Maps between non-commutative spaces.pdf

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TRANSACTIONS OF THEAMERICAN MATHEMATICAL SOCIETYVolume 356, Number 7, Pages 29272944S 0002-9947(03)03411-1Article electronically published on November 18, 2003

MAPS BETWEEN NON-COMMUTATIVE SPACES

S. PAUL SMITH

Abstract. Let Jbe a graded ideal in a not necessarily commutative gradedk-algebraA = A0A1 in which dimkAi < for alli. We show that themapA A/Jinduces a closed immersion i : ProjncA/J ProjncAbetween

the non-commutative projective spaces with homogeneous coordinate rings AandA/J. We also examine two other kinds of maps between non-commutative

spaces. First, a homomorphism : A B between not necessarily commu-tative N-graded rings induces an affine map ProjncB U ProjncA froma non-empty open subspace U ProjncB. Second, ifA is a right noetherian

connected graded algebra (not necessarily generated in degree one), and A(n)

is a Veronese subalgebra ofA, there is a map ProjncA ProjncA(n)

; weidentify open subspaces on which this map is an isomorphism. Applying thesegeneral results when A is (a quotient of) a weighted polynomial ring produces

a non-commutative resolution of (a closed subscheme of) a weighted projectivespace.

1. Introduction

This paper concerns maps between non-commutative projective spaces of theform Projnc A. Before summarizing our main results we define the relevant terms.

Following Rosenberg [8, p. 278] and Van den Bergh [13], a non-commutativespaceX is a Grothendieck category ModX. A map g : Y Xbetween two spaces

is an adjoint pair of functors (g

, g) with g : ModY ModX and g

left adjointto g. The mapg is affine [8, page 278] ifg is faithful and has a right adjoint.For example, a ring homomorphism : R S induces an affine map g : Y Xbetween the affine spaces defined by ModY :=ModSandModX:= ModR.

Letk be a field. An N-gradedk-algebraA is locally finiteif dimkAi < for alli. The non-commutative projective spaceX with homogeneous coordinate ring Ais defined by

ModX:= GrModA/FdimA

(see Section2), and

Projnc A:= (ModX, OX),

whereOX is the image ofA in ModX. Thus Projnc A is an enriched quasi-scheme

in the language of [13]. Let Y be another non-commutative projective space withhomogeneous coordinate ringB. A mapf: Projnc B Projnc Ais a mapf :Y Xsuch that fOX = OY.

Received by the editors September 18, 2002 and, in revised form, April 29, 2003.2000 Mathematics Subject Classification. Primary 14A22; Secondary 16S38.The author was supported by NSF grant DMS-0070560.

c2003 American Mathematical So ciety

2927


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2928 S. PAUL SMITH

When A is a commutative N-graded k-algebra we write Proj A for the usualprojective scheme. We will always view a quasi-separated, quasi-compact schemeXas a non-commutative space by associating to it the enriched space (QcohX, O

X).

The rule X (QcohX, OX) is a faithful functor.

Summary of results. The main results in this paper are Theorems 3.2,3.3,4.1,and Proposition4.8.

A map g : Y X is a closed immersion if it is affine and the essential image ofModY in ModXunder g is closed under submodules and quotients. Theorem3.2shows that a surjective homomorphism A A/Jof graded rings induces a closedimmersioni : Projnc A/J Projnc A. The functors i

andi are the obvious ones(see the proof of3.2). It seems to be a folklore result thati is left adjoint toi, butwe could not find a proof in the literature so we provide one here. Several peoplehave been aware for some time that this is the appropriate intuitive picture, but,as far as I know, no formal definition of a closed immersion has been given and sono explicit proof has been given.

IfA is a graded subalgebra ofB, commutative results suggest there should bea closed subspace Z of Y = Projnc B and an affine map g : Y\Z Projnc A.Theorem3.3establishes such a result under reasonable hypotheses onA and B. Infact, that result is set in a more general context, namely a homomorphism : A B of graded rings. Corollary3.4 then says that if : A B and B is a finitelypresented leftA-module, then there is an affine map g : Projnc B Projnc A. Thisis a (special case of a) non-commutative analogue of the commutative result that afinite morphism is affine.

If A is a quotient of a commutative polynomial ring, and A(n) is the gradedsubring with components (A(n))i = Ani, then there is an isomorphism of schemesProj A =Proj A(n). Verevkin [12] proved that Projnc A =Projnc A

(n) when A isno longer commutative, but is connected and generated in degree one. Theorem4.1shows that when A is not required to be generated in degree one, there is stilla map Projnc A Projnc A

(n), and Proposition4.8describes open subspaces onwhich this map is an isomorphism.

The results here are modelled on the commutative case, and none is a surprise.In large part the point of this paper is to make the appropriate definitions sothat results from commutative algebraic geometry carry over verbatim to the non-commutative setting. Thus we formalize and make precise some of the terminologyand intuition in papers like [2]and [7].

In Example4.9we show how our results apply to a quotient of a weighted poly-nomial ring to obtain a birational isomorphismg : Projnc A X= Proj A, whereXis a commutative subscheme of a weighted projective space. It can happen thatXis singular whereas Projnc Ais smooth. Thus we can view Projnc A Proj Aassomething like a non-commutative resolution of singularities. Furthermore, in this

situationgg

=id.We freely use basic notions and terminology for non-commutative spaces fromthe papers[9], [10], and [13].

2. Definitions and preliminaries

Throughout this paper we assume thatAis a locally finite N-graded algebraover afieldk. ThusA = A0 A1 , and dimk Ai < for alli. Theaugmentation idealm ofA isA1 A2 . IfA0 is finite dimensional andA is right noetherian, then it


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MAPS BETWEEN NON-COMMUTATIVE SPACES 2929

follows that dimkAi< for alli becauseAi/Ai+1 is a noetherianA/m-module.We write GrModA for the category ofZ-graded rightA-modules, and define

TailsA:= GrModA/FdimA,

whereFdimAis the full subcategory consisting of direct limits of finite dimensionalA-modules. Equivalently, FdimA consists of those modules in which every elementis annihilated by a suitably large power ofm. We write for the quotient functorGrModA TailsA and for its right adjoint.

The projective space with homogeneous coordinate ring A is the space Xdefinedby ModX := TailsA. We write Projnc A = (ModX, OX), where OX denotes theimage ofA in TailsA.

A closed subspace Zof a space X is a full subcategory ModZ ofModX that isclosed under submodules and quotient modules inModXand such that the inclusionfunctori: ModZ ModXhas both a left adjoint i and a right adjoint i!.

Two spaces are isomorphic if their module categories are equivalent. Hence a

mapY Xis a closed immersion if and only if it is an isomorphism from Y to aclosed subspace ofX.The complementX\Zto a closed subspace Zis defined by

ModX\Z:= ModX/T,

the quotient category ofModXby the localizing subcategory T consisting of thoseX-modulesMthat are the direct limit of modulesNwith the property thatN hasa finite filtration N = Nn Nn1 N1 N0 = 0 such that each Ni/Ni1is in ModZ. Because T is a localizing category, there is an exact quotient functor

j : ModX ModX\Z, and its right adjoint j : ModX\Z ModX. The pair(j, j) defines a map j : X\Z X. We call it an open immersion.

We sometimes write ModZX for the category T and call it the category ofX-modules supported on Z.

Letf :Y Xbe a map. Iff is faithful, then the counit idY f!f is monicand the unit ff idY is epic.

Watts Theorem for graded modules. Let A and B be Z-graded k-algebras.We recall the analogue of Watts Theorem proved by Del Rio [3, Proposition 3]that describes thek-linear functors GrModA GrModB that have a right adjoint.

A bigradedA-B-bimodule is an A-B-bimodule

M=

(p,q)Z2pMq

such that Ai.pMq.Bj i+pMq+j for all i ,j,p,q Z. Write for k. IfL is agraded rightA-module, we define

LA M :=

qZ(L

A M)q,

where (LA M)q is the image of

p(LppMq) under the canonical mapLML AM. This gives L A Mthe structure of a graded right B -module; it is a B-module direct summand of the usual tensor product L AM.

IfNis a graded rightB-module, we define

HomB

(M, N) := {f HomGrB(M, N) | f(pM) = 0 for almost all p}.


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2930 S. PAUL SMITH

This is made into a graded right A-module by declaring that deg f=piff(iM) = 0for alli = p. Hence Hom

B(M, N)pis naturally isomorphic to HomGrB(pM, N),

and there is a natural isomorphismHom

B(M, N) =

p

HomGrB(pM, N).

The usual adjoint isomorphism between Hom and then induces an isomorphism

HomGrB(LA M, N) =HomGrA(L, HomB(M, N)),(2-1)

showing that A M :GrModA GrModB is left adjoint to HomB(M, ).

Theorem 2.1(Del Rio [3]). LetA andB be gradedk-algebras, andF :GrModA GrModB ak-linear functor having a right adjoint. Then F = A M, whereMis the bigradedA-B-bimodule

M= pZ

F(A(p))

with homogeneous components pMq = F(A(p))q.IfF also commutes with the twists by degree, thenF is given by tensoring with

a graded A-B-bimodule, say V =

n Vn. The corresponding M in this case isM=

V(p) with pMq = V(p)q.

The leftA-action onM is given by declaring thatx Ai acts on pM asF(x),wherex: A(p) A(p+ i) denotes left multiplication byx.

3. Maps induced by graded ring homomorphisms

Throughout this section we assume that A and B are locally finite N-gradedalgebras over a field k .

We consider the problem of when a homomorphism : A B of graded ringsinduces a map g : Projnc B Projnc A and, if it does, how the properties ofg aredetermined by the properties of.

Associated to is an adjoint triple (f, f, f!) of functors between the categories

of graded modules. Explicitly, f = A B, f= BBA is the restriction map,and f! =

pZHomGrB(B(p), ). We wish to establish conditions on which

imply that these functors factor through the quotient categories in the followingdiagrams:

GrModB f GrModA

TailsB TailsA

GrModB f, f!

GrModA

TailsB TailsA

Lemma 3.1. LetA andB be Grothendieck categories with localizing subcategoriesS A andT B. Let : A A/S and :B B/T be the quotient functors, andlet and be their right adjoints. Consider the following diagram of functors:

A F B

A/S B/T.


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Suppose thatF(S) T.

(1) There is a unique functorG : A/S B/T such thatF =G.

(2) IfH: B A is a right adjoint to F, thenH is a right adjoint to G.(3) IfH is a right adjoint to F andG is a right adjoint to G, thenH(T) S

if and only ifG=H.

Proof. (1) The existence and uniqueness ofG is due to Gabriel [6,Corollaire 2, p.368].

(2) To show that G has a right adjoint it suffices to show that it is right exactand commutes with direct sums. IfM is a collection of objects in A/S, then eachis of the form M = M for some objectM in A. Both andFcommute withdirect sums because they have right adjoints, so G commutes with direct sums; also commutes with direct sums. Therefore

GM =G M=G M

= GM= GM.Thus G commutes with direct sums.

To see that G is right exact, consider an exact sequence

0 L M N 0(3-1)

in A/S. By Gabriel [6,Corollaire 1, p. 368], (3-1) is obtained by applying to anexact sequence 0 L M N 0 inA. Both andFare right exact becausethey have right adjoints, so F L F M F N 0 is exact. In other words,GL GM GN 0 is exact.

Hence G has a right adjoint, say G. It follows that G is a right adjoint toG. But G = F has H as a right adjoint, so G =H . Since =idA/S,G = H . Since a right adjoint is only determined up to natural equivalence,

H is a right adjoint to G.(3) IfH(T) S, then H vanishes on T so, by Gabriel [6, Corollaire 2, page

368], there is a functor V : B/T A/Ssuch thatV =H. ThusV =H , andthis is isomorphic toG by (2). Hence G =H. Conversely, ifG =H, thenH(T) = 0, so H(T) S.

Warning. The functor H in part (2) of Lemma 3.1 need not have the propertythat H(T) is contained in S. An explicit example of this is provided by takingB= A, F =H= idA, S = 0, and T = B.

Theorem 3.2. Let J be a graded ideal in an N-graded k-algebra A. Then thehomomorphismA A/J induces a closed immersion i: Projnc A/J Projnc A.

Proof. Write X = Projnc A and Z = Projnc A/J. Write m = A1 A2 .Thus ModX= GrModA/FdimA. We write : GrModA ModX for the quotientfunctor and for a right adjoint to it. Similarly, : GrModA/J ModZ is thequotient functor, and is a right adjoint to it. See [12]and [1,Section 2] for moreinformation about this.

Let f : GrModA/J GrModA be the inclusion functor. It has a left adjointf = A A/J, and a right adjointf! that sends a gradedA-module to the largestsubmodule of it that is annihilated by J.


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2932 S. PAUL SMITH

By [6,Corollaire 2, p. 368] (= part (1) of Lemma3.1), there is a unique functori such that

GrModA/J f GrModA

ModZ i

ModX

commutes, and i is exact because f is [6,Corollaire 3, p. 369]. Thus i =f.

By Lemma3.1(2), i has a right adjoint, namely i! := f!. It is clear that f!

sends FdimA to FdimA/J, so f!=i! by Lemma3.1(3).It is clear that f sends FdimA to FdimA/J, so by [6, Corollaire 2, p. 368],

there is a functor i : ModX ModZ such that f = i. Since f is rightadjoint tof, it follows from Lemma3.1(2) thatf is a right adjoint to i. Butf

=i=i. Hence i is left adjoint to i.

We now show that i is faithful. Since i has a left and a right adjoint, it isexact, so it suffices to show that ifiM = 0, then M = 0. Suppose thatiM= 0for someM GrModA/J. ThenfM= 0, and we conclude that M is in FdimA,and hence in FdimA/J; therefore M= 0. Hence i is faithful.

We will show that i is full after establishing the following fact.Claim. f=f.Proof. Let M GrModA/J, let Mdenote the largest submodule of M that

is in FdimA/J (equivalently, in FdimA), and set M = M/M. Then M = Mand fM = f M, so the two functors take the same value on Mif and only ifthey take the same value on M. Hence we can, and will, assume that M= M; i.e., M= 0.

We must show that M = M. By definitionMis the largest essentialextension 0 M M T 0 such thatT FdimA. The definition ofM

is analogous, althoughTis now required to belong to FdimA/J. It therefore sufficesto prove thatMis in GrModA/J. The top row in the diagram

MAJ (M) AJ TAJ 0 MJ (M)J

is exact, and the first vertical map is zero because M is an A/J-module, so thereis an induced map TAJ (M)J. This map is surjective. However, TAJ belongs to FdimA because T does, so (M)J FdimA. This implies thatM (M)J FdimA. But M = 0, so M (M)J= 0, and it follows that(M)J= 0 because M is essential in M. In other words, M GrModA/J.

This completes the proof of the claim.We haveff =idGrModA/J and=idModZ, so

ii=(f)(f

) =ff=idModZ.

It follows from this that i is full.To see thati is a closed immersion, it remains to check that i(ModZ) is closed

under submodules and quotients in ModX. Let M ModZ and suppose that0 L iM N 0 is an exact sequence inModX. There is an exact sequence


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0 L iM N R1L in GrModA. Let Ndenote the image of iMin N. ThenL = L and N = N, because is exact and R1L FdimA.Now M = M for some M GrModA/J, so iM = iM = fM

=fM, from which we conclude that iM is annihilated by J. ThereforeL

is also annihilated by J, so is of the form fL for some L GrModA/J; henceL= L= fL = iL i(ModZ). SinceN is a quotient ofiM it is alsoannihilated by J, and a similar argument shows that N i(ModZ).

SincefA= A/J, iOX = OZ.

We retain the notation of the theorem.Becausei is fully faithful, we often view ModZas a full subcategory ofModX

and speak ofZ= Projnc A/Jas a closed subspace ofX= Projnc A and call it thezero locus ofJ.

It is not the case that every closed subspace of Projnc A is the zero locus of atwo-sided ideal inA. For example, ifA = kq[x, y] is the ring defined by the relationyx = qxy where 0 = q k, then Proj

nc

A = P1, but the closed points of P1 arenot cut out by two-sided ideals when q = 1: for example, (x+ y)A is not atwo-sided ideal whenq= 1 and= 0. This is essentially due to the fact that theauto-equivalence M M(1) ofModXinduced by the degree shift on A does notgenerally sendZ-modules to Z-modules.

A more difficult question is whether every closed subspace of Projnc Ais the zerolocus of a two-sided ideal in somehomogeneous coordinate ring of Projnc A. Wedo not know the answer to this question.

Theorem 3.3. Suppose that : A B is a map of locally finite N-graded k-algebras. WriteX= Projnc A andY= Projnc B. Let m be the augmentation idealofA, and letI be the largest two-sided ideal ofB contained in(m)B. LetZ Ybe the zero locus ofI. IfB (m)n (m)B for some integern, then induces anaffine map

g: Y\Z X.

Proof. The category of modules over Y\Z is ModY /ModZY. This is equivalent tothe quotient category GrModB/T, where T consists of those modules Mwith theproperty that every element ofMis killed by some power ofI. Let :GrModBGrModB/T be the quotient functor. We have functors (f, f, f

!) between thegraded module categories and a diagram

GrModB f GrModA

Mod Y\Z ModX.

To check that f sends FdimA to T, it suffices to check that f(A/m) T,because f commutes with direct limits and with the degree twist (1). However,f(A/m) =B/(m)B is in T becauseI (m)B. Hence there is a unique functorg : TailsA (GrModB)/T satisfyingg= f.

To check thatf sendsT to FdimA, it suffices to check thatf(B/I) is inFdimA.However, (B/I).mn = B(m)n + I /I; the hypothesis that B(m)n (m)Bensures that B(m)n I, so (B/I).mn = 0. Hence there is an exact functorg: GrModB/T TailsA such thatg =f.


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2934 S. PAUL SMITH

By Lemma3.1(2),g has a right adjoint g! =f!.

To show that g is faithful we must show that ifM is a graded B -module suchthatg

M= 0, thenM T. SincefM= 0, as anA-moduleMis a direct limitlim M where eachM is a finite dimensional A-module. There is an epimorphism

lim(M AB)=(lim M) AB

=MAB M

of gradedB-modules. SinceMABequalsfM, it is inT; butT is closed underdirect limits and quotients, so M is in T. Thus g is faithful.

The following consequence of the theorem slightly extends a result of Van denBergh [13,Proposition 3.9.11].

Corollary 3.4. Let : A B be a homomorphism of graded rings such thatBbecomes a finitely presented graded leftA-module. Then induces an affine mapg: Projnc B Projnc A.

Proof. If we apply A/m A to a finite presentation ofB as a left A-module, wesee that B/(m)B has finite dimension. Thus, as a rightA-module B/(m)B isannihilated by mn for some n 0. Equivalently, B(m)n (m)B. Thus thehypotheses of the theorem are satisfied. It remains to show that Z is empty.

LetIdenote the right annihilator in B ofB/(m)B. We have already observedthat(m)n I. SinceA/mn is finite dimensional, so is A/mn AB =B/(m)nB.ThusB/I is finite dimensional. Hence the zero locus ofIin Projnc B is empty.

Remark. If, in Theorem 3.3, BA is finitely presented, then we have the usefultechnical fact that g != f!. This follows from Lemma3.1(3) once we show thatf! sendsFdimAtoFdimB. LetM= lim Mbe a direct limit of finite dimensional A-modules. IfB is a finitely presented rightA-module, then HomGrA(B, ) commuteswith direct limits, so HomGrA(N, lim M) = lim HomGrA(B, M); this is a direct

limit of finite dimensional B-modules, because BA is finitely generated. Hencef!(FdimA) FdimB.

4. The Veronese mapping

Throughout this section A is a locally finite N-graded k-algebra and nis a positiveinteger. Thenth VeronesesubalgebraA(n) is defined by

A(n)i :=Ani.

It is a classical result in algebraic geometry that ifA is a finitely generated commu-tative connected graded k-algebra generated in degree one, then Proj A =Proj A(n).This isomorphism is implemented by the Veronese embedding.

Verevkin proved a non-commutative version of this result when A is noetherian

and generated in degree one [12,Theorem 4.4].Theorem4.1 and Proposition4.8show what happens when A need not be com-

mutative and need not be generated in degree one.

Theorem 4.1. LetA be a left noetherian locally finiteN-gradedk-algebra. Fix apositive integern. There is a mapg : Projnc A Projnc A

(n). Furthermore, g isexact andgg=id. IfA is also right noetherian, theng has a right adjointg!.

We will use the notation X:= Projnc A(n) andX := Projnc A.


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We need two preliminary results before proving the theorem. First we explainhow the functors defining the map g : Projnc A Projnc A

(n) in the theorem areinduced by functors between the categories GrModA and GrModA(n).

IfL is a gradedA-module, we define the graded A(n)-moduleL(n) by

L(n)i :=Lni.

The rule L L(n) extends to give an exact functor

f : GrModA GrModA(n).

The functor f is not faithful when n 2, because f((A/m)(1)) = 0.

Proposition 4.2. LetA be a locally finiteN-gradedk-algebra. Fix a positive integern. LetW andW be the spaces with module categories

ModW =GrModA(n)

and

ModW =GrModA.

Then there is a map f : W Wwith direct image functor given byfL= L(n).Furthermore,f has a right adjointf

!.

Proof. It is clear that f is an exact functor commuting with direct sums. By thegraded version of Watts Theorem, f = A M, where

M :=pZ

A(p)(n)

with components pMq = (A(p)(n))q = A(p)nq. The right action ofA

(n) on M is

given by right multiplication, and each A(p)(n) is a rightA(n)-submodule. The leftaction ofA is given by left multiplication, whereby a Ai acts by sending A(p)nqtoA(p+ i)nq.

Define f : GrModA(n) GrModA by fN = NA(n) A with the usual rightaction ofA, and grading given by

(NA(n)A)s=

ni+j=s

Ni Aj .

It is not hard to show that f is a left adjoint to f. Thereforef = A(n)Q,where

Q=pZ

f(A(n)(p)) =pZ

A(np);

multiplication A(n)(p) A(n) A A(np) gives an isomorphism of graded right A-modules. Thus

pQ

=

A(np) with its usual grading. One can verify directly thatf=HomA(Q, ).

The right adjoint to f is the functor f! = Hom

A(n)(M, ). If N is a graded

rightA(n)-module, then

(f!N)i= HomGrA(n)(iM, N) = HomGrA(n)(A(i)(n), N).

IfNis a gradedA(n)-module, thenff(N) = N, so ff is naturally equivalentto idW.


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2936 S. PAUL SMITH

Let : GrModA TailsA and : GrModA(n) TailsA(n) be the quotientfunctors. To prove Theorem4.1, we must find functors g, g, and g

! making thefollowing diagrams commute:

GrModA f GrModA(n)

TailsA

gTailsA(n).

GrModA f,f!

GrModA(n)

TailsA g,g!

TailsA(n)

wheref, f, and f!, are the functors in the previous proposition.

Since f sends FdimA to FdimA(n), there is a functor g : TailsA TailsA(n)

such thatg =f. Becausef is exact, g is too [6].

To ensure that f and f! induce functors between the quotient categories, we

must impose a noetherian hypothesis. Although there is no noetherian hypothesisin Proposition4.2, in Theorem4.1 it is assumed that A is right noetherian. Thishypothesis ensures that f sendsFdimA(n) to FdimA.

Recall that GrModA denotes the category of graded rightA-modules.

Lemma 4.3. LetA be a locally finiteN-gradedk -algebra. Then:

(1) f sends right noetherianA-modules to right noetherianA(n)-modules;

(2) ifA is right noetherian so isA(n), andA is a finitely generated rightA(n)-module.

(3) ifA is left noetherian, thenf sendsFdimA(n) to FdimA;(4) ifA is left noetherian, then there is a functorg :TailsA(n) TailsA such

thatg= f;(5) ifA is right noetherian, thenf! sendsFdimA(n) to FdimA;(6) ifA is right noetherian, then there is a functorg ! :TailsA(n) TailsA such

thatg!= f!;

Proof. (1) Let Mbe a right noetherian graded A-module. IfN is a submodule ofM(n), then N=N A M(n). Hence any proper ascending chain of submodules inM(n) would give a proper ascending chain of submodules ofMby multiplying byA. SinceM contains no such chain, neither does M(n).

(2) Applying (1) to M=A shows that A(n) is right noetherian.Applying (1) toM :=AA(1)A(n1) gives the result, becauseM(n)=A

as a right A(n)-module.(3) Because A is left noetherian, the left module version of (2) implies that A

is a finitely generated left A(n)-module. Hence if N is a finite dimensional right

A(n)-module, NA(n) A is a finite dimensional A-module. Thus f sends finitedimensional rightA(n)-modules to finite dimensional right A-modules. Since f isa left adjoint, it commutes with direct limits. The result follows.

(4) Because A is left noetherian, we may invoke (3). The existence of such g

now exists by the universal property of the quotient functor .(5) First we show that it suffices to prove that f! send finite dimensional A(n)-

modules to finite dimensional A-modules. To this end, let N FdimA(n) andwrite N = lim N as a direct limit of finite dimensional modules. Recall that


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f!N= HomA(n)

(M, N), where

M= pZ

A(p)(n).

By the noetherian hypothesis, A(p)(n) is a finitely generated, and hence finitely pre-sented, right A(n)-module, so HomGrA(n)(A(p)

(n), ) commutes with direct limits.It follows that

f!(lim N) =p

HomGrA(n)(A(p)(n), lim N)

=p

lim HomGrA(n)(A(p)(n), N)

= lim

p

HomGrA(n)(A(p)(n), N)

= lim f

!

(N).Hence, if eachf!(N) is finite dimensional,f!Nis a direct limit of finite dimensionalmodules.

Now we show that f!N is finite dimensional when N is a finite dimensionalgraded right A(n)-module. It suffices to show that (f!N)p, which is equal to

HomGrA(n)(A(p)(n), N), is zero for almost all p and is finite dimensional for all p.

By the noetherian hypothesis, A(p)(n) is a finitely generated right A(n)-module,so HomGrA(n)(A(p)

(n), N) has finite dimension.We now show that HomGrA(n)(A(p)

(n), N) is zero if |p| is sufficiently large. Fixp. For every integer j we have

A(p + nj)(n)=A(p)(n)(j),

soHomGrA(n)(A(p+ nj)

(n), N) =HomGrA(n)(A(p)(n), N(j)).

Since A(p)(n) is finitely generated and Nis finite dimensional, when |j| is sufficientlylarge HomGrA(n)(A(p)

(n), N(j)) is zero. Hence HomGrA(n)(A(p)(n), N) is zero for

|p| sufficiently large. This completes the proof that f! sends finite dimensionalmodules to finite dimensional modules.

(6) follows from (5) in the same way that (4) follows fom (3).

Proof of Theorem4.1. By Lemma 4.3 there are functors g and g between thecategoriesMod Projnc A

(n) =TailsA(n) andMod Projnc A= TailsA satisfying

g= f, g =f, g

!= f!.

Applying Lemma 3.1 to f

, we see that g

has f

as a right adjoint. Butf =g

=g, sog is a right adjoint tog. We have already remarked that

g is exact because f is. Furthermore,

id = =ff= g

f= gg =gg

.

SincefA(n) =A, gOProjnc

A(n) = OProjncA.

Now suppose that A is also right noetherian. Theng has a right adjoint g! by

Lemma4.3 and g != f!. This completes the proof of Theorem 4.1.


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2938 S. PAUL SMITH

In the next result Proj A is the usual commutative scheme viewed as a non-commutative space with module category Qcoh(Proj A).

Corollary 4.4. IfA is a finitely generated commutative connected gradedk-algebra,there is a map g: Projnc A Proj A. Furthermore,g has a right adjointg

!.

Proof. For some sufficiently largen, A(n) is generated in degree one; so

TailsA(n) =Qcoh Proj A(n) =Qcoh Proj A.

Hence Projnc A(n)=Proj A. Therefore Theorem4.1 gives the result.

Remarks. 1. Suppose that A is both left and right noetherian, as in Theorem 4.1.Sinceg has both a left and a right adjoint, it is exact; its right adjointg

! thereforepreserves injectives. Hence there is a convergent spectral sequence

ExtpProjnc

A(M, Rqg!N) Extp+q

Projnc

A(n)(gM, N)

forM andNmodules over Projnc A and Projnc A(n) respectively.

2. IfJ is a two-sided ideal ofA, then the natural map A A/J induces anisomorphismA(n)/J(n) (A/J)(n), so there is a commutative diagram

Projnc A/J i Projnc A g

Projnc A(n)/J(n) Projnc A

(n)

where the horizontal maps are the natural closed immersions.IfA is prime, right noetherian, we define

FractgrA := {ab1 | a, b A are homogeneous and b is regular}.

Proposition 4.5. Let A be a right noetherian, locally finite, N-gradedk-algebra.Suppose thatA is prime andFractgrA contains a copy ofA(n) for alln Z. Then

(1) Projnc A andProjnc A(n) are integral spaces in the sense of[10], and

(2) g: Projnc A Projnc A(n) is a birational isomorphism in the sense that it

induces an isomorphism between the function fields.

Proof. That Projnc A is an integral space is proved in [10, Th. 4.5]. It is alsoshown there that the function field of Projnc A is isomorphic to (FractgrA)0. It is

clear that (FractgrA(n))0 (FractgrA)0, and the reverse inclusion follows from the

observation thatab1 =abn1bn.

Remarks. 1. IfA is prime noetherian and has a regular element of degree d for alld 0, then FractgrA contains a copy ofA(n) for all n Z, so the previous resultapplies.

2. Ifz is a normal regular element, then the complement in Projnc Ato the zero

locus ofz is the open subspace U := Projnc A[z1

]. Now ModU is equivalent toGrModA[z1]. Ifd is the smallest positive integer such that A[z1] has a unit ofdegreed, then Uis an affine space with coordinate ring

R0 R1 Rd1

R1 R0 Rd2...

...Rd+1 Rd+2 R0

,


13/18


where R = A[z1]. The reason for this is thatd1

i=0 A[z1](i) is a generator in

GrModA[z1] and the tiled matrix ring is the endomorphism of this generator.3. In the situation of Proposition4.5,ifs and t are homogeneous regular elements

of relatively prime degrees in A, and st is normal, meaning that stA=Ast, thenA[(st)1] has a unit of degree one; so, ifUis the open complement to the zero locusofst in Projnc A, then Uis the affine space with coordinate ring A[(st)

1]0. Thisring is equal to A(n)[(st)n]0, so the open complement is isomorphic to the opencomplement to the zero locus of (st)n in Projnc A

(n).4. If FractgrA fails to contain a copy of every A(n), the map Projnc A

Projnc A(n) need not be a birational isomorphism. For example, takeA = k[x] with

deg x= 2.

Example 4.6. IfA is not generated in degree one, then g need not be faithful.Let A = k[x, z] be the polynomial ring with deg x = 1 and deg z = n 2.

The image under ofM = A/(x) is a simple module Op in Projnc A. We haveOp(1) = 0, but (M(1))(n) = 0, so g(Op(1)) = 0.

One might anticipate that g : Projnc A Projnc A(n) is an isomorphism on

suitable open subspaces: in the previous example, g restricts to an isomorphismfrom the complement to the zero locus ofx in Projnc A to the complement to thezero locus ofx in Projnc A

(n). We prove a general result of this type in Proposition4.8. First we need a lemma.

For each integer r, define

A(n)+r :=jZ

Anj+r.

ObviouslyA(n)+rA(n)+s A(n)+r+s, so eachA(n)+r is anA(n)-A(n)-bimodule, andthese bimodules depend only on r (mod n). Define

Ir :=A(n)+r

A=jZ

Anj+rA

and

I:=rZ

Ir = I1 I2 In.

AlthoughIr is in general only a right ideal ofA, I(n)r is a two-sided ideal ofA(n).

SinceAqIr Iq+r, Iis a two-sided ideal ofA.

Notice thatA(n)+rA(n)r =I(n)r .

Lemma 4.7. With the above notation, I2n I(n)A.

Proof. From the containment

I2

IrI=A(n)+r

I A(n)+r

Inr = A(n)+r

A(n)+nr

A= I(n)r A,

it follows that

I2n I(n)1 I2n2 I(n)1 I

(n)2 I

2n4 I(n)1 I(n)n A.

But this last term is contained inI

(n)1 I

(n)n

A= I(n)A,

which completes the proof.


14/18

2940 S. PAUL SMITH

Proposition 4.8. With the hypotheses of Theorem4.1, the map g restricts to anisomorphismg : Projnc A\Z

Projnc A(n)\Z, whereZ andZare the zero loci of

I and

I

(n) respectively.

Proof. Write X = Projnc A, X= Projnc A(n), U =X\Z and U = X\Z. Write

: U X and : U X for the inclusions. We will use Lemma3.1to showthat there is an isomorphism h : U Usuch that the diagram

U

X = Projnc A

h

gU

X= Projnc A

(n)

commutes.Let T be the localizing subcategory of GrModA consisting of those modules L

such that every element ofL is killed by a suitably large power of I. Let S be

the localizing subcategory ofGrModA(n)

consisting of those modules N such thatevery element ofNis killed by a suitably large power ofI(n). These two localizingsubcategories contain all the finite dimensional modules. The spaces U andU aredefined by

ModU := (GrModA(n))/S and ModU := (GrModA)/T.

Let f and f be the functors defined in Proposition 4.2. We will show thatf(T) S and f(S) T. The first of these inclusions is obvious: if every elementof an A-module L is annihilated by a power of I, then every element of L(n) isannihilated by a power ofI(n). To show thatf(S) T, it suffices to show thatf(A(n)/I(n)) belongs to T. But f(A(n)/I(n)) = A/I(n)A, and by Lemma 4.7,A/I(n)A is annihilated by I2n so belongs to T.

We now use Lemma3.1 in the context of the following diagram:

GrModA(n) f GrModA

ModU ModU.

Becausef(S) T, there exists a functor h :ModU ModU such thath=f. Because f is right adjoint to f, h :=f is right adjoint to h.Thus,h and h define a map h : U U. Since g =f, a computation givesh =g. Thereforeh = g.

It remains to show that h is an isomorphism.The unit idGrModA(n) ff

is an isomorphism because the natural map L (L A(n) A)

(n) is an isomorphism for all L GrModA(n). Becausef(T) S, part(3) of Lemma3.1 gives h

=f. Therefore,

hh=hh

= hf=

ff=idU.

To show that the natural transformation hh idU is an isomorphism, wefirst consider the natural transformation ff idGrModA. For an A-module Mthis is the multiplication map

ffM=M(n) A(n) A M.

We claim that the kernel and cokernel of this map belong to T.


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Suppose that

mi ai M(n) A(n)A is in the kernel. Then

i miai = 0. Bytaking homogeneous components we can reduce to the case where each ai belongstoA

nr+ A

2nr

+ for somer {1, . . . , n}. Then, ifb Anj+r

for somej , theni

mi ai

b=

i

mi aib=i

miaib 1 = 0.

Thus

i miai

Ir = 0. Hence the kernel is annihilated by I, so belongs toT. The

cokernel offfM M is M /M(n)A. Ifr {1, . . . , n}, then MnjrIr M(n)A.HenceIannihilates the cokernel.

Since the kernel and cokernel offf id belong to T, there is an isomorphismff . Hence

hh = hf

=ff

=idU.

In Example4.6, I1 = (x) and I2 = A, so I= (x), whence Z is the zero locus of

x. This explains why we need to remove the zero locus ofx to get the isomorphism.If A is generated in degree one, then Ir = Ar for r {0, 1, . . . , n 1}, so

A/Ir FdimA, whence A/I FdimA. It follows that Z and Z are empty, andtherefore U = X and U = X. We therefore recover Verevkins result X = X

whenA is generated in degree one over A0.

Example 4.9. Let A be a weighted polynomial ring. That is, A = k [x0, . . . , xn],where deg xi = qi 1. Write Q = (q0, . . . , q n). Then PnQ := Proj A is called aweighted projective space. It is isomorphic to the quotient variety Pn/Q, whereQ = q0 qn. There is a sufficiently divisible integer d such that A

(d) isgenerated in degree one. Hence

PnQ= Proj A= Proj A

(d) =Projnc A(d).

By Theorem4.1,there is a map

g: Projnc A Projnc A(d)= PnQ.

This is an isomorphism on an open subspace by Proposition 4.8. Since A hasglobal homological dimension n + 1, Projnc A has global homological dimension n.We therefore think of Projnc Aas a smooth space of dimension n and the map g asa non-commutative resolution ofPnQ. LetX P

nQ be the closed subscheme cut out

by an ideal J inA. Then there is a commutative diagram

Projnc A/J i

Projnc A

f

gX PnQ

in which f is a birational isomorphism and i is a closed immersion. It can happenthat Projnc A/J is smooth even when X is singular. Thus Projnc A is a non-commutative resolution ofX. An interesting case to examine in some detail isthat where X is an orbifold of a Calabi-Yau three-fold.

IfA = k[x] with deg x= 2, andn = 2, then Projnc A=Spec k k and Proj A =

Spec k. Furthermore,Z =X andZ= X. This is a special case of the next result,the truth of which was suggested by Darin Stephenson.


16/18

2942 S. PAUL SMITH

Proposition 4.10. LetA be a locally finiteN-gradedk-algebra such thatAi = 0whenever i 0 (mod n). ThenProjnc A is isomorphic to the disjoint union ofncopies of

Projnc A

(n).

Proof. Letpr : GrModA GrModA be the functor defined by

pr(M) =iZ

Mr+in

on objects, and pr() =|pr(M) whenever HomGrA(M, N). The hypothesis onAensures that eachpr(M) is a gradedA-submodule ofM, so pr is indeed a functorfromGrModAto itself. It is clear that idGrModA= p0 pn1, where this directsum is taken in the abelian category of k-linear functors from GrModA to itself;essentially, this is the observation thatM=p0(M) pn1(M), and that anymap : M Nof graded A-modules respects this decomposition. Furthermore,eachpr is idempotent and the prs are mutually orthogonal. It follows from this thatthere is a decomposition of GrModA as a product of categories, each component

being the full subcategory on which pr is the identity.It is clear that the shift functor (1) cyclicly permutes these subcategories, so

they are all equivalent to one another and (n) is an autoequivalence of each compo-nent. However, any one of these categories together with its autoequivalence (n) isequivalent to GrModA(n) with its autoequivalence (1). Thus GrModA is equivalentto the product ofn copies ofGrModA(n).

Finally, this decomposition descends to the Tails categories.

5. An Ore extension and an example

The morphism

p: Pn\{(0, . . . , 0, 1)} Pn1,

(0, . . . , n) (0, . . . , n1),is called the projection with center (0, . . . , 0, 1). This section examines a non-commutative analogue of this basic operation.

Consider a connected gradedk-algebraR and a connected graded Ore extension

S= R[t; , ]

with respect to a graded automorphism and a graded -derivation of degreen 1. ThusS=

n=0 Rt

n andtr = rt + (r) for all r R. Since (Ri) Ri+nfor all i, by setting deg t= n, Sbecomes a connected graded algebra.

One expects that the inclusion mapR Sinduces a map Projnc S Projnc R.Indeed, the projection map above can be obtained as a special case of this.

Let m denote the augmentation ideal ofR. Since(m) m, mSis a two-sided

ideal ofS. Furthermore,S/m

S=k[t] as graded rings.

Proposition 5.1. With the above notation, letZ denote the zero locus of mS inProjnc S.

(1) Z=Spec kn.(2) There is an affine map g: Projnc S\Z Projnc R.

Proof. (2) The existence ofg is a special case of Theorem3.3. That theorem appliesbecause Sm mS.


17/18


(1) The quotient ringS/mSis isomorphic to the polynomial ringk[t] with deg t=n, so this follows from Proposition4.10.

We think of Projnc S as a cone over Projnc R with vertex Z. It would beinteresting to describe the fibers of the map g .

When deg t > 1, the Ore extension S= R[t; , ] is not generated by its elementsof degree one. This sometimes causes technical problems; however, ifRis generatedin degree one, then the nth VeroneseS(n) is generated in degree one. We can thencombine Theorems3.3 and 4.1 to analyze the space with homogeneous coordinateringSas follows.

Proposition 5.2. The inclusion of then-Veronese subalgebras ofSandS/mSgivesa commutative diagram of rings and an induced commutative diagram of spaces as

in the following diagrams:

k[t](n)

k[t] = S/Sm S(n) S R(n) R

Spec k =v Z=Spec kn

Projnc S

(n) g Projnc S Projnc S

(n)\{v} Projnc S\Z

Projnc R(n) Projnc R

An application. In [11], a family of three-dimensional Artin-Schelter regular alge-brasA is constructed and studied. Although the algebraic properties ofA are quitewell understood, our understanding of the corresponding geometric object Projnc A

is rudimentary. The algebras are of the formA = R[t; , ] with deg t= n and Ra two-dimensional Artin-Schelter regular algebra generated in degree one. It iswell-known thatR and its Veronese subalgebras are (not necessarily commutative)homogeneous coordinate rings of P1. By Proposition5.2, there is a commutativediagram of spaces and maps

v= Spec k Z=Spec kn Projnc A

(n) g Projnc A Projnc A

(n)\{v} g

Projnc A\Z

P1

= P1.

(5-1)

Acknowledgements

This work was stimulated by Darin Stephensons paper [11]. I would like to thankhim for explaining his results, and also suggesting that Proposition4.10should betrue. The final paragraph of this paper applies our results to Stephensons algebras.


18/18

2944 S. PAUL SMITH

I thank James Zhang for bringing [3] to my attention. I thank the referee forsuggesting several improvements to an earlier version of this paper.

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Department of Mathematics, Box 354350, University of Washington, Seattle, Wash-

ington 98195

E-mail address: [email protected]
http://www.ams.org/mathscinet-getitem?mr=96a:14004http://www.ams.org/mathscinet-getitem?mr=96a:14004http://www.ams.org/mathscinet-getitem?mr=92d:16011http://www.ams.org/mathscinet-getitem?mr=92d:16011http://www.ams.org/mathscinet-getitem?mr=88h:14052http://www.ams.org/mathscinet-getitem?mr=88h:14052http://www.ams.org/mathscinet-getitem?mr=85g:14060http://www.ams.org/mathscinet-getitem?mr=85g:14060http://www.ams.org/mathscinet-getitem?mr=38:1144http://www.ams.org/mathscinet-getitem?mr=38:1144http://www.ams.org/mathscinet-getitem?mr=2003f:58017http://www.ams.org/mathscinet-getitem?mr=2003f:58017http://www.ams.org/mathscinet-getitem?mr=97b:14004http://www.ams.org/mathscinet-getitem?mr=97b:14004http://www.ams.org/mathscinet-getitem?mr=2003f:14002http://www.ams.org/mathscinet-getitem?mr=2003f:14002http://www.ams.org/mathscinet-getitem?mr=2003d:16037http://www.ams.org/mathscinet-getitem?mr=2003d:16037http://www.ams.org/mathscinet-getitem?mr=2001a:16044http://www.ams.org/mathscinet-getitem?mr=2001a:16044http://www.ams.org/mathscinet-getitem?mr=93j:14002http://www.ams.org/mathscinet-getitem?mr=93j:14002http://www.ams.org/mathscinet-getitem?mr=2002k:16057http://www.ams.org/mathscinet-getitem?mr=2002k:16057http://www.ams.org/mathscinet-getitem?mr=2002k:16057http://www.ams.org/mathscinet-getitem?mr=93j:14002http://www.ams.org/mathscinet-getitem?mr=2001a:16044http://www.ams.org/mathscinet-getitem?mr=2003d:16037http://www.ams.org/mathscinet-getitem?mr=2003f:14002http://www.ams.org/mathscinet-getitem?mr=97b:14004http://www.ams.org/mathscinet-getitem?mr=2003f:58017http://www.ams.org/mathscinet-getitem?mr=38:1144http://www.ams.org/mathscinet-getitem?mr=85g:14060http://www.ams.org/mathscinet-getitem?mr=88h:14052http://www.ams.org/mathscinet-getitem?mr=92d:16011http://www.ams.org/mathscinet-getitem?mr=96a:14004

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