simple modules and primitive ideals of non-noetherian generalized down-up algebras

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Simple Modules and Primitive Ideals of Non-NoetherianGeneralized Down-Up AlgebrasIwan Praton aa Department of Mathematics , Franklin and Marshall College , Lancaster, Pennsylvania, USAPublished online: 04 Mar 2009.

To cite this article: Iwan Praton (2009) Simple Modules and Primitive Ideals of Non-Noetherian Generalized Down-UpAlgebras, Communications in Algebra, 37:3, 811-839, DOI: 10.1080/00927870802231304

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Communications in Algebra®, 37: 811–839, 2009Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927870802231304

SIMPLE MODULES AND PRIMITIVE IDEALSOF NON-NOETHERIAN GENERALIZED DOWN-UP ALGEBRAS

Iwan PratonDepartment of Mathematics, Franklin and Marshall College,Lancaster, Pennsylvania, USA

Generalized down-up algebras were first introduced in Cassidy and Shelton (2004).Their simple weight modules were classified in Cassidy and Shelton (2004) in thenoetherian case, and in Praton (2007) in the non-noetherian case. Here we concentrateon non-noetherian down-up algebras. We show that almost all simple modules areweight modules. We also classify the corresponding primitive ideals.

Key Words: Down-up algebras; Primitive ideals; Simple modules.

2000 Mathematics Subject Classification: 16D60; 16D70; 16D25.

1. INTRODUCTION AND PRELIMINARIES

Down-up algebras were introduced by Benkart and Roby (1999). Theywere motivated by combinatorial considerations, but down-up algebras turn outto be generalizations of low-dimensional versions of well-known algebras (e.g.,the enveloping algebra of semisimple Lie algebras). Cassidy and Shelton (2004)introduced a more general version of down-up algebras. They showed that many ofthe properties of the Benkart and Roby’s down-up algebras still hold true in theirgeneralized version.

Primitive ideals of down-up algebras were classified in Praton (2004) andPraton and May (2005). It is natural to wonder whether the classification carriesover to generalized down-up algebras. In this article we look at this question in thecase where the algebra is non-noetherian. Simple weight modules of non-noetheriangeneralized down-up algebras were classified in Praton (2007); we will see later thatin most cases, simple modules must be weight modules. Thus classifying primitiveideals is mostly a matter of finding the annihilators of the simple weight modulesdescribed in Praton (2007).

We begin by recalling the definition of generalized down-up algebras. We workover the complex numbers for simplicity, although the results below are valid forany uncountable algebraically closed field of characteristic zero. Fix r� s� � ∈ � and

Received August 16, 2007; Revised December 15, 2007. Communicated by S. Kleiman.Address correspondence to Iwan Praton, Department of Mathematics, Franklin and Marshall

College, Lancaster, PA 17604, USA; Fax: (717) 358-4507. E-mail: [email protected]

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812 PRATON

a polynomial f ∈ ��x�. The generalized down-up algebra A = A�f� r� s� �� is thealgebra generated by u, d, and h with relations

hu− ruh− �u = 0�

dh− rhd − �d = 0�

du− sud − f�h� = 0�

(In Cassidy and Shelton (2004) the definition is somewhat different—there are minussigns attached to � and f—but the difference is just notational. The sign conventionused here allows us to avoid leading minus signs in the formulas describing certainsimple modules.) The algebra A is noetherian if and only if rs �= 0. Since we areconcentrating on the non-noetherian case here, we assume throughout that rs = 0,i.e., either r = 0 or s = 0.

There is a useful �-grading on A defined by setting the degrees of u, h, andd as 1, 0, and −1, respectively. Thus the monomial uihjdk has degree i− k. Anyelement x ∈ A of degree m ≥ 0 can be written as umx0, where x0 has degree 0, andany element of degree −m ≤ 0 can be written as x0d

m, where x0 is again of degree 0.We now recall the definition of weight modules. If V is an A-module, an

element v ∈ V has weight � = �� ∈ �2 if hv = v and udv = v; we say then thatv is a weight vector. The set of all elements with weight � is called the weight spaceV�, and V is a weight module if it is the direct sum of its weight spaces.

In the first part of this article we show that most simple modules are weightmodules; when f �= 0, the annihilators of simple modules that are not weightmodules must be generated by h− � (when r = 0) or by f��h− ��/r�− ud (whens = 0). In the second part of the article we classify the annihilators of simple weightmodules and hence obtain a reasonably complete description of the primitive idealsof non-noetherian down-up algebras.

The methods we use are inevitably similar to the ones used in Praton andMay (2005) to tackle the standard, nongeneralized case. Some of the results andproofs are identical; we repeat them here for completeness. Other results are similar,but we give different proofs (e.g., the proof for the annihilator of Verma modulesis new). Roughly speaking, results about the standard case generalize without toomuch trouble to the present case.

Along the way we use the following version of Schur’s lemma.

Theorem 1.1. Suppose R is a finitely generated algebra over � (e.g., R could be ourgeneralized down-up algebra). If N is a simple module, then any element in the centerof R acts as a scalar on N .

This theorem can be found in McConnell and Robson (1987, Proposition9.1.7). It is used in Praton (2004) and Praton and May (2005).

2. SIMPLE MODULES

In this section we show that in most cases, simple modules are weight modules.Furthermore, even though there exist simple modules that are not weight modules,their annihilators are mostly the same. It is therefore not unreasonable to expect

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SIMPLE MODULES, PRIMITIVE IDEALS 813

that all primitive ideals are annihilators of weight modules. (Thus the faint surpriseexpressed in Praton and May (2005) is somewhat misplaced.)

Here is a more detailed summary of the situation.

Theorem 2.1. Let N be a simple module. If either:

(i) r = s = 0, or(ii) r = 0, s = 1, f�� = 0, or(iii) r = 0, s a primitive nth root of unity (n > 1), or(iv) s = 0, r = 1, � = 0, or(v) s = 0, r a primitive nth root of unity (n > 1),

then N must be a weight module. In all other cases, there exist simple modules that arenot weight modules.

Suppose now that N is a simple module that is not a weight module.

a) If r = 0, s �= 0, then AnnN = �h− ��.b) If s = 0, r �= 0, f �= 0, then AnnN = �f( h−�

r

)− ud�.c) If s = 0, r �= 0, f = 0, then AnnN = �u� when ker d = �0� and AnnN = �d� when

ker d �= �0�.

We will spend the rest of this section providing a proof the theorem.We start by constructing examples of simple modules that are not weight

modules. The underlying vector space is the same in all cases: V is the vector spacewhose basis is �vi � i ∈ ��. The action of A on this vector space depends on the valuesof r and s; we give descriptions below.

Example 2.2. Suppose r = 0 and s is not a root of unity, s �= 0. Define, for alli ∈ �,

hvi = �vi� uvi = vi+1� dvi = si−1vi−2 −f��

s − 1vi−1�

Example 2.3. Suppose r = 0 and s = 1, and f�� = 0. Define, for all i ∈ �,

hvi = �vi� uvi = vi+1� dvi = vi−2 + if��vi−1�

Example 2.4. Suppose s = 0 and r not a root of unity, r �= 0. Define, for all i ∈ �,

hvi = rivi+1 −�

r − 1vi� uvi = f

(h− �

r

)vi+1� dvi = vi−1�

Example 2.5. Suppose s = 0, r not a root of unity, r �= 0, and f = 0. Define, forall i ∈ �,

hvi = rivi+1 −�

r − 1vi� uvi = vi+1� dvi = 0�

Example 2.6. Suppose s = 0, r = 1, and � �= 0. Define, for all i ∈ �,

hvi = vi+1 + i�vi� uvi = f�h− ��vi+1� dvi = vi−1�

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Example 2.7. Suppose s = 0, r = 1, � �= 0, and f = 0. Define, for all i ∈ �,

hvi = vi+1 + i�vi� uvi = vi+1� dvi = 0�

It is straightforward to verify these claims. As a sample of the computationsinvolved, here is a verification of Example 2.6. In this case, the defining relations are

hu = u�h+ �� dh = �h+ ��d� du = f�h��

These relations imply that for any polynomial p,

p�h− ��u = up�h�� and dp�h− �� = p�h�d�

We want to show that V is an A-module. First we calculate that

�h+ ��dvi = �h+ ��vi−1

= vi + �i− 1��vi−1 + �vi−1

= dhvi�

This immediately implies that dp�h− ��vi = p�h�dvi for all polynomials p and i ∈ �.Next we observe that

duvi = d�f�h− ��vi+1� = f�h�dvi+1 = f�h�vi

while

u�h+ ��vi = u�vi+1 + �i+ 1��vi�

= f�h− ��vi+2 + �i+ 1��f�h− ��vi+1

= f�h− ��vi+2 + �i+ 1��vi+1�

= f�h− ��hvi+1

= hf�h− ��vi+1 = huvi

thus V is indeed an A-module.We now show that V is simple. This is a pretty standard argument. First

note that if a submodule contains vk (where k is an arbitrary integer), then it alsocontains dvk = vk−1. Similarly, it contains hvk = vk+1 + k�vk, and hence it containsvk+1. Therefore, any submodule that contains vk must be all of V .

Now suppose a submodule W contains a nonzero element w. We can writew = ∑n

i=m aivi, where an �= 0. We’ll show that W must contain a vk; this is enoughto show that W is all of V . We do this by induction on the length of w, i.e., on��w� = #�i � ai �= 0�.

Note that �dh− 1�vi = vi + i�vi−1 − vi = i�vi−1. Therefore, W contains

�dh− 1�w − n�dw =n∑

i=m

iai�vi−1 −n∑

i=m

nai�vi−1 =n−1∑i=m

�i− n�ai�vi−1�

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whose length is at least one less than ��w�. By induction, we see that W contains avk; as noted above, this shows that V is simple.

Finally, we show that h has no eigenvector in V . This is also quite standard.Suppose hw = �w. If w �= 0, then we can write w = ∑n

i=m aivi, where an �= 0. Then

0 = �h− ��w = anvn+1 +n∑

i=m+1

�ai−1 + i�ai − �ai�vi + �m�− ��amvm

which implies an = 0. This contradictions shows that w = 0. Thus h has no eigen-vector, as claimed. So V is not a weight module. This completes the verification ofExample 2.6. The other examples can be verified using basically similar computations.

We now take the first steps in proving Theorem 2.1. It is convenient to define

h1 ={h− � if r = 0

f��h− ��/r�− ud if r �= 0� s = 0�

The element h1 is crucial in showing that A has zero divisors. Specifically,

h1u = 0� dh1 = 0�

(When r = 0, it is trivial to verify the relations above. When r �= 0, s = 0, thecomputation is a bit less trivial: h1u = f��h− ��/r�u− udu = uf�h�− uf�h� = 0.The case dh1 = 0 is handled similarly.)

Now let N be a simple module. We consider two cases: ker d = �0� andker d �= �0�. It turns out that the case ker d �= �0� is less complicated, so we startwith that. We begin with a straightforward way of determining whether N is aweight module.

Lemma 2.8. Suppose N is a simple module where ker d �= �0�. If ker d contains aweight vector, then N is a weight module.

Proof. Suppose v ∈ ker d is a weight vector. Then hv = �v for some � ∈ � andudv = 0. Define vi = uiv for i ≥ 0. Then hv1 = huv0 = �ruh+ �u�v0 = �r�+ ��v1,and by an easy induction we have, for i ≥ 1,

hvi = �ivi� where �i = ri�+ �1+ r + · · · + ri−1��

In particular, vi is an eigenvector of h.Similarly, we have dv1 = duv0 = �sud + f�h��v0 = f��v0, and by another easy

induction, we have, for i ≥ 1,

dvi = �ivi−1� where �i = si−1f��+ si−2f��1�+ · · · + f��i−1��

Thus udvi = �ivi, and vi is an eigenvector of ud.

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From these calculations we conclude that: (i) the span of the vis is a (nonzero)submodule, and hence must be all of N ; (ii) each vi a weight vector. Therefore, N isa weight module. �

This lemma is strong enough to show that most simple modules withker d �= �0� must be weight modules. Specifically, we have the following result.

Proposition 2.9. Suppose N is simple and ker d �= �0�. If r = 0, then N is a weightmodule. If r �= 0, s = 0, and f is not the zero polynomial, then N is also a weightmodule.

Proof. Let v be any nonzero element of ker d. We first show that if there exists apolynomial p, of degree at least 1, such that p�h�v = 0, then N is a weight module.The proof is straightforward. First note that h preserves ker d: if dw = 0, thendhw = �rh+ ��dw = 0. Now since p�h�v = 0, the set �hiv � i ≥ 0� cannot be linearlyindependent. Therefore, the span V of this set is finite-dimensional, and since h actson V , we conclude that h has an eigenvector w (say with eigenvalue �) located inV ⊆ ker d. Then w is a weight vector with weight �� 0�. Thus by Lemma 2.8, N isa weight module.

So all we have to do is find a nontrivial polynomial p�h� that annihilatesa nonzero element in ker d. As before, pick a nonzero v ∈ ker d. There are twopossibilities: h1v = 0 or h1v �= 0. If h1v = 0 and r = 0, then �h− ��v = 0, and wecan take p�h� = h− �. If h1v = 0 and r �= 0, s = 0 and f �= 0, then we have f��h−��/r�v = 0, and we can take p�h� to be f��h− ��/r�. (Note that f in this case cannotbe a nonzero constant polynomial since v �= 0). Thus we can assume that h1v �= 0.Note that dh1v = 0, so h1v lies in ker d.

If hh1v = 0, then we can take p�h� to be h—it annihilates h1v. So assume thathh1v = h1hv �= 0. Since N is simple, we can find an element x ∈ A such that xh1hv =v. Then h1xh1hv = h1v. But x can be written in the form x = ∑

i�j uigi�j�h�d

j , wheregi�j�h� is a polynomial in h. Then

h1xh1h = ∑i�j

h1uigi�jd

jh1h = h1g0�0�h�h1h�

since uh1 = dh1 = 0. To conserve indices write g�h� for g0�0�h�. Thus we haveh1g�h�h1hv = h1v. If r = 0, this would imply that

��h− ��g�h�h− 1�h1v = 0�

and hence we can take p�h� to be �h− ��g�h�h− 1, certainly a polynomial of degreeat least 1. On the other hand, if r �= 0, s = 0, and f is not the zero polynomial, wewould have (

f

(h− �

r

)g�h�h− 1

)h1v = 0�

which means we can take p�h� to be f��h− ��/r�g�h�h− 1, which is a polynomialof degree at least 1. Thus in all cases we have found a nontrivial polynomial p�h�that annihilates a nonzero element of ker d, which is what we needed to show. �

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Thus the situation when ker d �= �0� is relatively straightforward. All simplemodules are weight modules in this case, with the exception of when r �= 0, s = 0,and f = 0. Examples 2.5 and 2.7 show that simple modules that are not weightmodules exist in these exceptional cases, so the result of the proposition is the bestwe can do.

We now take a look at the case where ker d = �0�. As mentioned above, thisis the case where most exceptional simple modules arise. At first, however, we willconcentrate on finding a way to tell whether a simple module is a weight module.We begin with a helpful preliminary result.

Lemma 2.10. Let V be an A-module and suppose ker d = �0�. If r = 0, then h actsas the scalar � on V . If s = 0, then hdv = �dv implies that udv = f��v.

Proof. If r = 0, then d�h− �� = 0, so for any v ∈ V , we have d�hv− �v� = 0. Sinceker d = �0�, it follows that hv = �v.

If s = 0, then du = f�h�. Suppose that hdv = �dv. Then dudv = f�h�dv =f��dv, so d�udv− f��v� = 0. Since ker d = �0�, we conclude that udv = f��v, asrequired. �

We note an immediate corollary that takes care of the first part ofTheorem 2.1.

Corollary 2.11. Suppose r = s = 0 and N is a simple module. Then N is a weightmodule.

Proof. If ker d = �0�, then Lemma 2.10 implies that all elements of N have weight�� f��. If ker d �= �0�, then the first part of Proposition 2.9 implies that N is aweight module. �

Lemma 2.10 is also helpful in establishing the following criterion fordetermining whether a simple module is a weight module.

Lemma 2.12. Suppose N is a simple module with ker d = �0�. If N contains a weightvector, then N is a weight module. More specifically, in the case r = 0, s �= 0, N is aweight module if ud has an eigenvector. In the case r �= 0, s = 0, N is a weight moduleif h has an eigenvector.

Proof. The previous corollary already takes care of the case r = s = 0. Supposenow that r = 0 but s �= 0. Then Lemma 2.10 implies that h acts as the scalar � on N .Thus any eigenvector of ud is automatically a weight vector. Say udv = �v, wherev is nonzero. Then it is easy to calculate that uv has weight �� s� + f��. Also, since

uddv = 1s�dudv− f�h�dv� = 1

s�� − f��dv�

it follows that dv has weight �� − f��/s�. Therefore, if v is a weight vector,then so are hv, uv, and dv. Thus the submodule generated by a weight vector mustbe a weight module. Since N is simple, it is generated by any weight vector it maycontain, so if N contains a weight vector, then N is a weight module.

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818 PRATON

We now tackle the case r �= 0, s = 0. Suppose v is an element of N withhv = �v. Then huv = �ruh+ �u�v = �r�+ ��uv and uduv = uf�h�v = f��uv, so uvis a weight vector with weight �r�+ �� f��. On the other hand,

hdv = 1r�dh− �d�v = 1

r��− ��dv�

so Lemma 2.10 implies that udv = f��− ��/r�v. Hence v is in fact a weight vector.The same reasoning can be applied to dv; we conclude that dv is also a weightvector. Thus whenever v is a weight vector, then hv, uv, and dv are also weightvectors. As in the previous paragraph, it follows that if N has a weight vector, thenN is a weight module. �

In the case where ker d is nontrivial, a similar sounding criterion (Lemma 2.8)is sufficient to determine the simple weight modules. Unfortunately, in the presentcase, the situation is more complicated. The problem, of course, is that in thiscase there do exist simple modules that are not weight modules. We will refineLemma 2.12 and at the same time determine the annihilators of those exceptionalsimple modules.

Let N be a simple module. We first note that dh1N = 0, so if ker d = �0�, weconclude immediately that h1N = 0. Thus in this case h1 must lie in the annihilatorof N . The ideal generated by h1 turns out to be the annihilator of most of theexceptional simple modules, so we will denote this important ideal by I .

Write a ≡ b to indicate that a− b lies in I . To let us treat the various casesthe same way, it is convenient to define

h′ =

du+ f��/�s − 1� if r = 0� s �= 0� s �= 1�

h+ �/�r − 1� if s = 0� r �= 0� r �= 1�

du if r = 0� s = 1�

h if s = 0� r = 1�

We then have the following commutation formulas.

Lemma 2.13. Define h′ as above and write = r + s. (Thus either = r or = s,depending on which one of s and r is zero.) If �= 1, then

h′ui ≡ iuih′� �h′�iu ≡ iu�h′�i� dih′ ≡ ih′di� d�h′�i ≡ i�h′�id

for i ≥ 0. On the other hand, if = 1, then

h′ui ≡ ui�h′ + i�� dih′ ≡ �h′ + i��di�

where � = f�� if r = 0, s = 1, and � = � if r = 1, s = 0.

Proof. This is straightforward computation. For example, when r = 0, s �= 0,s �= 1, we have h1 = h− �, and thus h ≡ �. In this case h′ = du+ f��/�s − 1� =

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sud + f�h�+ f��/�s − 1� ≡ sud + sf��/�s − 1�, so

h′u ≡ �sud + sf��/�s − 1��u ≡ su�du+ f��/�s − 1�� ≡ suh′�

as required. The other calculations are similar. �

The rather complicated definition of h′ is motivated by the followingrefinement of Lemma 2.12.

Lemma 2.14. Suppose N is a simple module. If there exist nonzero polynomial g anda nonzero element v ∈ N such that g�h′�v = 0, then N is a weight module.

Proof. Since g�h′�v = 0, the subspace generated by ��h′�iv � i ≥ 0� is finite-dimensional, and so h′ has an eigenvector. When r = 0, then ud is congruent to alinear function of h′, so ud also has an eigenvector, and thus N is a weight moduleby 2.12. On the other hand, when r �= 0 and s = 0, then h is a linear function of h′,and thus h has an eigenvector. By Lemma 2.12 again, N is a weight module. �

The result above is enough to dispose of a few special cases and prove a fewmore parts of Theorem 2.1.

Corollary 2.15. Suppose r = 0 and s = 1 or r = 1 and s = 0. Define � as inLemma 2.13, i.e., � = f�� if r = 0, s = 1; � = � if r = 1, s = 0. If � = 0, then anysimple module is a weight module.

Proof. Let N be a simple module. The hypotheses imply that h′ commutes witheverything in A as an operator on N . Thus h′ acts as a scalar on N . By Lemma 2.14,N is a weight module. �

Corollary 2.16. Suppose = r + s is a primitive nth root of unity, where n > 1. Thenany simple module is a weight module.

Proof. Let N be a simple module. Then Lemma 2.13 implies that �h′�n commuteswith u and d. When r = 0, we know that h acts as �, so �h′�n certainly alsocommutes with h. When s = 0, r �= 0, the definition of h′ shows that �h′�n alsocommutes with h. Therefore, �h′�n commutes with everything in A as an operatoron N . Thus �h′�n acts as a scalar on N . By Lemma 2.14, N is a weight module. �

These two corollaries finish off the claims of Theorem 2.1 about cases wheresimple modules must be weight modules. We now concentrate on finding theannihilators of simple modules that are not weight modules. We start with someformulas highlighting the relationship between u, d, and h′.

Lemma 2.17. If r �= 0, s = 0, and f = 0, then uidi ≡ 0 for all i ≥ 1. In all othercases, uidi is congruent to a nonzero polynomial in h′.

Proof. We have ud ≡ f��h− ��/r� = 0 if r �= 0, s = 0 and f = 0. Thus uidi =ui−1�ud�di−1 ≡ 0 for all i ≥ 1 in this case.

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820 PRATON

The other cases are taken care of by an induction argument. If r �= 0, s = 0and f �= 0, then we have ud ≡ f��h− ��/r�, which is a nonzero polynomial. Onthe other hand, if r = 0 and s �= 0, then du = sud + f�h� ≡ sud + f��, so ud ≡s−1�du− f��. This means ud ≡ s−1�h′ − f��/�s − 1�− f�� if s �= 1, or ud ≡ h′ −f�� if s = 1. In all cases ud is congruent to a nonzero polynomial in h′. Thus thestatement is true for i = 1.

Assume that uidi ≡ qi�h′� where qi is a nonzero polynomial. Then in the case

�= 1, we have, using Lemma 2.13,

ui+1di+1 ≡ uqi�h′�d ≡ qi�

−1h′�ud ≡ qi�−1h′�q1�h

′��

which is certainly a nonzero polynomial in h′. The case where = 1 is similar. �

Lemma 2.18. Any element in the algebra A is congruent to an element of the form∑ni=0 u

ipi�h′�+∑m

k=1 qk�h′�dk, where pi and qk are polynomials.

Proof. Suppose first that r = 0. Then h ≡ �. Any element of A can be written as alinear combination of uihjdk. If i ≤ k, then

uihjdk ≡ �juididk−i ≡ �jpi�h′�dk−i�

where pi�h′� is a polynomial in h′, by Lemma 2.17. If i ≥ k, then

uihjdk ≡ �jui−kukdk ≡ �jui−kpk�h′��

where pk is again a polynomial. The result of the lemma follows.On the other hand, if r �= 0 and s = 0, then either h = h′ (if r = 1) or h = h′ −

f��/�r − 1� (if r �= 1). Thus any element of A can be written as a linear combinationof ui�h′�jdk. By Lemma 2.13, we have ui�h′�j ≡ r−ij�h′�jui if r �= 1, and ui�h′�j ≡�h′ − i��jui if r = 1. In either case, ui�h′�j ≡ gi�j�h

′�ui where gi�j is a polynomial.Similarly, �h′�jdk = dkgk�j�h

′�. Thus if k ≥ i,

ui�h′�jdk ≡ gi�j�h′�uidk ≡ gi�j�h

′�uididk−i ≡ gi�j�h′�qi�h

′�dk−i

while if k ≤ i,

ui�h′�jdk ≡ uidkgk�j�h′� ≡ ui−kukdkgk�j�h

′� ≡ ui−kqk�h′�gk�j�h

′��

The conclusion of the lemma follows. �

We are now ready to determine the annihilators of most of the exceptionalsimple modules.

Proposition 2.19. Suppose N is a simple module that is not a weight module, withker d = �0�. Assume that f is not the zero polynomial. Then the annihilator of N isgenerated by h1.

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Proof. By Corollaries 2.15 and 2.16, we can assume that is not a primitive nthroot of unity, and if = 1, we can assume that � �= 0.

We have noted that the annihilator of N must contain h1. Suppose it is strictlylarger than �h1�. We will show that N must be a weight module in this case.

Let x be an element in the annihilator of N that is not in the ideal �h1�. ByLemma 2.18, x is congruent to an element of the form

∑ni=0 u

ipi�h′�+∑m

k=1 qk�h′�dk,

where not all of the polynomials pi and qk are zero. Then xdn is also in theannihilator, and

xdn =n∑

i=0

uipi�h′�dn +

m∑k=1

qk�h′�dk+n

=n∑

i=0

pi�−ih′�uidn +

m∑k=1

qk�h′�dk+n

=n∑

i=0

pi�−ih′�ai�h

′�dn−i +m∑

k=1

qk�h′�dk+n

=n+m∑i=0

gi�h′�di�

where gi�h� = pn−i�−n+ih′�an−i�h

′� if i ≤ n and gi�h′� = qi�h

′� if i > n. Note that notall of the gi�h

′� are zero.Thus we have found in the annihilator of N an element of the form∑m

i=0 gi�h′�di, where not all of the polynomials gi�h

′� are zero. Among all suchelements, choose one—call it y—with minimal m. If m = 0, then there exists a(nonzero) polynomial g0 such that g0�h

′� annihilates N . Then N is a weight moduleby Lemma 2.14. If m > 0, then

yh′ =m∑i=0

gi�h′�dih′ =

m∑i=0

igi�h′�h′di if �= 1

m∑i=0

gi�h′��h′ + i��di if = 1�

When �= 1, we calculate that

my − yh′ =m∑i=0

�m − i�gi�h′�h′di =

m−1∑i=0

�m − i�gi�h′�h′di

while if = 1, we calculate that

�h′ +m��y − yh′ =m∑i=0

�m− i��gi�h′�di =

m−1∑i=0

�m− i��gi�h′�di�

In both cases we have found an element in the annihilator of N of the form∑m−1i=0 qi�h

′�di; thus by the minimality of m, it must be true that qi is the zeropolynomial for 0 ≤ i ≤ m− 1. We assumed that is not a root of unity when �= 1,and we assumed that � �= 0 when = 1. We conclude that gi is the zero polynomial

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822 PRATON

for 0 ≤ i ≤ m− 1; thus y = gm�h′�dm. Since ker d = �0�, this implies that there is a

nonzero element v ∈ N annihilated by a polynomial in h′. By Lemma 2.14, N is aweight module and we are done. �

To complete the proof of Theorem 2.1, all that remains is to determine theannihilators of exceptional simple modules when f is the zero polynomial and s = 0,r �= 0.

Proposition 2.20. Suppose s = 0, r �= 0, and f = 0. Let N be a simple module thatis not a weight module. If ker d = �0�, then the annihilator of N is �u�; if ker d �= �0�,then the annihilator of N is �d�.

Proof. The proof is quite similar to the proof of Proposition 2.19; we just changethe base ideal to �u� or to �d�.

Assume first that ker d = �0�. We can assume that r is not a root of unity, orif r = 1, that � �= 0. Since duN = f�h�N = 0 and ker d = �0�, we see that uN = �0�.Thus AnnN includes u. Let J denote the ideal �u�, and for this proof write a ≡ bto mean a− b ∈ J . We will show that if AnnN is bigger than J , then N must be aweight module.

So suppose x is any element of AnnN that is not in J . Then x is congruentto an element of the form

∑mi=0 gi�h

′�di where not all of the polynomials gi�h′� are

zero. Among all such xs, choose one where m is minimal. Arguing exactly as inthe proof of Proposition 2.19, we see that x is actually congruent to gm�h

′�dm. Wethen conclude that gm�h

′� annihilates a nonzero vector in N , and thus N is a weightmodule, as required.

The proof when ker d �= �0� is similar, with u and d exchanging roles. �

Note. Examples 2.4–2.7 show that the simple modules described in the propositionabove really do exist.

3. SIMPLE WEIGHT MODULES

One of our goals is to determine the primitive ideals of A. The previoussections show that in order to achieve this goal, it remains to find the annihilatorsof simple weight modules. Simple weight modules were classified in Praton (2007);they can be grouped into the following categories:

a) Doubly-infinite modules W��;b) Lowest-weight infinite-dimensional modules V��, which we call Verma modules;c) Highest-weight infinite-dimensional modules V ′��, in a sense dual to the Verma

modules;d) One-dimensional modules;e) Finite-dimensional quotients L�� of Verma modules;f) Cyclic finite-dimensional modules F�� and F ′�� .

In the list above, � and � are complex numbers, and � is nonzero. We can defineW�� and V�� for any value of �, but they are irreducible only for certain valuesof �. (Similar caveats hold for the other categories.) W�� and V�� are universal in

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the sense that any simple weight module is necessarily a quotient of either W�� orV��.

There is some overlap among these categories (e.g., L�� could be one-dimensional). We will nevertheless find annihilators of the simple modules in eachcategory, thereby completing the classification of primitive ideals.

We recall here the operator � on weights described in Cassidy and Shelton(2004) and more particularly in Praton (2007). This operator � and its inverse �−1

basically move weights up and down; the orbit of weights under � has an importantrole in determining the annihilators of weight modules. As in Praton (2007), �actually acts on �×�×�×: for any �� b� t ∈ �, with t �= 0, we have

�� b� t� = t�+ b�

In what follows, we usually suppress b and t and write �� for �� b� t�. Asin Praton (2007), let �� denote the orbit ��i�� i ∈ ��, and let �� denote thecardinality of ��. If �� is an infinite set, we write �� = �.

We usually write �i to denote �i��. It is not hard to calculate that

�i =

�+ ib if t = 1

ti(�− b

1− t

)+ b

1− tif t �= 1�

Thus if t is a primitive nth root of unity, n > 1, then either �� = 1 or �� = n.For other values of t, either �� = 1 or �� = �. When �� = �, the values of�i are all different.

4. ANNIHILATORS OF SIMPLE WEIGHT MODULESWHEN r = 0, s �= 0

In this section we take up the case where r = 0 and s �= 0. We begin by statinga useful result about elements in A of degree 0. Recall that in this case, h1 is definedas h− �; thus h1u = dh1 = 0.

Lemma 4.1. Any element of degree 0 can be written as

p�du�+ q0�h1�h1 + uq1�h1�h1d + u2q2�h1�h1d2 + · · · + umqm�h1�h1d

m�

where p, q0� � � � � qm are polynomials.

Proof. Any element of degree 0 can be written as a sum of elements of the formukgk�h1�d

k, where gk is a polynomial. Since the polynomial gk can be written asgk�h1� = qk��h1�h1�+ ck, where ck is a constant, it suffices to show that ukdk can bewritten in the required form.

Before we show this, we first note the formula u�du�i = s−i�du− f��iu fori ≥ 0, which can be easily established by induction. We also note that �du�h1 =�sud + f�h��h1 = f�h�h1, i.e., �du�h1 is a polynomial in h1. It follows that theproduct of a polynomial in du with a polynomial in h1 is a sum of a polynomial indu with a polynomial in h1.

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824 PRATON

We now show that ukdk can be written in the required form. This is done byinduction; the statement is true for k = 1 because ud = s−1�du− f�h��. Assuming itis true for k, we have

uk+1dk+1 = u�ukdk�d = u

(p�du�+

k−1∑i=0

uiqi�h1�h1di

)d�

so we only need to show that up�du�d can be written in the required form. Ifp�du� = ∑m

i=0 ci�du�i, then

up�du�d =m∑i=0

ciu�du�id =

m∑i=0

s−ici�du− f��iud

=m∑i=0

s−i−1ci�du− f��i�du− f�h��

which is of the required form by the remarks in the previous paragraph. �

Note that since h1 = h− �, any polynomial in h1 is also a polynomial in h,and vice-versa. Thus any element of degree 0 can also be written in the form

p�du�+ q0�h�h1 + uq1�h�h1d + u2q2�h�h1d2 + · · · + umqm�h�h1d

m�

We now describe the doubly-infinite module W�� explicitly. For any complexnumber �, W�� has a basis �wi � i ∈ ��, and

hwi = �wi for i ∈ �

uwi = wi+1� uw−i−1 = �−iw−i for i ≥ 0

dw−i = w−i−1� dwi+1 = �i+1wi for i ≥ 0�

where �i = �i�� f�� s�. (We use the notation �i instead of �i to avoid confusionwith the �i that appears in the definition of Verma modules, below.)

In this case W�� is simple if and only if �� = � and �i �= 0 for all i ∈ �.If �� < �, then W�� is never simple, and the simple quotients of W�� are finite-dimensional. (In fact, if s is not a primitive nth root of unity with n > 1, then thesimple quotients of W�� are all one-dimensional.)

Note that although the actions of u and d seem rather complicated, for i ≥ 0we have duwi = �i+1wi and duw−i−1 = �−iw−i−1. Thus duwj = �j+1wj for all j ∈ �.

We have the following result about the annihilator of W��.

Theorem 4.2. If W�� is simple, then its annihilator is �h− ��. The annihilator ofsimple infinite-dimensional quotients of W�� is also �h− ��.

Proof. We note that h− � = h1 does indeed annihilate W�� and hence all itsquotients. We need to show that all elements that annihilate a simple infinite-dimensional quotient of W�� can be generated by h1.

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Since elements of degree k take wi to �wi+k, we can assume that x ishomogeneous, say of degree k ≥ 0. Then x = ukx0, where x0 has degree 0. ByLemma 4.1, we can write

x0 = p�du�+ q0�h�h1 + uq1�h�h1d + u2q2�h�h1d2 + · · · + umqm�h�h1d

m�

We need to show that p is the zero polynomial. Now xwi = ukx0wi = p��i+1�ukwi.

Since our simple module is infinite-dimensional, there are infinitely many values of isuch that ukwi �= 0. Thus xwi = 0 implies that p��i+1� = 0 for infinitely many valuesof i. Thus p has infinitely many different zeroes, and hence p must indeed be thezero polynomial, as required. �

This result automatically takes care of the simple infinite-dimensional highest-weight modules, because such modules are quotients of W��.

We now consider Verma modules. For any � ∈ �, the Verma module V�� hasbasis �vi � i ≥ 0�, with

hv0 = �v0� hvi = �vi� for i ≥ 1

uvi = vi+1� for i ≥ 0

dv0 = 0� dvi+1 = �ivi for i ≥ 0�

where �i = �i�f�� f�� s� for i ≥ 0.Recall that V�� is simple if and only if �i �= 0 for all i ≥ 0. We also recall that

if � = �, then V�� is a quotient of W��.

Theorem 4.3. Suppose V�� is an irreducible Verma module.

a) If � = �, then AnnV�� = �h− ��.b) If � �= � and ��f�� = n ≥ 1, then AnnV�� = ��h− ��h− �� du�n −��, where

� = ∏n−1i=0 �i.

c) If � �= � and ��f�� = �, then AnnV�� = ��h− ��h− ��.

Proof. If � = �, then as noted above, V�� is a simple quotient of W��, and henceits annihilator is �h− �� by Theorem 4.2.

It is straightforward to verify that �h− ��h− �� does annihilate V�� in allcases. In addition, if �� = n ≥ 1, then �i+n = �i for all i ≥ 0. Since �du�vi = �ivi,we see that in this case �du�n acts as the scalar �0�1 · · · �n−1 = �, and hence �du�n −� does indeed annihilate V��.

Now suppose x is an arbitrary element that annihilates V��. We need toshow that x lies in the ideal specified by the theorem. We can assume that x ishomogeneous; in fact, since on V�� d is a surjective operator and u is an injectiveoperator, we can assume that x has degree 0. Thus x can be written as

p�du�+ q0�h�h1 + uq1�h�h1d + u2q2�h�h1d2 + · · · + umqm�h�h1d

m�

When �� = n < � then �du�n −� is in the ideal, so we can assume that thepolynomial p has degree at most n− 1.

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When i > j, we have ujqj�h�h1dj · vi = �i−1 · · · �i−ju

jqj�h�h1vi−j = 0 becausei− j ≥ 1. Thus for all large enough values of i (it sufficies to take i > m), we have

0 = xvi = p��i�vi �⇒ p��i� = 0�

When �� = �, we conclude that p has infinitely many zeroes, and hence p has tobe the zero polynomial. When �� = n, then p has n distinct zeroes, but since weassumed that p has degree at most n− 1, we can also conclude that p is the zeropolynomial.

Now xv0 = 0 implies that ��− ��q0�� = 0, which in turn implies that q0��= 0.Thus q0 is a multiple of �h− ��. Similarly, xv1 = 0 implies that �0��− ��q1��= 0,which implies that q1 is a multiple of �h− ��. We continue in this manner,concluding that each qi (0 ≤ i ≤ m) is a multiple of �h− ��. Thus x is a multiple�h− ��h1, which finishes the proof. �

When s is a primitive nth root of unity (n > 1), it is possible to calculate �

explicitly. In this case we have �i = f��

1−s− si

(f��

1−s− f��

), so

� =n−1∏i=0

�i =(

f��

1− s

)n

−(

f��

1− s− f��

)n

�

We note here that in some special cases, the list of generators of AnnV�� canbe shortened. For example, in the standard case, we have f�h� = h, so du = h− sud.When ��f�� = 1, � �= �, and V�� is simple, then the theorem says that AnnV��is generated by �h− ��h− �� and du− �. But in this case du− � = h− sud − �

and so �h− ��du− �� = �h− ��h− ��− s�h− ��ud = �h− ��h− ��, so AnnV��is actually generated by just du− �. Unfortunately, du− � by itself does not seemto generate the whole annihilator when ��f�� = 1 but f is a polynomial of highdegree.

We now look at simple quotients L�� of Verma modules. We can describeL�� explicitly as follows. Pick � ∈ � so that �m = 0 for some m ≥ 0 but �i �= 0 for0 ≤ i ≤ m− 1. Then L�� is �m+ 1�-dimensional, with basis �vi � 0 ≤ i ≤ m�. Theaction of A on L�� is given by

hv0 = �v0� hvi+1 = �vi+1�

uvi = vi+1� uvm = 0�

dv0 = 0� dvi+1 = �ivi�

where 0 ≤ i ≤ m− 1. It is easy to verify that L�� is a quotient of V��.For standard (nongeneralized) down-up algebras, the annihilator of L�� was

described in Praton and May (2005). In the present case the description is virtuallyidentical, with the same proof. We describe the result and its proof here forcompleteness.

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First define y� as follows. Let ��0� �1� � � � � �m� be the solution to the equation

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

��

� � ��

1 �m−1 �m−1�m−2 · · · �m−1 · · · �0

�0�1�2��m

=

−�+ �

00��0

�

Note that there is a unique solution because the matrix is nonsingular. Then definey� as

y� = h1 + �0 + �1ud + �2u2d2 + · · · + �mu

mdm�

We can now describe the annihilator of L��.

Theorem 4.4. The annihilator of L�� is �um+1� dm+1� y��.

Proof. It is straightforward to verify that um+1, dm+1, and y� annihilate L��. LetI denote the ideal generated by these three elements. Congruences in this proof willbe modulo I .

We first claim that any element of A of degree 0 is congruent to a linearcombination of the monomials uidi, where 0 ≤ i ≤ m. Assuming this claim is true,suppose x annihilates L��. We can assume that x is homogeneous; for definiteness,let’s say that x has degree k ≥ 0. Certainly, k ≤ m. Then

x ≡ uk�a0 + a1ud + · · · + am−kum−kdm−k��

where ai ∈ �. Then a short calculation shows that xv0 = 0 implies a0 = 0, and xv1 =0 implies a1 = 0, and so on. We conclude that x ≡ 0, as required.

Thus we only need to prove the claim mentioned above. In this case theproof is straightforward and identical to the standard case. We know that h1 ≡−∑

�iuidi. Thus h2

1 ≡ −∑�ih1u

idi ≡ −�0h1 ≡∑

�0�iuidi. Similarly, any power of

h1 is congruent to a linear combination of the monomials uidi. Since any elementof degree 0 is a linear combination of the monomials uih

j1d

i, we see that the claimis true. This finishes the proof. �

Note that the proof does not use Lemma 4.1; it does not involve s−1 at all.Thus the proof works even when s = 0.

Next, we take a look at the cyclic simple modules F�� and their closerelatives F ′�� . These are quotients of W�� and arise only when s is a primitiventh root of unity, where n > 1. They can be described explicitly as follows. Pick � ∈�× and � ∈ � such that �i �= �j for 0 ≤ i < j ≤ n− 1; recall that �i = si�+ si−1

s−1 f��

in this case. (Consequently, any value of � will do, as long as � �= f��

1−s.) Then F��

is an n-dimensional module whose basis is �wi � i ∈ �/n��—note that we use thecyclic group �/n� as our index set. The action of A on F�� is

hwi = �wi�

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828 PRATON

uwi = �wi+1�

dwi+1 = �−1�i+1wi�

F ′�� is also n-dimensional with basis �wi � i ∈ �/n��, but the action of A is

hwi = �wi�

dwi = �wi−1�

uwi = �−1�i+1wi+1�

The annihilators of F�� and of F ′�� are determined in exactly the sameway as in the standard case.

Theorem 4.5. If F�� is simple, then its annihilator is

�h− �� un − �n� dn − �−n�′��where �′ = ∏n−1

i=0 �i. If F′�� is simple, then its annihilator is

�h− �� dn − �n� un − �−n�′��Proof. We repeat the proof in Praton and May (2005) since it works almostwithout change in the present case. First of all, it is easy to check that h1, u

n −�n, and dn − �−1�′ all annihilate F�� . Congruences will be modulo the idealgenerated by these three elements.

Suppose x annihilates F�� . We can assume, without loss of generality, thatx is homogeneous of (absolute) degree at most n− 1. Since x ≡ �−nunx, we can evenassume that x is homogeneous of degree k ≥ 0. Thus x = ukx0 where x0 has degree 0;since uk is injective, we see that x0 also annihilates F�� . By Lemma 4.1, we can write

x0 = p�du�+ q0�h1�h1 + · · · + umqm�h1�h1dm�

Since un ≡ dn ≡ 0 and h ≡ �, we can assume that p is a polynomial of degree atmost n− 1. Thus x0 ≡ p�du� and x0wi = 0 implies p��i+1� = 0, which means that phas n distinct zeroes. Thus p must be the zero polynomial and x0 ≡ 0, as required.

�

We can calculate �′ explicitly:

�′ =(

f��

1− s

)n

−(

f��

1− s− �

)n

�

We note also that for the purpose of listing primitive ideals, it is unnecessary tolist both Ann F�� and Ann F ′�� , since they have substantial overlap. (Indeed,F�� and F ′�� are isomorphic if �′ �= 0.) If we write

�h− �� un − zu� dn − zd� where zuzd �=

(f��

1− s

)n

�

then these ideals include all annihilators of F�� and F ′�� .

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Finally, we take a look at one-dimensional simple modules. Let’s say u, h, andd act as scalars cu, ch, and cd, respectively. From the defining relations of A, thesescalars must satisfy

cuch = �cu� cdch = �cd� �1− s�cucd = f�ch��

The annihilator of our one-dimensional module would then be

�u− cu� h− ch� d − cd��

5. ANNIHILATORS OF SIMPLE WEIGHT MODULESWHEN r �= 0, s = 0

We now treat the case where r �= 0 and s = 0. It has inevitable similarities withthe case where r = 0 and s �= 0, but there are also some differences that make itworthwhile to treat this case separately. Recall that when r �= 0 and s = 0, we defineh1 as f

(h−�

r

)− ud. The following result about elements of degree 0 is the analogueof Lemma 4.1.

Lemma 5.1. Any element of degree 0 can be written in the form

p�h�+ q0�h�h1 + uq1�h�h1d + · · · + umqm�h�h1dm�

where p, q0� q1� � � � � qm are polynomials.

Proof. Since every element of degree 0 can be written as a sum of elements of theform uig�h�di (where g is a polynomial), it suffices to prove that uig�h�di can bewritten in the stated form. Before we do this, we take a look first at the commutationrelation between h and u.

Using induction we can easily see that uhi = r−i�h− ��iu for i ≥ 0. Thusug�h� = g�r−1�h− ��u for any polynomial g.

We now show that uig�h�di can be written in the desired form. This is doneby induction on i. It is certainly true for i = 0. Assume it is true for i. Then

ui+1g�h�di+1 = u�uig�h�di�d

= up�h�d + uq0�h�h1d + u2q1�h�h1d2 + · · · + uiqi−1�h�h1d

i

= p��r−1�h− ��ud + uq0�h�h1d + · · · + uiqi−1�h�h1di

= p��r−1�h− ��f�r−1�h− ��− h1�

+ uq0�h�h1d + · · · + uiqi−1�h�h1di�

= p0��1�h��f��1�h��− p0��1�h��h1

+ uq0�h�h1d + u2q1�h�h1d2 + · · · + uiqi−1�h�h1d

i�

which is of the required form. �

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830 PRATON

We can use the lemma above to determine the annihilators of W�� and V��when they are simple. First, we recall the definition of W�� in this situation. Asbefore, � is a complex number, and W�� has basis �wi � i ∈ ��, where the action ofA is given by

hwi = �iwi for i ∈ �

uwi = wi+1� uw−i−1 = f��−i−1�w−i for i ≥ 0

dw−i = w−i−1� dwi+1 = f��i�wi for i ≥ 0�

Here �i = �i�� r�. In this case, W�� is simple if and only if �� = � and f��i� �=0 for all i ∈ �. We note as before that if �� < �, then W�� is not simple and itssimple quotients are finite-dimensional.

We have the following result about the annihilator of W��. It is basicallyidentical with the result for r = 0, s �= 0; the proof is also identical.

Theorem 5.2. Suppose W�� is simple. Then its annihilator is �h1�. The annihilatorof a simple infinite-dimensional quotient of W�� is also �h1�.

Proof. The proof is a rewrite of the proof of Theorem 4.2, except we useLemma 5.1 instead of Lemma 4.1, and with h now playing the role of du. �

As noted before, this result automatically takes care of the simple infinite-dimensional highest-weight modules, because such modules are quotients of W��.

We now look at Verma modules. As before, we start with an explicitdescription. For any � ∈ �, the Verma module V�� has basis �vi � i ≥ 0�, with

hvi = �ivi� for i ≥ 0

uvi = vi+1� for i ≥ 0

dv0 = 0� dvi+1 = f��i�vi for i ≥ 0�

where �i = �i�� r�. In this case V�� is simple if and only if f��i� �= 0 for all i ≥ 0.Although only �i, i ≥ 0, appear in the definition of V��, it turns out that �−1 =

r−1��− �� also play an important role in determining the annihilator of V��. Forexample, recall that if f��−1� = 0, then V�� is a quotient of W��, and hence if V��turns out to be simple, its annihilator is just �h1�. This is the first case described inthe following theorem.

Theorem 5.3. Suppose V�� is simple.

(i) If f��−1� = 0, then AnnV�� = �h1�.(ii) If f��−1� �= 0 and �� = n < � (where n ≥ 1), then AnnV�� = ��h− ��h1�

hn −��, where � = ∏n−1i=0 �i.

(iii) If f��−1� �= 0 and �� = �, then AnnV�� = ��h− ��h1�.

Proof. As noted above, the case where f��−1� = 0 follows from Theorem 5.2. Theother cases are treated exactly as in Theorem 4.3, where we use Lemma 5.1 instead

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of Lemma 4.1, and with h playing the role of du. The only other difference is thatin this case, we have h1v0 = f��−1�v0 instead of h1v0 = ��− ��v0, but this differencecauses no difficulty. �

As in the discussion after Theorem 4.3, in certain cases the list of generators ofAnnV�� can be shortened. For example, if f��−1� �= 0 and �� = 1, then accordingto the theorem, AnnV�� is generated by �h− ��h1 and h− �. But clearly h− � byitself suffices to generate this ideal. Compare this with the situation when r = 0 ands �= 0.

We now consider finite-dimensional modules. We look first at L��, the finite-dimensional quotients of Verma modules. L�� can be described explicitly as follows.Pick � ∈ � such that f��m� = 0 for some m ≥ 0 but f��i� �= 0 for 0 ≤ i ≤ m− 1.Then L�� is �m+ 1�-dimensional with basis �vi � 0 ≤ i ≤ m�; the action of A is

hvi = �ivi for 0 ≤ i ≤ m

uvi = vi+1� uvm = 0 where 0 ≤ i ≤ m− 1

dv0 = 0� dvi+1 = f��i�vi� where 0 ≤ i ≤ m− 1�

The annihilator of L�� is determined in the same way as for the case r = 0,s �= 0. We first define ��0� �1� � � � � �m� as the (unique) solution to the system ofequations

1 0 0 · · · 01 f��0� 0 · · · 01 f��1� f��1�f��0� · · · 0��

��

� � ��

1 f��m−1� f��m−1�f��m−2� · · · ∏m−1i=0 f��i�

�0�1�2��m

= −

�0�1�2��m

since the matrix is nonsingular, there is a (unique) solution. Then we define y� as

y� = h+ �0 + �1ud + �2u2d2 + · · · + �mu

mdm�

Now we can describe AnnL��.

Theorem 5.4. If L�� is simple, then its annihilator is

�um+1� dm+1� y��

Proof. The proof is similar to the proof of Theorem 4.4, with h taking the placeof h1 and f��i� taking the place of �i. The only difference occurs when provingthe claim that every element of degree 0 is congruent to a linear combination ofthe monomials ujdj , where 0 ≤ j ≤ m. It suffices to prove this claim for hi; this isdone by induction on i. It is certainly true for i = 0 and i = 1 (since h ≡ −∑

�jujdj).

Assume it is true for i; then

hi+1 ≡ h∑

cjujdj

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832 PRATON

≡ ∑cj�r

jujh+ �1+ r + · · · + rj−1��dj

≡ ∑cjr

jujhdj +∑cj�1+ r + · · · + rj−1��ujdj�

The first sum is a linear combination of the desired monomials (it follows from thecase i = 1) while the second sum is certainly a linear combination of the desiredsort. Thus the claim is proven. The rest of the proof follows the same way as inTheorem 4.4. �

We now take a look at the cyclic modules F�� and F ′�� . Here we assumethat r is a primitive nth root of unity, n > 1. As before, F�� and F ′�� arequotients of W��; the explicit description are as follows. F�� has basis �wi � i ∈�/n��, with

hwi = �iwi�

uwi = �wi+1�

dwi+1 = �−1f��i�wi�

where �i is defined as before, i.e., �i = ri�+ ri−1r−1 �. The module F�� is simple if

and only if �i �= �j whenever i and j are different elements of �/n�.The dual F ′�� also has basis �wi � i ∈ �/n��, but the action of A is as

follows:

hwi = �−iwi�

dwi = �wi−1�

uwi+1 = �−1f��−i�wi�

where �i is as above. The simplicity criterion for F ′�� is the same as the one forF�� .

Theorem 5.5. Let � = ∏n−1i=0 �i as before, and let � = �−n

∏n−1i=0 f��i�.

If F�� is simple, then its annihilator is

�h1� hn −�� un − �n� dn − ��

If F ′�� is simple, then its annihilator is

�h1� hn −��dn − �n� un − ��

Proof. Note that there are four elements generating the annihilator, instead ofthree as in the case where r = 0, s �= 0.

The proof proceeds exactly as before, using Lemma 5.1 instead of Lemma 4.1.Let x denote an element that annihilates F�� ; as before, we can assume that x hasdegree 0. So by Lemma 5.1

x = p�h�+ q0�h�h1 + uq1�h�h1d + · · · + umqm�h�h1dm�

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Since hn ≡ �, we can assume that p is a polynomial of degree at most n− 1; sinceh1 ≡ 0, we see that x ≡ p�h�. We can conclude that x ≡ 0, just as in the proof ofTheorem 4.5. �

In this situation, we can calculate � explicitly:

� =n−1∏i=0

�i =(

�

1− r

)n

−(

�

1− r− �

)n

�

Unfortunately, � resists such explicit display unless we know more about thepolynomial f . Thus to list succinctly the annihilators of all possible F�� andF ′�� , as was done after Theorem 4.5, is hard to do in this case.

Finally, we take a look at one-dimensional simple modules. Let’s say u, h, andd act as scalars cu, ch, and cd, respectively. From the defining relations of A, thesescalars must satisfy

�1− r�cuch = �cu� �1− r�cdch = �cd� cucd = f�ch��

The annihilator of our one-dimensional module would then be

�u− cu� h− ch� d − cd��

6. ANNIHILATORS OF SIMPLE WEIGHT MODULESWHEN r = s = 0

When r = s = 0, there are certain simplifications that arise. For example, W��is never simple, so we can ignore this module entirely. Simple quotients of V��are either one- or two-dimensional, so we can write down very explicit formulas.Unfortunately, there are also complications, chief among them being that we donot have the analogue of Lemmas 4.1 and 5.1. Thus we have to find other ways tocalculate the annihilator of simple Verma modules.

The Verma module V�� (where � ∈ �) has basis �vi � i ≥ 0�. The action of thealgebra is described as follows, for i ≥ 0:

hv0 = �v0� hvi+1 = �vi+1

uvi = vi+1

dv0 = 0� dv1 = f��v0� dvi+2 = f��vi+1�

Here V�� is simple if and only if f�� = 0 and f�� = 0.

Theorem 6.1. Suppose V�� is simple. If f�� = f��, then the annihilator of V�� isgenerated by

h− �+ �− �

f��ud�

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834 PRATON

On the other hand, if f�� = f��, then the annihilator of V�� is generated by theelements �h− ��h− �� and

uhd − f��f��

f��− f��h− �f��− �f��

f��− f��ud + �f��f��

f��− f��

Proof. Take the case where f�� = f�� = 0 first. As usual, let I be the idealgenerated by

⟨h− �+ �−�

f��ud

⟩. Congruences will be mod I . We have

h ≡ �+ �− �

f��ud�

and thus

h2 ≡ �h+ �− �

f��udh ≡ �

(�+ �− �

f��ud

)+ �− �

f��ud

≡ �2 + ��+ ��− �

f��ud�

Similarly, hk is congruent to a linear function of ud, for any k ≥ 1. As aconsequence, any element of degree 0 is congruent to an element of the form

a0 + a1ud + a2u2d2 + · · · + anu

ndn�

Now the annihilator of a Verma module is generated by elements of degree 0.Suppose x annihilates the Verma module. Then x is congruent to an element of theform above. Now xv0 = 0 implies that a0 = 0. Then xv1 = 0 in turn implies thata1f�� = 0, i.e., that a1 = 0. And so on and so on. We easily see that a2 = a3 =· · · = 0; i.e., that x is congruent to 0 mod I . Thus x is already in I , as expected. Thisfinishes the case where f�� = f��.

Now suppose that f�� = f��. It is straightforward to verify that the elementsmentioned in the theorem do indeed annihilate V��. Let I now denote the idealgenerated by these two elements. Congruences will be mod I .

Because �h− ��h− �� is in I , every polynomial in h is congruent to a linearpolynomial in h. Furthermore, we have

uhd ≡ f��f��

f��− f��h+ �f��− �f��

f��− f��ud − �f��f��

f��− f��

We conclude that every element of order 0 is congruent to an element of the form

x = b0h+ a0 + a1ud + a2u2d2 + · · · anu

ndn�

If x annihilates V��, then xvn = 0 implies that

b0�+ a0 + a1f��+ a2f��2 + · · · + anf��f��

n−1 = 0�

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But xvn+1 = 0 implies

b0�+ a0 + a1f��+ a2f��2 + · · · + anf��f��

n−1 = 0�

The two equations together imply that an = 0 (because f�� = f��). Similarly, wecan show that an−1 = an−2 = · · · = a1 = 0.

Thus x is congruent to b0h+ a0. But then xv0 = 0 implies that �b0 + a0 = 0,while xv1 = 0 implies that �b0 + a0 = 0. Together these two equations imply thatb0 = a0 = 0 (since � �= �). Therefore, x is congruent to 0, which means x is in theideal to begin with. This concludes the proof. �

The only other simple modules are either one- or two-dimensional. The one-dimensional modules are easy to take care of. If hv = chv, uv = cuv, and dv = cdv,then the annihilator is �h− ch� u− cu� d − cd�, where the scalars ch, cu, cd mustsatisfy

chcu = �cu� chcd = �cd� cdcu = f�ch��

The two-dimensional module exists only if f�� = 0. It can be described explicitly asfollows. If �v0� v1� form a basis for the module, then we have

hv0 = �v0� hv1 = �v1�

uv0 = v1� uv1 = 0�

dv0 = 0� dv1 = f��v0�

The module is simple if and only if f�� = 0. Note that we clearly have � �= �because f�� = 0 while f�� = 0.

Theorem 6.2. The annihilator of this simple two-dimensional module is⟨u2� d2� h− �+ �− �

f��ud

⟩�

Proof. The proof of Theorem 4.5 applies in this present instance. The statedtheorem is simply the explicit version of Theorem 4.5 in the case m = 1. �

7. A LIST OF PRIMITIVE IDEALS

We now give an explicit list of primitive ideals of A for various values of r, s,and f . Given the results in previous sections, the list is not hard to construct—wejust have to be careful in considering the various cases.

As in the standard case, primitive ideals can be divided into two classes. Thereare the annihilators of Verma modules and their quotients; these primitive idealsare parametrized by � ∈ �. We’ll denote them by I� below. Then there are otherprimitive ideals; they all contain h1, and they either annihilate (quotients of) W��,or simple modules that are not weight modules.

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836 PRATON

The list below has some duplication (e.g., in the list for r = s = 0� f�� = 0,the ideal �u− cu� d − cd� h− �� already appears as I� when cu = cd = 0). It does notseem worthwhile to make the list even more complicated than it already is by tryingto avoid such duplications.

The Case r = s = 0� f�� = 0. All simple modules are weight modules in thiscase; they are either one or two dimensional. Refer to Theorem 6.2. The primitiveideals are:

a) I� ={�u� d� h− �� if f�� = 0⟨u2� d2� h− �+ �−�

f��ud

⟩if f�� = 0

b) �u− cu� d − cd� h− �� where cucd = 0.

The Case r = s = 0� f�� = 0. All simple modules are weight modules. Referto Theorem 6.1. Write z� to indicate the element

uhd − f��f��

f��− f��h− �f��− �f��

f��− f��ud + �f��f��

f��− f��

Then the primitive ideals are:

a) I� =

�u� d� h− �� if f�� = 0⟨h− �+ �−�

f��ud

⟩if f�� = f��

��h− ��h− �� z�� if f�� = 0� f�� = f��

b) �u− cu� d − cd� h− �� where cucd = f��.

The Case r = 0� s = 1� f�� = 0. All simple modules are weight modules, sowe need to refer to Theorem 4.3 (as well as results about one-dimensional modules).The primitive ideals are:

a) I� ={�u� d� h− �� if f�� = 0

��h− ��h− �� du− f�� if f�� = 0

b) �h− �� u− cu� d − cd� where cu� cd ∈ � are arbitrary.

The Case r = 0� s = 1� f�� = 0. Once again, all simple modules are weightmodules. Here we need to to refer to Theorems 4.3 and 4.4; in particular, we needthe definition of y� from Theorem 4.4. We can also look at Theorem 4.2, but theprimitive ideal described there already appears in the list generated by Theorem 4.3.The primitive ideals are:

a) I� =

�h− �� if � = �

��h− ��h− �� if � �= �� f�� = −if�� for all i ≥ 0

�uk+1� dk+1� y�� if � �= �� f�� = −kf�� for a k ≥ 0�

The Case r = 0, s not a Root of Unity, f�� = 0. There are simple modulesthat are not weight modules, but their annihilators are all �h− ��. We need to refer

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to Theorems 4.2 and 4.3. The primitive ideals are:

a) I� ={�u� d� h− �� if f�� = 0

��h− ��h− �� if f�� = 0

b) �h− ��;c) �u− cu� d − cd� h− �� where cucd = 0.

The Case r = 0, s not a Root of Unity, f�� = 0. Once again, there are simplemodules that are not weight modules; their annihilators are all �h− ��. Refer toTheorems 4.2–4.4. The primitive ideals are:

a) I� =

�h− �� if � = �

��h− ��h− �� du− f��

1−s� if f�� = f��/�1− s�

�um+1� dm+1� y�� if f�� = s−m−1s−1 f��

��h− ��h− �� for other values of � b) �u− cu� d − cd� h− �� where cucd = f��/�1− s�.

The Case r = 0, s a Primitive nth Root of Unity, f�� = 0. All simple modulesare weight modules. We recall the definitions of � and �′ from Theorems 4.3and 4.5. In this case � = �−f��n and �′ = �−��n. The cyclic modules F�� andF ′�� are irreducible as long as � �= 0. The primitive ideals are:

a) I� ={�u� d� h− �� if f�� = 0

��h− ��h− �� du�n − �−f��n� if f�� = 0 b) �u− cu� d − cd� h− �� where cucd = 0;c) �un − zu� d

n − zd� h− �� where zuzd �= 0.

The Case r = 0, s a Primitive nth Root of Unity, f�� = 0. All simple modulesare weight modules. Here we mostly refer to Theorems 4.3 and 4.5, especially thediscussion after Theorem 4.5. The primitive ideals are:

a) I� =

��h− ��h− �� du− �� if f�� = f��/�1− s�

�um+1� dm+1� y�� if f�� = s−m−1s−1 f��

��h− ��h− �� du�n −�� for other values of � b) �u− cu� d − cd� h− �� where cucd = f��/�1− s�;c) �un − zu� d

n − zd� h− �� where zuzd �= (f��

1−s

)n.

The Case s = 0, r = 1, � = 0. All simple modules are weight modules. This isa simple case: we only need to refer to Theorem 5.3 (and the results about one-dimensional modules). The primitive ideals are:

a) I� ={�u� d� h− �� if f�� = 0

�h− �� if f�� = 0 b) �u− cu� d − cd� h− ch� where cucd = f�ch�.

The Case s = 0, r = 1, � �= 0, f �= 0. There are simple modules that are notweight modules, but their annihilators are all �h1�.

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838 PRATON

Recall that in the present case we have �i = �+ i�. It is convenient to makethe following definition. Suppose � ∈ � satisfy f��i� = 0 for some integer i ≥ 0. Letm �= m�� denote the smallest integer value m ≥ 0 such that f��m� = 0.

From Theorems 5.3 and 5.4, the primitive ideals are:

a) I� =

�um+1� dm+1� y�� if f��i� = 0 for some i ≥ 0

�h1� if f��−1� = 0 but f��i� �= 0 for i ≥ 0

�h1�h− �� if f��i� �= 0 for all i ≥ −1 b) �h1�.

The Case s = 0, r = 1, � �= 0, f = 0. Here there are simple modules that arenot weight modules. Their annihilators are either �u� or �d�. The primitive idealsare:

a) I� = �u� d� h− �� for all � ∈ �;b) �u�;c) �d�.

The Case s = 0, r not a Root of Unity, f �= 0, f��/�1− r�� = 0. There are simplemodules that are not weight modules, but their annihilators are all �h1�.

In this case we have �i = ri��− �

1−r�+ �

1−r. As before, it is convenient to define

m �= m�� to be the smallest integer m ≥ 0 such that f��m� = 0.Using Theorems 5.2, 5.3, and 5.4, the primitive ideals are:

a) I� =

�um+1� dm+1� y�� if f��i� = 0 for some i ≥ 0

�h1� if f��−1� = 0 but f��i� �= 0 for i ≥ 0

�h1�h− �� if f��i� �= 0 for all i ≥ −1 b) �h1�;c) �u− cu� d − cd� h− �

1−r� where cucd = 0.

The Case s = 0, r not a Root of Unity, f �= 0, f��/�1− r�� = 0. There are againsimple modules that are not weight modules, but their annihilators are all �h1�.

Once again, we define m �= m�� to be the smallest integer m ≥ 0 such thatf��m� = 0.

As before, we compute the primitive ideals using Theorems 5.2, 5.3, and 5.4.Let �1 denote

�

1−r. The primitive ideals are:

a) I� =

�h− �� if � = �1�um+1� dm+1� y�� if � �= �1� but f��i� = 0 for an i ≥ 0

�h1� if � �= �1� f��−1� = 0� and f��i� �= 0� for all i ≥ 0

�h1�h− �� if � �= �1� and f��i� �= 0 for i ≥ −1 b) �h1�;c) �u− cu� d − cd� h− �1� where cucd = f��1�.

The Case s = 0, r not a Root of Unity, f = 0. There are simple modules thatare not weight modules, and their annihilators are either �u� or �d�. The primitive

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ideals are:

a) I� = �u� d� h− �� where � ∈ �;b) �u− cu� d − cd� h− �

1−r� where cucd = 0;

c) �u�;d) �d�.

The Case s = 0, r an nth Root of Unity, f��/�1− r�� = 0. All simple modulesare weight modules. We define m�� as in the case s = 0, r not a root of unity. Notethat in this case, m�� is at most n− 1. Recall also the definitions of � and � fromTheorem 5.5.

The primitive ideals are:

a) I� ={�um+1� dm+1� y�� if f��i� = 0 for some i� 0 ≤ i ≤ n− 1

�h1�h− �� hn −�� if f��i� �= 0 for all i ≥ 0 b) �u− cu� d − cd� h− �

1−r� where cucd = 0;

c) �h1� hn −�� un − �n� dn − �� where � �= �

1−r;

d) �h1� hn −��dn − �n� un − �� where � �= �

1−r.

The Case s = 0, r an nth Root of Unity, f��/�1− r�� = 0. Again, all simplemodules are weight modules. We define �, �, m�� the same way as in the previouscase. Recall that �1 = �

1−r.

The primitive ideals are:

a) I� =

�h− �� if � = �1�um+1� dm+1� y�� if � �= �1� f��i� = 0 for an i ≥ 0

�h1� if � �= �1� f��−1� = 0 and f��i� �= 0 for all i ≥ 0

�h1�h− �� hn −�� if � �= �1 and f��i� �= 0 for all i ≥ −1 b) �u− cu� d − cd� h− �1� where cucd = f��1�;c) �h1� h

n −�� un − �n� dn − �� where � �= �1;d) �h1� h

n −��dn − �n� un − �� where � �= �1.

REFERENCES

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

Cassidy, T., Shelton, B. (2004). Basic properties of generalized down-up algebras. J. Algebra279:402–421.

McConnell, J., Robson, J. (1987). Noncommutative Noetherian Rings. New York: John Wileyand Sons.

Praton, I. (2004). Primitive ideals of noetherian down-up algebras. Comm. Algebra32:443–471.

Praton, I. (2007). Simple weight modules of non-noetherian generalized down-up algebras.Comm. Algebra 35:325–337.

Praton, I., May, S. (2005). Primitive ideals of non-noetherian down-up algebras. Comm.Algebra 33:605–622.

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simple modules and primitive ideals of non-noetherian generalized down-up algebras

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