that Kur a no a nd Sr iniv as has re cen tly constructed an example of a local ring R suc h t ha t the map G( R ) Q → G( ˆ R ) Q is not injectiv e (see [KS]). The ring in their example is not normal, a nd we do no t know if a norma l exa mple exists in that context (i.e. with r a tional co efficien ts). In sec tion 3 we will discuss so me q uestions on the k ernel of the map G( R ) → G( ˆ R ). W e would like to thank Anurag Singh for telling us ab out this question and for some inspiring conversations. W e thank Me lv in Ho chster for generous ly s ha ring his 1 2 HAILONG DA O unpublished note [Ho2], which provided the key ideas for our example. W e also thank the r eferee for many helpful comments. 2. Our example W e shall prove Pro pos ition 1.2. Fir st we nee d to reca ll some classica l r e sults: Corollary 2.1 . (Swan, [Sw ] , Cor ol lary 11.8) L et k b e a field of char acteristic not 2 , n > 1 an inte ger and R = k [ x 1 , ..., x n ] / ( f ) wher e f is a non-de gener ate quadr atic form in k [ x 1 , ..., x n ] . Then G( R ) = Z ⊕ Z / 2 Z if C 0 ( f ) , the even p art of the Cliffor d algebr a of f , is simple. Prop osition 2.2. (Samu el, se e [F o] , Pr op osition 11.5) L et k b e a field of char- acteristic not 2 and f b e a non-de gener ate quadr atic form in k [ x 1 , x 2 , x 3 ] . L et R = k [ x 1 , x 2 , x 3 ] / ( f ) . If f = 0 has no non-trivial solution in k then Cl( R ) = 0 . Prop osition 2.3. (Kamoi-Kur ano) L et S = ⊕ n ≥ 0 S n b e a gr ade d ring over a field S 0 and S + = ⊕ n> 0 S n . L et A = S S + . Then the map G( S ) → G( A ) induc e d by lo c alization is an isomorphism. Pr o of. See the pro of of Theorem 1.5 (ii) in [K K].  Prop osition 1.2 now follows from the following Pro positio ns (clearly , R is normal, since the s ing ular lo cus V ( x, y , z ) has co dimension 2): Prop osition 2.4. [ ˆ M ] = 0 in G( ˆ R ) . Pr o of. ˆ R = R [[ x, y , z , w ]] / ( x 2 + y 2 − ( w + 1 ) z 2 ). W e want to show tha t [ ˆ R/P ˆ R ] = 0 in G( ˆ R ). Let α = √ w + 1 which is a unit in ˆ R . Let Q = ( x, y − αz ) ˆ R . Then clearly Q is a height 1 prime in ˆ R and P ˆ R = Q + ( y + αz ) ˆ R . The s ho rt exact sequence : 0 → ˆ R/Q → ˆ R/Q → ˆ R/P ˆ R → 0 where the seco nd map is the multiplication by y + αz shows that [ ˆ R/P ˆ R ] = 0 in G( ˆ R ).  Prop osition 2.5. [ M ] 6 = 0 in G( R ) . Pr o of. It is enough to show that [ M P ] 6 = 0 in G( R P ). Let K = R ( w ) then R P ∼ = K [ x, y , z ] ( x,y ,z ) / ( f ) where f = x 2 + y 2 − ( w + 1 ) z 2 . Let S = K [ x, y , z ] / ( f ). Clearly f is a non-dege nerate quadratic form. Since the rank of f is 3 , an o dd num b er, C 0 ( f ) is a simple algebra o ver K (see, for exa mple, [L a], Chapter 5 , Theorem 2.4). B y 2.1 and 2 .3, G( R P ) = G( S ) = Z ⊕ Z / (2). W e claim that f has no non- trivial so lution in K . Suppos e it has. Then by clearing denominators , there are po lynomials a ( w ) , b ( w ) , c ( w ) ∈ R [ w ] such tha t a ( w ) 2 + b ( w ) 2 = ( w + 1) c ( w ) 2 . The deg ree o f a ( w ) 2 + b ( w ) 2 is alwa ys even. The degre e o f ( w + 1) c ( w ) 2 is o dd unless c ( w ) = 0. But then a ( w ) 2 + b ( w ) 2 = 0 which forces a ( w ) = b ( w ) = 0, a c o n tr adiction. By the claim a nd 2 .2, Cl( R P ) = Cl( S ) = 0. Th us [ R P ] and [ R P /P R P ] genera te G( R P ) = Z ⊕ Z / (2) (since the Grothendieck group is gener ated by { [ R P /Q ] , Q ∈ Sp ec R P } and dim R P = 2 ). Since Z ⊕ Z / (2) can not b e generated by one element, [ R P /P R P ] must be no nzero (it is ea s y to see that [ R P /P R P ] is 2-torsio n).  ON INJECTIVITY OF MAPS BETWEEN GROT HE NDIECK GROUPS INDUCED BY COMPLETION 3 3. On the kernel of the map G( R ) → G( ˆ R ) In this section we raise some questions ab out the kernel of the map G( R ) → G( ˆ R ). First we fix s ome notations. Thro ughout this section we will assume, for simplicity , that R is excellent, and is a homomorphic image of a regula r lo cal r ing T . Let d = dim R . Let A i ( R ) b e the i-th Chow group of R , i.e., A i ( R ) = M P ∈ Sp ec R, d im R/P = i Z · [Sp ec R/P ] h div( Q, x ) | Q ∈ Spec R , dim R/Q = i + 1 , x ∈ R \ Q i where div( Q, x ) = X P ∈ Min R R/ ( Q,x ) l R P ( R P / ( Q, x ) R P )[Spec R/P ] . F o r an ab elian gro up A , w e let A Q = A ⊗ Z Q . The Chow group of R is defined to b e A ∗ ( R ) = ⊕ d i =0 A i ( R ). It is well known that there is a Q -vector space isomo rphism: τ R/T : G( R ) Q → A ∗ ( R ) Q (It is unknown whether this is indep endent of T ). W e also remark that the Grothendieck group G( R ) admits a filtration b y F i G( R ) = h [ M ] ∈ G( R ) | dim M ≤ i i . The ex isting examples on the failur e of injectivity for the map G ( R ) → G( ˆ R ) and the a ffirmative results in [KK] motiv ate the following question: Question 3. 1. Supp ose t hat R satisfies (R n ) (i.e, r e gular in c o dimension n ). Then is ker(G( R ) → G ( ˆ R )) c ontaine d in F d − n − 1 G( R ) ? Question 3.1 is closely related to the following: Question 3.2 . Su pp ose that R satisfies (R n ) . Then is t he map A i ( R ) → A i ( ˆ R ) inje ctive for i ≥ d − n ? In fact, if we allow rational co efficien ts, then the previous questions are equiv a- lent . Let G i ( R ) = F i G( R ) /F i − 1 G( R ). Then clearly we hav e a decomp osition: G( R ) Q = d M i =0 G i ( R ) Q Also, the Riemann-Ro ch ma p deco mpose s in to iso morphisms τ i : G i ( R ) Q → A i ( R ) Q , which make the following diagram: G i ( R ) g i   τ i R/T / / A i ( R ) f i   G i ( ˆ R ) τ i ˆ R/ ˆ T / / A i ( ˆ R ) commutativ e. It follows that ker( G ( R ) Q → G( ˆ R ) Q ) ∼ = d M i ker( f i ) ∼ = d M i ker( g i ) . Thu s we have: 4 HAILONG DA O Prop osition 3.3. L et R b e an exc el lent lo c al ring which is a homomorph ic image of a r e gular lo c al ring. L et dim R = d and let 0 < l ≤ d b e an int e ger. Then the maps A i ( R ) Q → A i ( ˆ R ) Q ar e inje ct ive for i ≥ l if and only if ker(G( R ) Q → G( ˆ R ) Q ) ⊆ F l − 1 G( R ) Q . W e do no t know if 3.2 is true even if l = 1. Note that if R is nor mal, then b oth 3.1 and 3.2 are true for l = 1. In that situation A 1 ( R ) ∼ = Cl( R ), and the map betw een c lass groups of R and ˆ R is injectiv e. F urthermo re, it is w ell known that G( R ) /F d − 2 G( R ) ∼ = A d ( R ) ⊕ A d − 1 ( R ) (see, for example [Ch], Co rollary 1), so 3.1 is also tr ue for l = 1 . Finally , one could formulate a stronger version of 3.1 as follows. Note that in b oth Ho c hs ter’s example a nd the example presented her e, the supp ort of the mo dules given a ctually equal to the singula r lo cus of R . So one co uld ask: Question 3.4. L et R b e an exc el lent lo c al ring. L et X = Sp ec R , Y = Sing ( X ) , ˆ X = Sp ec ˆ R and ˆ Y = Sing ( ˆ X ) . One then has a c ommutative diagr am: G( Y )   f / / G( X ) g   G( ˆ Y ) / / G( ˆ X ) (Her e G( X ) denotes the Gr othendie ck gr oup of c oher ent O X -mo dules and the maps ar e natur al ly induc e d by close d immersions or flat morphisms). Is ker( g ) c ontaine d in im( f ) ? References [F o] R. F ossum, The Di v isor Class Gr oup of a K rul l Domain , Springer-V erlag, New Y ork 1973. [Ch] C.-Y. J. Chan, Filtr ations of mo dules, the Chow gr oup, and the Gr othendie ck gr oup , J. Algebra 219 (1999), 330–34 4. [F u] W. F ulton, Interse ction The ory , second edition, Springer-V erlag, Berlin 1998. [Ho1] M. Hochster, Thirte en op en questions in Commutative A lge br a , talk given at LipmanF est, July 2004, av ail able online at htt p: //www.math.lsa.umic h.edu/ ∼ hochste r/Lip.text.pdf . [Ho2] M. Hochster, N on-inje ctivity of a map to a Gr othendie ck gr oup of a co mpletion , unpub- lished note. [KK] Y. Kamoi, K.K urano, On maps of Gr othendie ck gr oups induc e d by co mpletion , J. Algebra 254 (2002), 21–43. [KS] K.Kurano, V. Sri niv as, A lo c al ring such that the map b etwe en Gr othendie ck gr oups with r ational c o efficient induc e d by c ompletion is not inje ctive , preprin t, arXiv mathA C/0707.0547. [La] T.Y. Lam, Algebr aic The ory of Quadr ati c F orms , W.A Benjamins, 1973. [Sw] R.G. Sw an, K - The ory of Q uadric Hyp ersurfac es , Ann. of Math. 122 (1985), 113–153. Dep ar tment of Ma themat ics, Un iversity o f Ut ah , 155 South 1400 East, Sal t Lake City, UT 84112-009 0, USA E-mail addr ess : hdao@math .utah.ed u