This article was downloaded by: [University of Auckland Library]On: 25 November 2014, At: 14:32Publisher: Taylor & FrancisInforma Ltd Registered in England and Wales Registered Number: 1072954 Registeredoffice: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK

Communications in AlgebraPublication details, including instructions for authors andsubscription information:http://www.tandfonline.com/loi/lagb20

PRIMITIVE IDEALS OF NON-NOETHERIANDOWN-UP ALGEBRASIwan Praton a & Stephen May ba Department of Mathematics and Computer Science , Franklin andMarshall College , Lancaster, Pennsylvania, USAb Department of Mathematics , North Carolina State University ,Raleigh North Carolina, USAPublished online: 01 Feb 2007.

To cite this article: Iwan Praton & Stephen May (2005) PRIMITIVE IDEALS OF NON-NOETHERIAN DOWN-UP ALGEBRAS, Communications in Algebra, 33:2, 605-622, DOI: 10.1081/AGB-200047448

To link to this article: http://dx.doi.org/10.1081/AGB-200047448

PLEASE SCROLL DOWN FOR ARTICLE

Taylor & Francis makes every effort to ensure the accuracy of all the information (the“Content”) contained in the publications on our platform. However, Taylor & Francis,our agents, and our licensors make no representations or warranties whatsoever as tothe accuracy, completeness, or suitability for any purpose of the Content. Any opinionsand views expressed in this publication are the opinions and views of the authors,and are not the views of or endorsed by Taylor & Francis. The accuracy of the Contentshould not be relied upon and should be independently verified with primary sourcesof information. Taylor and Francis shall not be liable for any losses, actions, claims,proceedings, demands, costs, expenses, damages, and other liabilities whatsoever orhowsoever caused arising directly or indirectly in connection with, in relation to or arisingout of the use of the Content.

This article may be used for research, teaching, and private study purposes. Anysubstantial or systematic reproduction, redistribution, reselling, loan, sub-licensing,systematic supply, or distribution in any form to anyone is expressly forbidden. Terms &Conditions of access and use can be found at http://www.tandfonline.com/page/terms-and-conditions

Communications in Algebra®, 33: 605–622, 2005Copyright © Taylor & Francis, Inc.ISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1081/AGB-200047448

PRIMITIVE IDEALS OF NON-NOETHERIAN DOWN-UPALGEBRAS#

Iwan PratonDepartment of Mathematics and Computer Science, Franklin and MarshallCollege, Lancaster, Pennsylvania, USA

Stephen MayDepartment of Mathematics, North Carolina State University, Raleigh,North Carolina, USA

We identify the primitive ideals of non-Noetherian down-up algebras by determiningspecific elements that generate them. Primitive ideals of Noetherian down-up algebrashave been previously identified, so in this work we complete the classification ofprimitive ideals in down-up algebras over �.

Key Words: Down-up algebras; Non-Noetherian rings; Primitive ideals.

Mathematics Subject Classification: Primary 16D25, 16D70, 17B10; Secondary 16S30, 17B35.

1. INTRODUCTION

For any complex numbers �� �� �, the down-up algebra A��� �� �� is the algebra(over �) generated by u and d subject to the relations

d2u = �dud + �ud2 + �d and du2 = �udu+ �u2d + �u�

It is known that if � = 0, then A��� 0� �� is not a domain and is a non-Noetherian ring. Further details about down-up algebras can be found inBenkart (1998) and Benkart and Roby (1998), where they were introduced,and other papers that followed, e.g., Kirkman et al. (1999), Jordan (2000),Carvalho and Musson (2000), Benkart and Witherspoon (2001), and Hildebrand(2002). Here we determine the primitive ideals of down-up algebras with� = 0. The primitive ideals of A��� �� �� with � �= 0 have been determined in Praton(2004), so with this work we complete the classification of primitive ideals of alldown-up algebras over �.

As in the Noetherian case, there are many possible ways of determiningthe primitive ideals of A��� 0� ��. We choose here to stay close to the original

Received August 2003; Revised July 2004#Communicated by S. Kleiman.Address correspondence to I. Praton, Department of Mathematics and Computer Science,

Franklin and Marshall College, Lancaster, PA 17604, USA; Email: iwan.praton@fandm.edu

605

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

606 PRATON AND MAY

approach to down-up algebras; thus we do not use skew polynomial rings orfiltrations of rings or other advanced techniques. Down-up algebras have smallenough dimensions that we can achieve our results by explicit calculations. Asa consequence of this approach, we describe our primitive ideals by giving areasonably explicit list of their generators.

Down-up algebras are similar to enveloping algebras of semisimple Liealgebras, so it is no surprise that their primitive ideals share certain similarities.It turns out that the primitive ideals of A��� �� �� are the annihilators of irreduciblehighest weight modules L���, but there is a “multiplicity” at � = �. At that value of�, we have not only the primitive ideal AnnL��� but certain other primitive idealsas well. (They are usually, but not always, annihilators of one-dimensional modules.See Theorem 4.1 for the details.)

We describe here the convention and notations we use. As usual, � and� denote the complex numbers and the integers, respectively. We write A forA��� �� ��—no confusion should result. We always set � = 0 in this paper. Ideals aretwo-sided; the ideal generated by r1� r2� � � � is denoted �r1� r2� � � � �. We write x ≡ y

mod I to mean x − y ∈ I , as usual.In Zhao (1999), Lemma 2.2, it is shown that the monomials ui�du− aud +

b�jdk, with i� j� k ≥ 0 and a� b ∈ �, form a basis for A. We will make frequent useof this basis. The algebra A becomes a �-graded algebra by taking u to have degree1 and d to have degree −1. For example, the monomial ui�du− aud + b�jdk hasdegree i− k. Any element of degree i > 0 can be written as uix0, where x0 is anelement of degree 0; similarly, any element of degree −k < 0 can be written as x0d

k,where x0 is again an element of degree 0.

It is convenient to define h1 = du− �ud − �. We then have h1u = dh1 =0—this relation will be very useful to us. It is also convenient to define h2 = du+�/��− 1� when � �= 1 and h2 = du when � = 1. (Note that h1 = h2 when � = 0.)The following commutation relations then hold for k ≥ 0:

h2uk =

{�kukh2 if � �= 1�

uk�h2 + k�� if � = 1dkh2 =

{�kh2d

k if � �= 1�

�h2 + k��dk if � = 1�

The monomials uihj1d

k form a basis for A. So do the monomials uihj2d

k.One more notational convention: as we shall see, the expression �/��− 1�

appears so often that it is convenient to use �1 to denote it.There are many classes of A-modules described in Benkart and Roby (1998),

but we only need a few of them here: Verma modules, quotients of Verma modules,and finite-dimensional modules. (In one case—when � = 1, � = 0—we need to bringin a doubly-infinite module, but that is its only appearance.) All these modules areweight modules (see Benkart and Roby, 1998, for the definition of weight modules).There are irreducible A-modules that are not weight modules (for example, when� = 0 and � is not a root of unity, the module whose basis is vi � i ∈ ��, whereuvk = vk+1 and dvk = �kvk−2, is irreducible). It is interesting that the primitive idealsof A are annihilators of such a restricted class of modules.

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

PRIMITIVE IDEALS OF NON-NOETHERIAN DOWN-UP ALGEBRAS 607

2. VERMA MODULES AND PRIMITIVE IDEALS

Verma modules play a central role in the calculation of primitive ideals of A,so we review here their definition and properties. See Benkart and Roby (1998), 2.2to 2.8, for more details and proofs.

Let � be an arbitrary complex number. Define a sequence �k�k≥0 as follows:�0 = �, and �k+1 = ��k + � for k ≥ 1. The Verma module with highest weight �,denoted by V���, is an A-module whose basis is v0� v1� � � � � vn� � � � �; the action of Ais defined by

uvk = vk+1� dvk+1 = �kvk for k ≥ 0 and dv0 = 0�

It is convenient to define �1 = �− � (for all values of �) and �2 = �+ �1 (when� �= 1). (Recall that �1 = �/��− 1�.) We then have h1v0 = �1v0 and h2v0 = �2v0 when� �= 1. Using the commutation relation between h1 and u, and also between h2 andu, we can easily figure out that

h1vk = 0 for k ≥ 1 and all values of ��

and for k ≥ 0,

h2vk ={�k�2vk if � �= 1�

��+ k��vk if � = 1�

In particular, h1�h1 − �1�vk = 0 for all k ≥ 0. Thus AnnV��� always containsh1�h1 − �1�; this is a useful fact for us when we calculate the annihilators of Vermamodules.

The Verma module V��� is irreducible if and only if �k �= 0 for all k ≥ 0. It istherefore useful to have a convenient formula for �k. In general, the formulas arecomplicated (see Benkart and Roby, 1998, Proposition 2.12), but since � = 0 in ourcase, the results are relatively straightforward:

�k ={�k�2 − �1 if � �= 1�

�+ k� if � = 1�

If V��� is not irreducible, then there exists a k ≥ 0 such that �k = 0. Pick theminimum such value of k. Then spanvi � i > k� is a maximal submodule of V���;this maximal submodule is denote by M��� in Benkart and Roby (1998). We willcall it the standard maximal submodule of V���. If V��� is irreducible, we define M���to be �0�. The quotient V���/M��� is an irreducible module denoted by L���; wewill call it the standard quotient of V���. Thus L��� = V��� if V��� is irreducible;otherwise L��� is a proper quotient of V���, and in fact, L��� is finite-dimensional.It is possible for V��� to have other irreducible quotients (i.e., V��� may have othermaximal submodules besides M���), but if the �is are distinct, then L��� is the onlyirreducible quotient. (Necessary and sufficient conditions for V��� to have only oneirreducible quotient are somewhat complicated. There is a full discussion in Benkartand Roby, 1998, Corollary 2.28, completed in the addendum to Benkart and Roby,1998).

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

608 PRATON AND MAY

Recall from Benkart and Roby (1998, Definition 2.7) that an A-module V is ahighest weight module of weight � if there is a v ∈ V such that Av = V , �du�v = �v,and dv = 0. In particular, the Verma module V��� is a highest weight module. Morethan that, V��� has the usual universal property: if V is any highest weight modulewith weight �, then V is a quotient of V���.

The following two results justify our claim that Verma modules play a crucialrole in determining the primitive ideals of A.

Proposition 2.1. Let M be an irreducible A-module. Then either h1M = 0 or thereexists a complex number c such that h1�h1 − c�M = 0.

Proof. The proof is a simplified version of Kirkman and Kuzmanovich (2000a,Theorem 3.4). We get a more precise conclusion because we are working over analgebraically closed field.

Suppose there is an element v ∈ M such that h1v �= 0. Because M isirreducible, there exists an element x ∈ A such that xh1v = v. We can write x =∑

i�k≥0 uipik�h1�d

k, where pik�h1� is a polynomial in h1. Since dh1 = 0, we can assumethat x has the form

∑i≥0 u

ipi�h1�. Thus we have the equation

∑i≥0

uipi�h1�h1v = v�

Multiply both sides on the left by h1, and recall that h1u = 0. We get

h1p0�h1�h1v = h1v�

Thus we have �1− p0�h1�h1�h1v = 0, where 1− p0�h1�h1 is a polynomial in h1 ofdegree at least 1. Factor it into linear factors, and we get �h1 − cn��h1 − cn−1� · · ·�h1 − c2��h1 − c1�h1v = 0. Let k be the first index such that

∏ki=1�h1 − ci�h1v �= 0

but �h1 − ck+1�∏k

i=1�h1 − ci�h1v = 0. Let v′ = ∏ki=1�h1 − ci�v; then v′ �= 0, h1v

′ �=0, and �h1 − ck+1�h1v

′ = 0. To avoid repeating the index, write c = ck+1 so that�h1 − c�h1v

′ = 0.We now claim that �h1 − c�h1M = 0. Since h1v

′ �= 0, any element in M can bewritten as

∑i≥0 u

iqi�h1�h1v′, where qi�h1� is a polynomial in h1. But then

�h1 − c�h1

∑i≥0

uiqi�h1�h1v′ = �h1 − c�h1q0�h1�h1v

′ �since h1u = 0�

= h1q0�h1��h1 − c�h1v′ = 0�

This proves the proposition. �

Theorem 2.2. Suppose M is an irreducible module. If h1M �= 0, then M is isomorphicto L��� for some � ∈ �.

Proof. Pick a nonzero element v ∈ M such that h1v �= 0. (Such an element existssince we are assuming that h1M �= 0.) Since M is irreducible, M is generated by h1v.

Now any x ∈ A can be written as∑

i�k≥0 uipik�h1�d

k. But then xh1v =∑i≥0 u

ipi0�h1�h1v. By Proposition 2.1, there is a c ∈ � such that h1�h1 − c�v = 0,

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

PRIMITIVE IDEALS OF NON-NOETHERIAN DOWN-UP ALGEBRAS 609

i.e., h21v = ch1v. So pi0�h1�h1v must be a multiple of h1v. Thus xh1v = ∑

i≥0 ciuih1v.

Since x was arbitrary, we see that M is spanned by uih1v � i ≥ 0�.Let vi = uih1v. Since M = Av0 and dv0 = dh1v = 0, we see that M is a highest

weight module. By the universal property of Verma modules, we conclude that Mis either isomorphic to a Verma module V��� or a quotient of V���. We are not yetdone; we still have to show that if M is isomorphic to a proper quotient of V���,then it must be the standard quotient.

We first figure out the value of �. Since �du�v0 = �h1 + �ud + ��h1v = �h21 +

�h1�v = �c + ��h1v = �c + ��v0. The highest weight is thus � = c + �.We now claim that c �= 0. If it is, then h1vi = h1�u

ih1v� = 0 for i > 0(since h1u = 0), and h1v0 = h1�h1v� = h2

1v = 0 (since h1�h1 − c� = h21 annihilates M).

Therefore h1 annihilates M , contrary to our original assumption.Now suppose M is not isomorphic to V���. Then V��� is not irreducible: it

has a proper submodule N . Suppose w = a0v0 + a1v1 + · · · + anvn is an elementin N , with a0 �= 0. Then h1w = ca0v0 ∈ N ; since c �= 0, we have v0 ∈ N , so N = V���,a contradiction. Therefore N ⊆ spanv1� v2� � � � �, so by Benkart and Roby (1998,2.4(c)), N must lie inside the standard maximal submodule. Thus M must beisomorphic to the standard irreducible quotient L���. �

Corollary 2.3. The primitive ideals of A��� 0� �� are

• AnnL���� � ∈ �,• primitive ideals containing h1.

The corollary gives us a good overall view of the primitive ideals of A. We wouldlike to get a more detailed view; specifically, we want to give a reasonably explicitlist of the generators of each primitive ideal. Unfortunately, the calculations forthe most part involve case-by-case considerations. Some are similar, but there areenough annoying local differences that it would be clearer to separate the cases. Westart, however, by gathering results that apply to all cases.

3. GENERAL RESULTS

We gather here some useful results that are valid for all (or most) values of �.

Annihilators of Some Verma Modules. As mentioned above, the annihilator ofa Verma module V��� must contain h1�h1 − �1�. In most cases, h1�h1 − �1� turnsout to generate the whole annihilator, but there are certain special cases where theannihilator is bigger. To calculate these large annihilators, we start with a usefullemma.

Lemma 3.1. Suppose V��� is an irreducible Verma module. Then AnnV��� isgenerated by elements of degree 0.

Proof. From the definition of the action of A, it is clear that an element of degreei takes vk to �vk+i. Therefore if x = ∑

xi annihilates V���, where xi is of degreei, then each xi also annihilates V���. Thus the annihilator of V��� is generated byhomogeneous elements.

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

610 PRATON AND MAY

Now suppose x is an element in AnnV��� with degree i > 0. Then we can writex = uix0, where x0 is an element of degree 0. Now ui, considered as an operator onV���, is injective: if ui

(∑j cjvj

) = 0, then∑

j cjvj+i = 0, which implies that cj = 0 forall j. Therefore uix0V��� = 0 implies that x0V��� = 0, i.e., x0 is already in AnnV���.Thus x is a multiple of a degree 0 element.

On the other hand, suppose x is an element in AnnV��� with degree −i < 0.Then we can write x = x0d

i, where x0 is an element of degree 0. In thiscase we need to show that di is a surjective operator. This is not hard: vk =di

(��k�k+1 · · · �k+i−1�

−1vk+i

)—note that none of the �j is zero since V��� is irreducible.

Thus x0diV��� = 0 implies that x0V��� = 0, i.e., x0 is also in AnnV���. Thus in this

case x is also a multiple of a degree 0 element. Thus to find the generators ofAnnV���, we only need to consider elements of degree 0. �

We use the lemma above to calculate the annihilator of some special Vermamodules.

Proposition 3.2. Let V��� be an irreducible Verma module. Suppose h = du+ aud +b (where a� b ∈ �) annihilates V���. Then the annihilator of V��� is generated by h.

Proof. Suppose x ∈ A annihilates V���; by the lemma above, we can assume thatx is an element of degree 0. Then we can write x = ∑n

i=0 uipi�h�d

i, where pi�h� is apolynomial in h. We want to show that pi�h� is a multiple of h.

From xv0 = 0 we conclude that �∑n

i=0 uipi�h�d

i�v0 = p0�0�v0 = 0 since div0 = 0for i > 0. Thus p0�0� = 0.

From xv1 = 0 we conclude (since div1 = 0 for i > 1) that �p0�0�+ �0p1�0��v1 = 0. Since we already know that p0�0� = 0 and �0 �= 0, we see that p1�0� = 0.

We can continue such calculations easily. In general, for i ≤ n, the equationxvi = 0 leads to

p0�0�+ �i−1p1�0�+ · · · + �i−1�i−2 · · · �0pi�0� = 0�

from which we conclude (by induction) that pi�0� = 0. Thus pi�h� is indeed amultiple of h, as required. �

Quotients of Verma Modules. We now calculate the annihilator of L��� whenL��� �= V���. Recall that if V��� is not irreducible, then there is a k ≥ 0 such that�k = 0. Choose k to be the smallest such possible value. Thus the images in L��� ofv0� v1� � � � � vk form a basis for L���: L��� is �k+ 1�-dimensional.

Proposition 3.3. Let L��� be a proper standard irreducible quotient of a Vermamodule as described above. If k = 0 (i.e., � = 0), let y� = 0; if k ≥ 1, let y� = h1 + 0 + 1ud + 2u

2d2 + · · · + kukdk, where � 0� 1� � � � � k� is the (unique) solution to

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

PRIMITIVE IDEALS OF NON-NOETHERIAN DOWN-UP ALGEBRAS 611

the system of equations

1 0 0 0 · · · 01 �0 0 0 · · · 01 �1 �1�0 0 · · · 01 �2 �2�1 �2�1�0 · · · 0���

������

���� � �

���1 �k−1 �k−1�k−2 �k−1�k−2�k−3 · · · �k−1�k−2 · · · �0

0 1 2 3��� k

=

−�+ �000���0

�

Then the annihilator of L��� is �uk+1� dk+1� y��.

Proof. If � = 0, then L�0� is one dimensional and uv0 = dv0 = 0, so AnnL�0� =�u� d�.

If k ≥ 1 it is clear that uk+1 and dk+1 annihilate L���. We need to verifythat y� also annihilates L���. Now y�v0 = ��− �+ 0�v0 = 0 by definition of 0.Furthermore, if 1 ≤ j ≤ k, then

y�vj = � 0 + �j−1 1 + �j−1�j−2 2 + · · · + �j−1 · · · �0 j�vj = 0

by definition of the is. Therefore y� does indeed annihilate L���.Let I denote the ideal generated by uj+1, dj+1, and y�. All congruences in this

proof will be mod I . First note that h1y� = h21 + 0h1 is in I , and hence h2

1 ≡ − 0h1.Furthermore, from y� ≡ 0 we get h1 ≡ −∑

iuidi. We also have uk+1 ≡ dk+1 ≡ 0.

Thus any element of A—which is a linear combination of the monomials uihn1d

j—iscongruent to a linear combination of the monomials uidj where i� j ≤ k.

Now let x be an element of degree j that annihilates L���. For definitenessassume that j ≥ 0. By the discussion above, x ≡ uj�a0 + a1ud + · · · + ak−ju

k−jdk−j�where ai ∈ �. Then xv0 = 0 implies that a0vj = 0, i.e., a0 = 0. Also, xv1 = 0 impliesthat a0 + �0a1 = 0, i.e., a1 = 0. We can continue doing this kind of calculation,concluding that ai = 0 for 0 ≤ i ≤ k− j. Thus x ≡ 0. The same result holds whenj ≤ 0. Hence x must already be in I , which concludes the proof. �

The Algebra A/�h1�. Suppose M is an irreducible A-module and the annihilator ofM contains h1. Let I denote the ideal generated by h1. Then M is also an irreducibleA/I-module, so to determine the primitive ideals of A containing I it suffices todetermine the primitive ideals of A/I . The algebras A/I are in many cases well-known (for example, when � = 1 and � �= 0, then A/I is the first Weyl algebra), butwe will give a unified treatment of these algebras for completeness.

A/I is generated by the elements d and u subject to the relation du = �ud + �.Any element of A/I can be written (uniquely) as a linear combination of themonomials uidk, i� k ≥ 0. Since the relation du− �ud − � = 0 is homogeneous ofdegree 0, the algebra A/I inherits the usual degree function from A. The monomialuidk has degree i− k.

Assume that � �= 0. Then ud is a first-degree polynomial in du. By inductionwe can establish that uidi is a polynomial in du of degree i, and hence it is alsoa polynomial in h2 of degree i. Thus any homogeneous element in A/I of degreek ≥ 0 can be written as ukp�h2�, where p�h2� is a polynomial in h2. Similarly, any

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

612 PRATON AND MAY

homogeneous element of degree −k ≤ 0 can be written as q�h2�dk, where q�h2� is a

polynomial in h2.Note also that if � �= 0, then any element of A/I can also be written uniquely

as a linear combination of the monomials dkui, i� k ≥ 0, as well as a linearcombination of the monomials uidk as mentioned above. Consequently, neither unor d are zero divisors in A/I .

One-dimensional Modules. The irreducible module L�0� is one-dimensional, butthere are other possible one-dimensional modules. They have all been classifiedin Benkart and Roby (1998), but we discuss them here since the classification isstraightforward.

Suppose �v is a one-dimensional A-module. Then uv = cuv and dv = cdv forsome cu� cd ∈ �. We must have

cdc2u = �cucdcu + �cu and c2dcu = �cdcucd + �cd�

It is then easy to figure out that either cu = cd = 0 (in which case our module is justL�0�) or �1− ��cdcu = �. Thus we have the following result.

Proposition 3.4. The annihilators of one-dimensional (irreducible) modules are

(1) (if � �= 1): �u− cu� d − cd� where cu� cd ∈ � with cucd = −�1,(2) (if � = 1 and � = 0): �u− cu� d − cd� where cu� cd ∈ �.

Schur's Lemma. The following result (the strong form of Schur’s lemma) will bevery useful to us.

Proposition 3.5. Suppose R is a finitely generated algebra over an uncountablealgebraically closed field, and M is a simple R-module. Then any element in the centerof R acts as a scalar. In particular, if R is commutative, then all elements, of R act asscalars, so M has to be one-dimensional.

A proof is implicit in McConnell and Robson (1987, Proposition 9.1.7). It is used inthe down-up algebra setting in Kirkman and Kuzmanovich (2000b) and in Praton(2004).

4. PRIMITIVE IDEALS

Here is a list of all primitive ideals of A��� 0� ��. Recall that h1 = du− �ud − �,h2 = du+ �1 if � �= 1, and h2 = du if � = 1. The element y� was defined inProposition 3.3.

Theorem 4.1. If � = 1 and � = 0, the primitive ideals of A�1� 0� 0� are

• AnnL��� ={�u� d� if � = 0�

�du− �� if � �= 0

• �d − cd� u− cu�, where cd� cu ∈ �, not both zero.

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

PRIMITIVE IDEALS OF NON-NOETHERIAN DOWN-UP ALGEBRAS 613

If � = 1 and � �= 0, the primitive ideals of A�1� 0� �� are

• AnnL��� =

�h1� if � = ��

�uk+1� dk+1� y�� if � = −k�� k ≥ 0�

�h1�h1 − �+ ��� for other values of ��

If � is not a root of unity and � �= 0, but � = 0, the primitive ideals of A��� 0� 0� are

• AnnL��� ={�u� d� if � = 0�

�h1�h1 − ��� if � �= 0• �h1�;• �d − cd� u− cu�, where cdcu = 0, but cd and cu are not both zero.

If � is not a root of unity and � �= 0, and � �= 0, the primitive ideals of A��� 0� �� are

• AnnL��� =

�uk+1� dk+1� y�� if � = ��−k − 1��1� k ≥ 0�

�h1� if � = ��

�h2� if � = −�1�

�h1�h1 − �+ ��� for other values of �

• �d − cd� u− cu�, where cu� cd ∈ � with cucd = −�1.

If � = 0 and � = 0, the primitive ideals of A�0� 0� 0� are

• AnnL��� ={�u� d� if � = 0�

�u2� d2� y�� if � �= 0

• �u− cu� d − cd� where cucd = 0, but cu and cd are not both zero.

If � = 0 and � �= 0, then the primitive ideals of A�0� 0� �� are

• AnnL��� =

�u� d� if � = 0�

�h1� if � = ��⟨h1�h1 − �+ ��� �ud − ���ud − ��− ��

�− �h1

⟩if � �= �� � �= 0

• �u− cu� d − cd� where cucd = �.

If � is a primitive nth root of unity �n > 1� and � = 0, the primitive ideals of A��� 0� 0�

are

• AnnL��� ={�u� d� if � = 0�

�h1�h1 − ��� hn2 − �n� if � �= 0

• �u− cu� d − cd� where cucd = 0, but cu and cd are not both zero,

• �h1� dn − zd� u

n − zu�, where zdzu �= 0.

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

614 PRATON AND MAY

If � is a primitive nth root of unity �n > 1� and � �= 0, the primitive ideals of A��� 0� ��are

• AnnL��� =

�uk+1� dk+1� y�� if � = ��−k − 1��1� 0 ≤ k ≤ n− 1�

�h2� if � = −�1�

�h1�h1 − �+ ��� hn2 − ��+ �1�

n� for other values of �

• �u− cu� d − cd� where cucd = −�1,• �h1� dn − zd� u

n − zu�, where zdzu �= �−�1�n, and zd and zu are not both zero.

We prove this theorem through case-by-case considerations. For each value of � and�, we need to (1) calculate the annihilators of irreducible Verma modules and (2)determine the primitive ideals containing h1.

Proof when � = 1. We start by determining the annihilators of L���. When � = 1,we have �k = �+ k� for k ≥ 0.

Assume first that � = 0. If � = 0, then V�0� is not irreducible and L�0� isone-dimensional. Its annihilator is �u� d� by Proposition 3.3. If � �= 0, then V���is irreducible. Its annihilator is �du− �� by Proposition 3.2. This completes thedetermination of AnnL��� when � = 1, � = 0.

Assume now � �= 0. If � = −k� for some nonnegative integer k, then �k = 0;furthermore, no earlier value of �i is zero. Thus L�−k�� is �k+ 1�-dimensional, andits annihilator is �uk+1� dk+1� y�� by Proposition 3.3.

It only remains to calculate the annihilators of irreducible Verma modules. If� = �, then AnnV��� = �h1� by Proposition 3.2. Thus we can assume that � �= −k�,k ≥ −1. We want to find AnnV���.

Suppose pi�h1� is a polynomial in h1. Recall that h1v0 = �1v0 �= 0 and h1vj = 0for j > 0. So for k > i, we have

�uipi�h1�di�vk = �k−1�k−2 · · · �k−ipi�0�vk

= pi�0���+ �k− 1�����+ �k− 2��� · · · ��+ �k− i���vk�

Define polynomials fi�t� ∈ ��t� by

fi�t� ={1 if i = 0∏i

j=1�t − j�� if i > 0�

The polynomial fi has degree i, so f0� f1� � � � � fn are linearly independent in thevector space ��t�. The calculations above show that, for k > i, �uipi�h1�d

i�vk =pi�0�fi��+ k��vk.

Now suppose x is an element of degree 0 that annihilates V���. (By Lemma3.1 we only need to consider elements of degree 0.) We want to show that xis in the ideal generated by h1�h1 − �1�. Since x has degree 0, we can write x =∑n

i=0 uipi�h1�d

i. For all large values of k (it suffices to take k > n), the equationxvk = 0 implies

∑ni=0 pi�0�fi��+ k�� = 0. Thus the polynomial

∑ni=0 pi�0�fi�t� has

infinitely many zeroes, namely, �+ k� for all k > n. Therefore it has to be the zeropolynomial. Since the fi are linearly independent, we can conclude that pi�0� = 0.

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

PRIMITIVE IDEALS OF NON-NOETHERIAN DOWN-UP ALGEBRAS 615

Now from xv0 = 0 we can conclude that p0��1� = 0. From xv1 = 0 weconclude that p0�0�+ �0p1��1� = 0; since we now know that p0�0� = 0, we concludethat p1��1� = 0. We can continue such calculations easily; we conclude that pi��1� =0 for 0 ≤ i ≤ n.

Thus pi�0� = pi��1� = 0, which implies that pi�h1� is a multiple of h1�h1 − �1�.Therefore x is inside the ideal generated by h1�h1 − �1�. So Ann V��� = �h1�h1 −�1��, as stated in the theorem. This concludes the calculation of AnnL��� when � =1 and � �= 0.

Ideals containing h1: We now calculate ideals containing h1. Recall that weuse the algebra A/I , where I = �h1�.

If � = 0, then du = ud in A/I , i.e., A/I is commutative. By the strong form ofSchur’s lemma, M is one-dimensional. This takes care of the case � = 0.

Now assume � �= 0. There are two possibilities to consider: (1) there exists anonzero v ∈ M such that dv = 0, and (2) dv �= 0 for all nonzero v ∈ M .

In the first case, we have �du�v = ��ud + ��v = �v. Since M is irreducible, wehave M = Av. Thus M is a highest weight module with highest weight �, so Mis a homomorphic image of V���. Since V��� is irreducible, M must actually beisomorphic to V���. By Proposition 3.2, the annihilator of M is �h1�.

The second case is more complicated. Assume there exists a nonzero x ∈ A/Isuch that xM = 0. We will derive a contradiction.

Multiplying by a high enough power of d, we can assume that x has the formx = ∑n

i=0 pi�h2�di. (Recall that h2 = du.) Choose x so that the highest power of d is

as small as possible, i.e., choose x so that n is minimal. Then (using the commutationrelation dih2 = �h2 + i��di),

�h2 + n��x − xh2 =n∑

i=0

�h2 + n��pi�h2�di −

n∑i=0

pi�h2��h2 + i��di =n∑

i=0

�n− i��pi�h2�di

is an element in the annihilator where the highest power of d is at most n− 1. Thusit must be zero; since �n− i�� �= 0 for i = 0� � � � � n− 1, this means pi�h2� = 0 for ibetween 0 and n− 1. Thus the nonzero element pn�h2�d

n is in the annihilator of M .Multiplying by un, we see that there is a polynomial in h2 that annihilates M .

Factor this polynomial, and we conclude that there is a nonzero element w ∈ Msuch that �h2 − c�w = 0, i.e., h2w = cw, where c ∈ �. For i ≥ 0, the element diwis nonzero since we are in case (2), and h2d

iw = di�h2 − i��w = �c − i��diw. Butpn�h2�d

n is in the annihilator of M , so

0 = pn�h2�dn�diw� = pn�c − �i+ n���di+nw�

which implies that pn�c − �i+ n��� = 0 for all i ≥ 0. Thus pn is a polynomial withinfinitely many roots, and hence it must be the zero polynomial. We have found ourcontradiction. Hence the annihilator of M in A must be just �h1�. This takes care ofthe case � �= 0 and finishes the proof when � = 1.

Proof when � is not a root of unity, � �= 0. The proof for this case is similar inlarge outline with the proof for � = 1, but the local details differ.

We start by determining the annihilators of L���. In this case �k = �k�+ ��k −1��1 = �k�2 − �1. So if � = 0, then V�0� is not irreducible, and its standard quotient

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

616 PRATON AND MAY

L�0� is a one-dimensional module whose annihilator is �u� d� by Proposition 3.3.On the other hand, suppose � �= 0. In this case, if �2 = �−k�1, i.e., if � = ��−k − 1��1,then �k = 0, and no earlier value of �i is zero. Thus L��� is �k+ 1�-dimensional, andits annihilator is �uk+1� dk+1� y�� by Proposition 3.3. Furthermore, if � = �, then V���is irreducible, and its annihilator is �h1� by Proposition 3.2. Similarly, if � = −�1,i.e., if �2 = 0, then V��� is irreducible; its annihilator is �h2�, also by Proposition 3.2.

It remains to show that AnnV��� = �h1�h1 − �1�� for other values of �.We allow the possibility that � = 0 here, but we can assume that �1 �= 0 and �2 �= 0.The calculation is similar to the case � = 1. This time we have

�k−1�k−2 · · · �k−i = ��k−1�2 − �1���k−2�2 − �1� · · · ��k−i�2 − �1��

and we define fi�t� ∈ ��t� by

fi�t� ={1 if i = 0∏i

j=1��−j�2t − �1� if i > 0�

These polynomials are also linearly independent. Then we have �uipi�h1�di�vk =

fi��k�pi�0�vk for i ≥ k.The rest of the proof is basically identical to the proof when � = 1. We give

a quick recap. Suppose x = ∑ni=0 u

ipidi is an element of degree 0 that annihilates

V���. We start from xvk = 0 where k is large, and we conclude that the polynomial∑ni=0 pi�0�fi�t� has infinitely many zeroes, namely, �k for all large enough values of k.

Thus pi�0� = 0 for 0 ≤ i ≤ n. Then from xv0 = 0� xv1 = 0� � � � � xvn = 0, we concludethat pi��1� = 0 for 0 ≤ i ≤ n. Thus pi�h1� is a multiple of h1�h1 − �1�, which is whatwe needed to show. This concludes the calculation of AnnL���.

Ideals containing h1: To investigate primitive ideals containing h1, wedistinguish two possibilities (similar to when � = 1): (1) there is a nonzero v ∈ Msuch that dv = 0, and (2) dv �= 0 for all nonzero v ∈ M . In the first case we deducejust as before that M is a homomorphic image of V���. Thus the annihilator of Mis already listed among the annihilators of L���.

Now assume that dv �= 0 for all nonzero v ∈ M . We again work in A/I .Suppose there is a nonzero element x = ∑n

i=0 pi�h2�di ∈ A/I that annihilates M . We

will show that M must then be one-dimensional. Choose x so that the highest powerof d that appears is as small as possible. Then

�nh2x − xh2 =n∑

i=0

�npi�h2�h2di −

n∑i=0

pi�h2�dih2 =

n∑i=0

��n − �i�pi�h2�h2di

is in the annihilator of M where the highest power of d that appears is at mostn− 1. By choice of x, it must be zero. Since �i �= �n, we must have pi�h2� = 0 fori = 0� 1� � � � � n− 1. Thus x = pn�h2�d

n.Multiply x by un and we see that there is a polynomial in h2 that annihilates M .

Factor this polynomial over �; we conclude that there is a nonzero element w ∈ Mand c ∈ � such that �h2 − c�w = 0. We now claim that c = 0. For a contradictionsuppose c �= 0. We proceed as in the proof for � = 1. For any i ≥ 0, we have

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

PRIMITIVE IDEALS OF NON-NOETHERIAN DOWN-UP ALGEBRAS 617

h2diw = �−idih2w = c�−iw, so diw is an eigenvector of h2 with eigenvalue c�−i. Since

pn�h2�dn annihilates M ,

0 = pn�h2�dndiw = pn�c�

−n−i�dn+iw�

which implies pn�c�−n−i� = 0 since dn+iw �= 0. Thus pn has infinitely many zeroes,

and hence pn is the zero polynomial, a contradiction.We conclude that c = 0, i.e., h2w = 0. (We didn’t have to deal with

this situation when � = 1.) Now h2uidjw = �i−juidjh2w = 0. Since M = Aw =∑

i�j �uidjw, we conclude that h2 ∈ Ann M . Now A/�h1� h2� is commutative (weeasily check that du = ud = −�1 in A/�h1� h2�), so M must be one-dimensional.

We have shown that if a primitive ideal is strictly larger than �h1�, then itmust be the annihilator of a one-dimensional module. When � �= 0 this finishes theproof, but when � = 0, we haven’t shown that �h1� is actually a primitive ideal!(Compare with the situation when � = 1 and � = 0: in that case, any primitive idealthat contains h1 must be the annihilator of a one-dimensional module, so in thatsituation, �h1� is definitely not a primitive ideal.) When � = 0, the ideal �h1� is notthe annihilator of an irreducible Verma module, so we have to look elsewhere forour irreducible module. Fortunately it is not hard to find. Let � be an arbitrarynonzero complex number. Define �k = �k� for all integers k (including negativevalues of k). Define an A-module M whose basis is vk � k ∈ �� and whose A-actionis defined by uvk = vk+1 and dvk = �k−1vk−1. It is easy to check that M is indeed anA-module—in fact, it is one of the doubly-infinite modules considered in Benkartand Roby (1998). By their Proposition 5.11, M is irreducible.

Since h1vk = �du− �ud�vk = ��k − ��k−1�vk = 0, we see that h1 is in AnnM ,so M is also an irreducible A/I-module. Now as in the proof of Lemma 3.1, ui isinjective and di is surjective, so AnnM is generated by elements of degree 0. Suchan element can be written as a polynomial f�du� in du. But then 0 = f�du�vk =f��k�vk implies that f has infinitely many zeroes (since the �ks are distinct), so f�du�is actually zero. This proves that the annihilator of M in A is �h1�, so �h1� is indeeda primitive ideal. This finishes the proof when � is not a root of unity, � �= 0.

Proof when � = 0. There is a different flavor to the proofs when � = 0 comparedto other values of �. We start, as usual, by determining the annihilators of L���.

When � = 0, we have �0 = � and �k = � for k ≥ 1. Therefore if � = 0,the Verma module V��� is never irreducible. The irreducible quotient L��� iseither one-dimensional (when � = 0) or two-dimensional (when � �= 0). Thus byProposition 3.3, the annihilators of L��� are as stated in the theorem. If � �= 0, thenV��� is irreducible if and only if � �= 0, and if � = 0, then the irreducible quotientL��� is one-dimensional. Its annihilator is �u� d�. If � = �, then AnnV��� = �h1� byProposition 3.2.

It remains to calculate AnnV��� when � �= �, � �= 0, and � �= 0. We can checkeasily that

h1�h1 − �+ �� and �ud − ���ud − ��− ��

�− �h1

are in the annihilator. Let J be the ideal generated by these two elements; weclaim that J is the whole annihilator. All congruences are mod J . Note that �ud − ��

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

618 PRATON AND MAY

�ud − �� = �ud�2 − ��+ ��ud+�� = u�h1 + ��d−��+ ��ud + �� = uh1d − �ud + ��,thus

uh1d ≡ �ud + ��

�− �h1 − ���

Therefore ukh1dk is congruent to an element of the form b1h1 + a0 + a1ud +

a2u2d2 + · · · + aku

kdk.Suppose x is an element of degree 0 that annihilates V���. The previous

paragraph shows that x is congruent to an element of the form b1h1 + a0 + a1ud +a2u

2d2 + · · · + akukdk. Assume that k ≥ 1. Apply this element to vk, and we get

0 = a0 + �a1 + �2a2 + · · · + �k−1ak−1 + ��k−1ak�

Apply the element to vk+1 and we get

0 = a0 + �a1 + �2a2 + · · · + �k−1ak−1 + �kak�

Since � �= �, we conclude that ak = 0. By backwards induction we conclude that ai =0 for all values of i between 1 and k. Thus x ≡ b1h1 + a0. Apply this to v1 and we geta0 = 0. Hence b1 = 0 too. Conclusion: x ≡ 0, and hence J is indeed the annihilatorof V���. This concludes the calculation of AnnL���.

Ideals containing h1: We want to show that when � = 0, all primitive idealscontaining h1 are annihilators of one-dimensional modules, while in the case � �= 0,all primitive ideals strictly containing �h1� are annihilators of one-dimensionalmodules. As usual we work with the algebra A/I , where I = �h1�. Suppose first that� = 0. Then in A/I , we have du = 0. Consider the subspace uM . We have duM =0 ⊆ M and uuM ⊆ uM , so uM is actually a submodule of M . Thus it is either 0or M . If uM = 0, then u is in the annihilator of M . Hence M is an irreducibleA/�u�-module, but A/�u� = ��d� is commutative, so by Proposition 3.5 M is one-dimensional. On the other hand, if uM = M , then duM = 0 implies that d is in theannihilator of M . The same argument as before, with u and d interchanged, showsthat M is also one-dimensional in this case.

Now suppose � �= 0. Then in A/I we have du = �, and by induction weestablish easily that dkuk = �k for all k ≥ 0. Now suppose there is a nonzero elementin A/I that annihilates M . It can be written in the form

∑i�j u

idj . Multiply by a largeenough power of u on the right, and we see that it can be written as a polynomialin u. Factoring the polynomial, we see that �u− c1� · · · �u− ck� is in AnnM . Thusthere is a nonzero v ∈ M such that �u− c�v = 0.

Now apply d to the equation �u− c�v = 0. We get duv = cdv, i.e., �v = cdv;since � �= 0, we conclude that c �= 0. Thus dv = c−1�v. Since uv = cv, we see that theone-dimensional subspace spanned by v is actually a submodule. Hence it must beall of M , i.e., M is one-dimensional. This finishes the proof in the case � = 0.

Proof when � is a primitive nth root of unity, n > 1. We now take care of thelast remaining case. Assume that �n = 1 and �k �= 1 for any 0 < k < n. Note thathn2u = �nuhn

2 = uhn2 and dhn

2 = �nhn2d = hn

2d, so hn2 is in the center of A. By Schur’s

lemma, hn2 must act as a scalar on any irreducible module M .

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

PRIMITIVE IDEALS OF NON-NOETHERIAN DOWN-UP ALGEBRAS 619

We start as usual by determining the annihilators of L���. We have �k =�k�2 − �1 for k ≥ 0. Note that �k = �k+n. Suppose first that � = 0. Then V��� isirreducible if and only if �2 = � �= 0; if � = 0, then L�0� is one-dimensional andits annihilator is �u� d� as stated in the theorem. Now suppose that � �= 0. ThenV��� is irreducible if and only if �2 �= �−k�1, i.e., � �= ��−k − 1��1 for any nonnegativeinteger k. (Note that V��� is not irreducible even when � �= 0, in contrast to thesituation for other values of �.) If � = ��−k − 1��1, where 0 ≤ k ≤ n− 1, then �k = 0,and furthermore, this is the earliest value of �i that is zero. Thus the annihilator ofL��� is indeed �uk+1� dk+1� y��, as stated in the theorem.

If � �= 0 and � = −�1 (i.e., �2 = 0), then AnnV��� = �h2�, by Proposition 3.2.It remains, then, to calculate AnnV��� where we can assume that �1 �= 0, �2 �= 0. It isstraightforward to check that h1�h1 − �1� and hn

2 − �n2 annihilates V���. Let J denote

the ideal generated by h1�h1 − �1� and hn2 − �n

2 . All congruences will be modJ .We have h2

1 ≡ �1h1. A long but straightforward calculation shows that

h21 = h2

2 − 2��+ �1�h2 + ��+ �1�2 + �2�1ud − �2uh2d�

and so from h21 ≡ �1h1 we get

�2uh2d ≡ h22 − �2�+ 2�1 + �1�h2 + ��+ �1���+ �1 + �1�+ ��2�1 + ��1�ud�

(The details are not important: what is crucial is that uh2d is congruent to apolynomial in h2 plus a constant times ud.)

Using this expression (and the commutation relation between h2 and u and d),any monomial of the form ujhk

2dj is congruent to a sum of a polynomial in h2 and

a linear combination of the monomials uidi.Now suppose x is an element of degree 0 and the annihilator of V���. We

want to show that x ≡ 0. x can be written as a linear combination of the monomialsujhk

2dj ; by the remarks above, x is congruent to an element of the form f�h2�+

a1ud + a2u2d2 + · · · + asu

sds, where f is a polynomial of degree at most n− 1.We want to show that x is congruent to 0.

Now xvs−1 = 0, so

f��s−1�+ a1�s−2 + a2�s−2�s−3 + · · · + as−1�s−2�s−3 · · · �0 = 0� (1)

Also, xvs−1+n = 0, so f��s−1+n�+ a1�s−2+n + a2�s−2+n�s−3+n+ · · · + as−1�s−2+n�s−3+n · · ·�n + as�s−2+n�s−3+n · · · �n�n−1 = 0. Because �k+n = �k, this equation can berewritten as

f��s−1�+ a1�s−2 + a2�s−2�s−3 + · · · + as−1�s−2�s−3 · · · �0 + as�s−2 · · · �0�n−1 = 0� (2)

Equations (1) and (2) imply that as = 0. Continue in the same manner; we get thatas−1 = 0, as−2 = 0, etc, until a1 = 0. Thus x is congruent to just a polynomial f�h2�of degree at most n− 1. But since xvk = 0 for k = 0� � � � � n− 1, we have f��0� =f��1� = · · · = f��n−1� = 0, i.e., f has n distinct roots. Thus it must be the zeropolynomial. Therefore x is congruent to 0, as required. This finishes the calculationof the annihilators of L���.

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

620 PRATON AND MAY

Ideals containing h1: We again work in the algebra A/I , where I = �h1�. InA/I , we have du = �ud + �, and by induction,

dku = �kudk + �k − 1�− 1

�dk−1 duk = �kukd + �k − 1�− 1

�uk−1�

In particular, dn and un commute with everything, and so by 3.5 they act as scalars,say cd and cu, on any irreducible module M . The elements dn − cd and un − cu arethus in the annihilator of M . Since M is isomorphic to a quotient of �A/I�/�dn −cd� u

n − cu�, and �A/I�/�dn − cd� un − cu� is already finite-dimensional, we conclude

that M must be finite-dimensional.Thus any primitive ideal containing h1 must be the annihilator of a finite-

dimensional module. Unfortunately, in this case the dimension is not necessarily 1(in contrast with other values of �). We need to look at higher dimensional modules.Now all finite-dimensional simple modules of A have been completely classified inCarvalho and Musson (2000), but we only need to consider a few special cases here.We therefore take a short digression to recall a construction described in Benkartand Roby (1998).

Let �/n denote the (finite) group of integers modn. Pick � ∈ � and � �= 0.For i ∈ �/n, define �i = �i�+ ��i − 1��1; since � is an nth root of unity, �i is welldefined. (Note that the formula for �i is the same as the one we encounter whendefining Verma modules. In particular, it can be rewritten as �i = �i�2 − �1, where�2 = �+ �1, as usual.) Define the module N��� �� as the one with basis vi � i ∈ �/n�under the following action of A:

dvi = �vi−1� uvi = �−1�ivi+1�

It is straightforward to check that N��� �� does become an A-module under thisaction. See (Benkart and Roby, 1998, 5.5) for a more general version.

Note that if �2 �= 0 (i.e., if � �= −�1), then �0� �1� � � � � �n−1 are all distinct. Thisimplies that N��� �� is irreducible—see the discussion preceding Proposition 5.11 inBenkart and Roby (1998).

We also need to look at the dual module N ′��� ��. The general definition ofdual modules is given in Benkart and Roby (1998) before Proposition 2.35, but inthis case we can provide an explicit description. Let vi � i ∈ �/n� be a basis forN ′��� ��; then the action of A on N ′��� �� is given by

uvi = �vi+1� dvi = �−1�i−1vi−1�

The dual N ′��� �� is also irreducible if � �= −�1. If �i �= 0 for 0 ≤ i ≤ n− 1, thenN ′��� �� is isomorphic to N��� ��.

We now calculate the annihilators of N��� �� and N ′��� ��.

Lemma 4.2. Suppose � �= −�1 and N��� �� and N ′��� �� are irreduciblen-dimensional modules as described above. Write � to denote �−1�n�−n ��n1 − �n

2�. Then

AnnN��� �� = �h1� dn − �n� un − ���

AnnN ′��� �� = �h1� un − �n� dn − ���

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

PRIMITIVE IDEALS OF NON-NOETHERIAN DOWN-UP ALGEBRAS 621

Proof. We will concentrate on AnnN ′��� ��, leaving the similar AnnN��� �� tothe reader. It is straightforward to check that h1 annihilate N ′��� ��. Also, unvi =�nvi+n = �nvi for all i ∈ �/n, so un − �n also annihilates N ′��� ��. Finally, dnvi =�−n�0�1 · · · �n−1vi+n = �−n�0�1 · · · �n−1vi for all i ∈ �/n, so dn acts as the scalar

�−nn−1∏i=0

��i�2 − �1� = �−1�n�−n��n1 − �n2� = ��

so dn − � also annihilates N ′��� ��.Let J be the ideal generated by h1� u

n − �n, and dn − �. All congruences willbe modJ . Note that since h1 ∈ J , we have du ≡ �ud − �, so ud can be written as afirst-degree polynomial in du. By induction we can easily show that uidi is congruentto a polynomial of degree at most i in du.

Now suppose x is an element that annihilates N ′��� ��. We can write x ≡∑i�j ai�ju

idj , where 0 ≤ i� j ≤ n− 1. Then

x ≡ �−nunx ≡ �−n∑i�j

ai�jun+i−jujdj ≡

n−1∑k=0

ukfk�du��

where fk�du� is a polynomial in du of degree at most n− 1. Now xvi = 0 becomes∑n−1k=0 fk��i�vi+k = 0; since vi� vi+1� � � � � vi+n−1� is linearly independent, we conclude

that fk��i� = 0 for 0 ≤ k� i ≤ n− 1. Thus for a given value of k, polynomial fk hasn distinct zeroes �0� �1� � � � � �n−1. Since fk has degree at most n− 1, it must be thezero polynomial. Hence x ≡ 0, as required. �

Now that we know about N��� ��, N ′��� ��, and their annihilators, we turnback to the algebra A. Suppose M is an irreducible A-module whose annihilatorcontains h1. We want to show that M is either one-dimensional or isomorphic toL���, N��� ��, or N ′��� ��.

As mentioned before, un and dn must act as scalars and M must be finite-dimensional. We first consider the case where un and dn both act as zero. We wantto show that M is isomorphic to L��� in this case. The proof is straightforward:since dnM = 0, there exists a nonzero v0 ∈ M such that dv0 = 0. Since h1 annihilatesM , we have duv0 = ��ud + ��v0 = �v0, so v0 has weight �. By the universal propertyof Verma modules, we conclude that M must be a quotient of V���. But we cannotconclude yet that M is L���, since in this case, V��� has more than one maximalsubmodule.

Let N be the maximal submodule of V��� such that M � V���/N . Since un

annihilates M , the element unv0 = vn ∈ V��� must lie in N . But �n−1 = 0, so vngenerates the standard maximal submodule of V���. Thus M is indeed isomorphicto L��� in this case.

We now go back to the general case. Since M is finite-dimensional, we can finda nonzero v0 ∈ M such that h2v0 = �2v0. (Here �2 is just an eigenvalue of h2.) Firstassume that �2 = 0. Then h2�u

idjv0� = �i−juidjh2v0 = 0, so h2 annihilates M . Thealgebra A/�h1� h2� is commutative (in A/�h1� h2�, we have du = ud = −�1). Thus Mis one-dimensional.

The only remaining case is when �2 �= 0, and additionally, not both of un anddn act as the zero scalar. For definiteness let us say that un acts as the scalar �n,

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014

622 PRATON AND MAY

where � �= 0. Define vk = �−kukv0 for k = 1� 2� � � � � n− 1. Then uvk = �vk+1; sinceun = �n, we also see that uvn−1 = �v0. Thus we can use �/n as our index set; theformula uvk = �vk+1 would still be valid. Now h2vk = �−kh2u

kv0 = �−k�kukh2v0 =�k�2vk and dvk = �−1�du�vk−1 = �−1 �h2 − �1� vk−1 = �−1

(�k−1�2 − �1

)vk−1. Thus the

span of v0� v1� � � � � vn−1� is a submodule, and hence all of M .None of the vk is zero since un acts as a nonzero scalar. Since each vk belongs

to a different eigenspace of h2, they are linearly independent. Thus v0� v1� � � � � vn−1�is a basis for M . Comparing this with the definition of N ′��� ��, where � = �2 − �1,we see that M is isomorphic to N ′��� ��. Similarly, if dn acts as a nonzero scalar �n,then M is isomorphic to N��� ��.

To finish the proof of Theorem 4.1, we only need to verify that the set ofannihilators of N��� �� and N ′��� �� can be written in the form stated in the theorem.In Lemma 4.2 we have �n� = �−1�n��n1 − �n

2� �= �−�1�n, since �2 �= 0. Conversely,

suppose zu and zd are any complex numbers, not both zero, with zuzd �= �−�1�n. For

definiteness say zu �= 0. Then we can find a nonzero � such that �n = zu. We can alsofind �2 such that �−1�n�−n��n1 − �n

2� = zd; note that �2 �= 0, since otherwise we wouldhave zuzd = �−�1�

n. Let � = �2 − �1. Then AnnN ′��� �� = �h1� un − zu� d

n − zd�, asstated in the theorem.

REFERENCES

Benkart, G. (1998). Down-up algebras and Witten’s deformations of the universalenveloping algebra of ��2. In Recent Progress in Algebra (Hahn, S. G., Myung, H.C., Zelmanov, E. eds.). Contemporary Mathematics 224. Providence, RI: AmericanMathematical Society, pp. 29–45.

Benkart, G., Roby, T. (1998). Down-up algebras. J. Alg. 209:305–344; Addendum, (1999).213:378.

Benkart, G., Witherspoon, S. (2001). A Hopf structure for down-up algebras. Math. Z.238:523–553.

Carvalho, P., Musson, I. (2000). Down-up algebras and their representation theory. J. Alg.228:286–310.

Hildebrand, J. (2002). Centers of down-up algebras over fields of prime characteristic.Comm. Alg. 30:171–191.

Jordan, D. A. (2000). Down-up algebras and ambiskew polynomial rings. J. Algebra228:311–346.

Kirkman, E., Kuzmanovich, J. (2000a). Non-Noetherian down-up algebras. Comm. Algebra28:5255–5268.

Kirkman, E., Kuzmanovich, J. (2000b). Primitivity of Noetherian down-up algebras. Comm.Algebra 28:2983–2997.

Kirkman, E., Musson, I., Passman, D. (1999). Noetherian down-up algebras. Proc. Amer.Math. Soc. 127:3161–3167.

McConnell, J., Robson, J. (1987). Noncommutative Noetherian Rings. New York: John Wiley.Praton, I. (2004). Primitive ideals of Noetherian down-up algebras. Comm. Algebra

32:443–471.Zhao, K. (1999). Centers of down-up algebras. J. Algebra 214:103–121.

Dow

nloa

ded

by [

Uni

vers

ity o

f A

uckl

and

Lib

rary

] at

14:

32 2

5 N

ovem

ber

2014