ZARISKI CANCELLATION PROBLEM FOR NONCOMMUTATIVE ALGEBRAS 6 1 JASONBELLANDJAMESJ.ZHANG 0 2 n Abstract. A noncommutative analogue of the Zariski cancellation problem ∼ ∼ a asks whether A[x]=B[x]impliesA=B whenAandB arenoncommutative J algebras. WeresolvethisaffirmativelyinthecasewhenAisanoncommutative finitelygenerateddomainoverthecomplexfieldofGelfand-Kirillovdimension 8 two. Inaddition,weresolvetheZariskicancellationproblemforseveralclasses 1 ofArtin-SchelterregularalgebrasofhigherGelfand-Kirillovdimension. ] A R 0. Introduction . h t Kraft said in his 1995 survey [Kr] that “there is no doubt that complex affine a n-space An = An is one of the basic objects in algebraic geometry. It is therefore m C surprising how little is known about its geometry and its symmetries. Although [ there has been some remarkable progress in the last few years, many basic problems 1 remain open.” His remark still applies even today—20 years later. Let us start v with one of the famous questions in commutative affine geometry. Throughoutthe 5 introduction, we let k be an algebraicallyclosed field of characteristic zero (except 2 for some results mentioned below). 6 4 Question 0.1 (Zariski Cancellation Problem). DoesanisomorphismY ×A∼= 0 An+1 imply that Y is isomorphic to An? Or equivalently, does an isomorphism . 1 B[t]∼=k[t1,··· ,tn+1] of algebras implies that B is isomorphic to k[t1,··· ,tn]? 0 6 For simplicity, let ZCP denote the ZariskiCancellationProblem. An algebraA 1 is called cancellative if A[t] ∼= B[t] for any algebra B implies that A ∼= B. So the : ZCP asks if the commutative polynomial ring k[x ,··· ,x ] is cancellative. Recall v 1 n i that k[x1] is cancellative by a result of Abhyankar-Eakin-Heinzer [AEH], k[x1,x2] X iscancellativebyFujita[Fu]andMiyanishi-Sugie[MS]incharacteristiczeroandby r Russell[Ru]inpositivecharacteristic. TheZCPwasopenformanyyears. In2013, a a remarkable development was made by Gupta [Gu1, Gu2] who completely settled theZCPnegativelyinpositivecharacteristicforn≥3. TheZCPincharacteristic zero remains open for n ≥ 3. We give a list of open questions and problems that are closely related to the ZCP. Question 0.2. For the following, let k× be k\{0}. (ChP:=Characterization Problem) Find an algebro-geometriccharac- terization of An. (EP:=Embedding Problem) Is every closed embedding Aa ֒→ Aa+n equivalent to the standard embedding? 2000 Mathematics Subject Classification. Primary16W70, 16W25, 16S38. Key words and phrases. Zariskicancellation problem, noncommutative algebra, effective and dominatingdiscriminant,locallynilpotentderivation,skewpolynomialring. 1 2 JASONBELLANDJAMESJ.ZHANG (AP:=Automorphism Problem) Describe the group of polynomial au- tomorphisms of An. (LP:=Linearization Problem) Is every automorphism of An of finite order linearizable? (JC:=Jacobian Conjecture)Iseverypolynomialmorphismφ:An →An with detJφ∈k× an isomorphism? There are some known relationship between these problems. For example, a positive solution of the LP would imply a positive solution of the ZCP. When n ≤ 2, most of these questions (except for the JC) were resolved and there is a diagram of implications EP=⇒AP=⇒LP=⇒ZCP alongwithapossible“missinglink”JC=⇒ZCP(see[vdE]). NotethattheEP(in dimension2) was solvedby Abyhyankar-Moh[AM] and Suzuki [Su]. Gupta’s work [Gu1, Gu2] would suggest a negative solution to the ZCP, even in characteristic zero. If the “missing link” could be established and if the ZCP had a negative solution,thentheJCcouldbesettlednegatively. Manyauthorshavebeenworking on these questions—see the references in [Kr, vdE]. Some naive and direct translations of these questions into the noncommutative setting are easily seen to have negative solutions. So it is important to carefully formulatenoncommutativeversionsofthesequestionsandtounderstandforwhich classesof(commutativeornoncommutative)algebrasthese questionshavepositive ornegativeanswers. Hopefully new ideas willemergevia the study ofthe noncom- mutative versions of these open questions. In this paper we mainly consider the following noncommutative formulation of the ZCP. Question 0.3. Let A be a noncommutative noetherian Artin-Schelter regular al- gebra [AS]. When is A cancellative? Since Artin-Schelter regular algebras are considered as a noncommutative gen- eralization of the commutative polynomial ring, the above question can be viewed as a noncommutative version of ZCP. In this paper we present two ideas to deal with the ZCP for some families of noncommutative algebras. One is to use the Makar-Limanov invariant and the other is to use discriminants. Let us first review the Makar-Limanov invariant. Let A be an algebra and let LND(A) be the collection of locally nilpotent k-derivations of A. The Makar- Limanov invariant of A is defined to be ML(A)= ker(δ). δ∈L\ND(A) The Makar-Limanov invariant was originally introduced by Makar-Limanov [Ma1] and has become a very useful invariant in commutative algebra. We say that A is LND-rigid if ML(A) = A, or equivalently if LND(A) = {0}. One of our main results(seeTheorem3.6fortheprecisestatementandproof)isthefollowing,which shows that rigidity controls cancellation. Theorem 0.4. Let A be a finitely generated domain of finite Gelfand-Kirillov di- mension. If A is LND-rigid, then A is cancellative. ZARISKI CANCELLATION PROBLEM FOR NONCOMMUTATIVE ALGEBRAS 3 By the abovetheorem,wewouldliketo showthatvariousclassesofnoncommu- tative algebras are LND-rigid. Here is one of the consequences [Corollary 3.7]. Theorem 0.5. Let A be a finitely generated domain of Gelfand-Kirillov dimension two. If A is not commutative, then A is cancellative. By [AEH], every commutative domain of Gelfand-Kirillov dimension (or GK- dimension, for short) one is cancellative. By [Da, Fi] there are commutative do- mains of GK-dimension two that are not cancellative. Theorem 0.5 ensures that every non-commutative domain of GK-dimension two is cancellative. Crachiola [Cr]showedthatcommutativeUFDs ofGK-dimensiontwoarealwayscancellative. Next let us talk about the discriminant method. The discriminant method was introduced in [CPWZ1, CPWZ2] to answer the AP for a class of noncommutative algebras. The definition of the discriminant in the noncommutative setting will be reviewed in Section 4. Suppose that A is finitely generated by Y = ⊕d kx as i=1 i an algebra over a base commutative ring k. An element f ∈ A is called effective, if for every testing N-filtered k-algebra T with grT := F T/F T being an N- i i−1 graded domain, and for every testing subset {y ,...,y } ⊂ T satisfying (a) it is 1 d L linearlyindependentinthequotientk-moduleT/k1 and(b)somey isnotinF T, T i 0 thereisapresentationoff ofthe formf(x ,...,x )whenliftedinthe freealgebra 1 d khx ,...,x isuchthatf(y ,··· ,y )iseitherzeroornotinF T. Forexample,any 1 d 1 d 0 monomial xa1···xad, for some positive integers a ,...,a , is effective. Note that 1 d 1 d therearenon-monomialeffective discriminants(see Examples5.5and5.6). Here is oneofourmainresults by usingthe discriminant,whichprovidesa uniformwayof showing the rigidity for some classes of noncommutative algebras. Theorem 0.6. Suppose that A is a domain which is a finitely generated module over its affine center C and that the discriminant d(A/C) is effective. Then A is cancellative. TheabovetheoremdoesnotsolvetheoriginalZCPas,whenAiscommutative, the discriminant over its center is trivial and not effective. However, Theorem 0.6 applies to a large family of noncommutative algebras. One can check, for example, that the skew polynomial ring k [x ,...,x ] where n is even and 1 6= q q 1 n is a root of unity has effective discriminant. Then, by Theorem 0.6, k [x ,··· ,x ] q 1 n is cancellative. The next result shows a connection between the noncommutative ZCP and the noncommutative AP. Let C denote the center of the algebra A and wereferto Definition4.4for the definitionof“dominating”. We havethe following result (see Theorem 5.7 for an expanded version). Theorem 0.7. Let A be a skew polynomial ring k [x ,··· ,x ] where each p is pij 1 n ij a root of unity. The following are equivalent. (1) The full automorphism group Aut(A) is affine [CPWZ1, Definition 2.5]. (2) The discriminant d(A/C) is dominating. (3) The discriminant d(A/C) is effective. (4) A is LND-rigid. Consequently, under any of these equivalent conditions, A is cancellative. In general, by using the Makar-Limanov invariant and Theorem 0.4, we show that if d(A/C) is dominating, then A is cancellative, see Theorem 4.6(2). As an example, we have the following result. 4 JASONBELLANDJAMESJ.ZHANG Theorem 0.8. Let A be any finite tensor product of the algebras of the form (a) k [x ,··· ,x ] where 16=p∈k× and n is even; p 1 n (b) khx,yi/(x2y−yx2,y2x+xy2); (c) khx,yi/(yx−qxy−1) where 16=q ∈k×. Then A is LND-rigid. As a consequence, A is cancellative. Remark 0.9. Suppose that n is odd and that q 6= 1 is a root of unity. It is an open question whether k [x ,··· ,x ] is cancellative. There are two results related q 1 n to this. (1) The following weak cancellative property holds as a consequence of [BZ, Theorem 3.1]: Let B be a connected graded algebra generated in degree one. If kq[x1,··· ,xn][t] ∼= B[t] as algebras, then kq[x1,··· ,xn] ∼= B as graded algebras. (2) A result of [CYZ2] says that Veronese subrings of k [x ,··· ,x ](v) is can- q 1 n cellative when m and v are not coprime, where m is the order of q. Acknowledgments. J. Bell was supported by NSERC grant 31-611456 and J.J. Zhangbythe USNationalScience Foundation(NSFgrantNos. DMS-0855743and DMS-1402863). 1. Trivial center vs. cancellation Throughout the rest of the paper let k be a base commutative domain. Some- times we further assume that k is a field. Everything is taken over k, for example, ⊗ stands for ⊗ . We sometimes consider k-flat algebras. If k is a field, then every k k-module is flat. First we recall the definition of cancellative. Definition 1.1. Let A be an algebra. (a) WecallAcancellativeifA[t]∼=B[t]forsomealgebraB impliesthatA∼=B. (b) We call Astrongly cancellative if, foranyd≥1, A[t1,...,td]∼=B[t1,...,td] for some algebra B implies that A∼=B. (c) WecallAuniversallycancellativeif,foranyk-flatfinitelygeneratedcommu- tativedomainRsuchthatR/I =k forsomeidealI ⊂Randanyk-algebra B, A⊗R∼=B⊗R implies that A∼=B. Remark 1.2. By definition, it is easy to see that universally cancellative =⇒ strongly cancellative =⇒ cancellative. But, it is not obvious to us whether any two of them are equivalent. Wehaveanimmediateobservationforthenoncommutativecancellationproblem. Let C(A) denote the center of A. Proposition1.3. Letk beafieldandAbean algebra with centerC(A)=k. Then A is universally cancellative. Proof. Let R be any affine commutative domain such that R/I =k for some ideal I ⊂ R and suppose that φ : A⊗R → B⊗R is an algebra isomorphism for some algebra B. Since C(A)=k, we have C(A⊗R)=R. Since C(B⊗R)=C(B)⊗R and since φ induces an isomorphism between the centers, we have (E1.3.1) R∼=C(B)⊗R. Consequently, C(B) is a commutative domain. By considering the GK-dimension of both sides of (E1.3.1), one sees that GKdimC(B) = 0, when regarded as a ZARISKI CANCELLATION PROBLEM FOR NONCOMMUTATIVE ALGEBRAS 5 k-algebra. Hence C(B) is a field. Since there is an ideal I such that R/I = k, C(B)=k. Consequently,C(B⊗R)=R. Nowtheinducedmapφisanisomorphism betweenC(A⊗R)=RtoC(B⊗R)=R,sowehaveR/φ(I)=k. Finally,φinduces an automorphism from A∼=A⊗R/(I)∼=B⊗R/(φ(I))∼=B. (cid:3) There are easy consequences. Example 1.4. We have the following results. (1) Let k be a field of characteristic zero and A the nth Weyl algebra. Then n C(A )=k. So A is universally cancellative. n n (2) Let k be a field and q ∈ k×. Let k [x ,··· ,x ] be the skew polynomial q 1 n ring generated by x ,...,x subject to the relations x x = qx x for all 1 n j i i j 1 ≤ i< j ≤n. If n≥ 2 and q is not a root of unity, then C(A) =k. So A is universally cancellative. 2. Higher derivations and Makar-Limanov invariant The Makar-Limanovinvariant is a very useful invariant to deal with the cancel- lationproblem. We will also use a modified versionof Makar-Limanovinvariantto better control the cancellation in positive characteristic. Given a k-algebra A, let Der(A)denotethecollectionofk-derivationsofAandLND(A)denotethecollection of locally nilpotent k-derivations of A. For a sequence of k-linear endomorphisms ∂ := {∂ } of A (satisfying certain i i≥0 finiteness conditions) and for any c∈k, define ∞ (E2.0.1) G :A→A by a→ ci∂ (a) c∂ i i=0 X and ∞ (E2.0.2) G :A[t]→A[t] by a→ ∂ (a)ti,t→t, ∂,t i i=0 X for all a∈A. Definition 2.1. Let A be an algebra. (1) A higher derivation (or Hasse-Schmidt derivation) [HS] on A is a sequence of k-linear endomorphisms ∂ :=(∂ ) such that: i i≥0 n ∂ =id , and ∂ (ab)= ∂ (a)∂ (b) 0 A n i n−i i=0 X for all a,b ∈ A and for all n ≥ 0. The collection of higher derivations is denoted by DerH(A). (2) A higher derivation is called iterative if ∂ ∂ = i+j ∂ for all i,j ≥0. i j i i+j (3) A higher derivation is called locally nilpotent if (cid:0) (cid:1) (a) for all a∈A there exists n≥0 such that ∂ (a)=0 for all i≥n, i (b) the map G defined in (E2.0.2) is an algebra automorphism of A[t]. ∂,t ThecollectionoflocallynilpotenthigherderivationsisdenotedbyLNDH(A) andthecollectionoflocallynilpotentiterativehigherderivationsisdenoted by LNDI(A). 6 JASONBELLANDJAMESJ.ZHANG (4) For any ∂ ∈DerH(A), then the kernel of ∂ is defined to be ker∂ = ker∂ . i i≥1 \ Givena higher derivation∂ =(∂ ) , ∂ is necessarilya derivationofA. Hence i i≥0 1 thereis amapDerH(A)→Der(A) bysending∂ to∂ . Incharacteristic0,the only 1 iterative higher derivation ∂ =(∂ ) on A such that ∂ =δ is given by: i 1 δn (E2.1.1) ∂ = n n! for all n≥0. This iterative higher derivation is called the canonical higher deriva- tion associated to δ. In this case, we have a map Der(A)→DerH(A) sending δ to (∂ ) as defined by (E2.1.1). Hence the mapDer(A)→DerH(A) is the rightinverse i of the map DerH(A)→Der(A). The following lemma is easy. Lemma 2.2. Let ∂ :=(∂ ) be a higher derivation. i i≥0 (1) Suppose ∂ is locally nilpotent. For any c ∈ k, G is an algebra automor- c∂ phism of A. (2) If ∂ is iterative and satisfies Definition 2.1(3a), then G is an algebra ∂,t automorphism of A. As a consequence, ∂ is locally nilpotent. (3) If G : A[t] → A[t] be a k[t]-algebra automorphism and if G(a) ≡ a mod t for all a∈A, then G=G for some ∂ ∈LNDH(A). ∂,t Proof. (3) Write G(a) = ∂ (a)ti for all a ∈ A. Then it is easy to show that i≥0 i ∂ :=(∂ ) is in LNDH(A). (cid:3) i P We now recall the definition of the Makar-Limanov invariant. Definition 2.3. Let A be an algebra over k. Let ∗ be either blank, or H, or I. (1) The Makar-Limanov∗ invariant [Ma1] of A is defined to be (E2.3.1) ML∗(A) = ker(δ). δ∈LN\D∗(A) ThismeansthatwehaveoriginalML(A), aswellas,MLH(A)andMLI(A). (2) We say that A is LND∗-rigid if ML∗(A)=A, or LND∗(A)={0}. (3) A is called strongly LND∗-rigid if ML∗(A[t ,...,t ])=A, for all d≥1. 1 d Remark 2.4. The following hold. (a) If k contains Q, then the induced map LNDH(A) → LND(A) is surjective and the induced map LND(A)→LNDH(A) is injective. As a consequence, MLH(A)⊂ML(A). Inparticular,ifA isLNDH-rigid,then itis LND-rigid. (b) Suppose k contains Q. Since LND(A) = LNDI(A), we have ML(A) = MLI(A). (c) IfkcontainsQ,itisnotobvioustouswhetherMLH(A)=ML(A)ingeneral. Inparticular,wedon’tknowifLND-rigidityisequivalenttoLNDH-rigidity. (d) If the prime field of k is finite, then there are examples so that MLI(A)=MLH(A))ML(A). In particular, the LNDH-rigidity is not equivalent to the LND-rigidity. In fact, [CPWZ1, Example 3.9] is such an example. ZARISKI CANCELLATION PROBLEM FOR NONCOMMUTATIVE ALGEBRAS 7 Example 2.5. Suppose k contains Z. Define ∂ = 1(d)n. Then ∂ := (∂ ) n n! dt n is in LNDI(k[t]) and LNDH(k[t]). One sees that ML(k[t]) = k = MLI(k[t]) = MLH(k[t]). A similar result holds for k[t ,...,t ]. 1 d Remark 2.6. Suppose A contains Z. Let ∗ be either blank, or H or I. (1) It is clear that ML∗(A[t ,...,t ]) ⊆ ML∗(A), but, it is not obvious to us 1 d whether ML∗(A[t ,...,t ])=ML∗(A). 1 d (2) Makar-Limanov made the following conjecture in [Ma2]: If A is a commu- tative domain over a field of characteristic zero, then ML(A[t ,...,t ]) = 1 d ML(A). AndheprovedthattheconjectureholdswhenGKdimA=1[Ma2]. 3. Rigidity controls cancellation We shall investigate the relationship between LND-rigidity (respectively, strong LND-rigidity) and cancellation (respectively, strong cancellation). We needthe followinglemma whichis [KL, Proposition3.11]. If Ais analgebra overa commutative base ring k (which is not a field in general), then the Gelfand- Kirillov dimension (or GK-dimension, for short) of A is defined to be GKdimA=sup lim log rank (Vn) n k V n→∞ (cid:16) (cid:17) where V varies over all finitely generated k-submodules of A. Lemma 3.1. Let k be a commutative domain. Let A be a k-algebra and R be an affine commutative k-algebra. (1) GKdimA = GKdim A ⊗ Q where Q is the field of fractions of k. In Q particular, if A is affine and commutative, GKdimA is an integer. (2) GKdimA⊗R=GKdimA+GKdimR. Proof. Left to the reader. (cid:3) Lemma 3.2. Let Y := ∞ Y be an N-graded domain. If Z is a subalgebra of Y i=0 i containing Y such that GKdimZ =GKdimY <∞, then Z =Y . 0 0 0 L Proof. Let X denote the subalgebra Y . Suppose Z strictly contains X as a sub- 0 algebra. Since Y is a graded algebra, Z is an N-filtered algebra with F Z = X. 0 By (a non-field version of) [KL, Lemma 6.5], GKdimZ ≥ GKdimgrZ. Since grZ is an N-graded sub-domain of Y that strictly contains X as the degree zero part of grZ, one can easily see that GKdimgrZ ≥GKdim(grZ) +1=GKdimX +1. 0 Combining these inequalities, one obtains that GKdimZ ≥ GKdimX +1. This contradicts the hypothesis that GKdimZ =GKdimX. Therefore Z =X. (cid:3) It is well-known that a domain of finite GK-dimension is an Ore domain. Here is the main result of this section. Theorem 3.3. Let A be a finitely generated domain of finite GK-dimension. Let ∗ be either blank, or H, or I. When ∗ is blank we further assume A contains Z. (1) If A is strongly LND∗-rigid, then A is strongly cancellative. (2) If ML∗(A[t])=A, then A is cancellative. Proof. We prove (1) and note that the proof of (2) is similar. Letφ:A[t ,...,t ]→B[t ,...,t ]beanisomorphismforsomealgebraB. Since 1 d 1 d AhasfiniteGK-dimension,soisB andGKdimB =GKdimA[Lemma3.1(b)]. For 8 JASONBELLANDJAMESJ.ZHANG eachi,let∂ := ∂ when∗isblankand∂ :={1(∂ )n}∞ when∗iseitherH orI. WehavethaitML∂t∗i(B[t ,...,t ])iscontaiinedinnB! ∂stiincedn=iff0erentiationwithrespect 1 d to t , namely ∂ , gives rise to a locally nilpotent derivation of B[t ,...,t ] and the i i 1 d intersection of the kernels of these maps is exactly B. On the other hand, we have that ML∗(A)[t ,...,t ]) = A by hypothesis. If ∂ is a locally nilpotent derivation 1 d of B then ∂◦φ is a locally nilpotent derivation of B and similarly if ∂′ is a locally nilpotent derivation of A then ∂′◦φ−1 is a locally nilpotent derivation of A. Thus φ induces an isomorphism between ML∗(A[t ,...,t ]) and ML∗(B[t ,...,t ]). In 1 d 1 d particular φ maps A into B. Let Y = A[t ,...,t ] with degt = 1 and Y = A 1 d i 0 and Z = φ−1(B). Then Lemma 3.2 implies that φ−1(B) = A. So A and B are isomorphic. The result follows. (cid:3) Fortherestofthissectionwegivesomecorollaries. Webeginwithawell-known result (see [BS, Lemma 3.2] or [Ba, Lemma 2.1] for related results). If A is an Ore domain, let Q(A) denote the fraction division ring of A. Lemma 3.4. Let A be an Ore domain containing Z. Suppose that A is endowed with a nonzero locally nilpotent derivation δ. Then the following hold. (1) A is embedded in the Ore extension E[x;δ ] and E[x;δ ] is embedded in 0 0 Q(A), where E ={a∈Q(A)|δ(a)=0} and δ is a derivation of E. 0 (2) Q(A)=Q(E[x;δ ]). 0 (3) δ can be extended to a locally nilpotent derivation of E[x;δ ] by declaring 0 that δ(E)=0 and δ(x)=1. Proof. (1)LetE denotethekernelofthe uniqueextensionofδ toQ(A). ThenE is adivisionsubalgebraofQ(A). Sinceδ isnonzeroandlocallynilpotent,wecanfind x∈A\E suchthatδ(x)∈E. Byreplacingxbyαxforsomeα∈E wemayassume that δ(x)=1. Now for every a∈E we have δ([x,a])=[δ(x),a]=[1,a]=0. Thus [x,a]∈E for all a∈E. In particular, [x,−] induces a derivation δ of E. 0 Let W ={a∈Q(A)|δn(a)=0,for some n≥0}. We claim that W is a subset of the subalgebra of Q(A) generated by E and x. Since [x,E]⊆E, we have that this subalgebra is just Exi. i≥0 X Toseethe claim,weleta∈W. Thenthereissomesmallestnforwhichδn(a)=0. We prove the claim by induction on n. When n = 0 we have a ∈ E and so the result follows. Now suppose that the claim holds whenever δj(a) = 0 for some j < n and consider the case that δn(a) = 0 but δj(a) 6= 0 for j < n. Then δn−1(a) = α ∈ E with α 6= 0. Since δn−1(αxn−1/(n − 1)!) = α, we see that δn−1(a−αxn−1/(n−1)!)=0 and so by the inductive hypothesis a∈ Exi. The claim follows. It is clear that Exi ⊆ W. So W = Exi. Since δ is in LND(AP), A ⊂ W. Thus A embeds in the subalgebra W generated by E and x. Since [α,x] = δ (α) 0 P P for α∈E, we see that W is isomorphic to a homomorphic image of E[t;δ ]. If W 0 is in fact isomorphic to a proper homomorphic image of E[t;δ ], then A embeds in 0 a finitely generated free E-module and since δ is zero on E, we see that it must be zero on the algebra generated by E and x and hence it must be identically zero ZARISKI CANCELLATION PROBLEM FOR NONCOMMUTATIVE ALGEBRAS 9 on A, a contradiction. Thus we see that A embeds in W which is isomorphic to E[x;δ ] as required. 0 Both (2) and (3) are clear. (cid:3) The following result was proved in [Ma2] in the commutative case. Lemma 3.5. Let A be a finitely generated Ore domain over k that contains Z. If A is LND-rigid, then ML(A[x])=A. Proof. Let C =ML(A[x]). Note that C ⊆A since differentiation with respect to x gives a locally nilpotent derivation of A[x] and the kernel of this map is exactly A. ItsufficestoshowthatC ⊇A. Supposethatthereisalocallynilpotentderivationδ ofA[x] thatdoesnotsendAto zero. Supposea ,...,a generateAasak-algebra. 1 s Then δ(A)⊆Aδ(a )A+···Aδ(a )A and so there exists some smallest m≥0 such 1 s that δ(A)⊆A+Ax+···+Axm. If m=0, then δ(A)⊆A. Since A is LND-rigid, δ(A)=0. This yields a contradiction and therefore m≥1. We write δ(a)=µ(a)xm+lower degree terms for some derivation µ of A. We now consider the following three cases. Case I: δ(x)∈A+Ax+···+Axm. In this case we have δ(xi) ⊆ i+m−1Axn and δ(Axi) ⊆ i+mAxn for all i. n=0 n=0 Thus P P δ2(a)=µ2(a)x2m+lower degree terms. More generally, we see that δj(a)=µj(a)xmj +lower degree terms. Thus µ is a locally nilpotent derivation and so µ(A) = 0, contradicting the mini- mality of m. Thus δ(A)=0 in this case. Case II: δ(x)=bxm+1+lower degree terms for some b6=0 in A. Applying δ to the equation [x,a] = 0, one sees that b commutes with a, or b is in the center of A. Now we define a new derivation δ′ of A[x] by declaring that δ′(a) = µ(a)xm for a ∈ A and δ′(x) = bxm+1. Then we see that δ′ sends Axi to Axi+m foreveryi≥0. Wecanviewδ′ asaassociatedgradedderivationofδ. Since δ is locally nilpotent, δ′ is a locally nilpotent derivation of A[x] [CPWZ2, Lemma 4.11]. Applying Lemma3.4to the algebraA[x], A[x] embedsinE[y;δ ]whereδ is 0 0 a derivation of E. Moreover,δ′ extends to a locally nilpotent derivation of E[y;δ ] 0 by declaring that δ′(E) = 0 and δ′(y) = 1. Under this embedding x = p(y) for some nonzero polynomial p. Let d denote the degree of this polynomial. Then bxm+1 gets sent to q(y)p(y)m+1 for some nonzero polynomial q(y). But since δ′(x) is nonzero, it has degree exactly d−1 and so we have (m+1)d+degq(y)=d−1, which is impossible. CaseIII:δ(x)=bxi+lower degree termsforsomeb6=0inAandsomei>m+1. In this case we see that, for any n≥2, n−1 δn(x)= ((i−1)s+1) bnx(i−1)n+1+lower degree terms, ( ) s=1 Y so δ cannot be locally nilpotent, which contradicts the hypothesis. 10 JASONBELLANDJAMESJ.ZHANG Combining these cases, δ(A)=0. The result follows. (cid:3) We next give the proof of Theorem 0.4. Theorem 3.6. Let A be a finitely generated domain containing Z. Suppose it has finite GK-dimension. If A is LND-rigid, then A is cancellative. Proof. SinceAisadomainoffiniteGK-dimension,itisanOredomain. ByLemma 3.5, ML(A[x])=A. Then the assertion follows from Theorem 3.3(2). (cid:3) WenowproveTheorem0.5. WesayanalgebraAisPIifitsatisfiesapolynomial identity. Corollary 3.7. Let A be a domain of GK-dimension two over an algebraically closed field k of characteristic zero. (1) If A is PI and not commutative, then A is LND-rigid. As a consequence, if further A is finitely generated over k, then A is cancellative. (2) If A is not PI, then A is universally cancellative. Proof. (1)IfAisnotLND-rigid,thenthereisanonzerolocallynilpotentderivation δ of A. So the kernel of δ is not equal to A. As in Lemma 3.4 let E denote the set of elements a ∈ Q(A) such that δ(a) = 0. By Lemma 3.4, A embeds in W :=E[x;δ ] for some derivation δ of E. Since W is a subalgebra Q(A), Q(A) is 0 0 infinite dimensionalasaleftandrightE-vectorspace. Hence E hasGK-dimension one [Be, Theorem 1.3]. By the Small-Warfield theorem [SW] and Tsen’s theorem, everydomainofGK-dimensiononeiscommutative,whenceE isafield. ByLemma 3.4(3), W := E[x;δ ] is a subring of Q(A). Since A is PI, Q(A) and then E[x;δ ] 0 0 are PI. This implies that δ = 0 and W is commutative. So A is commutative, 0 yielding a contradiction. The result follows. The consequence follows from the main assertion and Theorem 3.6. (2)IfAisnotPIandhasGK-dimensiontwo,then,by[SZ,Corollary2],C(A)= k. The assertion follows from Proposition 1.3. (cid:3) Definition3.8. AnOredomainAiscalledbirationallyaffine-ruledifQ(A)=D(x) for some division algebra D and birationally Weyl-ruled if Q(A) = Q(E[x;δ ]) for 0 some nonzero derivation δ of E. 0 By Lemma 3.4, if A is endoweda nonzerolocally nilpotent derivation,then A is either birationally affine-ruled or birationally Weyl-ruled. Corollary 3.9. Let A be a finitely generated PI domain containing Z with finite GK-dimension. If A is not birationally affine-ruled, then A is LND-rigid and can- cellative. Proof. By Theorem 3.6, it suffices to show that A is LND-rigid. IfAisnotLND-rigid,thenAisendowedwithanonzerolocallynilpotentderiva- tion. By Lemma 3.4, A ⊂ E[x;δ ] ⊂ Q(A) where E is a division subring of Q(A). 0 SinceAisPI,soareQ(A)andE[x;δ ]. ThenthecenterofE[x;δ ]isnotasubring 0 0 ofE. Let f =a xn+a xn−1+···+a be a centralelement inE[x;δ ]for some n n−1 0 0 n≥1 and a 6=0. Since f is central, 0 n n 0=[x,f]= [x,a ]xi = δ (a )xi, i 0 i i=0 i=0 X X