primitive ideals of noetherian down-up algebras
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Primitive Ideals of Noetherian Down-Up AlgebrasIwan Praton a ba Department of Mathematics and Computer Science , Franklin and Marshall College ,Lancaster , Pennsylvania , USAb Department of Mathematics and Computer Science , Franklin and Marshall College, P.O.Box 3003 , Lancaster , PA , 17604 , USAPublished online: 01 Feb 2007.
To cite this article: Iwan Praton (2004) Primitive Ideals of Noetherian Down-Up Algebras, Communications in Algebra,32:2, 443-471, DOI: 10.1081/AGB-120027906
To link to this article: http://dx.doi.org/10.1081/AGB-120027906
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Primitive Ideals of Noetherian Down-Up Algebras#
Iwan Praton*
Department of Mathematics and Computer Science, Franklin and MarshallCollege, Lancaster, Pennsylvania, USA
ABSTRACT
We determine the primitive spectrum of Noetherian down-up algebras bydescribing explicit generators for their primitive ideals.
Key Words: Down-up algebras; Primitive ideals; Noetherian rings.
1. INTRODUCTION AND NOTATIONS
1.1. Motivated by combinatorial considerations, Benkart and Roby (1998) intro-duced a class of algebras called down-up algebras, parametrized by three complexnumbers a; b; g. A down-up algebra Aða; b; gÞ is an algebra over the complexnumbers C generated by d and u subject to the relations
d2u ¼ adudþ bud2 þ gd
du2 ¼ auduþ bu2dþ gu:
Many familiar algebras are examples of down-up algebras. The enveloping algebraof the Lie algebra sl2ðCÞ and of the Heisenberg algebra H, for instance, are
#Communicated by S. Kleiman.*Correspondence: Iwan Praton, Department of Mathematics and Computer Science, Franklin
and Marshall College, P.O. Box 3003, Lancaster, PA 17604, USA; E-mail: [email protected].
COMMUNICATIONS IN ALGEBRA�
Vol. 32, No. 2, pp. 443–471, 2004
443
DOI: 10.1081/AGB-120027906 0092-7872 (Print); 1532-4125 (Online)
Copyright # 2004 by Marcel Dekker, Inc. www.dekker.com
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isomorphic to Að2;�1; 1Þ and to Að2;�1; 0Þ, respectively. It turns out that the theoryof down-up algebras has many similarities with the theory of enveloping algebras,e.g., there is a PBW basis and well-behaved Verma modules associated with anyAða; b; gÞ.
Other people soon took up the task of investigating these algebras. The papers(Carvalho and Musson, 2000; Jordan, 2000; Kirkman and Kuzmanovich, 2000a,b;Kirkman et al. 1999; Zhao, 1999) investigate, among other things, the centerof Aða; b; gÞ, its finite-dimensional representation theory, and criterions for when itis a domain, Noetherian, and primitive. We also know that theory of down-upalgebras are closely related to the theory of generalized Weyl algebras and skew poly-nomial rings. This paper continues the investigation of Aða;b; gÞ, answering one ofthe questions posed at the end of Benkart and Roby (1998). It is in a sense an exten-sion of Kirkman and Kuzmanovich (2000a), which determines which ones of theNoetherian down-up algebras are primitive, i.e., which of these algebras have 0 asa primitive ideal. Here we describe all the primitive ideals of a Noetherian down-up algebra.
There are many possible approaches to calculating the primitive ideals ofA :¼ Aða; b; gÞ. We could start with the theory of skew polynomial rings and=or gen-eralized Weyl algebras and then specialize to A. This is the approach taken inKirkman et al. (1999) and especially Jordan (2000). Another possibility: Jordan(2000) has determined most, but not all, the prime ideals in A; since primitive idealsare automatically prime, we could try to determine which of the prime ideals are pri-mitive. In this paper, though, we avoid these approaches and use only elementarytechniques in the spirit of Benkart and Roby (1998) and Zhao (1999). Indeed, thetechniques used here are similar to the ones in Kirkman and Kuzmanovich(2000a), especially in some parts of Sec. 3. We try to give quite explicit descriptionsof the primitive ideals by specifying their generators.
The next four sections of the article describe the annihilators of some standardmodules (Verma modules, their quotients, the doubly-infinite modules and also mod-ules NðF ; rÞ and their duals described in Benkart and Roby (1998, Theorem 5.5). Weshow that these annihilators are primitive ideals. In the last sections of the paper, weshow that these primitive ideals exhaust the supply, thus completely determining theprimitive spectrum of A when A is Noetherian. The list of primitive ideals—the mainresult of this paper—is given in Theorem 8.1.
1.2. We describe some of the notation we use. As usual, N;Z;C;C� denote thenonnegative integers, the integers, the complex numbers, and the nonzero complexnumbers, respectively. We will never have to refer explicitly to the square root of�1, so the letter i will be used freely as a summation index. We use Pn to denotethe set of primitive roots of order n. The two-sided ideal generated by elementsx; y; . . . is denoted by hx; y; . . .i. If I is an ideal, we use the standard notation x � y
mod I to mean x� y 2 I.
1.3. We assume that b 6¼ 0 throughout this paper. In Kirkman et al. (1999) it isshown that this is equivalent to A being Noetherian and to A being a domain. Itis also shown that du and ud are commuting algebraically independent elements,
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i.e., C½du; ud� is (isomorphic to) a polynomial ring in two variables. A basis forA consists of the elements ðudÞiðduÞjdk, ðudÞiðduÞjukþ1, where i; j; k 2 N. Thusany element x 2 A can be written as x ¼ P
i¼0 piui þP
j¼1 qjdj where pi and qj
are polynomials in C½du; ud�. We can define a grading of A by Z, i.e., A ¼Pi2Z Ai, where Ai ¼ C½du; ud�ui if i � 0 and Aj ¼ C½du; ud�dj if j � 0. We have
AiAj � Aiþj.
1.4. Following Benkart and Roby (1998), we define r1 and r2 to be the roots ofx2 � ax� b ¼ 0. Then r1 þ r2 ¼ a and r1r2 ¼ �b. Since b 6¼ 0, neither r1 nor r2 iszero. There are several cases to consider, and in each case we define two new genera-tors of C½du; ud� that will turn out to be more convenient than du and ud. (These arethe cases described in Carvalho and Musson (2000, 1.4), our definitions of h1 and h2
below are similar to their w1 and w2 but the normalization follows Zhao (1999,Section 1.3).
Case A. r1 6¼ r2 and neither is 1. In this case define
h1 ¼ du� r2udþ g1
h2 ¼ du� r1udþ g2;
where gi ¼g
ri � 1. Define G ¼ g
ðr1 � 1Þðr2 � 1Þ.Then h1 and h2 commute and C½du; ud� ¼ C½h1;h2�. We need to know the com-
mutation relations between h1;h2; u, and d. An easy calculation shows thathiu
k ¼ rki ukhi and dkhi ¼ rki hid
k for i ¼ 1; 2 and any positive integer k. We also havepðh1;h2Þu ¼ upðr1h1; r2h2Þ and dpðh1;h2Þ ¼ pðr1h1; r2h2Þd for any polynomialp 2 C½h1;h2�.
The classification of the primitive ideals of A depends on whether ri is a root ofunity. We thus distinguish the following subcases:
(A1) One ri is a root of unity and the other one is not. In this case we adopt theconvention that r1 is not a root of unity but r2 2 Pn.
(A2) Neither r1 nor r2 are roots of unity, but there exist positive integers i and j
such that ri1 ¼ rj
2 . In this case let n1 be the smallest possible value for i; tobe precise, let n1 be the minimal i > 0 with ri1 ¼ rk2 for some k 2 Z. Simi-larly define n2 to be the minimal j > 0 such that rj2 ¼ rk1 for some k 2 Z.It is an easy exercise in elementary number theory to show that rn11 ¼ r
n22 ,
and if ri1 ¼ rj
2 , then i ¼ qn1 and j ¼ qn2 for some q 2 Z.(A3) Neither r1 nor r2 are roots of unity, but there exist positive integers i and j
such that ri1rj
2 ¼ 1. Again, it is easy to show that there existðn1; n2Þ 2 N�N such that i must be a multiple of n1 (say i ¼ qn1 forsome q 2 Z), and j ¼ qn2.
(A4) Neither r1 nor r2 are roots of unity, and the only pair of integers ði; jÞ withri1r
j
2 ¼ 1 is ði; jÞ ¼ ð0; 0Þ.(A5) r1 2 Pn1 and r2 2 Pn2 . In this situation we define n to be the least
common multiple of n1 and n2.
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Case B. r1 ¼ r2 ¼: r 6¼ 1. In this case define
h1 ¼ du� rudþ g1
h2 ¼ r
r � 1du� udð Þ;
where, as before, g1 ¼g
r � 1. In this case G ¼ g
ðr � 1Þ2.
Again, C½du; ud� ¼ C½h1;h2�. The commutation relations between h1;h2; uand d are as follows: h1u
k ¼ rkukh1;dkh1 ¼ rkh1d
k;h2uk ¼ rkukðkh1 þ h2Þ, and
dkh2 ¼ rkðkh1 þ h2Þdk. If pðh1;h2Þ is any polynomial in C½h1;h2�, thenpðh1;h2Þu ¼ upðrh1; rh1 þ rh2Þ and dpðh1;h2Þ ¼ pðrh1; rh1 þ rh2Þd.
The classification of the primitive ideals of A again depends on whether r is aroot of unity. Thus we distinguish two subcases:
(B1) r 2 Pn.(B2) r not a root of unity.
Case C. One of the ri is 1 but the other is not. For definiteness say that r1 6¼ 1 butr2 ¼ 1. In this case define
h1 ¼ du� udþ g1h2 ¼ du� r1ud;
where again g1 ¼g
r1 � 1. In this case G is undefined.
Once again we have C½du; ud� ¼ C½h1;h2�. The commutation relations betweenh1;h2; u and d are as follows: h1u
k ¼ rk1ukh1;d
kh1 ¼ r1h1dk;h2u
k ¼ ukðh2 þ kgÞ, anddkh2 ¼ ðh2 þ kgÞdk. If pðh1;h2Þ is any polynomial in C½h1;h2�, then pðh1;h2Þu ¼upðr1h1;h2 þ gÞ and dpðh1;h2Þ ¼ pðr1h1;h2 þ gÞd.
We again distinguish two subcases:
(C1) r1 2 Pn.(C2) r1 not a root of unity.
Case D. r1 ¼ r2 ¼ 1. In this case our down-up algebra is isomorphic to Uðsl2Þ ifg 6¼ 0 and to UðHÞ if g ¼ 0, i.e., they are isomorphic to enveloping algebras of Liealgebras. See Benkart and Roby (1998). Since these are well known algebras, we willnot consider Case (D) in this article.
Note that in cases (A), (B), and (C), we always have r1 6¼ 1 and h1 is defined thesame way.
2. DOUBLY-INFINITE MODULES
2.1. We now consider a set of modules whose annihilators turn out to be primitiveideals. In this section we look at infinite-dimensional modules; in Secs. 4 and 5 we
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will look at finite-dimensional ones. Most of the infinite-dimensional modules thatwe need are Verma modules, but there are primitive ideals that are not annihilatorsof Verma modules, so we have to look at a somewhat more general class of modules.The construction here is taken from Benkart and Roby (1998, Proposition 2.33).
Let flkgk2Z be a sequence of complex numbers (indexed by the integers) suchthat lk ¼ alk�1 þ blk�2 þ g for all k 2 Z. Then define a doubly-infinite A-modulewhose basis is fvkgk2Z, and A acts as follows:
u � vk ¼ vkþ1; d � vk ¼ lk�1vk�1:
It is a straightforward matter to verify that this does define an A-module. SeeBenkart and Roby (1998, Proposition 2.33).
2.2. To get further—in particular, to calculate annihilators of these modules—weneed to solve the second-order recurrence relation lk ¼ alk�1 þ blk�2 þ g givenabove. The results were worked out in Carvalho and Musson (2000, Lemma 2.2);here we rewrite them slightly. In the formulas below, L1 and L2 are arbitrarycomplex numbers—they are the parameters we get when we solve a second-orderrecurrence relation.In Case (A),
lk�1 ¼ G� rk1L1 � rk2L2: ð2:2:AÞIn Case (B),
lk�1 ¼ G� krkL1 � rkL2: ð2:2:BÞIn Case (C),
lk�1 ¼ L2 � rk1L1 � kg1: ð2:2:CÞWe write V ðL1;L2Þ to denote the doubly-infinite A-module described above. Notethat this differs slightly from the notation in Benkart and Roby (1998, Proposition2.33).
2.3. Some special cases of the doubly-infinite modules are of particular interest.One special case occurs when l�1 ¼ 0. In that case, the subspace spanned byfvkgk�0 is a submodule, as is easily checked. This submodule is a Verma module,denoted by V ðlÞ, where l ¼ l0. Annihilators of Verma modules are a rich sourceof primitive ideals; we will investigate them in the next section.
2.4. Another special case is when one of the parameters L1;L2 is zero. In most (butnot all) cases this assumption leads to infinite dimensional modules with straightfor-ward structures, allowing us to find their annihilators easily. We start with a generalresult about annihilators of V ðL1;L2Þ.
Lemma. Suppose lk 6¼ 0 for all k 2 Z. Then the annihilator of V ðL1;L2Þ isgenerated by elements in C½h1;h2�.
Proof. Suppose x annihilates VðL1;L2Þ. Recall (Sec. 1.3) that x ¼ Pi2Z xi where
xi 2 Ai. Since Ai � vk � Cvkþi and the set fvj j j 2 Zg is linearly independent, each xi
Primitive Ideals of Noetherian Down-Up Algebras 447
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must also annihilate V ðL1;L2Þ. Now if i � 0, then xi ¼ pui where p is a polynomialin C½h1;h2�; while if i � 0, then xi ¼ qdi where q is a polynomial in C½h1;h2�. Wetake each of these possibilities in turn.
If pui annihilates VðL1;L2Þ, then for all k 2 Z, we have 0 ¼ pui � vk ¼ p � vkþi.Thus p annihilates vj for all j 2 Z, i.e., p annihilates V ðL1;L2Þ.
On the other hand, if qdi annihilates VðL1;L2Þ, then for all k 2 Z, we have0 ¼ qdi � vk ¼ lk�1lk�2 � � � lk�iq � vk�i. Since none of the lj is zero, we can concludethat q � vk�i ¼ 0, i.e., q annihilates V ðL1;L2Þ.
In both cases xi is a multiple of an element in C½h1;h2� that annihilatesVðL1;L2Þ. The lemma follows. &
2.5. We now investigate what happens when one of the parameters L1;L2 is zero.First suppose L1 ¼ 0. Then we have
lk�1 ¼ G� rk2L2 in cases (A) and (B)L2 � kg1 in case (C)
�
for all k 2 Z. In cases (A2), (A3), (A4), and (B2) (i.e., in cases where r2 is not a root ofunity), the formula shows that we can choose L2 so that lk�1 is nonzero and distinctfor all k 2 Z. Similarly in case (C) with g 6¼ 0, we can choose L2 so that lk�1 is non-zero and distinct for all k 2 Z.
On the other hand, suppose L2 ¼ 0. In this situation, we focus only on cases (A)and (C) with g ¼ 0. Then
lk�1 ¼ G� rk1L1 in case (A)�rk1L1 in case (C) with g ¼ 0
�
for all k 2 Z. In cases (A1), (A2), (A3), (A4), and (C2) with g ¼ 0 (i.e., where r1 is nota root of unity), we can choose L1 so that lk�1 is nonzero and distinct for all k 2 Z.
Lemma. In cases (A2), (A3), (A4), (B2), and (C) with g 6¼ 0, choose L2 asdescribed above. Then AnnV ð0;L2Þ ¼ hh1i.
Similarly, in cases (A1), (A2), (A3), (A4), and (C2) with g ¼ 0, choose L1 asdescribed above. Then AnnV ðL1; 0Þ ¼ hh2i.
Proof. We’ll provide details for the first assertion. The second is quite similar.To show that h1 2 AnnVð0;L2Þ, we only need to do a straightforward
calculation. For example, in cases (A) and (B), we have
h1 � vk ¼ ðdu� r2udþ g1Þ � vk¼ ðlk � r2lk�1 þ g1Þvk¼ ðG� rkþ1
2 L2 � r2Gþ rkþ12 L2 þ g1Þvk
¼ ð1� r2ÞGþ g1ð Þvk ¼ ð�g1 þ g1Þvk¼ 0:
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The calculation in case (C) is similar. Thus hh1i � AnnVð0;L2Þ; we need to show theconverse inclusion. By Lemma 2.4 we only need to find elements in C½h1;h2� thatannihilate Vð0;L2Þ. Now any polynomial p in C½h1;h2� can be written as a polyno-mial in h1 and du, and such a polynomial can be written in the form pðh1;duÞ ¼Pm
j¼0 gjðduÞhj
1, where gj is a polynomial in du. Suppose p 2 AnnVð0;L2Þ, i.e., pannihilates vk for all k 2 Z. Since h1 � vk ¼ 0, we conclude that g0ðduÞ � vk ¼ 0; sincedu � vk ¼ lkvk, this is equivalent to g0ðlkÞvk ¼ 0. Thus g0ðlkÞ ¼ 0 for all k 2 Z. Sincethe lks are all distinct, the polynomial g0 has infinitely many roots, and hence mustbe the zero polynomial. Therefore pðh1;duÞ ¼
Pmj¼1 gjðduÞhj
1, i.e., p 2 hh1i, which iswhat we want. &
2.6. The result above gives us a list of primitive ideals.
Theorem. In cases (A2)–(A4), (B2), and (C) with g 6¼ 0, the ideal hh1i is aprimitive ideal.
In cases (A1)–(A4) and (C2) with g ¼ 0, the ideal hh2i is a primitive ideal.
Proof. We only need to show that the modules Vð0;L2Þ and V ðL1; 0Þ described inLemma 2.5 are irreducible. This follows from Benkart and Roby (1998, Proposition5.11). &
3. ANNIHILATORS OF VERMA MODULES
3.1. In formulas ð2:2:AÞ; ð2:2:BÞ, and ð2:2:CÞ, choose L1 and L2 so that l�1 ¼ 0.As mentioned at the beginning of Sec. 2.3, the Verma module VðlÞ of highest weight lis the submodule of VðL1;L2Þ spanned by fvkgk�0; here the parameter l is definedto be l0. In this section we determine the annihilators of certain Verma modules.We are most interested in the irreducible ones, but we consider a somewhat wider classof Verma modules since it requires no more effort to determine their annihilators.Kirkman and Kuzmanovich (2000a) have determined the Verma modules whoseannihilators is h0i; the methods used here are similar to theirs.
In Benkart and Roby (1998, Proposition 2.4a) it is shown that a Verma moduleV ðlÞ is irreducible if and only if lk 6¼ 0 for all k � 0. We won’t require V ðlÞ to beirreducible here, but we do require that only finitely many lk are zero. In this case,let n0 be the largest integer such that ln0�1 ¼ 0. Then M ¼ spanfvk j k � n0g is anirreducible submodule of VðlÞ; it is isomorphic to Vðln0Þ. It will turn out thatAnnVðlÞ is the same as AnnM, so even though VðlÞ might not be irreducible,AnnVðlÞ is a primitive ideal anyway.
Suppose x 2 AnnVðlÞ. We can write x ¼ Pi2Z xi where xi 2 Ai. As in the proof
of Lemma 2.4, each xi also annihilates VðlÞ. If i � 0, then xi ¼ uip where p is a poly-nomial in C½h1;h2�; while if i � 0, then xi ¼ diq where q is a polynomial in C½h1;h2�.We consider each case separately.
Define m1;k and m2;k by h1 � vk ¼ m1;kvk and h2 � vk ¼ m2;kvk. If uip annihilatesV ðlÞ, then 0 ¼ uip � vk ¼ pðm1;k; m2;kÞvkþi for all k � 0, so
uipðh1;h2Þ 2 AnnV ðlÞ¼) pðm1;k; m2;kÞ ¼ 0 for all k � 0: ð3:1:1Þ
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On the other hand, if djp annihates VðlÞ, then 0 ¼ lk�1lk�2 � � � lk�jpðm1;k; m2;kÞvk�j.For all sufficiently large values of k, the product lk�1lk�2 � � � lk�j is nonzero.(It suffices to take k > j þ n0.) For these values of k, we have pðm1;k; m2;kÞ ¼ 0.Therefore
djpðh1;h2Þ 2 AnnV ðlÞ¼) pðm1;k; m2;kÞ ¼ 0 for all large values of k: ð3:1:2ÞTo find AnnV ðlÞ, therefore, it is sufficient to find polynomials pðh1;h2Þ withpðm1;k; m2;kÞ ¼ 0 for all large values of k. It will turn out that the restriction on k issuperfluous: if pðm1;k; m2;kÞ ¼ 0 for all k > N0, then pðm1;k; m2;kÞ ¼ 0 for all k � 0anyway. This implies that AnnVðlÞ ¼ AnnV ðln0Þ, so AnnVðlÞ is a primitive ideal.
3.2. In order to calculate the annihilators of V ðlÞ, we need to find explicit expres-sions for lk and for m1;k and m2;k. We already worked out expressions for lk in Sec. 2.2(they are also given in Benkart and Roby, 1998, Proposition 2.12), but we rewritethem in a slightly more convenient form here.
Case A. Here l�1 ¼ 0 implies that G ¼ L1 þ L2. Then l ¼ l0 ¼ G� r1L1 � r2L2.Define m1 ¼ ðr2 � r1ÞL1 and m2 ¼ ðr1 � r2ÞL2. In terms of the parameter l, we havem1 ¼ lþ g1 and m2 ¼ lþ g2. Then a straightforward calculation shows thathi � v0 ¼ miv0ði ¼ 1; 2Þ, and thus hi � vk ¼ rki mivk. The formula for lk in terms of m1and m2 is
ðr1 � r2Þlk�1 ¼ rk1 � 1� �
m1 � rk2 � 1� �
m2: ð3:2:AÞ
Case B. Here l�1 ¼ 0 implies that L2 ¼ G. Then l ¼ l0 ¼ �g1 � rL1. Definem1 ¼ �rL1 ¼ lþ g1 and m2 ¼ rl=ðr � 1Þ. Then hi � v0 ¼ miv0 ði ¼ 1; 2Þ. Thereforeh1 � vk ¼ rkm1vk and h2 � vk ¼ rkðkm1 þ m2Þvk. The formula for lk in terms of m1 is
lk�1 ¼ krk�1m1 þ ð1� rkÞG: ð3:2:BÞ
Case C. Here l�1 ¼ 0 implies that L1 ¼ L2. Then l ¼ l0 ¼ ð1� r1ÞL1 � g1. Definem1 ¼ ð1� r1ÞL1 ¼ lþ g1. Then h1 � v0 ¼ m1v0 and h2 � v0 ¼ lv0. (Thus we could havedefined m2 ¼ l here.) Therefore h1 � vk ¼ rk1m1vk and h2 � vk ¼ ðlþ kgÞvk. We alsohave
ðr1 � 1Þlk�1 ¼ ðrk1 � 1Þm1 � kg: ð3:2:CÞ
3.3. We now give a list of the annihilators of V ðlÞ.
Theorem. Suppose VðlÞ is a Verma module where only finitely many lk are zero.
(A) Suppose l 6¼ �g1 and l 6¼ �g2. (Thus m1m2 6¼ 0.)
(1) In case (A1), AnnVðlÞ ¼ hhn2 � mn2i.
(2) In case (A2), AnnVðlÞ ¼ hmn22 hn11 � mn11 h
n22 i.
(3) In case (A3), AnnVðlÞ ¼ hhn11 h
n22 � mn11 mn22 i.
(4) In case (A4), AnnVðlÞ ¼ h0i.
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(B) Suppose l 6¼ �g1, i.e., m1 6¼ 0.
(1) In case (B1), AnnVðlÞ ¼ hhn1 � mn1i.
(2) In case (B2), AnnVðlÞ ¼ h0i.
(C) Suppose l 6¼ �g1, i.e., m1 6¼ 0.
(1) In case (C1) with g 6¼ 0, AnnVðlÞ ¼ hhn1 � mn1i.
(2) In case (C2) with g 6¼ 0, AnnVðlÞ ¼ h0i.(3) In case (C2) with g ¼ 0, AnnVðlÞ ¼ hh2 � li.
(Case (A5) and case (C1) with g ¼ 0 do not appear in the list because in thesecases there are always an infinite number of lks equal to zero.)
We will spend the rest of this section providing the proof of the theorem above.
3.4. Before tackling the proof, we first set a definition of the degree of a two-variable polynomial. Put a lexicographic order on N�N: ði; jÞ > ði0; j0Þ if and onlyif either j > j0 or j ¼ j0 and i > i0—similar to alphabetizing by last names. Anypolynomial of two variables can be written as pðh1;h2Þ ¼
Pði;jÞ2S aijh
i1h
j
2 whereS is a finite subset of N�N. The degree of p is the largest pair ðm1;m2Þ 2 S
(according to the lexicographical order) such that am1;m26¼ 0.
3.5. We will also need the following definition to deal with case (A). A finite subsetS of N�N is said to be distinctive if it has the following property: whenever ði; jÞand ði0; j0Þ are in S, then ri1r
j
2 6¼ ri01 r
j02 . (Thus we can distinguish ordered pairs in S sim-
ply by looking at them as exponents of r1 and r2.) A polynomialP
ði;jÞ2S aijhi1h
j
2 isdistinctive if S is a distinctive set.
3.6. Distinctive sets are important because of the following result. Write Il for theideals on the right side of Theorem 3.3, part (A). For example, in case (A1),Il ¼ hhn
2 � mn2i, in case (A2), Il ¼ hmn22 hn11 � mn11 h
n22 i, etc. We will also deal with case
(A5) because it will be used later. In case (A5), then, let Il denote the ideal generatedby fhi
1hj
2 � mi1mj
2 j allði; jÞ with ri1rj
2 ¼ 1g; here m1 and m2 are still nonzero.
Lemma. In case (A), any polynomial pðh1;h2Þ is congruent mod Il to a distinctivepolynomial.
Proof. We can write pðh1;h2Þ ¼P
ði;jÞ2T aijhi1h
j
2 where T is a finite set. Suppose
ði; jÞ; ði0; j0Þ 2 T with ri1rj
2 ¼ ri01 r
j02 . We claim that this implies that mi
01m
j02 h
i1h
j
2�mi1m
j
2hi01h
j02 is in Il. Assuming this (we’ll prove it below), then hi0
1hj02 � mi
0�i1 mj
0�j
2 hi1h
j
2
mod Il, so we can write p � Pði;jÞ2T 0 a
0ijh
i1h
j
2 mod Il, where T0 is the set T nfði0; j0Þg.
Continuing this way, we see that p � Pði;jÞ2S bijh
i1h
j
2 mod Il, where S is a distinctive
set.It remains to prove that whenever ri1r
j
2 ¼ ri01 r
j02 , then mi
01m
j02h
i1h
j
2 � mi1mj
2hi01h
j02 is
in Il.
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In case (A1), ri1rj
2 ¼ ri01 r
j02 is equivalent to i ¼ i0 and j � j0 mod n. Say j ¼ j0 þ tn
for some t 2 Z. Then mj02h
j
2 � mj02h
j02h
tn2 � mj
02h
j02 m
tn2 � mj2h
j02 mod Il. Thus indeed,
mi01m
j02h
i1h
j
2 � mi1mj
2hi01h
j02 mod Il.
In case (A2), we have ri�i01 ¼ r
j0�j
2 , so i ¼ i0 þ tn1 and j0 ¼ j þ tn2 for some t 2 Z.
Using mn22 hn11 � mn11 h
n22 mod Il, it is straightforward to show that mi
01m
j02 h
i1h
j
2 �mi1m
j
2hi01h
j02 mod Il.
In case (A3), we have ri�i01 r
j�j02 ¼ 1, so i ¼ i0 þ tn1 and j ¼ j 0 þ tn2 for some
t 2 Z. Using hn11 h
n22 � mn11 mn22 mod Il, it is straightforward to show that mi
01m
j02h
i1
hj
2 � mi1mj
2hi01h
j02 mod Il.
In case (A4), ri1rj
2 ¼ ri01 r
j 02 if and only if i ¼ i0 and j ¼ j0, so clearly mi
01m
j02h
i1h
j
2 ¼mi1m
j
2hi01h
j02 .
In case (A5), we can assume that i� i0 � 0. Then ri�i01 r
j�j02 ¼ 1. If j � j0 � 0, then
hi�i01 h
j�j02 � mi�i0
1 mj�j02 mod Il and we get the desired result. If j � j0 < 0, then there is
a positive integer t such that j � j0 þ tn2 � 0. Since rn22 ¼ 1, we have hi�i0
1 hj�j0þtn22 �
mi�i01 mj�j0þtn2
2 mod Il. Using hn22 � mn22 mod Il, we get the desired result. &
3.7. The following lemma is the key part in our approach to calculating AnnVðlÞin case (A).
Lemma. Suppose pðh1;h2Þ is a distinctive polynomial and pðrk1m1; rk2m2Þ ¼ 0 for allsufficiently large k, say for all k � n0. Then pðh1;h2Þ is the zero polynomial.
Proof. Write pðh1;h2Þ ¼P
ði;jÞ2S aijhi1h
j
2 where S is distinctive. Then for all k � n0,Xði;jÞ2S
ðri1rj2Þkaijmi1mj2 ¼ 0 for all k � n0. ð3:7:1Þ
Suppose jSj ¼ m. We now switch from using S as our index system to usingf1; 2; . . . ;mg. Write b1; b2; . . . ; bm for the elements of fri1rj2 j ði; jÞ 2 Sg in some order,and write x1; x2; . . . ; xm for the corresponding elements of faijmi1mj2g. Note that theb‘s are nonzero and distinct because S is distinctive. Equation (3.7.1) in just the firstm cases then becomes
Xm‘¼1
bk‘x‘ ¼ 0 for k ¼ n0;n0 þ 1; . . . ; n0 þm� 1:
This is an m�m linear system. The coefficient matrix of this system is aVandermonde matrix whose determinant is a nonzero multiple of
Qk 6¼‘ðbk � b‘Þ,
i.e., the determinant is nonzero. Thus xk ¼ 0 for k ¼ 1; . . . ;m, which means (sincem1m2 6¼ 0) aij ¼ 0 for all ði; jÞ 2 S. Thus pðh1;h2Þ ¼ 0, as promised. &
3.8. We are now ready to prove Theorem 3.3 in case (A).
Proof of Theorem 3.3, Part (A). It is straightforward to calculate that Il annihilatesVðlÞ, so all we need to do is show that AnnVðlÞ Il.
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According to (3.1.1) and (3.1.2), we need to find a polynomial fðh1;h2Þ withfðrk1m1; rk2m2Þ ¼ 0 for all sufficiently large k. By Lemma 3.6 we can write f � p
mod Il, where p is a distinctive polynomial. By Lemma 3.7, p ¼ 0. Thus f 2 Il, asrequired. &
3.9. We state a result about distinctive polynomials which we will use later(in Sec. 7.2).
Lemma. Suppose pðh1;h2Þ is a distinctive polynomial of degree ðm1;m2Þ andpðr1h1; r2h2Þ ¼ r
m1
1 rm2
2 pðh1;h2Þ. Then pðh1;h2Þ ¼ chm1
1 hm2
2 for some c 2 C.
Proof. This is straightforward. Write pðh1;h2Þ ¼P
ði;jÞ2S aijhi1h
j
2. Then pðr1h1;r2h2Þ ¼
Pði;jÞ2S aijr
i1r
j
2hi1h
j
2, so the hypothesis of the lemma implies
Xði;jÞ2S
aijðri1rj2 � rm1
1 rm2
2 Þhi1h
j
2 ¼ 0:
Since S is distinctive, we have ri1rj
2 � rm1
1 rm2
2 6¼ 0 unless ði; jÞ ¼ ðm1;m2Þ. Thus aij ¼ 0whenever ði; jÞ 6¼ ðm1;m2Þ. &
3.10. We now consider case (B). We start by stating a lemma about polynomialsin two variables; it is the analogue of Lemma 3.9 in case (B).
Lemma. Suppose pðh1;h2Þ is a polynomial with degree ðm1;m2Þ and
pðrh1; rh1 þ rh2Þ ¼ rm1þm2pðh1;h2Þ: ð3:10:1Þ
If r is not a root of unity, or if r 2 Pn but m1 < n, then pðh1;h2Þ ¼ chm1
1 for somec 2 C.
Proof. Suppose first that r is not a root of unity. We use induction on ðm1;m2Þ. Wecan write pðh1;h2Þ ¼
Pm1
j¼0 gjðh2Þhj
1 where gj is a single-variable polynomial. Thenpð0;h2Þ ¼ g0ðh2Þ. Now pð0; rh2Þ ¼ rm1þm2pð0;h2Þ, so g0ðrh2Þ ¼ rm1þm2g0ðh2Þ. Thusif a is a zero of g0, then so are ra; r2a; . . .. Since r is not a root of unity, we concludethat either the only zero of g0 is 0 (i.e., g0ðh2Þ ¼ chk
2 for some complex number c andinteger k � 0), or that g0 has no zero (i.e., g0 is a constant). We take each of thesepossibilities in turn.
If g0ðh2Þ ¼ chk2, then g0ðrh2Þ ¼ rm1þm2g0ðh2Þ leads to crkhk
2 ¼ crm1þm2hk2, i.e.,
g0ðh2Þ ¼ chm1þm2
2 . Since the degree of p is ðm1;m2Þ, we conclude that m1 ¼ 0 andthat pðh1;h2Þ ¼ ch
m2
2 . Then Eq. (3.10.1) implies crm2ðh1 þ h2Þm2 ¼ crm2hm2
2 , whichis impossible unless m2 ¼ 0 or c ¼ 0. In either case pðh1;h2Þ is constant and weare done.
If g0 is a constant, then g0ðrh2Þ ¼ rm1þm2g0ðh2Þ implies that either m1 þm2 ¼ 0or g0 is not only constant but identically zero. If m1 þm2 ¼ 0 then pðh1;h2Þ is con-stant and we are done. So we can proceed with the assumption that g0ðh2Þ ¼ 0.
In this case define qðh1;h2Þ ¼Pm1
j¼1 gjðh2Þhj�11 . Then pðh1;h2Þ ¼ h1qðh1;h2Þ,
and so using Eq. (3.10.1), rh1qðrh1; rh1 þ rh2Þ ¼ rm1þm2h1qðh1;h2Þ. Thus q has the
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same property as p, but it has lower degree. By the induction hypothesis,qðh1;h2Þ ¼ ch
m1�11 , and so pðh1;h2Þ ¼ h1qðh1;h2Þ ¼ ch
m1
1 , as required.Now suppose r 2 Pn with m1 < n. Using Eq. (3.10.1) repeatedly, we get
pðrjh1; rjjh1 þ rjh2Þ ¼ rðm1þm2Þjpðh1;h2Þ for all j � 0. In particular, with j ¼ n, we
get pðh1; nh1 þ h2Þ ¼ pðh1;h2Þ.We can write pðh1;h2Þ ¼
Pm2
i¼0 fiðh1Þhi2 where fi is a single-variable polynomial.
Let a 2 C� be fixed but arbitrary. Then pða;h2Þ ¼Pm2
i¼0 fiðaÞhi2 is a polynomial in
h2. Call it gaðh2Þ. Because pða;h2Þ ¼ pða;h2 þ naÞ, if b is a zero of ga, then so arebþ na, bþ 2na, . . .. We conclude that ga must be the constant polynomial, hencefiðaÞ ¼ 0 for all i > 0. Since a was arbitrary, the polynomials fi must be the zeropolynomial for i > 0. Thus pðh1;h2Þ ¼ f0ðh1Þ. Since pðh1;h2Þ has degree ðm1;m2Þ,we conclude that m2 ¼ 0 and f0 has degree m1.
Now f0ðrh1Þ ¼ rm1f0ðh1Þ by Eq. (3.10.1). So if z is a zero of f0, then z; rz; . . . ;rn�1z are also zeroes of f0. Since m1 < n, the only zero of f0 must be z ¼ 0. Thusf0ðh1Þ ¼ ch
m1
1 , as required. &
3.11. We can now prove Theorem 3.3 in case (B). Write Il for the ideal on the rightside of Theorem 3.3.
Proof of Theorem 3.3, Part (B). It is again straightforward to check that Il anni-hilates VðlÞ. (In case (B2) there is nothing to verify!) Thus we only need to show thatAnnVðlÞ � Il. Let pðh1;h2Þ be a polynomial that annihilates vk for all large valuesof k, say k � n0. We have
pðrkm1; rkðkm1 þ m2ÞÞ ¼ 0 for all k � n0: ð3:11:1ÞAccording to (3.1.1) and (3.1.2), it suffices to show that pðh1;h2Þ 2 Il.
For a contradiction, let pðh1;h2Þ be the polynomial in C½h1;h2�nIl with thesmallest degree satisfying Eq. (3.11.1). Let’s say that the degree of p is ðm1;m2Þ.In case (B2) we must have m1 < n; in case (B1) m1 can be arbitrary.
The polynomial qðh1;h2Þ ¼ pðrh1; rh1 þ rh2Þ � rm1þm2pðh1;h2Þ also satisfiesEq. (3.11.1), and it has smaller degree than p. By the minimality of ðm1;m2Þ, we musthave q 2 Il; since m1 < n, this implies q ¼ 0, i.e., pðrh1; rh1 þ rh2Þ ¼ rm1þm2
pðh1;h2Þ. By Lemma 3.10, pðh1;h2Þ ¼ chm1
1 . Eq. (3.11.1) then implies that c ¼ 0,as required. &
3.12. We now tackle case (C). As for (B), we begin by stating a lemma aboutpolynomials in two variables; this is the analogue of Lemma 3.9 for case (C).
Lemma. Suppose g 6¼ 0 and pðh1;h2Þ ¼Pm
j¼0 fjðh2Þhj
1 is a polynomial with
pðrh1;h2 þ gÞ ¼ rmpðh1;h2Þ: ð3:12:1Þ
If r is not a root of unity, or r 2 Pn and m < n, then pðh1;h2Þ ¼ chm1 for some
c 2 C.
Proof. We use induction on m. First note that f0ðh2Þ ¼ pð0;h2Þ. By (3.12.1),pð0;h2 þ gÞ ¼ rmpð0;h2Þ, so f0ðh2 þ gÞ ¼ rmf0ðh2Þ. As before, we conclude that
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f0 is the constant polynomial. But then f0ðh2 þ gÞ ¼ rmf0ðh2Þ implies that eitherm ¼ 0 or f0 is the zero polynomial. If m ¼ 0 then pðh1;h2Þ is constant and we aredone. So suppose f0ðh2Þ ¼ 0.
In this case pðh1;h2Þ ¼ h1
Pm�1j¼0 fjðh2Þhj
1 ¼ h1qðh1;h2Þ. From (3.12.1) we getrh1qðrh1;h2 þ gÞ ¼ rmh1qðh1;h2Þ, and thus qðrh1;h2 þ gÞ ¼ rm�1qðh1;h2Þ. Thereforeq satisfies the same property as p, but the parameter m has decreased to m� 1.By the induction hypothesis, qðh1;h2Þ ¼ chm�1
1 . Thus pðh1;h2Þ ¼ h1qðh1;h2Þ ¼ chm1 ,
as required. &
3.13. We can now provide a proof of Theorem 3.3 for case (C). As in (A) and (B),write Il for ideal on the right side in Theorem 3.3.
Proof of Theorem 3.3, Part (C). As in the proof for parts (A) and (B), we need tofind polynomials p 2 C½h1;h2� such that
pðrk1m1; lþ kgÞ ¼ 0 for all k � n0. ð3:13:1Þ
Write pðh1;h2Þ ¼Pm
j¼0 fjðh2Þhj
1. Suppose g ¼ 0. Then by our assumption r1 is not aroot of unity. The polynomial gðh1Þ :¼
Pm
j¼0 fjðlÞhj
1 therefore has infinitely manyroots rk1m1, k � n0. It must be the zero polynomial. Thus fjðlÞ ¼ 0 for all j,0 � j � m. Hence fjðh2Þ is a multiple of h2 � l, which implies that pðh1;h2Þ 2hh2 � li, as required.
Suppose now that g 6¼ 0. Choose pðh1;h2Þ ¼Pm
j¼0 fjðh2Þhj
1 to be a polynomialin C½h1;h2� n Il of the smallest degree satisfying Eq. (3.13.1). In case (C1) we havem < n; in case (C2) m can be arbitrary. Then the polynomial qðh1;h2Þ ¼pðr1h1;h2 þ gÞ � rm1 pðh1;h2Þ also satisfies (3.13.1) and has smaller degree than p.Thus q 2 Il and hence q ¼ 0; therefore pðr1h1;h2 þ gÞ ¼ rm1 pðh1;h2Þ. By Lemma3.12 we conclude that pðh1;h2Þ ¼ chm
1 . But then cðrk1m1Þm ¼ 0, so c ¼ 0 andpðh1;h2Þ ¼ 0, as required. &
3.14. Theorem 3.3 assumes that only finitely many of the lks are zero. We nowdetermine when this happens. In most cases this is an easy matter, but cases (A2)and (A3) need a little care.
Lemma. In case (A), suppose m1m2 6¼ 0. Then in cases (A1) and (A3), only finitelymany of the lks are zero. This is also true for case (A2) except when g ¼ 0 andn1 ¼ n2.
In case (B1), suppose m1 6¼ 0. Then only finitely many of the lks are zero.In case (C1), suppose g 6¼ 0. Then only finitely many of the lks are zero.In case (C2) with g ¼ 0, suppose m1 6¼ 0. Then only finitely many of the lks arezero.In cases (A4), (B2), and (C2) with g 6¼ 0, we can choose l so that none of thelks are zero.
Proof. From Eqs. ð3:2:AÞ, ð3:2:BÞ, and ð3:2:CÞ, we see that there are at most acountably infinite number of values for l that lead to lk ¼ 0 for some k. The claimin cases (A4), (B2), and (C2) then follows. Most of the other claims also follow
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immediately from Eqs. ð3:2:AÞ, ð3:2:BÞ, and ð3:2:CÞ. The only trouble comes fromcases (A2) and (A3). In those cases, let s1 and s2 denote an n2th root of r1 and ann1th root of r2, respectively, i.e., r1 ¼ s
n21 and r2 ¼ s
n12 . Then in case (A2) we have
ðs1s�12 Þn1n2 ¼ 1 and in case (A3) we have ðs1s2Þn1n2 ¼ 1. In case (A2) let
y ¼ ðs1s�12 Þn2 . Note that y is a root of unity. Then r1 ¼ s
n21 s
�n22 s
n22 ¼ ysn22 . Now from
Eq. ð3:2:AÞ we have that
lk�1 ¼ 0() ykm1sn2k2 � m2s
n1k2 þ m2 � m1 ¼ 0:
Since y is a root of unity, there are only finitely many polynomials fðxÞ ¼ykm1x
n2 � m2xn1 þ m2 � m1 ¼ 0. Now if lk�1 ¼ 0 then sk2 is a root of one of these poly-
nomials. If none of these polynomials are identically zero, then there are only finitelymany possibilities for sk2, and hence only finitely many values of k with lk�1 ¼ 0. Thepolynomial fðxÞ is never zero identically unless n1 ¼ n2 and m1 ¼ m2. This proves thelemma in case (A2).
In case (A3) we define y ¼ ðs1s2Þn2 . Note that y is again a root of unity. In thiscase,
lk�1 ¼ 0() m2ðsk2Þn1þn2 þ ðm1 � m2Þðsk2Þn2 � m1yk ¼ 0:
So if lk�1 ¼ 0, then sk2 is a root of one the polynomials fðxÞ ¼ m2xn1þn2 þ
ðm1 � m2Þxn2 � m1yk. In contrast to case (A2), these polynomials are never identically
zero. Since y is a root of unity, there are only finitely many of these polynomials,hence only finitely many possible values of sk2, hence only finitely many values of kwith lk�1 ¼ 0. &
The Lemma seems to indicate that case (A2) with g ¼ 0 and n1 ¼ n2 is anexceptional case. We will deal with this possibility starting in Sec. 8.2.
3.15. Theorem 3.3 gives us the following list of primitive ideals.
Theorem. Let c be any nonzero complex number. Then the following are primitiveideals.
(1) In case (A1): hhn2 � ci.
(2) In case (A2) with either g 6¼ 0 or n1 6¼ n2: hhn11 � ch
n22 i.
(3) In case (A3): hhn11 h
n22 � ci.
(4) In case (A4): h0i.(5) In case (B1): hhn
1 � ci.(6) In case (B2): h0i.(7) In case (C1) with g 6¼ 0: hhn
1 � ci.(8) In case (C2) with g 6¼ 0: h0i.(9) In case (C2) with g ¼ 0: hh2 � ci.
Proof. We take case (A3) as an example; the other cases are similar. Given anyc 2 C�, we can find a l 2 C such that ðlþ g1Þn1ðlþ g2Þn2 ¼ c. Clearly l 6¼ �g1and l 6¼ �g2. By Theorem 3.3, the ideal hhn1
1 hn22 � ci is then primitive. &
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4. IRREDUCIBLE QUOTIENTS OF VERMA MODULES
4.1. In the previous two sections we investigated the annihilators of certain infinite-dimensional modules. We now look at finite-dimensional modules. There are twotypes of finite-dimensional modules that are of interest: irreducible quotients ofVerma modules and the modules NðF ; rÞ and N 0ðF ; rÞ introduced in Benkart andRoby (1998, Theorem 5.5). In this section we concentrate on quotients of Vermamodules; we’ll consider NðF ; rÞ and N 0ðF ; rÞ in the next section.
4.2. Suppose VðlÞ is a nonsimple Verma module. Then lm�1 ¼ 0 for some m > 0;choose m to be minimal. The module MðlÞ :¼ spanfvk j k � mg is a maximalsubmoduleofV ðlÞandLðlÞ :¼VðlÞ=MðlÞ is an irreducible (m-dimensional)A-module.
We now describe the annihilator of LðlÞ.
Theorem. The annihilator of LðlÞ is generated by um, dm, and polynomialspðdu; udÞ where pðlk; lk�1Þ ¼ 0 for k ¼ 0; 1; . . . ;m� 1.
Proof. Let J be the ideal in C½du; ud� consisting of polynomials pðdu; udÞ withpðlk; lk�1Þ ¼ 0 for k ¼ 0; 1; . . . ;m� 1. Let I be the ideal in A generated by J , um,and dm. We want to show that Ann LðlÞ ¼ I.
The module LðlÞ has a basis v0; v1; . . . ; vm�1, where vk is the coset vk þMðlÞ.We first show that I � Ann LðlÞ. For 0 � k � m� 1, we have um � vk ¼
vkþm 2 MðlÞ and dm � vk ¼ 0, so um � vk ¼ 0 and dm � vk ¼ 0, i.e., um and dm are inthe annihilator of LðlÞ. Furthermore, since ðduÞ � vk ¼ lkvk and ðudÞ � vk ¼ lk�1vk,we have pðdu; udÞ � vk ¼ pðlk; lk�1Þvk. Therefore pðdu; udÞ 2 Ann LðlÞ is equivalentto pðlk; lk�1Þ ¼ 0 for all k ¼ 0; 1; . . . ;m� 1. Thus J is in Ann LðlÞ. We concludethat I � Ann LðlÞ.
To prove the other inclusion, first note that the same calculation shows that theannihilator of LðlÞ in C½du; ud� is just J . Now take a look at the elementy ¼ uþ dm�1. By induction we can show that yk � uk þ qkd
m�k mod I for 1 � k �m, where qk 2 C½du; ud�. In particular, ym � qm mod I, where qm is an element ofC½du; ud�. We now determine what qm is.
Note that y � vk ¼ vkþ1 if 0 � k � m� 2, while y � vm�1 ¼ ðl0l1 � � � lm�2Þv0. Writex for the product l0l1 � � � lm�2; because LðlÞ is irreducible, we know that x 6¼ 0. Wenow easily calculate that ym � vk ¼ xvk for all k ¼ 0; 1; . . . ;m� 1. In other words,ym acts as the scalar x on LðlÞ. From the previous paragraph, however, we know thatym acts as the polynomial qm on LðlÞ. The annihilator in C½du; ud� of LðlÞ is J , sowe conclude that qm � x mod J and hence ym � x mod I.
Now suppose x is an arbitrary element in Ann LðlÞ. We want to show that x 2 I.Without loss of generality we can write x ¼ Pm�1
k¼0 fkuk þ gkd
k� �
, where fk and gk arepolynomials in C½du; ud�. Since C½du; ud�ukvi � C vkþi and C½du; ud�dkvi � C vk�i,we can assume without any harm that x ¼ fku
k or x ¼ gkdk. For definiteness let’s
say that x ¼ gkdk; the other case is similar. Now xyk ¼ gkd
k� �
uk þ qkdm�k
� � � p
mod I, where p 2 C½du; ud�. Because p annihilates LðlÞ and the annihilator inC½du; ud� of LðlÞ is just J , we conclude that p � 0 mod I. Therefore xykym�k � 0
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mod I, i.e., xx � 0 mod I. Since x 6¼ 0, we conclude that x 2 I. Thus Ann LðlÞ � I,as required. &
5. THE MODULES NðF ; qÞ AND N 0ðF ; qÞ
5.1. We now consider another set of finite-dimensional modules whose annihilatorswill turn out to be the primitive ideals we need. First recall a construction given inBenkart and Roby (1998, Theorem 5.5). A weight n is an ordered pair of complexnumbers n ¼ ðn0; n00Þ. A element m in an A-module M is said to have weight n ifðduÞ �m ¼ n0m and ðudÞ �m ¼ n00m. There is a pair of operators on weights, d andm, whose actions are as follows: dðnÞ ¼ ðn 00; b�1ðn0 � an00 � gÞÞ, and mðnÞ ¼ðan 0 þ bn00 þ g; n0Þ. The operators m and d are inverses of each other.
For our purposes it is enough to concentrate on the following special case.Suppose lk (where k 2 Z) is the sequence of complex numbers described in Sec. 2.1(i.e., lk satisfies the recurrence lk ¼ alk�1 þ blk�2 þ g for all k 2 Z). Define theweights ok ¼ ðlk; lk�1Þ and let F denote the set of distinct weights consisting of theseok: F ¼ fok j k 2 Zg. Then F is closed under m and d, and F is generated under m andd by o0.
Given any r 2 C�, we define two A-modules NðF ; rÞ and N 0ðF ; rÞ. As vectorspaces they are identical. Both have a basis consisting of vectors indexed by F : sayfvk jok 2 Fg. (Note that vk ¼ vj if ok ¼ oj.) On NðF ; rÞ the action of A is given by
d � vk ¼ rvk�1 and u � vk ¼ r�1lkvkþ1 ð5:1:1Þ
while on N 0ðF ;rÞ the action of A is given by
d � vk ¼ r�1lk�1vk�1 and u � vk ¼ rvkþ1: ð5:1:2Þ
It follows from Benkart and Roby (1998, Theorem 5.5) that NðF ; rÞ and N 0ðF ; rÞ areA-modules. They are irreducible if lk 6¼ 0 for any k 2 Z.
5.2. We are most interested in NðF ; rÞ and N 0ðF ; rÞ when they are finitedimensional, i.e., when F is a finite set. We now determine when this occurs.
In Case (A), recall that lk�1 ¼ G� rk1L1 � rk2L2: From this, we easily deduce thatok ¼ oj if and only if ðrk1 � r
j
1ÞL1 ¼ ðrk2 � rj
2ÞL2 ¼ 0. Thus
(1) In case (A1), jF j ¼ 1 if L1 ¼ L2 ¼ 0; and jF j ¼ n if L1 ¼ 0, L2 6¼ 0.(2) In cases (A2), (A3), (A4), jF j ¼ 1 if L1 ¼ L2 ¼ 0.(3) In case (A5), jF j ¼ 1 if L1 ¼ L2 ¼ 0, jF j ¼ n1 if L1 6¼ 0;L2 ¼ 0; jF j ¼ n2 if
L1 ¼ 0;L2 6¼ 0; and jF j ¼ n if L1L2 6¼ 0.
In Case (B), lk�1 ¼ G� krkL1 � rkL2. From this, we easily see that
(1) In case (B1), jF j ¼ 1 if L1 ¼ L2 ¼ 0; and jF j ¼ n when L1 ¼ 0, L2 6¼ 0.(2) In case (B2), jF j ¼ 1 if L1 ¼ L2 ¼ 0.
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In case (C), lk�1 ¼ L2 � rk1L1 � kg1. We deduce easily that jF j is finite only wheng ¼ 0. In that case,
(1) In case (C1), jF j ¼ 1 if L1 ¼ 0; and jF j ¼ n when L1 6¼ 0.(2) In case (C2), jF j ¼ 1 if L1 ¼ 0.
Note that when jF j ¼ n, we can use Z=nZ as the index set of our basis vectors:thus fvk j k 2 Z=nZg is a basis for NðF ; rÞ, and the action of A is given by Eq. (5.1.1),with additions in the index performed in Z=nZ.
5.3. In the situation above, write Iðr;L1;L2Þ and I 0ðr;L1;L2Þ for the annihilator ofNðF ; rÞ and N 0ðF ; rÞ, respectively. We want to determine generators for Iðr;L1;L2Þand I 0ðr;L1;L2Þ.
Theorem. In the formulas below, assume Li 6¼ 0 unless specified otherwise. In case(C) assume that g ¼ 0. Then the annihilators of NðF ; rÞ when NðF ; rÞ are finite-dimensional are as follows:
(1) In all cases (A) and (B), Iðr; 0; 0Þ ¼ hd� r; u� r�1Gi.(2) In cases (A1) and (B1), Iðr; 0;L2Þ ¼ h1;d
n � rn; un � r�nðGn � Ln2Þ
� �.
(3) In case (C), Iðr; 0;L2Þ ¼ hd� r; u� r�1L2i. Here L2 could be zero.(4) In case (C1), Iðr;L1;L2Þ ¼ h2 � ð1� r1ÞL2;d
n � rn; un � r�nðLn2 � Ln
1Þ� �
.Here L2 could be zero.
(5) In case (A5), Iðr; 0;L2Þ ¼ h1;dn2 � rn2 ; un2 � r�n2ðGn2 � Ln2
2 Þ� �, and
similarly, Iðr;L1; 0Þ ¼ h2;dn1 � rn1 ; un1 � r�n1ðGn1 � Ln2
2 Þ� �.
(6) In case (A5), write L for the product l0l1 � � � ln�1, and write m1 and m2 forðr2r�1
1 � 1ÞL1 and ðr1r�12 � 1ÞL2, respectively. Let Il denote the ideal in
C½h1;h2� generated by hi1h
j
2 � mi1mj
2 for all ði; jÞ with ri1rj
2 ¼ 1. (See (3.6).)Then
Iðr;L1;L2Þ ¼ dn � rn; un � r�nL; Ilh i:
We get corresponding formulas for I 0 by interchanging u and d.
Proof. We’ll concentrate on NðF ; rÞ since the proof for N 0ðF ;rÞ is very similar.If L1 ¼ L2 ¼ 0, then F ¼ fðG;GÞg in cases (A) and (B); and F ¼ fðL2;L2Þg in
case C. So NðF ; rÞ is a one-dimensional module and the action of A is given byd � v ¼ rv, u � v ¼ r�1cv where c ¼ G in cases (A) and (B), and c ¼ L2 in case (C).Parts (1) and (3) of the theorem follows.
In part (2), we have lk�1 ¼ G� rk2L2. Let I ¼ h1;dn � rn; un � r�nðGn � Ln
2Þ� �
.We first show that I annihilates NðF ; rÞ.
Wehaveh1 �vk¼ðdu� r2udþ g1Þ�vk¼ðlk � r2lk�1 þ g1Þvk¼ðG� r2Gþ g1Þvk ¼ 0,so h1 annihilates NðF ; rÞ. Furthermore, since d � vk ¼ rvk�1, we have dn � vk ¼rnvk�n ¼ rnvk, so dn � rn also annihilates NðF ; rÞ. Finally, from u � vk ¼ r�1lkvkþ1
we get un � vk ¼ r�nlklkþ1 � � � lkþn�1vkþn ¼ r�nl0l1 � � � ln�1vk, so un�r�nl0l1 � � � ln�1
annihilates NðF ; rÞ. NowQn�1
i¼0 li ¼Qn�1
i¼0 G� ri2L2
� � ¼ Gn � Ln2. Thus I does indeed
annihilates NðF ; rÞ.
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We need to show the converse. Now A can be written as a direct sumA ¼ P
i2Z=nZ Bi, where Bi ¼P0
j C½du; ud�dj þP0k C½du; ud�uk; the first sum is over
all j with j � i mod n, and the second sum is over all k with k � �i mod n. We haveBi � vk � Cvk�i. Any x 2 A can be written as x ¼ P
i2Z=nZ xi with xi 2 Bi; if x annihi-lates NðF ;rÞ, then each xi also annihilates NðF ; rÞ. It does no harm, therefore, toassume that x ¼ xi for some i 2 Z=nZ.
Let y ¼ xdn�i. Then y 2 B0. We can write y ¼ gðh1; udÞ þ y0 where y0 2 I and g isa polynomial in h1 and ud. Thus gðh1; udÞ also annihilates NðF ; rÞ. Now gðh1; udÞcan be written as g0ðudÞ þ
Pmj¼1 gjðudÞhj
1 where gj are polynomials in ud only. Sinceh1 2 I, we conclude that g0ðudÞ also annihilates NðF ; rÞ. We want to show that g0 isin I.
We first show by induction that ðudÞk � ckukdk þ fkðudÞ mod I where ck 2 C�
and fkðudÞ is a polynomial in ud of degree at most k� 1. Thus we haveg0ðudÞ � fðudÞ mod I where f is a polynomial of degree at most n� 1. Sinceg0ðudÞ annihilates NðF ; rÞ, we conclude that fðudÞ also annihilates NðF ; rÞ. But thenfðudÞ � vk ¼ 0 implies that fðlk�1Þ ¼ 0 for all k 2 Z=nZ; since l0; . . . ; ln�1 are alldistinct, this means that f has n different zeroes. We conclude that f is the zeropolynomial. Tracing back all the way to y, we conclude that y 2 I.
Hence ydi ¼ xdn is also an element of I. But xdn � rnx mod I. Since r is non-zero, we conclude that x 2 I, as required.
The proofs for parts (4) and (5) are almost identical to the proof above, so weleave the details for the reader.
The proof for part (6) is similar to the proof of Theorem 3.3, part (A). Note thathi � v0 ¼ miv0ði ¼ 1; 2Þ and that m1m2 6¼ 0. Write I for dn � rn; un � r�nL; Ilh i. Astraightforward calculation shows that I � Iðr;L1;L2Þ. We need to show the reverseinclusion.
Let x 2 Iðr;L1;L2Þ. As in the proof of part (2), it suffices to assume thatx 2 C½h1;h2�. By Lemma 3.6 x � gðh1;h2Þ mod Il, where gðh1;h2Þ is distinctive.We need to show that gðh1;h2Þ ¼ 0.
Now the multiplicative subgroup of C� generated by r1 and r2 is a finite group ofn elements. Thus a distinctive set S has at most n elements. Since gðh1;h2Þ 2Iðr;L1;L2Þ, we have gðh1;h2Þ � vk ¼ 0 for all k 2 Zn, i.e.,
Pði;jÞ2S aijðri1rj2Þkmi1mj2 ¼ 0
for k ¼ 0; 1; . . . ; n� 1. We can regard this as a linear system of n equations in n
unknowns aijmi1mj
2; its coefficient matrix has determinant equal to the nonzeroVandermonde determinant
Qði;jÞ6¼ði0;j0Þ2Sðri1rj2 � ri
01 r
j02 Þ. Thus aij ¼ 0, as required.
&
5.4. We have expended some effort calculating the annihilators of NðF ; rÞ andN 0ðF ; rÞ in the previous section. Since our ultimate aim is to produce primitive ideals,the effort would have been wasted if NðF ; rÞ or N 0ðF ; rÞ is not irreducible.Fortunately they are.
Lemma. The finite-dimensional modules NðF ; rÞ and N 0ðF ;rÞ described above aresimple modules.
Proof. The proof is basically the same as the one in Benkart and Roby (1998,Theorem 5.5); the only difference is that we allow one of lk to be zero. Let M be
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a nonzero submodule of N 0ðF ; rÞ and let v ¼ Pakvk be a nonzero element in M.
Since vk has weight ok and the weights are all distinct, we have vk 2 M if ak 6¼ 0.Now whenever we have vk 2 M, we also have u � vk ¼ rvkþ1 2 M; since r 6¼ 0, weget vkþ1 2 M. Because our index set is cyclic, this means that vi 2 M for alli 2 Z=nZ, i.e., M ¼ N 0ðF ; rÞ and N 0ðF ; rÞ is irreducible. A very similar proof worksto show that NðF ; rÞ is also irreducible. &
5.5. We can now enlarge our list of primitive ideals.
Theorem. (a) In cases (A) and (B), hu� cu;d� cdi is a primitive ideal, wherecu; cd are any complex numbers with cucd ¼ G. In case (C) with g ¼ 0,hu� c0u;d� c0di is a primitive ideal, where c0u; c
0d are arbitrary complex numbers.
(b) Let zu; zd 2 C with zuzd 6¼ Gn. Then hh1; un � zu;d
n � zdi is a primitiveideal in cases (A1) and (B1).
Similarly, hh2 � z0h; un � z0u;d
n � z0di is a primitive ideal in case (C1) with g ¼ 0,
where z0uz0d 6¼ z0h
1� r1
� �n
.
Proof. Part (a) follows immediately from the description of Iðr; 0; 0Þ given inTheorem 5.3. Alternatively, we can check that the one-dimensional vector space Cv
with A-action given by u � v ¼ cuv and d � v ¼ cdv (in cases (A) and (B)), or by u � v ¼c0uv and d � v ¼ c0dv (in case (C) with g ¼ 0), does give rise to an A-module. Thisis straightforward.
For part (b), we’ll do the first claim. If zuzd 6¼ Gn and zu 6¼ 0, we can findr;L2 2 C� such that zu ¼ rn and zd ¼ r�nðGn � Ln
2Þ. By Theorem 5.3 the idealhh1;d
n � zd; un � zui is the annihilator of NðF ; rÞ and hence primitive. If zd 6¼ 0, then
we similarly conclude that hh1;dn � cnd; u
n � cndi is a primitive ideal since it annihi-lates N 0ðF ; rÞ. If g 6¼ 0, there is still the possibility that zu ¼ zd ¼ 0. In this case,hh1;d
n; uni is not the annihilator of NðF ; rÞ or N 0ðF ; rÞ, so we need to find anotherirreducible module whose annihilator is hh1;d
n; uni. It turns out that an irreduciblequotient of a Verma module does the trick, as we show below.
In Eq. (4.3.1) with r2 2 Pn, let L2 ¼ r2G. Then l�1 ¼ 0 and lk ¼ ð1� rkþ12 ÞG.
We can form the Verma module V ðlÞ where l ¼ l0. Note that ln�1 ¼ 0 and soLðlÞ is an n-dimensional irreducible A-module. To show that its annihilator ishh1;d
n; uni, we basically repeat the proof of Theorem 5.3. Here is a quick sketch.According to Theorem 2.2, Ann LðlÞ is generated by un, dn, and polynomialsgðdu; udÞ such that gðlk; lk�1Þ ¼ 0. Now h1 is clearly in the annihilator; if the anni-hilator contains polynomials other than those generated by h1, then we can find oneof the form fðduÞ, where f has degree at most n� 1. But then fðlkÞ ¼ 0 for allk ¼ 0; 1; . . . ; n� 1, so f has n distinct zeroes and hence must be zero. &
6. INTERLUDE
6.1. In the preceding sections we have constructed a number of primitive ideals,based on four different classes of irreducible modules. We now switch gears to showthat our list of primitive ideals is complete (with one exception). There are no more.
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One of our main tools to prove this completeness is the following proposition. Itis a well-known version of Schur’s lemma (see, e.g., McConnell and Robson, 1987,Proposition 9.1.7, or Kirkman and Kuzmanovich, 2000a, Proposition 3.2).
Proposition. Let R be a finitely-generated C-algebra. Suppose M is an irreducibleR-module and a is an element in the center of R. Then there is a za 2 C such thata �m ¼ zam for all m 2 M, i.e., a acts as a scalar on M.
6.2. Another of our tool is the following simple lemma describing the action of h1
and h2 on simple modules.
Lemma. Let M be an irreducible A-module.
(a) In cases (A), (B), (C), either h1M ¼ M or h1M ¼ 0. More precisely, themap h1 : M ! M is either the zero map or invertible.
(b) In case (A), the map h2 : M ! M is also either the zero map orinvertible.
Proof. h1M is a submodule of M because uh1M ¼ r�11 h1uM � h1M and dh1M ¼
r1h1dM � h1M. The result follows because M is irreducible. To prove the more pre-cise statement, we need to show that Ker h1 is also submodule of M. This is straight-forward: if m 2 Ker h1, then h1um ¼ r1uh1m ¼ 0 and h1dm ¼ r�1
1 dh1m ¼ 0, so um
and dm are both also in Ker h1. Thus Ker h1 is indeed submodule of M.Part (b) has the same proof. &
6.3. We now give an outline of how Proposition 6.1 and Lemma 6.2 help us showthat our list of primitive ideals is complete. We’ll use case (A1) as a typical example;other cases are similar and will be treated in detail later.
In case (A1), the element hn2 is in the center of A. By Proposition 6.1, hn
2 must actas a scalar on any irreducible module, and therefore any primitive ideal must containhn2 � c for some complex number c. There are two possibilities to consider: c ¼ 0 or
c 6¼ 0.If c ¼ 0, then h2 is not invertible and hence must act as 0 by Lemma 6.2. So in
this case the primitive ideal must contain h2. Now hh2i by itself is already a primitiveideal (Theorem 2.6). We will show that if I is a primitive ideal strictly larger thanhh2i, then it must be the annihilator of LðlÞ, or NðF ; rÞ, or N 0ðF ; rÞ. Therefore wealready know all primitive ideals containing h2.
If c 6¼ 0, then the ideal hhn2 � ci by itself is already a primitive ideal. Let I be any
primitive ideal strictly larger than hhn2 � ci. Either I contains h1 or it doesn’t. We will
show that any primitive ideal containing h1 must be the annihilator of LðlÞ,or NðF ; rÞ, or N 0ðF ; rÞ, so if I contains h1, we already know what the possibilitiesare for I. If I does not contain h1, then we will show that any primitive ideal strictlycontaining hn
2 � c (and where h1 is invertible) must be the annihilator of LðlÞ. Thusin either case, we have already found all primitive ideals containing hn
2 � c.We therefore have two tasks ahead of us: show, roughly speaking, that any
primitive ideal strictly containing the annihilator of a Verma module must be theannihilator of an irreducible quotient of a Verma module, and show also that any
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primitive ideal strictly containing h1 (or h2 in case (A)) must be the annihilator ofLðlÞ, or NðF ; rÞ, or N 0ðF ; rÞ. We do this in the next section.
7. LARGER IDEALS
7.1. SupposeM is an irreducible A-module. We begin by showing that if a power ofu and a power of d both act as scalars on M, then M is isomorphic to one of thefinite-dimensional modules we considered in Secs. (4) and (5). The irreduciblefinite-dimensional modules have been completely determined in Carvalho andMusson (2000) and Jordan (2000) so we give only sketchy proofs here.
Proposition. Suppose M is an irreducible A-module whose annihilator I containsum � tu and dn � td for some m; n 2 N and tu; td 2 C. If tu ¼ td ¼ 0, then M is iso-morphic to LðlÞ for some l 2 C. Otherwise, M is isomorphic to some NðF ; rÞ orN 0ðF ; rÞ.Proof. Since um � tu and dn � td are in I, the quotient A=I is finite-dimensional.The module M, being a (left) quotient of A=I, must also be finite-dimensional.
If tu ¼ td ¼ 0, choose n so that dnM ¼ 0 but dn�1M 6¼ 0. Then V ¼fv 2 M jdv ¼ 0g is a nonzero subspace of M. We can find an eigenvector of du inV : say ðduÞ � v0 ¼ lv0. Then v0 is a highest weight vector of weight ðl; 0Þ; since M
is irreducible, we have M ¼ Av0. Thus M is a quotient of VðlÞ. Using umM ¼ 0 wecan then deduce that M ¼ LðlÞ.
Now suppose td 6¼ 0. Choose n so that dn acts as a scalar on M but dn�1 doesnot. Since M is finite-dimensional and du and ud commute, there is a nonzero vectorvn�1 2 M such that ðduÞ � vn�1 ¼ ln�1vn�1 and ðudÞ � vn�1 ¼ ln�2vn�1 for some ln�1;ln�2 2 C. Let r be an nth root of td, and define vn�k�1, k ¼ 1; 2; . . . ; n� 1, byvn�k�1 ¼ r�kdk � vn�1. Because dnM ¼ rnM, we can conclude that dv0 ¼ rvn�1.A straightforward calculation shows that dn�k is weight vector, with weightðln�k; ln�k�1Þ, where lk ¼ alk�1 þ blk�2 þ g. We can also conclude that u � vk ¼r�1lkvkþ1, so spanfvkg is a submodule of M. Using the minimality of n then allowsus to conclude that M is indeed isomorphic to NðF ; rÞ, where F ¼ fðlk; lk�1Þ j 0 �k � n� 1g.
A similar calculation when tu 6¼ 0 lets us conclude in this case that M is iso-morphic to N 0ðF ; rÞ. &
7.2. We now state a lemma describing the primitive ideals which containsannihilators of Verma modules.
Lemma. SupposeM is an irreducible module whose annihilator strictly contains Il,where Il is one of the ideals in Theorem 3.3. In case (A) assume thath1M ¼ h2M ¼ M; in cases (B) and (C) it suffices to assume that h1M ¼ M. ThenM is isomorphic to an irreducible quotient of a Verma module, i.e., M ¼ Lðl0Þ forsome l0 2 C.
Proof. Let I denote the annihilator of M. For a change of pace we divide the proofinto a formal sequence of claims.
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(1) If xdk 2 Il or xuk 2 Il for some k � 0, then x 2 Il. The hypothesis implies
that x � vj ¼ 0 for all sufficiently large value of j, so the claim follows from the proofof Theorem 3.3.
(2) There is an element in InIl of the formPm
i¼0 fidi where fi 2 C½h1;h2�.
Any element in A can be written asPm
i¼0ðfidi þ giþ1uiþ1Þ, where fi; giþ1 2 C½h1;h2�.
Multiply this on the right by dmþ1 and we get an element of the required form.The claim then follows from (1).
(3) There is an element in InIl of the form dmf where f 2 C½h1;h2� andm � 0. Choose an element x 2 InIl of the form
Pm
i¼0 fidi (where fi 2 C½h1;h2�), with
m as small as possible. Then xh1 ¼Pm
i¼0 ri1fih1d
i. Therefore the elementxh1 � rm1 h1x ¼ Pm�1
i¼0 ðri1 � rm1 Þfih1di is also in I, but the highest power of d that
appears is at most m� 1. By minimality of m, we conclude thatPm�1
i¼0 ðri1 � rm1 Þfih1d
i 2 Il. Because di � vk 2 Cvk�i, we must have ðri1 � rm1 Þ fih1di ¼ ðri1 � rm1 Þ
r�ii fid
ih1 2 Il for all i ¼ 0; 1; . . . ;m� 1.If r1 is not a root of unity, then ri1 6¼ rm1 and we can conclude that fid
ih1 2 Il andhence fid
i 2 Il. ThereforePm�1
i¼0 fidi 2 Il, and hence fmðh1;h2Þdm 2 I n Il. Using the
commutation relations between h1;h2 and dm, we can write fmðh1;h2Þdm ¼dmfðh1;h2Þ, proving the claim.
If r1 2 Pn, then we are in case (B1) or (C1). Take (B1); (C1) is similar. Thecalculations above imply that fid
i 2 Il whenever i 6� m mod n. Thusy ¼ P
i�m fidi 2 Il. Now yh2 ¼
Pi�m fir
iðih1 þ h2Þdi, so yh2 � rmðmh1 þ h2Þy ¼Pi�m rmði�mÞfih1d
i is also in Il. For i < m we conclude that fih1di ¼
r�ifidih1 2 Il, so fid
i 2 Il. ThereforePm�1
i¼0 fidi 2 Il, and hence fmðh1;h2Þ
dm 2 I n Il. Using the commutation relations between h1;h2 and dm as before, wecan write fmðh1;h2Þdm ¼ dmfðh1;h2Þ, proving the claim.
(4) There exists an element in InIl of the form dmhm1
1 hm2
2 . Choose a polyno-mial pðh1;h2Þ of the smallest possible nonzero degree such that there exists anm � 0 with x ¼ dmpðh1;h2Þ 2 InIl. (It is possible to find such a polynomial by theprevious claim.) Suppose that the degree of pðh1;h2Þ is ðm1;m2Þ. Note that in case(A), we can choose p to be distinctive (see Lemma 3.6); in case (B1) we havem1 < n; and in case (C1) with g 6¼ 0 we have m1 < n.
Now in case (A), xd ¼ dmpðh1;h2Þd ¼ dmþ1pðr1h1; r2h2Þ. Thus xd� rm1
1 rm2
2 dx ¼dmþ1 pðr1h1; r2h2Þ � r
m1
1 rm2
2 pðh1;h2Þ� � ¼ dmþ1qðh1;h2Þ. This element is still in I, but q
has smaller degree than p. By minimality of ðm1;m2Þ, q 2 Il. Since q can be chosen tobe distinctive, we conclude by Lemma 3.7 that q ¼ 0. Then Lemma 3.9 implies thatpðh1;h2Þ ¼ ch
m1
1 hm2
2 , proving the claim.The proofs for cases (B) and (C) are similar to the above. In case (B),
xd� rm1þm2 dx ¼ dmþ1 pðrh1; rh1 þ rh2Þ � rm1þm2pðh1;h2Þð Þ; as above, we concludethat pðrh1; rh1 þ rh2Þ ¼ rm1þm2pðh1;h2Þ. Then Lemma 3.10 implies pðh1;h2Þ ¼ch
m1
1 , again proving the claim.In case (C), we have to consider g ¼ 0 and g 6¼ 0. If g 6¼ 0, we have xd� r
m1
1
dx ¼ dmþ1 pðr1h1;h2 þ gÞ � rm1
1 pðh1;h2Þ� �
; as above, we also conclude thatpðr1h1;h2 þ gÞ ¼ r
m1
1 pðh1;h2Þ. Lemma 3.12 then implies that pðh1;h2Þ ¼ chm1
1 ,proving the claim. If g ¼ 0, then Il ¼ hh2 � li, so m2 ¼ 0. Hence pðh1;h2Þ is actuallya polynomial in h1 only; we can write pðh1;h2Þ ¼ pðh1Þ. Therefore xd� r
m1
1
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dx ¼ dmþ1 pðr1h1Þ � rm1
1 pðh1Þ� �
; we conclude that pðr1h1Þ ¼ rm1
1 pðh1Þ, and thereforepðh1Þ ¼ ch
m1
1 , finishing the claim.
(5) There exists an element in I n Il of the form dm. This follows from the pre-vious claim because h1M ¼ h2M ¼ M in case (A), and h1M ¼ M in cases (B) and (C).
(6) I is the annihilator of Lðl0Þ for some l0 2 C. The same technique allowsus to show that I contains uk for some k � 0. The result then follows fromProposition 7.1. &
7.3. We now investigate what happens if a primitive ideal contains the annihilatorof a doubly-infinite module as in Theorem 2.6.
Lemma. Suppose M is an irreducible module whose annihilator I contains h1. Ifr2 2 Pn (n > 1) or r2 ¼ 1 with g ¼ 0, thenM is finite-dimensional and is isomorphicto either LðlÞ, NðF ;rÞ, or N 0ðF ; rÞ. On the other hand, if r2 is not a root of unity orr2 ¼ 1 with g 6¼ 0, then either I ¼ hh1i, or M is finite-dimensional and is isomorphicto LðlÞ, NðF ; rÞ, or N 0ðF ; rÞ.
Similarly, in cases (A1)–(A4) and (C2) with g ¼ 0, suppose I contains h2. Theneither I ¼ hh2i, or M is finite-dimensional and isomorpic to LðlÞ, NðF ; rÞ, orN 0ðF ; rÞ.
Proof. Let I1 ¼ hh1i. Suppose I contains I1. Note that M is still irreducibleconsidered as an A=I1-module.
If r2 ¼ 1 and g ¼ 0, then h1 ¼ du� ud and so du � ud mod I1. Thus A=I1 iscommutative and the result follows from Proposition 7.1.
If r2 6¼ 1, we have du � r2ud� g1 mod I1, and by induction
duk � rk2ukd� ðrk2 � 1ÞGuk�1 mod I1
dku � rk2udk � ðrk2 � 1ÞGdk�1 mod I1
If r2 2 Pn, this implies that dnu � udn and dun � und mod I1, so by Proposition 6.1,dn and un act as scalars on M. The result follows from Proposition 7.1.
We now consider the case where r2 is not a root of unity. By Theorem 2.6, I1 isthe annihilator of a doubly-infinite module which is quite similar to a Verma module.The proof is therefore very similar to the proof of Theorem 7.2, with I1 replacingIl, h2 replacing h1, and m1 ¼ 0 in step (4). In case (A), everything goes through prettymuch unchanged. There are, however, some minor changes we have to make in cases(B) and (C).
In case (B), we have dh2 � rðh1 þ h2Þd � rh2d, so step (3) in the proof ofTheorem 7.2 goes through in this situation. In step (4) we conclude as before thatthere is an element in I1 of the form dmh
m2
2 . Now h2u � ruh2 and dh2 � rh2d mod I1,so as in Lemma 6.2 we have either h2M ¼ 0 or h2M ¼ M. If h2M ¼ M, then I
contains dm, which means step (5) of the proof is valid. If h2M ¼ 0, then h2 2 I,so du � ud mod I. This implies that A=I is commutative and our conclusion followsfrom Proposition 7.1.
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In case (C), we have h2u ¼ uðh2 þ gÞ so we have to change step (3) a little bit: welook at xh2 � ðh2 þmgÞx instead of xh1 � rm1 h1x. The conclusion is the same: there isan element in InI1 of the form fdm where f 2 C½h1;h2�. Of course, we can choosef to be a polynomial in h2 only. In step (4), choose pðh2Þ of the smallest degree suchthat there exists an m � 0 with x ¼ pðh2Þdm 2 InI1. Then dx� xd ¼ ðpðh2 þ gÞ�pðh2ÞÞdmþ1 is also in I; since the degree of ðh2 þ gÞ � pðh2Þ is smaller than the degreeof pðh2Þ, we conclude that pðh2 þ gÞ ¼ pðh2Þ. But this implies that p has infi-nitely many roots, i.e., p is a constant. Thus dm 2 InIl, which is the conclusion wewant. &
8. PRIMITIVE IDEALS
8.1. We are now in position to give a complete list of the primitive ideals of A.Here they are.
Theorem. In the following statements,
c is any nonzero complex number.
c0 is a complex number not equal to 0 or 1 (used only in case (A2)).
cu and cd are arbitrary complex numbers satisfying cucd ¼ G.
c0u and c0d are arbitrary complex numbers.
zu and zd are any complex numbers satisfying zuzd 6¼ Gn.
z0h, z0u and z0d are complex numbers with z0uz
0d 6¼ z0
h
1�r1
� n
.
In case (A2) we need to consider two subcases: (A2a) is when either n1 6¼ n2 org 6¼ 0, and (A2b) is when n1 ¼ n2 and g ¼ 0.
Then the primitive ideals of the down-up algebra A are as follows:
Case A.
(A1): hhn2 � ci, hh2i, hh1; u
n � zu;dn � zdi, hu� cu;d� cdi, and Ann LðlÞ.
(A2a): hhn11 � ch
n22 i, hh1i, hh2i, hu� cu;d� cdi, and Ann LðlÞ.
(A2b): hhn1 � c0hn
2i, hhn1 � hn
2 ;dni, hhn
1 � hn2 ; u
ni, hh1i, hh2i, hu� cu;d� cdi, andAnn LðlÞ.
(A3): hhn11 h
n22 � ci, hh1i, hh2i, hu� cu;d� cdi, and Ann LðlÞ.
(A4): h0i, hh1i, hh2i, hu� cu;d� cdi, and Ann LðlÞ.(A5): Iðr;L1;L2Þ, I 0ðr;L1;L2Þ (described in Theorem 5.3), and Ann LðlÞ.
Case B.
(B1): hhn1 � ci, hh1; u
n � zu;dn � zdi, hu� cu;d� cdi, and Ann LðlÞ.
(B2): h0i, hh1i, hu� cu;d� cdi, and Ann LðlÞ.
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Case C.
(C1) if g 6¼ 0: hhn1 � ci, hh1i, and Ann LðlÞ.
(C1) if g ¼ 0: hh2 � z0h; un � z0u;d
n � z0di, hu� c0u;d� c0di, and Ann LðlÞ.(C2) if g 6¼ 0: h0i, hh1i, and Ann LðlÞ.(C2) if g ¼ 0: hh2 � ci and hu� c0u;d� c0di.
The generators of Ann LðlÞ are described in Theorem 4.2.
Proof. Except in case (A2) with n1 ¼ n2 and g ¼ 0, we only need to show that theprimitive ideals listed in Theorems 2.6, 3.15, 4.2, and 5.5 are complete.
In cases (A5) and (C1) with g ¼ 0, the elements un and dn are in the center of A,so by Proposition 6.1 they act as scalars on any irreducible module M. By Lemma7.1, M must be isomorphic to LðlÞ, NðF ; rÞ, or N 0ðF ; rÞ. Their annihilators aredetermined in Theorems 4.2 and 5.3.
We now consider the remaining cases. LetM be an irreducible module with anni-hilator I. We have already outlined case (A1) in Sec. 6.3. Now take (B2) for anotherexample. We first note that h0i is a primitive ideal; in fact, h0i is the annihilator of aVerma module. From Lemma 6.2, either h1M ¼ M or h1M ¼ 0. If h1M ¼ M, then wecan apply Lemma 7.2 and conclude that we already know all primitive ideals in thiscase. But if h1M ¼ 0, then I contains hh1i. All primitive ideals containing it aredescribed by Lemma 7.3. So in all cases we have already determined the primitiveideals.
The reasoning in the other cases are similar. Only (A2) is somewhat different.Consider first the case where either n1 6¼ n2 or g 6¼ 0. In this case, either I containshh2i or it doesn’t. If it does, then the possibilities for I are described in Lemma7.3 (and Theorems 4.2 and 5.5). If it doesn’t, then h2M ¼ M and in this case we needto look at the slightly larger ring A0 which we get by adjoining h�1
2 to A. Note that theoperator M!M given by v 7!h2 � v is invertible, and hence M can be considered anA0-module. As such it is still irreducible. Now the element hn1
1 h�n22 is central in A0, so
by Proposition 6.1 it must act as a scalar c. Thus the annihilator in A of M must con-tain hhn1
1 � chn22 i. If c ¼ 0, then h1M ¼ 0 the possibilities for I are given in Lemma
7.3. If c 6¼ 0 then I must contain hhn11 � ch
n22 i. By Theorem 3.15, hhn1
1 � chn22 i is
already a primitive ideal which annihilates a Verma module. Primitive idealscontaining it are described in Lemma 7.2.
The case (A2) with n1 ¼ n2 and g ¼ 0 will be treated in the subsequentsections. &
8.2. In case (A2) with either n1 6¼ n2 or g 6¼ 0, we have seen that the idealshhn1
1 � chn22 i are primitive for any c 2 C�. (See Theorem 3.15.) In fact, these ideals
are annihilators of irreducible Verma modules. Unfortunately, if n1 ¼ n2 andg ¼ 0, then the corresponding Verma modules are not irreducible—infinitely manyof the lks are zero in this case. Thus it is not certain that the ideals hhn1
1 � chn22 i
are primitive. But it turns out that there are irreducible doubly-infinite moduleswhose annihilators are those ideals (at least when c 6¼ 1). We start by constructingthese modules.
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So for the rest of this section, assume that we are in case (A2) with g ¼ 0 andn1 ¼ n2 ¼: n. Thus rn1 ¼ rn2 , so we can write r1 ¼ yr2 where y 2 Pn. Write r for r2;then Eq. ð4:2:AÞ becomes
lk�1 ¼ �rkðykL1 þ L2Þ:
If L2 6¼ �ykL1 for k ¼ 0; 1; . . . ; n� 1, then lk is never zero and lk 6¼ lj for any k 6¼ j.Thus the doubly-infinite module VðL1;L2Þ is irreducible (Benkart and Roby, 1998,Theorem 5.5). We now find its annihilator.
As in 3.2 case (A), define m1 ¼ ðr � ryÞL1 and m2 ¼ ðry� rÞL2. A straight-forward calculation shows that
h1 � vk ¼ ðryÞkm1vk h2 � vk ¼ rkm2vk: ð8:2:1Þ
Lemma. Suppose L1 6¼ 0, L2 6¼ 0, and L2 6¼ �ykL1 for k ¼ 0; 1; . . . ; n� 1. Thenthe annihilator of VðL1;L2Þ described above is hðm2h1Þn � ðm1h2Þni.
Proof. Let I denote the ideal hðm2h1Þn � ðm1h2Þni. Equation (8.2.1) shows that I
annihilates V ðL1;L2Þ. We need to show the converse.By Lemma 2.4, the annihilator of V ðL1;L2Þ is generated by elements in
C½h1;h2�. Thus we need to find p 2 C½h1;h2� such that p rjyjm1; rjm2
� � ¼ 0 for allj 2 Z. By Lemma 3.6 p is congruent mod I to a distinctive polynomial q, and byLemma 3.7, q ¼ 0. Thus p 2 I and we are done. &
Corollary. hhn1 � chn2i is a primitive ideal whenever c 6¼ 0; 1.
Proof. According to the lemma, the ideal hhn1 �
mn1mn2
hn2i is primitive as long as
L1 6¼ 0, L2 6¼ 0, and L2 6¼ �ykL1 for k ¼ 0; 1; . . . ; n� 1. Nowm1m2 ¼
L1�L2; thus we
can always find L1 6¼ 0, L2 6¼ 0 with L2 6¼ �ykL1 such thatmn1mn2
¼ c. &
8.3. We now investigate the situation when c ¼ 1, i.e., we look at the idealhhn
1 � hn2i. It turns out, rather surprisingly, that it is not a primitive ideal. We provide
a proof below.
From the definitions of h1 and h2 we get ðr1 � r2Þud ¼ h1 � h2, and then usinginduction we get, for k � 1,
ðr1 � r2Þkukdk ¼Yk�1
i¼0
ðr�i1 h1 � r�i
2 h2Þ ¼Yk�1
i¼0
r�iðy�ih1 � h2Þ ð8:3:1Þ
In particular, ðr1 � r2Þnundn ¼ ð�1Þnrnðn�1Þ=2ðhn1 � hn
2Þ. Thus any ideal containinghn1 � hn
2 must also contain undn.
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Now let M be an irreducible module whose annihilator contains hn1 � hn
2. Byinduction, we can show that for k � 1,
ðr1 � r2Þdku ¼ ðrk1 � rk2Þdudk�1 � ðrk1r2 � r1rk2Þudk;
and in particular, dnu ¼ rnudn. Using the method of proof in Lemma 6.1, we con-clude that either dnM ¼ M or dnM ¼ 0. Similarly, either unM ¼ M or unM ¼ 0. Sincewe know that undnM ¼ 0, it cannot be the case that unM ¼ dnM ¼ M. Hence theannhilator of M must contain either un or dn.
Thus any primitive ideal containing hhn1 � hn
2i must also contain either dn or un.It turns out this is sufficient; adding either dn or un produces a primitive ideal. Weshow this in the next section.
8.4. In ð2:2:AÞ, let L1 ¼ �L2. Then lk�1 ¼ rkðyk � 1ÞL2. Now define an A-moduleV 0 as follows: as a vector space, it has a basis fvk j k 2 Zg, and the action of A is givenby u � vk ¼ vkþ1 and d � vk ¼ lk�n�1vk�n�1. It is not immediately obvious that V 0 isindeed an A-module, but the verification is straightforward: d2u � vk ¼ lk�n
lk�2n�1vk�2n�1, and ðadudþ bud2Þ � vk ¼ lk�n�1ðalk�2n�1 þ blk�2n�2Þvk�2n�1 ¼lk�n�1 lk�2nvk�2n�1. It is now easy to check that lk�nlk�2n�1 ¼ lk�n�1lk�2n, thus ver-ifying that d2u � vk ¼ ðadudþ bud2Þ � vk. A similar calculation shows thatdu2 � vk ¼ ðauduþ bu2dÞ � vk. Thus V 0 is indeed an A-module. V 0 is similar to aVerma module, but it has some important differences.
Note that whenever k � 0 mod n, we have lk�1 ¼ 0. Hence dn acts as the scalar 0on V 0. Furthermore, it is easy to verify that
h1 � vk ¼ ðyrÞk�nrðy� 1ÞL2vk�n h2 � vk ¼ rk�nrðy� 1ÞL2vk�n;
and hence hn1 � hn
2 annihilates V0. Note that V 0 is not a weight module in the sense of
4.1, unlike all the other modules we have looked at. But if we define h ¼ unh2, thenh � vk ¼ rk�nrðy� 1ÞL2vk, so vk is an eigenvector of h with eigenvaluerk�nrðy� 1ÞL2. Since r is not a root of unity, these eigenvalues are all distinct. Thusif M is a nonzero submodule containing
Pk2T akvk, then akvk 2 M for each k 2 T .
Since each vk generate V 0 under u and h2, we conclude that V 0 is irreducible.We would now like to find the annihilator of V 0. To make the calculations look
nicer, set L2 ¼ r�1ðy� 1Þ�1; then h1 � vk ¼ ðyrÞk�nvk�n and h2 � vk ¼ rk�nvk�n.
Lemma. The annihilator of V 0 described above is hdn;hn1 � hn
2i. Thus hdn;hn1 � hn
2iis a primitive ideal.
Proof. Let I denote the ideal hdn;hn1 � hn
2i. We need to show that if x annihilatesV 0, then x 2 I. We first consider the case when x is of the form x ¼ P
i fiui, where
f 2 C½h1;h2�.Note that the operator h2 : V
0 ! V 0 is invertible. Write h for h1h�12 ; then
h � vk ¼ ykvk. Since hi1h
j
2 ¼ ðh1h�12 Þihiþj
2 ¼ hihiþj
2 , any polynomial fðh1;h2Þ can bewritten as
Pj gjðhÞhj
2 where gj is a one-variable polynomial.
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Now hj
2ui � vk 2 Cvkþi�nj. Since the basis vectors fv‘g are independent, we can
assume, without loss of generality, that x is of the form x ¼ Pj gjðhÞhj
2unjþi. In fact,
since uiV 0 ¼ V 0, we can even assume that x is of the form x ¼ Pj gjðhÞhj
2unj.
In this case, x � vk ¼ 0 implies thatP
j gjðykÞrjðj�1Þn=2rjk ¼ 0. Since y is an nthroot of unity, we can write
Pj gjðykÞrjðj�1Þn=2rjðkþ‘nÞ ¼ 0 for all integer values of ‘.
Therefore, for any fixed k, the polynomialP
j gjðykÞrjðj�1Þn=2hj
2 ¼ 0 has infinitelymany roots. We conclude that gjðykÞ ¼ 0 for each value of j. Therefore gj is amultiple of the polynomial
Qn�1k¼0ðh� ykÞ ¼ hn � 1 ¼ ðh1h
�12 Þn � 1.
We conclude: if x ¼ Pi fiu
i annihilates V 0, where fi 2 C½h1;h2�, then fi is amultiple of the polynomial hn
1 � hn2, and hence x 2 I.
Now consider the general case, i.e., x is of the form x ¼ Pi fiu
i þPj qjd
j, wherefi; qj 2 C½h1;h2�. Then xum also annihilates V 0, and for a large enough value of m,xum can be written in the form
Pi fiu
iþm þPj q
0ju
j, where q0j 2 C½h1;h2�. By ourprevious conclusion, we can infer that fi is a multiple of hn
1 � hn2. ThereforeP
i fiui is already a member of I, and so
Pj qjd
j must by itself annihilate V 0.It remains to show that y ¼ P
j qjdj is in I. We can of course assume that all
powers of d are at most n. As before, the polynomial qj can be written as a poly-nomial in h and h2. Note that hi
2dj sends vk to Cvk�jðnþ1Þ�in, and k�
jðnþ 1Þ � in ¼ k� j0ðnþ 1Þ � i0n implies that j � j0 mod n, which, since both j
and j0 are between 0 and n� 1, implies that j ¼ j0. It follows that i ¼ i0. Hence eachterm in the sum qjd
j annihilates V 0. Therefore we can assume, without loss of gen-erality, that y ¼ qjðh1;h2Þdj for some integer j between 0 and n� 1; we can furtherassume that qjðh1;h2Þ ¼ gðhÞhi
2 for some integer i � 0. In this case, sincehi2d
j ¼ r�ijdjhi2 and hi
2V0 ¼ V 0, we can conclude that gðhÞdj � vk ¼ 0 for all k 2 Z.
Now gðhÞdj � vk ¼ lk�n�1lk�2ðnþ1Þ � � � lk�jðnþ1Þgðyk�jÞvk�jðnþ1Þ. Since li ¼ 0whenever i � �1 mod n but li 6¼ 0 otherwise, we can infer that gðyk�jÞ ¼ 0 for allk ¼ j; j þ 1; . . . ; n� 1, i.e., that gðykÞ ¼ 0 for k ¼ 0; 1; . . . ; n� j � 1. Therefore g isa multiple of the polynomial
Qn�j�1i¼0 ðh1h
�12 � yiÞ; thus qjðh� 1;h2Þ is a multiple
of fj ¼Qn�j�1
i¼0 ðh1 � yih2Þ.We now show that qjd
j 2 I; this implies that y 2 I, finishing the proof. Putk ¼ n� j in Eq. (8.3.1); we get ðr1 � r2Þn�j
un�jdn�j ¼ Qn�j�1i¼0 ðryÞ�iðh1 � yih2Þ,
which implies that un�jdn�j ¼ cQn�j�1
i¼0 ðh1 � yih2Þ ¼ cfj, where c is nonzero. There-fore un�jdn ¼ cfjd
j. The left side is a member of I, showing that fjdj is also in I,
as required. &
Note that hun;hn1 � hn
2i is also a primitive ideal; we simply exchange u and d inthe definitions above.
8.5. We can now complete the proof of Theorem 8.1. For clarity we restate theresult here.
Theorem. In case (A2) with g ¼ 0 and n1 ¼ n2 ¼ n, the primitive ideals arehhn
1 � chn2i (where c 6¼ 0; 1), hhn
1 � hn2 ;d
ni, hhn1 � hn
2 ; uni, hh1i, hh2i, hu� cu;d� cdi,
and Ann LðlÞ.
Proof. It only remains to show that if a primitive ideal contains hun;hn1 � hn
2ior hdn;hn
1 � hn2i, then it is the annihilator of a finite-dimensional module. This is
470 Praton
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ORDER REPRINTS
done as in Lemma 7.2, with V 0 taking the place of the Verma module. The proofthere is easily adapted to the present situation. We leave the (repetitive) details to thereader. &
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Received September 2001
Primitive Ideals of Noetherian Down-Up Algebras 471
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