# Affinoid Dixmier modules and the deformed Dixmier-Moeglin equivalence

aa r X i v : . [ m a t h . R T ] F e b Aﬃnoid Dixmier Modules and the deformedDixmier-Moeglin equivalence

Adam JonesFebruary 8, 2021

Abstract

The aﬃnoid envelope [ U ( L ) K of a free, ﬁnitely generated Z p -Lie algebra L , hasproven to be useful within the representation theory of compact p -adic Lie groups.Our aim is to further understand the algebraic structure of [ U ( L ) K , and to thisend, we will deﬁne a Dixmier module over [ U ( L ) K , and prove that this object isgenerally irreducible in case where L is nilpotent. Ultimately, we will prove that allprimitive ideals in the aﬃnoid envelope can be described in terms of the annihilatorsof Dixmier modules, and using this, we aim towards proving that these algebrassatisfy a version of the classical Dixmier-Moeglin equivalence. Contents ’ U ( L ) K on ’ D ( λ ) Locally Closed Ideals 28 ’ U ( L ) K . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285.2 Towards the Dixmier-Moeglin equivalence . . . . . . . . . . . . . . . . . 305.3 Classiﬁcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 Throughout, ﬁx p a prime, K \ Q p a ﬁnite extension, O the valuation ring of K , π ∈ O auniformiser. This research is partially inspired by a problem in classical non-commutative algebra.Let k be any ﬁeld of characteristic 0, and let R be a Noetherian k -algebra. We say thata prime ideal P of R is: • Primitive if P = Ann R M = { r ∈ R : rM = 0 } for some irreducible R -module M . • Weakly rational if Z ( R/P ) is an algebraic ﬁeld extension of k . • Rational if Z ( Q ( R/P )) is an algebraic ﬁeld extension of k , where Q ( R/P ) is theGoldie ring of quotients of

R/P , in the sense of [16, Theorem 2.3.6]. • Locally closed if P = ∩{ Q E R : Q prime, P ( Q } , i.e. if { P } is a locally closedsubset of Spec R with respect to the Zariski topology.It is not diﬃcult to see that if P is rational then it is weakly rational. Deﬁnition 1.1.

Let R be a Noetherian k -algebra. We say that R satisﬁed the Dixmier-Moeglin equivalence if for all prime ideals P of R , we have: P is primitive ⇐⇒ P is rational ⇐⇒ P is locally closed. Note that if R is commutative and k is algebraically closed, then this condition is essen-tially a strong version of Hilbert’s Nullstellensatz, i.e. that all prime ideals P of R ariseas an intersection of maximal ideals m of R of codimension 1.2his condition is known to be satisﬁed for numerous examples of Noetherian k -algebras,most notably the enveloping algebra U ( h ) for a ﬁnite dimensional k -Lie algebra h asproved independently by Dixmier and Moeglin in [7] and [15]. An ongoing project isto classify algebras satisfying the equivalence, and in [3], a detailed analysis of what isknown to date on this subject is given. In this paper, however, we will not be concernedwith the classical picture, but with a p -adic analogue.Speciﬁcally, let K be our p -adic ﬁeld, and let b R be a Noetherian, Banach K -algebra. Un-fortunately, to prove the full Dixmier-Moeglin equivalence for such algebras may prove tobe impractical, since the techniques classically used typically involve discrete generatingsets, which are absent here. However, there is a weaker statement that might be moreapproachable in this setting.For each n ∈ N , in Deﬁnition 2.3 below we will deﬁne the level n deformation b R n of b R ,which is a dense Banach subalgebra of b R , and using this we make the following deﬁnition: Deﬁnition 1.2.

We say that the Noetherian, Banach K -algebra b R satisﬁes the deformedDixmier-Moeglin equivalence if for all prime ideals P of R , there exists N ∈ N such thatfor all n ≥ N : P ∩ b R n is primitive ⇐⇒ P ∩ b R n is weakly rational ⇐⇒ P ∩ b R n is locally closed. Key example:

If we take R = U ( g ) and let R := U ( L ) for some O -Lie lattice L in g , then the π -adic completion b R = [ U ( L ) K of R with respect to R is called the aﬃnoidenveloping algebra (or aﬃnoid envelope ) of L , and b R n is just \ U ( π n L ) K .We believe that the aﬃnoid enveloping algebra [ U ( L ) K should satisfy the deformedDixmier-Moeglin equivalence, but to date, this is only known in the case where g isnilpotent and contains an abelian ideal of codimension 1, as shown in [13, Corollary1.6.2]. The other main motivation for this research lies within representation theory of compact p -adic Lie groups.Fix G a uniform pro- p group in the sense of [8, Deﬁnition 4.1]. When studying the p -adic representation theory of G , there are several avenues we can consider. If we wereinterested in general, abstract K -linear representations of G , naturally we would studymodules over the standard group algebra K [ G ], but since G is not discrete, this objectis of little practical use. Instead, deﬁne the rational Iwasawa algebra : KG := lim ←− U E o G O [ G/U ] ⊗ O K ,this is a Noetherian, topological K -algebra whose module structure completely describesthe continuous representations of G , a class which includes all ﬁnite dimensional, smoothand locally analytic representations. 3et L := p log( G ) be the Z p -Lie algebra of G , as deﬁned in [2, Section 10], and let [ U ( L ) K be the aﬃnoid envelope of L with coeﬃcients in K , as deﬁned above.Using [2, Theorem 10.4], we see that there exists a dense embedding of topological K -algebras KG → [ U ( L ) K , i.e. [ U ( L ) K is a Banach completion of KG , and thus we see thatrepresentations of the aﬃnoid enveloping algebra [ U ( L ) K naturally have the structure of KG -modules, which allows us to explore the representation theory of G via the represen-tation theory of L .In this paper, we will study the algebraic structure of the aﬃnoid envelope, and in aforthcoming work [12], we will show how we can use our results to deduce informationabout primitive ideals in KG . The key aim of this paper is to describe the primitive ideal structure of the aﬃnoid en-veloping algebra [ U ( L ) K , focusing on the case where L is nilpotent. Speciﬁcally, we aim toprove that under suitable conditions this algebra satisﬁes the deformed Dixmier-Moeglinequivalence.To this end, we will follow the representation theoretic approach outlined by JaquesDixmier in [7] when studying the classical enveloping algebra U ( g ). This approach wasto deﬁne a class of irreducible induced representations of U ( g ) whose annihilator idealscompletely describe the primitive ideals in U ( g ). In section 2, we will adapt this approachto the aﬃnoid envelope.Speciﬁcally, to each linear form λ ∈ Hom Z p ( L , O ), we will associate an induced mod-ule [ D ( λ ) over [ U ( L ) K , which we call a Dixmier module .We will prove in section 3 that Dixmier modules are generally irreducible, and thus theideal I ( λ ) := Ann \ U ( L ) K [ D ( λ ) is primitive in [ U ( L ) K . We call ideals of this form Dixmierannihilators . Our main result, which we will prove in section 6, describes all weaklyrational ideals of [ U ( L ) K in terms of Dixmier annihilators: Theorem A.

Let L be an O -Lie lattice in a nilpotent Lie algebra g , and let P be a weaklyrational ideal in [ U ( L ) K . Then there exists N ∈ N , λ ∈ Hom O ( π N L , O K ) such that forall n ≥ N : P ∩ \ U ( π n L ) K = I ( λ ) = Ann \ U ( π n L ) K [ D ( λ ) Note:

This makes sense because if λ takes values in the algebraic closure K of K , thenit must take values in F for some ﬁnite extension F of K , and we can consider the actionof [ U ( L ) K on [ D ( λ ) via the embedding of [ U ( L ) K into [ U ( L ) F .This result aims towards a correspondence between orbits of linear forms in Hom Z p ( L , O K )and primitive ideals in [ U ( L ) K , in line with the classical result of Dixmier [7, Theorem4.2.4]. In a subsequent paper [12], Theorem A will become a key step in classifying prim-itive ideals in KG for G nilpotent.Finally, in section 7, we will explore the properties of Dixmier modules and Dixmierannihilators further, and prove the deformed Dixmier-Moeglin equivalence in a speciﬁccase: Theorem B.

Let g be a nilpotent K -Lie algebra such that [ g , g ] is abelian, and let L bean O -Lie lattice in g . Then [ U ( L ) K satisﬁes the deformed Dixmier-Moeglin equivalence.In fact, for all weakly rational ideals P of [ U ( L ) K , n ∈ N suﬃciently high, \ U ( π n L ) K P ∩ \ U ( π n L ) K is asimple domain. This result provides an insight into the deeper algebraic structure of the aﬃnoid enve-lope, and we hope it will be of independent interest within the ﬁeld of non-commutativealgebra, and that it will advance the Dixmier-Moeglin project into the p -adic setting. Acknowledgments:

I am very grateful to Konstantin Ardakov and Ioan Stanciu formany helpful conversations and discussions. I would also like to thank EPSRC and theHeilbronn Institute for Mathematical Research for supporting and funding this research.

In this section, we sometimes want to consider the classical picture. So throughout, take k to be any ﬁeld of characteristic 0. First, we will recap some notions from rigid geometry. For a more detailed discussion,see [5].Recall from [5, Deﬁnition 2.2.2] that if R is a ring carrying a complete, separated ﬁltration w , the Tate algebra in d variables t , · · · , t d over R is the algebra: R h t , · · · , t d i := { P α ∈ N d λ α t α · · · t α d d : w ( λ α ) → ∞ as α → ∞} .In other words, the Tate algebra is the ring of power series with coeﬃcients in R thatconverge on the unit disc R d + , we call these Tate power series . This ring carries a sepa-rated ﬁltration given by w inf ( P α ∈ N d λ α t α · · · t α d d ) := inf { w ( λ α ) : α ∈ N d } .Normally, R is assumed to be commutative, and in our case, we will usually take R = K ,in which case the Tate algebra is Noetherian, and the ﬁltration w inf is Zariskian in thesense of [14, Ch. II Deﬁnition 2.1.1]. Recall from [5, Deﬁnition 3.1.1] that we deﬁne anaﬃnoid algebra to be any quotient of a Tate algebra over a complete, discretely valuedﬁeld. Clearly any aﬃnoid algebra A will carry a complete, Zariskian ﬁltration w A givenby the quotient ﬁltration with respect to w inf .Aﬃnoid algebras play a similar role in rigid geometry as commutative algebras play instandard algebraic geometry. Speciﬁcally, if A is an aﬃnoid algebra, we deﬁne Sp A to5e the space of maximal ideals of A . We call this an aﬃnoid variety , and we can realise A as a ring of K -valued functions on Sp A , where a ( p ) := a + p . This takes values in K since any maximal ideal in the Tate algebra has ﬁnite codimension by [5, Corollary2.2.12].We say that A is the ring of analytic functions on the aﬃnoid variety Sp A . Notethat for any ring homomorphism φ : A → B between aﬃnoid algebras induces a map φ : Sp B → Sp A, q φ − ( q ), continuous with respect to the Zariski topology, and wecall this a morphism of aﬃnoid varieties . Therefore, we can realise aﬃnoid varieties asa category, equivalent to the category of aﬃnoid algebras, via an equivalence where eachvariety is sent to its ring of analytic functions.Aﬃnoid varieties are useful in p -adic analysis, since they can indeed be realised as non-archimedean spaces. Recall that for each ǫ ∈ R , we deﬁne the d -dimensional polydisc ofradius ǫ to be the space D dǫ := { α ∈ K d : v π ( α i ) ≥ ǫ for each i } .When ǫ = 0 we call this the unit disc . We can consider this disc an aﬃnoid space, iso-morphic to Sp K h u , · · · , u d i , and thus all discs are isomorphic, regardless of the radius.Note that the Tate algebra K h u , · · · , u d i is precisely the set of power series convergingon D d , so we can indeed realise the Tate algebra as the ring of analytic functions on theunit disc. Moreover, for each n ∈ N , the subalgebra K h π n u , · · · , π n u d i is precisely thosefunctions which converge on D d − n .Now, recall the following deﬁnition ([5, Deﬁnition 3.3.9]): Deﬁnition 2.1.

Let X be an aﬃnoid variety. A subset U of X is called an aﬃnoidsubdomain if there exists an aﬃnoid variety Y with a morphism α : X → Y such that α ( X ) ⊆ U , and the following universal property is satisﬁed: If β : Z → X is anymorphism of aﬃnoid varieties such that β ( Z ) ⊆ U then there exists a unique morphism γ : Z → Y such that β = α ◦ γ . Note:

1. In the above deﬁnition, it follows from [5, Lemma 3.3.10] that the map α is anembedding of aﬃnoid spaces, and its image is equal to U , so we may sometimes identify Y with U .2. If X is an aﬃnoid variety, then X carries a Grothendieck topology, in the sense of [5,Deﬁnition 5.1.1], where the admissible open subsets are the aﬃnoid subdomains U of X .Roughly speaking, we deﬁne a rigid space to be a space carrying a Grothendieck topology,that is locally equivalent to an aﬃnoid variety. For example. we can consider the aﬃneplane A d ( K ) = K d to be a rigid variety, where its admissible open subsets consist of theunit discs D dǫ and their aﬃnoid subdomains. Indeed, for any aﬃne variety X , a similarprocedure known as analytiﬁcation allows us to realise X as a rigid variety. Recall the following deﬁnition [2, Deﬁnition 2.7]:6 eﬁnition 2.2.

Let V be a K -vector space. • An O -lattice in V is an O -submodule N of V such that T n ∈ N π n N = 0 and N ⊗ O K = V . • The completion of N is the O -module “ N := lim ←− n ∈ N Nπ n N . • The completion of V with respect to N is the K -vector space “ V = “ V N := “ N ⊗ O K . Note that if V is ﬁnite dimensional then “ V = V for any choice of lattice. More generally,every O -lattice N in V induces an exhaustive, separated ﬁltration w N on V given by w N ( u ) := sup { n ∈ Z : u ∈ π n N } .This is the π -adic ﬁltration associated to N , and the topological completion of V withrespect to w N is precisely the completion “ V N .Now, let R be a Noetherian K -algebra, and let R be a Noetherian O -lattice subaglebraof R . Then the completion b R of R with respect to R is a Noetherian, Banach K -algebra. Deﬁnition 2.3.

Let V ⊆ R be a K -vector subspace, and suppose that the lattice M := V ∩ R generates R as an O -algebra. Then R n := Oh π n M i ⊆ R is an O -subalgebralattice in R , and we deﬁne the level n deformation of b R to be b R n , the completion of R with respect to R n . Example: If R is the polynomial ring K [ t , · · · , t r ] and R = O [ t , · · · , t r ], then thecompletion b R of R with respect to R is the Tate algebra K h t , · · · , t r i . Moreover, wecan take the deformed Tate algebra K h π n t , · · · , π n t r i to be the level n deformation of b R .Now, let g be a ﬁnite dimensional K -Lie algebra, and let L be an O -Lie lattice in g , i.e. L is an O -lattice in g and it is closed under the Lie bracket. Note that the envelopingalgebra U ( L ) is an O -lattice in U ( g ). Deﬁnition 2.4.

Deﬁne the aﬃnoid enveloping algebra of L with coeﬃcients in O to be [ U ( L ) , the completion of U ( L ) with respect to its π -adic ﬁltration.Also, deﬁne the aﬃnoid enveloping algebra of L with coeﬃcients in K to be [ U ( L ) K := [ U ( L ) ⊗ O K .This is the completion of U ( g ) with respect to the π -adic ﬁltration associated to U ( L ) . Note:

We can take the level n deformation of [ U ( L ) K to be \ U ( π n L ) K .The following lemma is a straightforward consequence of the Poincarr´e-Birkoﬀ Witt the-orem, see e.g. [13, Proposition 2.5.1] for the proof. Lemma 2.1.

If we let { x , · · · , x d } be a K -basis for g which forms an O -basis for L ,then [ U ( L ) is isomorphic as an O -module to the Tate algebra Oh x , · · · , x d i , and hence [ U ( L ) K is isomorphic to K h x , · · · , x d i as a K -vector space. M be a g -representation, i.e. a U ( g )-module, and let N be an O -lattice in M such that L · N ⊆ N , and suppose that N is π -adically complete. Then it follows that M has the structure of a [ U ( L ) K -module. Unless otherwise stated, we will assume thatall modules are left modules. Proposition 2.2.

Let M be a ﬁnitely generated [ U ( L ) K -module. Then M contains an O -lattice N such that M is complete with respect to N and L · N ⊆ N .Proof. M = [ U ( L ) K m + · · · + [ U ( L ) K m s , so let N := [ U ( L ) m + · · · + [ U ( L ) m s , clearly N is an O -lattice in M .Note that since [ U ( L ) is π -adically complete and gr [ U ( L ) ∼ = O π O [ t, u , · · · , u d ] is Noethe-rian, it follows from [14, Ch.II, Theorem 2.1.2] that the π -adic ﬁltration on [ U ( L ) isZariskian, and hence any left submodule of [ U ( L ) s is closed in [ U ( L ) s .Since N ∼ = \ U ( L ) s J for some left submodule J of [ U ( L ), it follows that N is π -adicallycomplete. Let h be a ﬁnite dimensional Lie algebra over k . First recall the following deﬁnition [7,1.12.8]: Deﬁnition 2.5.

Given λ ∈ h ∗ , deﬁne a polarisation of h at λ to be a solvable subalgebra b of h such that for all u ∈ h , λ ([ u, b ]) = 0 if and only if u ∈ b . In particular, if b is a polarisation of h at λ , then λ ([ b , b ]) = 0, i.e. λ restricts to acharacter of b . Note that polarisations need not always exist. Lemma 2.3.

Given λ ∈ h ∗ , let h λ be the subalgebra { u ∈ h : λ ([ u, h ]) = 0 } of g . If b isa polarisation of h at λ , then: • h λ ⊆ b . • dim F b = (dim F h + dim F h λ ) . • If b ′ is a subalgebra of h such that λ ([ b ′ , b ′ ]) = 0 and dim F b ′ = (dim F h +dim F h λ ) ,then b ′ is a polarisation of h at λ . • b contains every ideal a of h such that λ ([ h , a ]) = 0 . In particular, b contains Z ( h ) and every ideal a such that λ ( a ) = 0 .Proof. The ﬁrst statement follows from the deﬁnition of a polarisation, since λ ([ b , h λ ]) =0, while the second and third follow from [7, 1.12.1]. The ﬁnal statement follows fromthe ﬁrst since if λ ([ a , h ]) = 0 then a ⊆ h λ ⊆ b . Examples:

1. If λ is a character of h , i.e. λ ([ h , h ]) = 0, then h is a polarisation of h at λ . In particular, if h is abelian, then for any λ ∈ h ∗ , h is a polarisation of h at λ .8. If h = a ⋊ kx for some abelian subalgebra a of h , x ∈ h , then for any λ ∈ h ∗ , if λ ([ h , h ]) = 0 then a is a polarisation of h at λ .3. If k is algebraically closed, and h is semisimple, then λ ∈ h ∗ has a polarisation b if andonly if λ is regular in the sense of [7, 1.11.6]. In this case b is a Borel subalgebra of h by[7, Proposition 1.12.18], and hence λ is a character of a Cartan subalgebra.In our case, we will be interested in the case where h is solvable or nilpotent. Recall that h is completely solvable if there exists a chain of ideals 0 = h ⊆ h ⊆ · · · ⊆ h d = h suchthat dim k h i = i for each i , e.g. if h is nilpotent. Note that for any solvable Lie algebra h , we can choose a ﬁnite extension F/k such that h ⊗ k F is completely solvable. Proposition 2.4.

Suppose that h is completely solvable. Then given λ ∈ h ∗ , and an ideal a of g such that λ ([ a , a ]) = 0 , there exists a polarisation b of h at λ such that a ⊆ b .Proof. Choose a chain of ideals 0 = h ⊆ h ⊆ · · · ⊆ h d = h such that dim k h i = i for each i , and we may choose this chain such that h j = a for some j .Setting λ i := λ | h i for each i , the subalgebra b := h λ + · · · + h λ d d is a polarisation of h at λ by [7, Proposition 1.12.18]. Moreover, since λ ([ a , a ]) = 0, it follows that h λ j j = a λ j = a ,and hence a ⊆ b as required.In particular, taking a = 0, this result ensures that polarisations always exist if h iscompletely solvable. In this paper, we mainly focus on the case where h is nilpotent.Now, for any k -Lie algebra h , λ ∈ h ∗ , and any polarisation b of h at λ , since λ restricts toa character of b , it follows that k λ := kv is a U ( b )-module via the b -action x · v := λ ( x ) v .This gives us the following deﬁnition: Deﬁnition 2.6.

Let λ ∈ h ∗ , and let b be a polarisation of h at λ . We deﬁne the b - Dixmiermodule of h at λ to be the U ( h ) -module: D ( λ ) = D ( λ ) b := U ( h ) ⊗ U ( b ) k λ (1)Note that D ( λ ) is a cyclic U ( h ) module, generated by a vector v λ on which U ( b ) acts byscalars.This deﬁnition is useful, because in the case where k is algebraically closed and h issemisimple, these Dixmier modules are precisely the well-known Verma modules, whichare fundamental within the representation theory of semisimple Lie algebras. So we maythink of Dixmier modules as a generalisation of Verma modules. Examples:

1. If h is semisimple, k is algebraically closed, the Verma module D ( λ ) hasa unique simple quotient L ( λ ), known as a simple highest weight module with weight λ .2. If h is abelian, or more generally if λ is a character of h , then D ( λ ) = k always.3. If h = a ⋊ kx , for a abelian, then if λ is not a character of h , D ( f ) ∼ = k [ t ], where x acts by t , and each u ∈ a acts by a polynomial in k [ ddt ].9 heorem 2.5 ([7, Theorem 6.1.1]) . Let h be solvable, and let λ ∈ h ∗ . Then there existsa polarisation b of h at λ such that D ( λ ) b is an irreducible U ( h ) -module. The following result yields a complete description of primitive ideals in U ( h ): Theorem 2.6 ([7, Theorem 6.1.7]) . Let h be solvable, and let P be a primitive ideal of h . Then there exists a ﬁnite extension F/k , λ ∈ h ∗ F and a polarisation b of h F at λ suchthat P = Ann U ( h ) D ( λ ) b . Now we will return to the p -adic case, i.e. K/ Q p is a ﬁnite extension, g is a K -Lie algebraand L is an O -lattice in g . We want to extend the conclusion of Theorems 2.5 and 2.6 tothe aﬃnoid setting.In this case, when we choose λ ∈ g ∗ , we will always assume further that λ ( L ) ⊆ O ,or in other words λ ∈ L ∗ = Hom O ( L , O ).Firstly, note that for any polarisation b of g at λ , if we set B := b ∩ L , then B is an O -Lie lattice in b . Furthermore, if we let K λ := Kv be the one dimensional U ( b )-moduleinduced by λ , then since π n U ( B ) v ⊆ p n O v , it follows that and K λ carries the structureof a [ U ( B ) K -module. Deﬁnition 2.7.

Let λ ∈ g ∗ such that λ ( L ) ⊆ O , and let b be a polarisation of g at λ .Deﬁne the b - aﬃnoid Dixmier module of L at λ to be the [ U ( L ) K -module deﬁned by: [ D ( λ ) = [ D ( λ ) b := [ U ( L ) K ⊗ \ U ( B ) K K λ (2) Notation:

If it is unclear what the ground ﬁeld K is, we may sometimes write [ D ( λ ) K for [ D ( λ ). Also, if it is unclear which lattice L we are considering, we may sometimeswrite [ D ( λ ) B instead of [ D ( λ ) b .Note that as in the classical case, [ D ( λ ) is a cyclic [ U ( L ) K -module, so [ D ( λ ) b = [ U ( L ) K v λ ,and [ U ( B ) K acts by scalars on v λ .In particular, using Proposition 2.2, we see that [ D ( λ ) is π -adically complete withrespect to some lattice. In fact it is a π -adic completion of the classical Dixmier module D ( λ ). Examples:

1. If g is split semisimple with Borel subalgebra b , the aﬃnoid Verma mod-ule [ V ( λ ) arises as a Dixmier module. This still has a unique simple quotient [ L ( λ ).2. If g is abelian, or more generally if λ is a character of g , then [ D ( λ ) = K always.3. If g = a ⋊ Kx , for a abelian, then if λ is not a character of g , [ D ( f ) ∼ = K h t i , where x acts by t , and each u ∈ a acts by a polynomial in K [ ddt ]. The following deﬁnition ([7, 4.7.7])) will be very useful to us throughout.

Deﬁnition 2.8.

Let h be a k -lie algebra. A reducing quadruple of h is a 4-tuple ( x, y, z, h ′ ) where: = x, y, z ∈ h and h ′ is an ideal of h of codimension 1, • y, z ∈ h ′ and x / ∈ h ′ , • z is central in h and y is central in h ′ , • [ x, y ] = αz for some = α ∈ k . The following results link reducing quadruples to polarisations, and they can be found inthe proof of [7, Theorem 6.1.1] and [7, Theorem 6.1.4]:

Proposition 2.7.

Suppose that h is nilpotent, dim( h ) > , and let λ ∈ h ∗ . If we assumethat λ ( a ) = 0 for all non-zero ideals a of g , then: ( i ) There exist x, y, z ∈ g and g ′ E g such that ( x, y, z, g ′ ) forms a reducing quadruple for g . ( ii ) If µ := λ | g ′ and b ⊆ g ′ is a polarisation of g ′ at µ , then b is a polarisation of g at λ .Proof. ( i ) Firstly, for any 0 = z in the centre of g , Kz is an ideal of g , and hence λ ( z ) = 0.Therefore, λ : Z ( g ) → K is injective, meaning that Z ( g ) must have dimension 0 or 1.But since g is nilpotent, the centre cannot be 0, hence Z ( g ) = Kz for some z ∈ Z ( g ).Thus note that g is non-abelian since dim( g ) > Z ( g ).Also using nilpotency of g we may choose y ∈ g such that y / ∈ Z ( g ) and [ y, g ] ⊆ Z ( g ) = Kz . Thus the linear map ad( y ) : g → g has rank 1, and hence it must have kernel ofdimension dim( g ) −

1. Let g ′ := ker(ad( y )), and it follows that g ′ is an abelian ideal in g of codimension 1.So, g = g ′ ⊕ Kx for some x ∈ g such that [ x, y ] = 0, and hence [ x, y ] = αz for some α ∈ K with α = 0. Hence ( x, y, z, g ′ ) is a reducing quadruple as required.( ii ) Since b is a polarisation of g ′ at µ , it must contain Z ( g ′ ) by Lemma 2.3, hence y, z ∈ b .Clearly b ⊆ g ′ is a solvable subalgebra of g , so suppose that b ⊆ V ⊆ g with λ ([ V, V ]) = 0.We will prove that V ⊆ g ′ , and it will follow that V = b by the deﬁnition of a polarisation,thus proving that b is a polarisation of g at λ .Suppose that βx + u ∈ V for some u ∈ g ′ , β ∈ K . Then since y ∈ b ⊆ V , it follows that λ ([ βx + u, y ]) = 0, and hence βαλ ( z ) = 0. But since α, λ ( z ) = 0, it follows that β = 0and hence V ⊆ g ′ as required. Proposition 2.8.

Suppose that h is nilpotent with n := dim F h > , and suppose that h has a reducing quadruple ( x, y, z, h ′ ) . We assume further that λ ∈ h ∗ and λ ( a ) = 0 for allnon-zero ideals a of h . Then for any polarisation b of h at λ , there exists a polarisation b ′ at λ , contained in h ′ , and a proper subalgebra t ( h such that b , b ′ ⊆ t .Proof. Firstly, note that h λ = { u ∈ h : λ ([ u, h ]) = 0 } is contained in b , otherwise b ( b + h λ and λ ([ b + h λ , b + h λ ]) = 0, contradicting the deﬁnition of a polarisation.11f b ⊆ h ′ , then taking b ′ = b , t = h ′ , the statement is trivially true, so we may assumethat b h ′ .Since y is central in h ′ but not in h , it is clear that h ′ = ker(ad( y )) and F z = im(ad( y )).So since there exists u ∈ b \ h ′ , we have that [ u, y ] = 0, i.e. [ u, y ] = βz for some0 = β ∈ F . But since F z is a non-zero ideal of h , λ ( z ) = 0, and thus λ ([ u, y ]) = 0.So since λ ([ b , b ]) = 0 and u ∈ b , this means that y / ∈ b .Let b ′ := ( b ∩ h ′ ) ⊕ F y . This is a subalgebra of h , and clearly it is contained in h ′ .Also, b ∩ h ′ has codimension 1 in b , therefore dim F b ′ = dim F b ∩ h ′ + 1 = dim F b . But itis clear that λ ([ b ′ , b ′ ]) = 0, so by Lemma 2.3, this means that b ′ is a polarisation of h at λ .Now, let t := b ⊕ F y . Since

F z = [ y, h ] ⊆ b , this is a subalgebra of h , and clearlyit contains b and b ′ , so we only need to prove that t = h , so assume for contradic-tion that t = h . This means that b has codimension 1 in h , so dim F b = n −

1. Butdim F b = ( n + dim F h λ ) by Lemma 2.3, and thus dim F h λ = n − βx + γy ∈ h λ , then βλ ([ x, y ]) = γλ ([ y, x ]) = 0, which is only possibleif β = γ = 0 since λ ( z ) = λ ( α − [ x, y ]) = 0. So Span F { x, y } ∩ h λ = 0, and therefore, h = F x ⊕ F y ⊕ h λ .Let a := ker( λ ) ∩ h λ . Then since z ∈ h λ and λ ( z ) = 0, it follows that a has codimension1 in h λ , which means that dim F a = n − λ ( a ) = 0, so we will ﬁnish by proving that a is an ideal of h , and this willimply that a = 0, and hence n − n = dim F h = 3 – contradicting our assumption.By the deﬁnition of h λ , it is clear that λ ([ h λ , h ]) = 0 and so [ h λ , a ] ⊆ h λ ∩ ker( λ ) = a . Sosince h = F x ⊕ F y ⊕ h λ , it remains to prove that [ y, a ] ⊆ a and [ x, a ] ⊆ a .Since a ⊆ h ′ , we have that [ y, a ] = 0 ⊆ a , and if we choose u ∈ b such that u / ∈ h ′ , then b = h λ ⊕ F u , so since h is nilpotent and b is a subalgebra, it follows that [ u, h λ ] ⊆ h λ , andhence [ u, a ] ⊆ h λ . Also, since λ ([ b , b ]) = 0, it follows that [ u, a ] ⊆ ker( λ ), hence [ u, a ] ⊆ a .But since u / ∈ h ′ , we have that u = βx + γy , where β = 0, so it follows immediatelythat [ x, a ] ⊆ a as required.Reducing quadruples will play an important role in many of the proofs in this paper,since they allow us to use an inductive strategy commonly employed by Dixmier in [7],which we outline below, very roughly: Dixmier’s Induction Strategy : We have a statement P involving a nilpotent k -Liealgebra h and a linear form λ : h → k . • Step 1: The base case is where dim k h = 1, this case should be straightforward.So we can assume that dim k h > P is true for all nilpotent Lie algebras ofdimension less than dim k h . • Step 2: If there exists a non-zero ideal a of h such that λ ( a ) = 0, then we can replace h by ha and apply induction. So we may assume that λ ( a ) = 0 for all non-zero ideals a of g . 12 Step 3: Applying Proposition 2.7, we can choose a reducing quadruple ( x, y, z, h ′ )of h . Since we know that P holds for h ′ and λ | h ′ by induction, we can hopefullyinduce up to h . Let us ﬁrst recall the deﬁnition of the Weyl algebra, given in [16, Section 1.3]:

Deﬁnition 2.9.

Let d ∈ N . We deﬁne the d ’th Weyl algebra A d ( k ) to be the k -algebrain d -variables x , · · · , x d , y , · · · , y d satisfying the following relations for ≤ i, j ≤ d , i = j : x i x j − x j x i = 0 , y i y j − y j y i = 0 , x i y j − y j x i = 0 , x i y i − y i x i = 1 . Lemma 2.9. A d ( k ) is isomorphic as a k -vector space to the polynomial ring k [ x , · · · , x d , ∂ , · · · , ∂ d ] ,subject to the relations that x , · · · , x d commute, ∂ , · · · , ∂ d commute, x i ∂ j = ∂ j x j if i = j and x i ∂ i = ∂ i x i + 1 . Note:

It is well known and quite straightforward to prove that A n ( k ) is a simple k -algebra domain.The Weyl algebra is of great importance within algebraic and diﬀerential geometry, sinceit can be realised as the ring of diﬀerential operators on the aﬃne plane A dk . The followingresult highlights the usefulness of this object within representation theory, since it canbe explicitly realised as a ring of endomorphisms: Lemma 2.10.

Let k [ t , · · · , t d ] be a polynomial ring in d -variables, and for each i =1 , · · · , d , let x i ∈ End k k [ t , · · · , t d ] be left multiplication by t i , and let ∂ i := ddt i ∈ End k k [ t , · · · , t d ] . Then the natural map A d ( k ) → End k k [ t , · · · , t d ] sending x i to x i and ∂ i to ∂ i is an injective ring homomorphism. Now, if we ﬁx h a nilpotent k -Lie algebra, the following result of Dixmier ([7, Theorem4.7.9]) shows us that all weakly rational ideals in U ( h ) are maximal, which completes theDixmier-Moeglin equivalence in the nilpotent case: Theorem 2.11.

Let I be a two-sided of U ( h ) such that Z Ä U ( h ) I ä = k , then there exists d ∈ N such that U ( h ) I ∼ = A d ( k ) . The integer d is called the weight of I . The proof of this Theorem is inductive, not constructive, so we cannot usually explicitlywrite down this isomorphism. However, using Theorem 2.6, we know that all weaklyrational ideals of U ( h ) arise as annihilators of Dixmier modules, and using this notionthere are cases where we can construct an isomorphism.Speciﬁcally, if I = Ann U ( h ) D ( λ ) for some linear form λ of h , with polarisation b ofcodimension d , then we know that U ( h ) I is a ring of k -linear endomorphisms of D ( λ ) ∼ = k [ u , · · · , u d ]. Since we know from Lemma 2.10 that A d ( k ) can be realised as a ring ofendomorphisms of k [ u , · · · , u d ], we want to show that these to rings can be identiﬁed.Roughly speaking, we want to show that elements in the polarisation b act by poly-nomials in ∂ , · · · , ∂ d , and that each basis vector u i of g / b acts by x i . In the next section,we will see how to do this explicitly.Now, we want to try and generalise Theorem 2.11 to the aﬃnoid case, so we need todeﬁne an aﬃnoid version of the Weyl algebra:13 eﬁnition 2.10. Let A d ( O ) be the O -lattice subalgebra of A d ( K ) consisting of polynomi-als in x , · · · , x d , ∂ , · · · , ∂ d with coeﬃcients in O . We deﬁne the d ’th Tate-Weyl algebra ,denoted \ A d ( K ) to be the completion of A d ( K ) with respect to A d ( O )Similarly to A d ( K ), the Tate-Weyl algebra \ A d ( K ) is also a simple domain. Lemma 2.12.

The d ’th Tate-Weyl algebra \ A d ( K ) is isomorphic as a K -vector space to K h x , · · · , x d , ∂ , · · · , ∂ d i , where x , · · · , x d commute, ∂ , · · · , ∂ d commute, x i ∂ j = ∂ j x j if i = j and x i ∂ i = ∂ i x i + 1 .Moreover, the level n deformation \ A d ( K ) n of \ A d ( K ) is the subalgebra consisting theelements of K h π n x , · · · , π n x d , π n ∂ , · · · , π n ∂ d i . Using this deﬁnition, we have the following aﬃnoid version of Lemma 2.10:

Lemma 2.13.

Let K h t , · · · , t d i be a Tate algebra in d -variables, and for each i =1 , · · · , d , let x i ∈ End K K h t , · · · , t d i be left multiplication by t i , and let ∂ i := ddt i ∈ End K K h t , · · · , t d i . Note that if λ α ∈ K for each α ∈ N d , and λ α → as α → ∞ ,then the series P α ∈ N d λ α x α · · · x α d d ∂ α d +1 · · · ∂ α d d is a well deﬁned K -linear endomorphism of K h t , · · · , t d i .Then the natural map A d ( k ) → End K K h t , · · · , t d i sending x i to x i and ∂ i to ∂ i is aninjective ring homomorphism. In a similar vein to Theorem 2.11, we would like to prove that all weakly rationalquotients of the aﬃnoid enveloping algebra [ U ( L ) K are isomorphic to Tate-Weyl algebras.However, it was shown in [13] that this need not always be the case. In section 7 we willexplore how to prove a related, weaker statement. Now we will recall some Lie theory. Assume h is nilpotent Lie algebra, and note that forevery u ∈ h , the map ad( u ) is a nilpotent derivation of h . So we can deﬁne:exp(ad( u )) := X n ≥ n ! ad( u ) n : h → h (3)Since ad( u ) is a derivation, it follows that exp(ad( u )) is a Lie-automorphism of h . Deﬁnition 2.11.

Deﬁne the adjoint algebraic group of h to be H ( h ) := { exp(ad( u )) ∈ Aut ( h ) : u ∈ h } Then H ( h ) is a subgroup of Aut ( h ), and if we deﬁne the functor H : k -Alg → Grp, R H ( h ⊗ k R ), then H is an aﬃne algebraic group in the sense of [10, Deﬁnition I.2.1], andit is unipotent. Note:

If we view the space ad( h ) ⊆ End k ( h ) as an aﬃne variety over k , then the mapexp : ad( h ) → H is an isomorphism of varieties, with inverse log.Now, let h ∗ : k -Alg → Set be the linear dual of h , i.e. h ∗ ( R ) = Hom R ( h ⊗ k R, R ) ∼ = h ∗ ( k ) ⊗ k R . Then h ∗ is an aﬃne scheme in the sense of [10, Deﬁnition I.1.3].14 eﬁnition 2.12. Deﬁne an action of H on g ∗ , i.e. a morphism of varieties H × h ∗ → h ∗ ,by ( g · f )( u ) = f ( g − u ) . This is the coadjoint action , and the orbits of this action in h ∗ are called coadjoint orbits . Given λ ∈ h ∗ ( k ), let X be the coadjoint orbit of λ in h ∗ , and let S be the stabiliser of λ in H , i.e. S ( R ) := { g ∈ H ( R ) : g · λ = λ } , an aﬃne algebraic subgroup of H . Lemma 2.14.

There exists an isomorphism of varieties H ∼ = S × X such that the map H → X, g g · λ is just the natural projection S × X → X .Proof. Since H is aﬃne and S is closed in H , we see using [4, Theorem II.6.8] that thequotient variety H /S exists. Since the orbit map H → X is surjective and k has char-acteristic 0, it follows that this map is separable in the sense of [4, I.8.2], and henceusing [4, Theorem I.17.3] and [4, Proposition II.6.7] it follows that X = H /S and thatthe map H → X is the quotient map. Since X is closed in h ∗ , it follows that H /S is aﬃne.Now, using [10, I.5.6(1)] we see that H × S ∼ = H × H /S H as varieties, and using thisisomorphism, it follows that the natural map H → H /S is an S -torsor in the sense of [6,Ch.III Deﬁnition 4.1.3].Therefore, since S is unipotent and H /S is aﬃne, it follows from [6, Ch.IV Proposition3.7(b)] that the torsor H → H /S is trivial, i.e. H ∼ = S × H /S = S × X as varieties andthe map H → X is just the projection to the second factor. ’ U ( L ) K on ’ D ( λ ) In this section, we will prove our aﬃnoid version of Theorem 2.5, at least in the casewhere g is nilpotent. Throughout, assume that g is a ﬁnite dimensional K -Lie algebra,with O -Lie lattice L . Since the aﬃnoid Dixmier module [ D ( λ ) is an induced [ U ( L ) K -module, we will ﬁrst exploresome general properties of induced modules. Lemma 3.1.

Let h be a subalgebra of g , let a be an ideal of g such that a ⊆ h . Let g := g / a , h := h / a . Also, set H := L ∩ h , A := L ∩ a , L := L / A , H := H / A , whichare Lie lattices in h , a , g and h respectively. Then: ( i ) There is a continuous surjection of K -algebras [ U ( L ) K ։ \ U ( L ) K induced by the sur-jection L ։ L . The kernel of this surjection is a [ U ( L ) K ( ii ) If M is a ﬁnitely generated \ U ( H ) K -module, then M has the structure of a \ U ( H ) K -module via the surjection in ( i ) , and [ U ( L ) K ⊗ \ U ( H ) K M ∼ = \ U ( L ) K ⊗ \ U ( H ) K M as [ U ( L ) K -modules.Proof. ( i ) It is clear that the surjection L ։ L induces a surjection U ( L ) ։ U ( L )sending π n U ( L ) to π n U ( L ), so this yields a continuous map [ U ( L ) → \ U ( L ).15f we ﬁx a basis { x , · · · , x d } for L such that { x r +1 , · · · , x d } is a basis for A , then usingLemma 2.1, we see that every element of \ U ( L ) K has the form P α ∈ N r λ α ( x + A ) α · · · ( x r + A ) α r ,where λ α ∈ O → α → ∞ . Clearly under the map [ U ( L ) → \ U ( L ), x i maps to x i + A for each i , and hence the map is surjective.Moreover, we can write any element of [ U ( L ) as P α ∈ N r x α · · · x α r r c α for some c α ∈ [ U ( A )converging to zero, and this maps to 0 if and only if c α maps to 0 for each α . But each c α has the form P β ∈ N d − r µ β x β r +1 r +1 · · · x β d d , and this maps to zero if and only if µ = 0, i.e. c α ∈ [ U ( A ) A . Hence the kernel of the surjection is [ U ( L ) A and part ( i ) follows.( ii ) Let φ : [ U ( L ) K → \ U ( L ) K be the surjection from part ( i ), and deﬁne a map:Θ : [ U ( L ) K ⊗ \ U ( H ) K M → \ U ( L ) K ⊗ \ U ( H ) K M, r ⊗ m φ ( r ) ⊗ m .It is clear that this is a well deﬁned map of [ U ( L ) K -modules, we want to prove that it isan isomorphism.Every element s ∈ \ U ( L ) K can be written uniquely in the form s = P α ∈ N r λ α ( x + A ) α · · · ( x r + A ) α r = φ ( P α ∈ N r λ α x α · · · x α r r ),so there is a unique element in K h x , · · · , x r i that maps onto s under φ . We call this ele-ment φ − ( s ), and it is clear that this deﬁnes a K -linear map φ − : \ U ( L ) → K h x , · · · , x r i .Therefore, we can deﬁne a K -linear map Ψ : \ U ( L ) K ⊗ \ U ( H ) K M → [ U ( L ) K ⊗ \ U ( H ) K M sending s ⊗ m to φ − ( s ) ⊗ m . We can show that this is well deﬁned by choosing anappropriate basis for H that extends to a basis for L , and clearly it is a right inverseto Θ.Using the fact that [ U ( L ) K is isomorphic as a K -vector space to K h x , · · · , x r ih x r +1 , · · · , x d i ,and x r +1 , · · · , x d ∈ A ⊆ H , we see that every simple tensor s ⊗ m ∈ [ U ( L ) K ⊗ \ U ( H ) K M can be written as an inﬁnite sum of simple tensors s n ⊗ m n converging to zero as n → ∞ ,with s n ∈ K h x , · · · , x r i . We know this sum converges by Proposition 2.2.Therefore, for any s ∈ [ U ( L ) K , m ∈ M , ΨΘ( s ⊗ m ) = P n ∈ N ΨΘ( s n ⊗ m n ) = P n ∈ N Ψ( φ ( s n ) ⊗ m n ),and since s n ∈ K h x , · · · , x s i for each n , φ − ( φ ( s n )) = s n , and hence ΨΘ( s ⊗ m ) = s ⊗ m .Thus Ψ and Θ are mutually inverse bijections.Now, recall from [1] that if A is a Banach K -algebra, and M is a left A -module, π -adicallycomplete with respect to some lattice N ⊆ M , then we may deﬁne the Tate module : M h t , · · · , t d i := { P α ∈ N d t α · · · t α d d s α : s α ∈ M, s α → | α | → ∞} .Note that we don’t necessarily give M h t , · · · , t d i the structure of an A -module, a prioriit is just a K -vector space. 16 roposition 3.2. Let h be a subalgebra of g , and let H := h ∩ L , so H is a Lie latticein h . Suppose that M is a ﬁnitely generated \ U ( H ) K -module. Then if r = dim K g / h ,there is an isomorphism of K -vector spaces [ U ( L ) K ⊗ \ U ( H ) K M ∼ = M h t , · · · , t r i , where t i v corresponds to x i ⊗ v for some O -basis { x , · · · , x r } for L / H . Thus M h t , · · · , t r i carriesthe structure of a [ U ( L ) K module.Moreover, if r = 1 , so L = H ⊕ O x for some x ∈ L , then we can choose this isomorphism [ U ( L ) K ⊗ \ U ( H ) K M ∼ = M h t i such that: ( i ) x acts by t on M h t i . ( ii ) If y, z ∈ H act on M by scalars β y , β z ∈ O , [ x, z ] = 0 and [ y, x ] = αz for some α ∈ K , then y acts on M h t i by αβ z ddt + β y . ( iii ) If α, β z = 0 and M is irreducible over \ U ( H ) K , then M h t i is irreducible over [ U ( L ) K .Proof. Let { x , · · · , x d } be an O -basis for L such that { x r +1 , · · · , x d } is a basis for H .Then by Lemma 2.1, writing x α := x α · · · x α d d , we have: [ U ( L ) K = { P α ∈ N d λ α x α : λ α ∈ O , λ α → | α | → ∞} .Deﬁne a map:Θ : [ U ( L ) K ⊗ \ U ( H ) K M → M h t , · · · , t r i , P α ∈ N d λ α x α ⊗ v P β ∈ N r t β · · · t β r r ( P γ ∈ N d − r λ ( β,γ ) x γ v ).Note that here ( β, γ ) refers to the d -tuple whose ﬁrst r terms are the terms of β , and thelast d − r terms are the terms of γ . It is straightforward but technical to show that thisis a well deﬁned K -linear map, so we need to prove that it is an isomorphism.Firstly, M = \ U ( H ) K v + · · · + \ U ( H ) K v t , so any element of M h t , · · · , t r i will have theform P β ∈ N r t β · · · t β r r ( a ,β v + · · · + a t,β v t ) for some a i,β ∈ \ U ( H ) K . This is the image of P β ∈ N r x β · · · x β r r a ,β ⊗ v + · · · + P β ∈ N r x β · · · x β r r a t,β ⊗ v t , so Θ is surjective.Furthermore, if P β ∈ N r t β · · · t β r r ( a ,β v + · · · + a t,β v t ) = 0 then a ,β v + · · · + a t,β v t = 0 for all β . Since [ U ( L ) K ⊗ \ U ( H ) K M is ﬁnitely generated, it is complete by Proposition 2.2, thus X β ∈ N r x β · · · x β r r a ,β ⊗ v + · · · + X β ∈ N r x β · · · x β r r a t,β ⊗ v t = X β ∈ N r x β · · · x β r r ⊗ ( a ,β v + · · · + a t,β v t ) = 0 , (4)hence Θ is injective. 17ence Θ is an isomorphism of K -vector spaces, and Θ( x i ⊗ v ) = t i v for all i ≤ r , v ∈ M .So clearly we can deﬁne an action of [ U ( L ) K on M h t , · · · , t r i making Θ into an isomor-phism of [ U ( L ) K -modules.( i ) Since Θ( x n v ) = t n v for all v ∈ M , n ∈ N , it is clear that the action of x on M h t i isgiven by multiplication by t .( ii ) Since y and z act by scalars on M , their action on M h t i is determined entirely bytheir action on the powers of t .Since [ x, z ] = 0, it follows that z commutes with all powers of x , and hence z · t n v = z Θ( x n ⊗ v ) = Θ( x n ⊗ z · v ) = Θ( β z ( x n ⊗ v )) = β z t n v , so z acts on M h t i via β z .Clearly y · v = β y v , so we will assume that for some n ≥ y · t n v = nαβ z t n − v + β y t n v and show that y · t n +1 v = ( n + 1) αβ z t n v + β y t n +1 v , and it will follow using induction that y acts by αβ z ddt + β y : y · t n +1 v = y Θ( x n +1 ⊗ v ) = Θ( yx n +1 ⊗ v ) = Θ(([ y, x ] x n + xyx n ) ⊗ v )= αz Θ( x n ⊗ v ) + xy Θ( x n ⊗ v ) = αzt n v + xyt n v = αβ z t n v + xnαβ z t n − v + xβ y t n v = ( n + 1) αβ z t n v + β y t n +1 v (5)( iii ) Let ∂ := ddt , and let ρ : [ U ( L ) K → End K ( M h t i ) be the action, then by part ( ii ), ρ ( y ) = αβ z ∂ + β y , so ∂ = ( αβ z ) − ( ρ ( y ) − β y ) = ρ (( αβ z ) − ( y − β y )) ∈ im( ρ ).Hence for each n ∈ N , ∂ [ n ] = n ! ∂ n ∈ im( ρ ).So, suppose that 0 = T ≤ M h t i is a submodule, i.e there exists P m ≥ t m s m ∈ T , s m ∈ M , s m not all zero, s m → m → ∞ .Then since ∂ [ n ] ( T ) ⊆ T for all n , it follows that P m ≥ n (cid:0) mn (cid:1) t m − n s m ∈ T , hence we may assumethat s = 0.Set s := s ∈ M \{ } , and deﬁne a sequence of elements in T by r := P m ≥ t m s m , and for i > r i := r i − − t i ∂ [ i ] ( r i − ).Now, if r i = s + P m>i t m s i,m , then t i ∂ [ i ] ( r i ) = P m>i t m (cid:0) mi (cid:1) s i,m , so r i +1 = r i − t i ∂ [ i ] ( r i ) = s + P m>i +1 t m ( s i,m − (cid:0) mi (cid:1) s i,m ).So inductively, we get that for each i ∈ N , r i = s + P m>i s i,m t m for some s i,m ∈ M with v ( s i,m ) ≥ v ( s i − ,m ). It follows easily that r i → s in M h t i as i → ∞ .But since M h t i is ﬁnitely generated over [ U ( L ) K , which is Noetherian, it follows that T is ﬁnitely generated over [ U ( L ) K , and hence T is π -adically complete by Proposition 2.2.18herefore, since each r i ∈ T , this means that s ∈ T . So 0 = s ∈ M ∩ T , and thus M ∩ T = 0. But since M is irreducible, M ∩ T = 0 or M , hence M ∩ T = M .It follows that M h t i = [ U ( L ) K ⊗ \ U ( L ′ ) K M ⊆ T , and hence T = M h t i . Since our choice of T was arbitrary, this implies that M h t i is irreducible as required. From now on, we will assume that g is nilpotent. We will now examine the action of [ U ( L ) K on [ D ( λ ) more closely, and write down an explicit formula for the action of ele-ments of g .First, let λ : g → K be a linear form such that λ ( L ) ⊆ O , and let b be a polarisation of g at λ with B = b ∩ L . Fix a basis { u , · · · , u r } for L / B , and it follows from Proposition3.2 that [ D ( λ ) b ∼ = K h u , · · · , u r i as a K -vector space.So, for each i = 1 , · · · , r , recall from Section 2.4 that we can deﬁne the endomorphisms x i , ∂ i ∈ End K [ D ( λ ) by x i ( f ) := u i f and ∂ i ( f ) := dfdu i . Note that for each u ∈ g , we canwrite u = v u + α ,u u + · · · + α r,u u r for some unique v u ∈ b , α i,u ∈ K . Using this, wedeﬁne: µ : g → End K [ D ( λ ) b , u λ ( v u ) + α ,u x + · · · + α r,u x r . (6)Naively, one might think that this deﬁnes the action of g on [ D ( λ ), but this is not true.However, the following result gives the explicit formula which allows us to completelydescribe this action: Proposition 3.3.

Let ρ : [ U ( L ) K → End K [ D ( λ ) b be the natural action of [ U ( L ) K on [ D ( λ ) b , then we may choose a basis { u , · · · , u r } for L / B such that for every u ∈ g , theaction of u is given by: ρ ( u ) = P α ∈ N r α ! · · · α r ! µ (ad( u r ) α r · · · ad( u ) α ( u )) ∂ α · · · ∂ α r r Note:

This is a ﬁnite sum, since g is nilpotent. Also, we deﬁne ad( x )( y ) := [ y, x ] , asopposed to the more conventional ad( x )( y ) = [ x, y ] .Moreover, if u ∈ b and ad( g ) n ( u ) ⊆ b for all n ∈ N , then this formula holds for anychoice of basis.Proof. Since g is nilpotent, we can choose a basis { u , · · · , u r } for L / B such that [ u i , g ] ⊆ b ⊕ Span K { u i + 1 , · · · , u r } for all i , and we will ﬁx such basis throughout the proof.So, deﬁne ρ ′ : g → End K [ D ( λ ) , u P α ∈ N r α ! · · · α r ! µ (ad( u r ) α r · · · ad( u ) α ( u )) ∂ α · · · ∂ α r r .We want to prove that ρ ( u ) = ρ ′ ( u ) for all u ∈ g , i.e. that ρ ( u )( f ) = ρ ′ ( u )( f ) for all f ∈ [ D ( λ ).Firstly, since by Lemma 3.2, every element of [ D ( λ ) can be written as a Tate power seriesin the variables u β · · · u β r r , it suﬃces to prove that ρ ( u )( u β · · · u β r r ) = ρ ′ ( u )( u β · · · u β r r )19or every β , · · · , β d ∈ N .Note that ρ ′ ( u )( u β · · · u β r r ) = P α ≤ β (cid:0) βα (cid:1) µ (ad( u r ) α r · · · ad( u ) α ( u )) u β − α · · · u β r − α r r , where α ≤ β means that α i ≤ β i for all i , and (cid:0) βα (cid:1) := (cid:0) β α (cid:1) · · · (cid:0) β r α r (cid:1) . So, we will prove by inductionon | β | := β + · · · + β r that ρ ( u β · · · u β r r ) = P α ≤ β (cid:0) βα (cid:1) µ (ad( u r ) α r · · · ad( u ) α ( u )) u β − α · · · u β r − α r r .First, suppose that β = 0, so u β · · · u β r r = 1, and hence ρ ′ ( u )( u β · · · u β r r ) = µ ( u )(1).Note that under the isomorphism [ D ( λ ) → K h u , · · · , u r i given by Lemma 3.2, theinverse image of 1 is 1 ⊗ v , where v generates the one dimensional subspace K λ of [ D ( λ ) = [ U ( L ) K ⊗ \ U ( B ) K K λ . Write u = v u + α ,u u + · · · + α r,u u r for v u ∈ b and α i,u ∈ K ,so µ ( u ) = λ ( v u ) + α ,u x + · · · + α r,u x r by deﬁnition. Then: ρ ( u )(1) = ρ ( u )(1 ⊗ v ) = u ⊗ v = v u ⊗ v + α ,u u ⊗ v + · · · + α r,u u r ⊗ v = λ ( v u ) ⊗ v + α ,u u ⊗ v + · · · + α r,u u r ⊗ v ,and the image of this under the isomorphism D ( λ ) → K h u , · · · , u r i is λ ( v u ) + α ,u u + · · · + α r,u u r by Lemma 3.2, and clearly this is equal to µ ( u )(1) = ρ ′ ( u )(1), so ρ ( u )(1) = ρ ′ ( u )(1) as required.So, now assume that n > | β | < n . Choose γ with | γ | = n ,and choose i ≥ γ i = 0, so u γ · · · u γ r r = u γ i i · · · u γ r r . Let β i := γ i − β j = γ j for all j > i , so that u γ i i · · · u γ r r = u β i +1 i u β i +1 i +1 · · · u β r r .Then ρ ( u )( u γ i i · · · u γ r r ) = ρ ( u )( u β i +1 i · · · u β r r ⊗ v ) = uu β i +1 i · · · u β r r ⊗ v = [ u, u i ] u β i i · · · u β r r ⊗ v + u i uu β i i · · · u β r r ⊗ v = ρ ([ u, u i ]) u β i i · · · u β r r + ρ ( u i ) ρ ( u ) u β i i · · · u β r r .Since | β | = | γ | − < n , applying induction gives that ρ ([ u, u i ]) u β i i · · · u β r r = P α ≤ β (cid:0) βα (cid:1) µ (ad( u r ) α r · · · ad( u i ) α i ([ u, u i ])) u β i − α i i · · · u β r − α r r = P α ≤ β (cid:0) βα (cid:1) µ (ad( u r ) α r · · · ad( u i ) α i +1 ( u )) u β i − α i i · · · u β r − α r r = P α j ≤ β j (cid:0) β i α i (cid:1) · · · (cid:0) β r α r (cid:1) µ (ad( u r ) α r · · · ad( u i ) α i +1 ( u )) u β i − α i i · · · u β r − α r r .and ρ ( u i ) ρ ( u ) u β i i · · · u β r r = ρ ( u i ) P α ≤ β (cid:0) βα (cid:1) µ (ad( u r ) α r · · · ad( u i ) α i ( u )) u β i − α i i · · · u β r − α r r .By our choice of basis, ad( u r ) α r · · · ad( u i ) α i ( u ) ∈ b ⊕ Span K { u i +1 , · · · , u r } for all α , andthus µ (ad( u r ) α r · · · ad( u i ) α i ( u )) ∈ K ⊕ Span K { x r , · · · , x i +1 } .This means that for all α , µ (ad( u r ) α r · · · ad( u i ) α i ( u )) u β i − α i i · · · u β r − α r r is linear combinationof monomials of the form u δ i i · · · u δ r r . 20o since ρ ( u i )( u δ i i · · · u δ r r ) = u δ i +1 i u δ i +1 i +1 · · · u δ r r = x i ( u δ i i · · · u δ r r ) for all δ , and x i com-mutes with x r , · · · , x i +1 , it follows that: ρ ( u i ) P α ≤ β (cid:0) βα (cid:1) µ (ad( u r ) α r · · · ad( u i ) α i ( u )) u β i − α i i · · · u β r − α r r = P α ≤ β (cid:0) βα (cid:1) µ (ad( u r ) α r · · · ad( u i ) α i ( u )) u β i − α i +1 i · · · u β r − α r r = P α j ≤ β j (cid:0) β i α i (cid:1) · · · (cid:0) β r α r (cid:1) µ (ad( u r ) α r · · · ad( u i ) α i ( u )) u β i − α i +1 i · · · u β r − α r r .Note that this is the only point in the proof where we use our particular choice of basis.Whereas if we assume that ad( g ) n ( u ) ∈ b for all n then, µ (ad( u r ) α r · · · ad( u i ) α i ( u )) ∈ K for all α , and clearly this commutes with ρ ( u i ) regardless of the choice of basis.Now, collecting terms gives us: ρ ( u )( u β i +1 i · · · u β r r ) = P α j ≤ β j (cid:0) β i α i (cid:1) · · · (cid:0) β r α r (cid:1) µ (ad( u r ) α r · · · ad( u i ) α i +1 ( u )) u β i − α i i · · · u β r − α r r + P α j ≤ β j (cid:0) β i α i (cid:1) · · · (cid:0) β r α r (cid:1) µ (ad( u r ) α r · · · ad( u i ) α i ( u )) u β i − α i +1 i · · · u β r − α r r = u β i +1 i · · · u β r r + P ≤ α i ≤ β i Ä (cid:0) β i α i (cid:1) + (cid:0) β i α i − (cid:1) ä (cid:0) β i +1 α i +1 (cid:1) · · · (cid:0) β r α r (cid:1) µ (ad( u r ) α r · · · ad( u i ) α i ( u )) u β i +1 − α i i · · · u β r − α r r + P α i = β i (cid:0) β i +1 α i +1 (cid:1) · · · (cid:0) β r α r (cid:1) µ (ad( u r ) α r · · · ad( u i ) β i +1 ( u )) u β i +1 − α i +1 i +1 · · · u β r − α r r = P α ≤ γ (cid:0) γα (cid:1) µ (ad( u r ) α r · · · ad( u i ) α i ( u )) u γ i − α i i · · · u γ r − α r r The last equality follows since (cid:0) β i α i (cid:1) + (cid:0) β i α i − (cid:1) = (cid:0) β i +1 α i (cid:1) = (cid:0) γ i α i (cid:1) . Note:

1. This proof is purely classical, and deﬁnes a formula for the action of g onthe classical Dixmier module D ( λ ). In fact, this result gives us a ring homomorphism U ( g ) → A r ( K ) whose kernel is Ann U ( g ) D ( λ ). We suspect this map is surjective in gen-eral, since we know from [7, Proposition 6.2.2] that U ( g ) / Ann D ( λ ) ∼ = A r ( K ).2. If u ∈ b and ad( g ) n ( u ) ⊆ b for all n ∈ N , i.e. u ∈ a for some ideal a of g with a ⊆ b ,then µ (ad( u r ) α r · · · ad( u ) α ( u )) = λ (ad( u r ) α r · · · ad( u ) α ( u )) for all α ∈ N r , so it followsfrom this proposition that u acts on [ D ( λ ) by a polynomial in K [ ∂ , · · · , ∂ r ]. Corollary 3.4.

Assume that L is a powerful Lie lattice in g , i.e. [ L , L ] ⊆ p L . Then forany linear form λ : g → K with λ ( L ) ⊆ O , if g has a polarisation b at λ of codimension r ,and I := Ann \ U ( L ) K [ D ( λ ) b , then there exists an injective ring homomorphism [ U ( L ) K /I → \ A r ( K ) , and thus [ U ( L ) K /I is a domain.Proof. The natural action ρ : [ U ( L ) K → End K [ D ( λ ) has kernel I , and since [ D ( λ ) ∼ = K h u , · · · , u r i by Lemma 3.2, the Tate-Weyl algebra \ A r ( K ) embeds as a subalgebra intoEnd K [ D ( λ ) by Lemma 2.13. Therefore, it remains to prove that the image of ρ is con-tained in \ A r ( K ). 21sing Proposition 3.3, we know that we can ﬁx a basis { u , · · · , u r } for L / B such thatfor each u ∈ g , ρ ( u ) = P α ∈ N r α ! · · · α r ! µ (ad( u r ) α r · · · ad( u ) α ( u )) ∂ α · · · ∂ α r r .Since µ ( g ) ⊆ K ⊕ Kx ⊕ · · · ⊕ Kx r ⊆ A r ( K ) and ∂ , · · · , ∂ r ∈ A r ( K ), it follows that ρ ( g ) ⊆ A r ( K ), and hence ρ ( U ( g )) ⊆ A r ( K ).Therefore, it remains to prove that ρ : U ( g ) → A r ( K ) is continuous with respect to the π -adic topologies on U ( g ) and A r ( K ) respectively. And since ρ is linear, this just meansproving that the image of U ( L ) under ρ is contained in A r ( O ) = O [ x , · · · , x r , ∂ , · · · , ∂ r ].Since [ L , L ] ⊆ p L , it follows that for all u ∈ L , α ∈ N r , ad( u r ) α r · · · ad( u ) α ( u ) ∈ p | α | L .And since µ : L → O ⊕ O x ⊕ · · · O x r is linear, µ (ad( u r ) α r · · · ad( u ) α ( u )) ∈ p | α | A r ( O ).Therefore, since α ! · · · α r ! p | α | = p α α ! · · · p αr α r ! ∈ O , it follows that: ρ ( u ) = P α ∈ N r α ! · · · α r ! µ (ad( u r ) α r · · · ad( u ) α ( u )) ∂ α · · · ∂ α r r ∈ A r ( O ).Therefore, the image of L under ρ is contained in A r ( O ), and therefore so is the imageof U ( L ) as required.Now, ﬁx an ideal a of g with a ⊆ b , and deﬁne a ⊥ := { u ∈ g : λ ([ u, a ]) = 0 } ,then a ⊥ is a subalgebra of g with b ⊆ a ⊥ , so we can ﬁx a basis { u , · · · , u r } for L / B suchthat { u s +1 , · · · , u r } is a basis for ( a ⊥ ∩ L ) / B for some s ≤ r . Proposition 3.5.

Every element of a acts on [ D ( λ ) by a polynomial in K [ ∂ , · · · , ∂ s ] , andthe image of U ( a ) under the action ρ : [ U ( L ) K → End K [ D ( λ ) is precisely K [ ∂ , · · · , ∂ s ] .Proof. Firstly, using Proposition 3.3, we know that for every u ∈ a , ρ ( u ) = P α ∈ N r α ! · · · α r ! λ (ad( u r ) α r · · · ad( u ) α ( u )) ∂ α · · · ∂ α r r ∈ K [ ∂ , · · · , ∂ r ].Also, if α i = 0 for any r ≥ i > s , then assuming i is maximal such that α i = 0, λ (ad( u r ) α r · · · ad( u ) α ( u )) = λ (ad( u i ) α i · · · ad( u ) α ( u )) = 0 since u i ∈ a ⊥ .Therefore, ρ ( u ) = P α ∈ N s α ! · · · α s ! λ (ad( u s ) α s · · · ad( u ) α ( u )) ∂ α · · · ∂ α s s ∈ K [ ∂ , · · · , ∂ s ]as required.For the second statement, clearly ρ ( U ( a )) ⊆ K [ ∂ , · · · , ∂ s ], so we just need to show that ∂ , · · · , ∂ s ∈ ρ ( U ( a )).We will ﬁrst need to construct our basis appropriately. For convenience, set A = a ∩ L and A ⊥ = a ⊥ ∩ L . First, write L / B = ( A ⊥ / B ) ⊕ V , for some complement V of A ⊥ / B in L / B .We deﬁne the upper central series by Z ( L ) := Z ( L ) and Z i ( L ) := { u ∈ L : [ u, L ] ⊆ Z i − ( L ) } for all i >

1. Since L is nilpotent, Z c ( L ) = L for some c ≥

1. Therefore,since V ∩ a ⊥ = 0, and hence λ ([ a , V ]) = 0, we can choose m > ([ A ∩ Z m ( L ) , V ]) = 0.So, choose y ∈ A ∩ Z m ( L ) such that λ ([ y , V ]) = 0. Then λ ◦ ad( y ) : V ⊗ K → K is a non-zero linear form, so we may write V = O u ⊕ V where λ ([ y , V ]) = 0 and λ ([ y , u ]) = 0.Now, let us suppose for induction, that for some i < s = dim K g / a ⊥ , we have the followingdata: • Integers 1 < m ≤ m ≤ · · · ≤ m i ≤ c . • Subspaces V i ⊆ V i − ⊆ · · · ⊆ V ⊆ V = V with λ ([ V j − , a ∩ Z m j − ( L )]) = 0 for each j ≥ • Elements u j ∈ V j − such that V j − = O u j ⊕ V j for each j ≥ • Elements y j ∈ A ∩ Z m j ( L ) such that λ ([ y j , V j ]) = 0 and λ ([ y j , u j ]) = 0.Note that for each j , dim K ( V j ⊗ K ) = s − j . So since i < s , V i ∩ a ⊥ = 0, so λ ([ V i , a ]) = 0.But since V i ⊆ V i − , we know that [ V i , a ∩ Z m i − ( L )] = 0, so choose m i +1 ≥ m i min-imal such that λ ([ V i , a ∩ Z m i +1 ( L )]) = 0, and choose y i +1 ∈ A ∩ Z m i +1 ( L ) such that λ ([ y i +1 , V i ]) = 0.Again, λ ◦ ad( y i +1 ) : V i ⊗ K → K is a non-zero linear form, so V i = O u i +1 ⊕ V i +1 for some u i +1 ∈ V i , where λ ([ y i +1 , V i +1 ]) = 0 and λ ([ y i +1 , u i +1 ]) = 0.Therefore, applying induction gives us a basis { u , · · · , u s } for V such that for each i , λ ([ u i , A ∩ Z m i − ( L )]) = 0, and elements y , · · · , y s ∈ A with y i ∈ Z m i ( L ) such that λ ([ y i , u j ]) = 0 for all j < i and λ ([ y i , u i ]) = 0.So, applying our explicit formula with the basis { u , · · · , u s } , we get that ρ ( y i ) = P α ∈ N s α ! · · · α s ! λ (ad( u s ) α s · · · ad( u ) α ( y i )) ∂ α · · · ∂ α s s for each i . Let us suppose that forall 0 ≤ j < i , ∂ j ∈ ρ ( U ( g )).Then since [ L , y i ] ⊆ Z m i − ( L ) and λ ([ u j , Z m i − ( L )]) = 0 for all j ≥ i , it follows that if λ (ad( u s ) α s · · · ad( u ) α ( y i +1 )) = 0 then either α j = 0 for all j ≥ i , or α j = 0 for all j < i and α i + · · · + α s = 1. Therefore: ρ ( y i ) = f ( ∂ , · · · , ∂ i − ) + λ ([ y i , u i ]) ∂ i + · · · + λ ([ y i , u s ]) ∂ s for some polynomial f .But since ∂ , · · · , ∂ i − ∈ ρ ( U ( a )), λ ([ y i , u j ]) = 0 for all j > i , and λ ([ y i , u i ]) = 0, it followsthat ∂ i ∈ ρ ( U ( a )). So applying induction, we see that ∂ , · · · , ∂ s ∈ ρ ( U ( a )) as required.Finally, if { u ′ , · · · , u ′ s } is any other basis for V , then since each u ′ i is a non-zero linearcombination in u , · · · , u s , it follows from the chain rule that ∂ u ′ i = β i ∂ u i for some non-zero β i ∈ K , and hence ∂ u ′ , · · · , ∂ u ′ s ∈ ρ ( U ( a )). 23 .3 Irreducibility Now we will prove the main theorem of this section.

Theorem 3.6.

Suppose that g is nilpotent, λ ∈ g ∗ with λ ( L ) ⊆ O . Then there existsa polarisation b of g at λ such that the aﬃnoid Dixmier module [ D ( λ ) b of L at λ withrespect to b is irreducible as a [ U ( L ) K -module.Proof. We will use induction on n = dim K g , so ﬁrst suppose that n = 1. Then g isabelian, so λ is a character of g , and [ D ( λ ) = K , which is clearly irreducible.For the inductive step, we will assume that the result holds for all m < n :Suppose ﬁrst that there exists a non-zero ideal a E g such that λ ( a ) = 0. Let g := g / a ,so since a = 0, dim K g < n , so we may apply the inductive hypothesis to g .Let A := L ∩ a , and let L := LA , then A , L are lattices in a , g respectively.Now, let λ be the linear form of g induced by λ , and clearly λ ( L ) ⊆ O . So by theinductive hypothesis, there exists a polarisation b of g at λ such that \ D ( λ ) b is irre-ducible over \ U ( L ) K .Since b = ba for some subalgebra b of g , it follows that b is a polarisation of g at λ , solet B := b ∩ L , B := b ∩ L , and let M := \ D ( λ ) b = \ U ( L ) K ⊗ \ U ( B ) K K λ .Using the surjection [ U ( L ) K ։ \ U ( L ) K given by Lemma 3.1( i ), we see that M has thestructure of an irreducible [ U ( L ) K -module, and using the fact that B = BA and Lemma3.1( ii ) we see that M ∼ = [ U ( L ) K ⊗ \ U ( B ) K K λ = [ D ( λ ) b , and hence [ D ( λ ) b is irreducible asrequired.So from now on, we may assume that λ ( a ) = 0 for all non-zero ideals a of g .Using Proposition 2.7, we can ﬁnd a reducing quadruple ( x, y, z, g ′ ) of g such that anypolarisation b ⊆ g ′ of g ′ at µ := λ | g ′ is in fact a polarisation of g at λ . Note that since y, z are central in g ′ , y, z ∈ b by Lemma 2.3.Set L ′ := L ∩ g ′ . Then since dim K g ′ = n − < n , using the inductive hypothesis wecan choose a polarisation b of g ′ at µ such that [ D ( µ ) = \ U ( L ′ ) K ⊗ B K µ is irreducible over \ U ( L ′ ) K .Now, since [ U ( L ) K = [ U ( L ) K ⊗ \ U ( L ′ ) K \ U ( L ′ ) K , it follows that [ D ( λ ) b = [ U ( L ) K ⊗ \ U ( L ′ ) K [ D ( µ ) b .Therefore, setting M := [ D ( µ ) b , we have that [ D ( λ ) b ∼ = M h t i as a K -vector space byProposition 3.2( i ), where x acts on M h t i by t .Also, y, z are central in g ′ and [ y, x ] = αz for some 0 = α ∈ K . Therefore, since y and z act on M by scalars λ ( y ), λ ( z ) respectively, we see using Proposition 3.2( ii ) that y acts24n M h t i by αλ ( z ) ddt + λ ( y ).So, ﬁnally, since Kz is an ideal of g , λ ( z ) = 0, so since α, λ ( z ) = 0, it follows fromProposition 3.2( iii ) that [ D ( λ ) b = M h t i is irreducible over [ U ( L ) K as required. From now on, we will always assume that g is nilpotent. We are interested in the anni-hilators inside [ U ( L ) K of aﬃnoid Dixmier modules.Since for any λ ∈ g ∗ with λ ( L ) ⊆ O , there exists a polarisation b of g at λ such that [ D ( λ ) b is irreducible by Theorem 3.6, it follows by deﬁnition that Ann \ U ( L ) K [ D ( λ ) b is aprimitive ideal of [ U ( L ) K .Ideally, we want to prove that all primitive ideals arise as annihilators of aﬃnoid Dixmiermodules. But in this section, we will ﬁrst show that these Dixmier annihilators are alwaysprimitive, regardless of the choice of polarisation. First we need some preliminary results. The ﬁrst is an aﬃnoid version of [7, Proposition5.1.7]:

Lemma 4.1.

Let h ≤ g be a subalgebra, let H := h ∩ L , and let M be a ﬁnitely gener-ated \ U ( H ) K -module, with J := Ann \ U ( H ) K M . Then Ann \ U ( L ) K (cid:16) [ U ( L ) K ⊗ \ U ( H ) K M (cid:17) is thelargest two-sided ideal in [ U ( L ) K contained in [ U ( L ) K J . It follows that if M, N are ﬁnitelygenerated \ U ( H ) K -modules such that Ann \ U ( H ) K M = Ann \ U ( H ) K N then Ann \ U ( L ) K ( [ U ( L ) K ⊗ \ U ( H ) K M ) = Ann \ U ( L ) K ( [ U ( L ) K ⊗ \ U ( H ) K N ) .Proof. Fix an O -basis { x , · · · , x d } for L such that { x , · · · , x r } is a basis for H . Thenby Lemma 2.1, every element r ∈ [ U ( L ) K can be written as r = P α ∈ N d λ α x α · · · x α d d , forsome λ α ∈ K converging to zero as α → ∞ , i.e. r = P α ∈ N r x α s α for some s α ∈ \ U ( H ) K suchthat s α → α → ∞ .Using Proposition 3.2, we see that [ U ( L ) K ⊗ \ U ( H ) K M is isomorphic as a K -vector space tothe Tate module M h t , · · · , t r i = { P α ∈ N r t α · · · t α r r s α : s α ∈ M, s α → | α | → ∞} via anisomorphism Ψ sending x α ⊗ m to t α · · · t α r r m . It is clear that the set of all elements in [ U ( L ) K that annihilate the set M inside M h t , · · · , t r i on the left contains the left ideal [ U ( L ) K J .Moreover, if rM = 0 for some r = P α ∈ N r x α s α ∈ [ U ( L ) K , then for all m ∈ M :0 = rm = P α ∈ N r x α s α m = Ψ − ( P α ∈ N r t α · · · t α r r s α m ), and hence s α m = 0 for all α ∈ N r .25hus s α ∈ J for all α , and hence r ∈ [ U ( L ) K J . Therefore the right ideal [ U ( L ) K J is theset of all elements of [ U ( L ) K that annihilate the set M .It follows that if r [ U ( L ) K ⊗ \ U ( H ) K M = 0, then r [ U ( L ) K annihilates M , so the two-sidedideal generated by r is contained in [ U ( L ) K J . Since our choice of r was arbitrary, itfollows that the annihilator of [ U ( L ) K ⊗ \ U ( H ) K M is contained in [ U ( L ) K J .Furthermore, if I ⊆ [ U ( L ) K J is a two-sided ideal of [ U ( L ) K , then I annihilates M , sosince I [ U ( L ) K = [ U ( L ) K I , it must also annihilate the submodule generated by M inside [ U ( L ) K ⊗ \ U ( H ) K M , which is clearly the whole module, and the result follows.The next result will be essential to several of our proofs, since it allows us to safely passto and from a reducing quadruple when studying two-sided ideals in [ U ( L ) K . Theorem 4.2.

Suppose that g has a reducing quadruple ( x, y, z, g ′ ) with x, y, z ∈ L , andlet L ′ := L ∩ g ′ . Then if I is a two-sided ideal of [ U ( L ) K such that z + I is not a zerodivisor in [ U ( L ) K /I , then I is controlled by L ′ , i.e.: I = (cid:16) I ∩ \ U ( L ′ ) K (cid:17) [ U ( L ) K = [ U ( L ) K (cid:16) I ∩ \ U ( L ′ ) K (cid:17) .Proof. Using Lemma 2.1, we see that every element of [ U ( L ) K can be written as g ( x ) forsome Tate power series g with coeﬃcients in \ U ( L ′ ) K . We will prove that if g ( x ) ∈ I thenthe coeﬃcients of g all lie in I , and the result follows.It will suﬃce to show that if g ( x ) = c + c x + c x + · · · ∈ I then the formal derivative g ′ ( x ) = c + 2 c x + 3 c x + · · · also lies in I . Then using an argument similar to the proofof Proposition 3.2( iii ), we can construct a sequence of elements in I converging to c . Byclosure of I in [ U ( L ) K , it follows that c ∈ I , so repeating the argument for n ! g ( n ) ( x ) foreach x , it follows that all coeﬃcients of g ( x ) lie in I as required.To prove that g ′ ( x ) lies in I , consider the action of y on [ U ( L ) K /I :Since y is central in L ′ , y commutes with everything in \ U ( L ′ ) K . Also, since [ x, y ] = αz ,clearly y · x = xy − αz , and an easy induction shows that y · x n = x n y − nαx n − z . Soif l y is the left action of y on [ U ( L ) K /I , r y is the right action, then l y − r y = − αz ddx .Therefore, since z is not a zero divisor modulo I , and α = 0, it follows that if g ( x ) ∈ I then ddx ( g ( x )) ∈ I as required. We will now prove the main result of this section, namely that Dixmier annihilator idealsare independent of the choice of polarisation. First, we deal with a special case.

Lemma 4.3.

Suppose that L has an O -basis { x, y, z } such that z is central and [ x, y ] = αz for some = α ∈ O . Then for any = β ∈ O , the ideal ( z − β ) [ U ( L ) K is a maximaltwo-sided ideal of [ U ( L ) K . roof. Let I be a proper ideal of [ U ( L ) K containing z − β . Then since β = 0, z + I is nota zero divisor in [ U ( L ) K /I . So setting g ′ := Span K { y, z } , since ( x, y, z, g ′ ) is a reducingquadruple, it follows from Theorem 4.2 that I is controlled by L ′ = g ′ ∩ L . Therefore, ifwe can prove that I ∩ \ U ( L ′ ) K = ( z − β ) \ U ( L ′ ) K then it follows that I = ( z − β ) [ U ( L ) K .Let ¯ y be the image of y in \ U ( L ′ ) K / ( z − β ) \ U ( L ′ ) K , then given r ∈ \ U ( L ′ ) K , by Lemma2.1, the image of r in \ U ( L ′ ) K / ( z − β ) \ U ( L ′ ) K has the form ¯ r = P n ≥ λ n ¯ y n for some λ n ∈ K , λ n → n → ∞ .Since [ x, y ] = αz , we have that x · ¯ y = ¯ yx + αβ , and an easy induction shows that x · ¯ y n = ¯ y n x + αβ ¯ y n − for all n , i.e. if l x and r x are the respective left and right actionsof x on \ U ( L ′ ) K / ( z − β ), then l x − r x = αβ dd ¯ y . Since α, β = 0 and I is a two-sided ideal, dd ¯ y preserves I ∩ \ U ( L ′ ) K / ( z − β ).So if g (¯ y ) = λ + λ ¯ y + λ ¯ y + · · · ∈ I ∩ \ U ( L ′ ) K / ( z − β ) \ U ( L ′ ) K , it follows that n ! g ( n ) (¯ y ) ∈ I for all n ∈ N , and using an argument similar to the proof of Proposition 3.2( iii ), we canconstruct a sequence of elements in I ∩ \ U ( L ′ ) K converging to λ ∈ K .By closure of I ∩ [ U ( L ) K , this implies that λ ∈ I , and hence λ = 0, and it follows afterreplacing g (¯ y ) by n ! g ( n ) (¯ y ) that λ n = 0 for all n , i.e. g (¯ y ) = 0. Therefore I ∩ \ U ( L ′ ) K =( z − β ) \ U ( L ′ ) K , and I = ( z − β ) [ U ( L ) K . So since our choice of I was arbitrary, ( z − β ) [ U ( L ) K is maximal as required. Theorem 4.4.

Suppose g is nilpotent, and let λ ∈ L ∗ . Then for any polarisations b , b of g at λ , Ann \ U ( L ) K [ D ( λ ) b = Ann \ U ( L ) K [ D ( λ ) b .Proof. If g is abelian then b = b = g so the statement is obvious. Since all nilpotentLie algebras of dimension 1 and 2 are abelian, we may assume that dim K g ≥ g is non-abelian and dim K g = 3 then it is straightforward to show that L has basis { x, y, z } with z central and [ x, y ] = αz for some α ∈ O\

0. If λ ( z ) = 0 then λ is a characterof g , so g is the only polarisation and the statement is trivially true. If λ ( z ) = 0, then forany polarisation b , z acts on [ D ( λ ) b by λ ( z ), and so the [ U ( L ) K -annihilator must contain( z − λ ( z )), which is a maximal ideal by Lemma 4.3, hence this must be the annihilatorin all cases as we require.So from now on, we may assume that n = dim K g ≥ n :Suppose ﬁrst that there exists a non-zero ideal a E g such that λ ( a ) = 0, so λ inducesa linear form λ of g := g / a . Setting A := a ∩ L , L := LA , it is clear that L is a Lielattice in g and λ ( L ) ⊆ O .Note that a ⊆ b , b by Lemma 2.3, so set b i, := b i / a for i = 1 ,

2, and b , , b , arepolarisations of g at λ . 27ince dim K g < n , it follows from induction that Ann \ U ( L ) K \ D ( λ ) b , = Ann \ U ( L ) K \ D ( λ ) b , .Using Lemma 3.1, we see that for i = 1 , \ D ( λ ) b i, = \ U ( L ) K ⊗ \ U ( B i, ) K K λ is naturallya [ U ( L ) K -module, and that it is isomorphic to [ U ( L ) K ⊗ \ U ( B i ) K K λ = [ D ( λ ) b i .If φ : [ U ( L ) K ։ \ U ( L ) K is the natural surjection, then it is clear that Ann \ U ( L ) K \ D ( λ ) b i, = Ann \ U ( L ) K \ D ( λ ) b i, ker( φ ) for i = 1 ,

2. Thus Ann \ U ( L ) K \ D ( λ ) b , = Ann \ U ( L ) K \ D ( λ ) b , , and hence:Ann \ U ( L ) K [ D ( λ ) b = Ann \ U ( L ) K \ D ( λ ) b , = Ann \ U ( L ) K \ D ( λ ) b , = Ann \ U ( L ) K [ D ( λ ) b as re-quired.So from now on, we may assume that λ ( a ) = 0 for all non-zero ideals a of g . UsingProposition 2.7, we see that this means that there exists a reducing quadruple ( x, y, z, g ′ )for g . Since we are assuming dim K g >

3, we may apply Proposition 2.8 to get that foreach i = 1 , h i of g containing b i , and a polarisation b ′ i of g at λ contained in h i and g ′ .By induction, since dim K g ′ < n , we get that Ann \ U ( L ′ ) K \ D ( λ | g ′ ) b ′ = Ann \ U ( L ′ ) K \ D ( λ | g ′ ) b ′ ,so by Lemma 4.1, Ann \ U ( L ) K [ D ( λ ) b ′ = Ann \ U ( L ) K [ D ( λ ) b ′ .Similarly, since h , h are proper subalgebras of g , we also have that Ann \ U ( H i ) K \ D ( λ | h i ) b i =Ann \ U ( H i ) K \ D ( λ | h i ) b ′ i for i = 1 , \ U ( L ) K [ D ( λ ) b i = Ann \ U ( L ) K [ D ( λ ) b ′ i , and it follows that Ann \ U ( L ) K [ D ( λ ) b = Ann \ U ( L ) K [ D ( λ ) b as required.Now that we have established that the annihilator of a Dixmier module does not dependon the choice of polarisation, we can unambiguously make the following deﬁnition: Deﬁnition 4.1.

Let

F/K be a ﬁnite extension, and let λ ∈ L ∗ F := ( L ⊗ O O F ) ∗ be a linearform. Deﬁne the Dixmier annihilator in [ U ( L ) K associated to λ to be the two sided ideal I ( λ ) := Ann \ U ( L ) K [ D ( λ ) F (or I ( λ ) F if it is unclear what the base ﬁeld is). Note:

This deﬁnition makes sense because there is a natural embedding [ U ( L ) K → \ U ( L F ) F for any ﬁnite extension F/K . Using Theorem 3.6, we see that I ( λ ) is a primitiveideal of [ U ( L ) K whenever F = K . Now we will study some general ring theoretic properties of the aﬃnoid enveloping alge-bra. ’ U ( L ) K Proposition 5.1.

Let P be a prime ideal of [ U ( L ) K , and let J be a two-sided ideal of R := [ U ( L ) K /P . Then if J = 0 there exists an element a ∈ J such that a n does not onverge to 0 as n → ∞ .Proof. Let w be the π -adic ﬁltration on [ U ( L ) K corresponding to the lattice [ U ( L ), and let¯ w be the quotient ﬁltration on R := [ U ( L ) K /P . Then since R is complete with respect to¯ w and gr ¯ w R ∼ = gr w \ U ( L ) K gr w P , it follows from [14, Ch II Theorem 2.2.1] that ¯ w is a Zariskianﬁltration on R .Also, since gr w [ U ( L ) K ∼ = U ( L /π L )[ t, t − ] by [2, Lemma 3.1], and it is well known that U ( L /π L ) is ﬁnitely generated over its centre, it follows that gr ¯ w R is also ﬁnitely gener-ated over its centre.Furthermore, t = gr π is central of positive degree in gr R , and it is non-nilpotent, soit follows that gr R is ﬁnitely generated over a central Noetherian subring whose positivepart is non-nilpotent. Hence after applying [11, Theorem 3.3], we can ﬁnd a ﬁltration v on the ring of quotients Q ( R ) such that v restricts to a valuation of the centre, and theinclusion ( R, ¯ w ) → ( Q ( R ) , v ) is continuous.Suppose for contradiction that for every element a ∈ J , a n → n → ∞ . Choosean arbitrary a ∈ J , and following [11, Deﬁnition 3.2], deﬁne the growth rate function ρ : Q ( R ) → R ∪ {∞} , q lim n →∞ v ( q n ) n , and let m := ⌈ ρ ( a ) ⌉ .If we assume that m < ∞ , then set b := π − ( m +1) a , and since π is central in R , we seeusing [11, Lemma 3.7( v )] that ρ ( b ) = ρ ( a ) − ( m + 1) v ( π ) ≤ ρ ( a ) − ( ρ ( a ) + 1) v ( π ) < b n does not converge to 0 as n → ∞ – contradiction since b ∈ J .Therefore m = ρ ( a ) = ∞ , so since Q ( R ) is simple and artinian, it follows from [11,Lemma 3.7( iv )] that a is nilpotent.Since our choice of a was arbitrary, this means that every element of J is nilpotent, andusing [7, Lemma 3.1.14] it follows that J is a nilpotent ideal of R . Since R is prime, thismeans that J = 0 as required.The following result is the aﬃnoid version of [7, Proposition 3.1.15]: Theorem 5.2.

Let I be a two sided ideal of [ U ( L ) K . Then I is semiprime if and only if I is an intersection of primitive ideals.Proof. Clearly if I is an intersection of primitive ideals, then it is semiprime, so it re-mains only to prove the converse, i.e. that if I is semiprime then it is the intersection ofprimitive ideals.Since semiprime ideals arise as an intersection of primes, we can assume that I is primein [ U ( L ) K , and we will show that J ( [ U ( L ) K /I ) = 0, from which the result follows.Assume for contradiction that J := J ( [ U ( L ) K /I ) = 0, then since I is prime it followsfrom Proposition 5.1 that we can choose an element a ∈ J such that a n does not convergeto zero as n → ∞ . 29et R := \ U ( L ) K I , and let C := R h X i be the Tate algebra in one variable over R . Then ifwe set g := g × K , L := L × O , it is clear that L is a Lie lattice in g and it followsfrom Lemma 2.1 that C ∼ = \ U ( L ) K /I \ U ( L ) K .Consider the element 1 − aX ∈ C . If this element is a unit, its inverse must have theform a + a X + a X + · · · for some a i ∈ [ U ( L ) K /I with a n → n → ∞ . But since1 = (1 − aX )( a + a X + a X + · · · ), it follows that a = 1, a = a , a = a , · · · , a n = a n ,and hence a n → n → ∞ – contradiction.Therefore 1 − aX is not a unit in C , so there exists a maximal left ideal of C containing1 − aX , i.e. there exists an irreducible C -module M and an element 0 = m ∈ M suchthat (1 − aX ) m = 0.Now, X does not act by zero on M , otherwise 1 − aX would act by 1, and we would have(1 − aX ) m = m = 0. So using Schur’s Lemma, the action of X is invertible, and using[17, Theorem 6.4.6] we see that the action of X − is algebraic over K , i.e. there exists f ( t ) = a + a t + · · · + a n t n for some a i ∈ K such that f ( X − ) = 0, and we may assumethat a = 0. So let g ( t ) := a − f ( t ) = 1 + b t + · · · + b n t n .Since aXm = m , we have that am = X − aXm = X − m , hence a r m = X − r m for all r ∈ N , and thus g ( a ) m = g ( X − ) m = 0.But g ( a ) = 1 + ( b + b a + · · · + b n a n − ) a , so since a ∈ J ( [ U ( L ) K /I ), this means that g ( a )is a unit in [ U ( L ) K /I . Therefore, since m = 0, g ( a ) m = 0 – contradiction.Therefore J ( [ U ( L ) K /I ) = 0 as we require. Now, recall from the introduction that a prime ideal P of [ U ( L ) K is locally closed if P = ∩{ Q E [ U ( L ) K : Q prime, P ( Q } . Proposition 5.3.

Let P be a prime ideal of [ U ( L ) K . Then: P is locally closed = ⇒ P is primitive = ⇒ P is weakly rational.Proof. First, suppose that P is primitive, i.e. P = Ann \ U ( L ) K M for some simple [ U ( L ) K -module M . Then it follows from [17, Theorem 6.4.6] that every element of End \ U ( L ) K M is algebraic over K . So since Z ( [ U ( L ) K /P ) is a domain, and embeds into End \ U ( L ) K M ,it follows that Z ( [ U ( L ) K /P ) is an algebraic ﬁeld extension of K , and hence P is weaklyrational.Using Theorem 5.2, we see that if P is equal to an intersection of primitive ideals. So if P is locally closed, i.e. not equal to the intersection of all prime ideals properly containingit, then P must be primitive. 30 ote: These implications also hold for primes P in U ( g ), and in the case where g isnilpotent, P weakly rational = ⇒ P maximal [7, Proposition 4.7.4]. We suspect thatthis is also true in the aﬃnoid case, but this is only known in the case where g containsan abelian ideal of codimension 1 [13, Corollary 1.6.2].Using this result, we see that to prove the Dixmier-Moeglin equivalence for [ U ( L ) K , itremains to show that weakly rational ideals are locally closed. Taking steps in thisdirection, the aim of this section is to prove that all locally closed prime ideals P of [ U ( L ) K have the form of a Dixmier annihilator I ( λ ) F for some ﬁnite extension F of K . Proposition 5.4.

Let g be a nilpotent K -Lie algebra, and let L be an O -Lie lattice in g such that every locally closed prime ideal in [ U ( L ) K has the form I ( λ ) F for some ﬁniteextension F/K and some K -linear map λ : g → F such that λ ( L ) ⊆ O F .Then given any prime ideal P in [ U ( L ) K , P arises as an intersection of Dixmier annihi-lators.Proof. Note that since gr [ U ( L ) K ∼ = U ( L /π L )[ t, t − ] has ﬁnite left and right Krull dimen-sion, it follows from [14, Ch.I Theorem 7.1.3] that [ U ( L ) K has ﬁnite left and right Krulldimension. Therefore, using [16, Lemma 6.4.5], it follows that [ U ( L ) K has ﬁnite classicalKrull dimension , i.e. there is a ﬁnite upper bound on the length of chains of prime idealsin [ U ( L ) K .So, given a prime ideal P of R , deﬁne the dimension dim( P ) of P to be the largest integer n ≥ P = P ( P ( · · · ( P n of [ U ( L ) K .We will proceed by induction on dim( P ).If dim( P ) = 0, then P is maximal, and hence locally closed, so P = I ( λ ) F for some ﬁniteextension F , λ : g → F as required. So suppose the result holds whenever dim( P ) < n .If dim( P ) = n then for every prime ideal Q of [ U ( L ) K with P ( Q , Q arises as an inter-section of Dixmier annihilators by the inductive hypothesis.If P is locally closed, then P is a Dixmier annihilator by assumption, otherwise P isequal to the intersection of all prime ideals properly containing it, and hence it is anintersection of Dixmier annihilators as required. Now we will complete the classiﬁcation of locally closed ideals of [ U ( L ) K in terms ofDixmier annihilators. First, we need some technical results: Lemma 5.5.

Suppose that dim K g > , and let I be an ideal of [ U ( L ) K . Suppose furtherthat Z ( [ U ( L ) K /I ) = K , and I ∩ g = 0 . Then g has a reducing quadruple ( x, y, z, g ′ ) .Proof. Firstly, suppose u, v ∈ g are central, then u + I, v + I ∈ Z ( [ U ( L ) K /I ) = K , hencethey are K -linearly dependent. So there exist non-zero α, β ∈ K such that αu + βv ∈ I ∩ g = 0, hence u, v are K -linearly dependent in g .31ince g is nilpotent, Z ( g ) = 0, so it follows that Z ( g ) has dimension 1, i.e. Z ( g ) = Kz for some z ∈ Z ( g ).So, since dim K g > g is non-abelian, and again using nilpotence of g , we can ﬁnd y ∈ g such that y is not central, but [ y, g ] ⊆ Z ( g ) = Kz .Let g ′ := ker(ad( y )). Then g ′ is an ideal of g , and since ad ( y ) : g → Kz is non-zero, g ′ must have codimension 1 in g , so g = g ′ ⊕ Kx , and it is clear that ( x, y, z, g ′ ) is a reducingquadruple for g . Proposition 5.6.

Let I be a two sided ideal of [ U ( L ) K such that F = Z ( [ U ( L ) K /I ) isa ﬁnite ﬁeld extension of K . Then \ U ( L ⊗ O O F ) F ∼ = [ U ( L ) K ⊗ K F , and there exists asurjection \ U ( L ⊗ O O F ) F → [ U ( L ) K /I of F -algebras with kernel containing I ⊗ K F .Proof. To see that \ U ( L ⊗ O O F ) F ∼ = [ U ( L ) K ⊗ K F , note that: U ( L ) ⊗ O O F ↔ U ( L ⊗ O O F ) a ⊗ α αau ⊗ α ← [ u ⊗ α Are isomorphisms of O F algebras, preserving the π -adic ﬁltration. Hence they induce anisomorphism \ U ( L ⊗ O O F ) ∼ = [ U ( L ) ⊗ O O F , and the result follows.Since F = Z ( [ U ( L ) K /I ) ⊆ [ U ( L ) K /I , it is clear that [ U ( L ) K /I is an F -algebra, and themap [ U ( L ) K ⊗ K F → [ U ( L ) K /I, r ⊗ ( α + I ) αr + I is clearly a surjection of F -algebrassending I ⊗ F to 0 as required. Lemma 5.7.

Let a be an ideal of g nilpotent, let A := a ∩ L and let L := L / A . Let P bea prime ideal of [ U ( L ) K , containing a , such that the image P E \ U ( L ) K of P under thesurjection [ U ( L ) K → \ U ( L ) K is a Dixmier annihilator. Then P is a Dixmier annihilator.Proof. We know that P = Ann \ U ( L ) K [ D ( µ ) F for some ﬁnite extension F/K , µ ∈ ( L / A ) ∗ F .Clearly µ is induced from a linear form λ of g ⊗ K F such that λ ( L ) ⊆ O F and λ ( a ) = 0.We will prove that P = Ann \ U ( L ) K [ D ( λ ) F .Choose a polarisation b of g ⊗ K F at λ , and since the annihilator is independent of thechoice of polarisation by Theorem 4.4, we may assume that a ⊆ b , i.e. b / a is a polar-isation of g / a at µ . Using Lemma 3.1( iii ), we see that [ D ( λ ) F = [ U ( L ) F ⊗ \ U ( B ) F F ∼ = \ U ( L / A ) F ⊗ \ U ( B / A ) F F = [ D ( µ ) F .Using Lemma 3.1( i ), we know that [ U ( L ) / \ a U ( L ) ∼ = \ U ( L / A ) K , and hence P = P/ a [ U ( L ) K .Therefore, since P = Ann \ U ( L ) K [ D ( µ ) F , and hence P [ D ( µ ) F = 0, it follows that P [ D ( λ ) F =0, i.e. P ⊆ Ann \ U ( L ) K [ D ( λ ) F .Moreover, if x [ D ( λ ) F = 0 then x [ D ( µ ) F = 0 so x + a [ U ( L ) K ∈ P and hence x ∈ P .Therefore P = Ann \ U ( L ) K [ D ( λ ) F as required.32ow we can prove the main theorem of this section, classifying locally closed ideals interms of Dixmier annihilators. Theorem 5.8.

Let g be a nilpotent K -lie algebra, with O -Lie lattice L , and let P bea locally closed prime ideal of [ U ( L ) K . Then there exists a ﬁnite extension F \ K and a K -linear map λ : g → F with λ ( L ) ⊆ O F such that P = I ( λ ) F .Proof. We will use induction on n = dim K g :First suppose that n = 1, and hence [ U ( L ) K ∼ = K h u i by Lemma 2.1. So if P is a locallyclosed ideal, then it is primitive, and hence maximal since [ U ( L ) K is commutative. So let F := [ U ( L ) K /P , then F is a ﬁeld.Furthermore, using [5, Corollary 2.2.12], we see that F is a ﬁnite extension of K , sodeﬁne λ : g → F, x x + P , and clearly this map is K -linear. Also, [ U ( L ) /P ∩ [ U ( L ) = Oh u i /P ∩ Oh u i is a lattice in F = K h u i /P . Thus [ U ( L ) /P ∩ [ U ( L ) ⊆ O F so clearly λ ( L ) ⊆ O F .So [ D ( λ ) F = F , where x ∈ [ U ( L ) K acts by zero if and only if λ ( x ) = 0, i.e. if and only if x ∈ P , so P = Ann \ U ( L ) K [ D ( λ ) F = I ( λ ) F as required.So now suppose that the result holds whenever dim K g < n .Again, suppose that P is a locally closed ideal of [ U ( L ) K , and let a := P ∩ g , A := a ∩ L .Clearly a is an ideal of g , contained in P , and A is a Lie lattice in a . We will supposeﬁrst that a = 0.Let P be the image of P under the surjection [ U ( L ) K → \ U ( L / A ) K , then P is a lo-cally closed ideal of \ U ( L / A ) K . Since dim K g / a < n , it follows from induction that P is aDixmier annihilator. Therefore, using Lemma 5.7, P is a Dixmier annihilator as required.So from now on we may assume that a = P ∩ g = 0.Since we know by Proposition 5.3 that P is primitive, it follows from [17, Theorem 6.4.6]that F = Z ( [ U ( L ) K /P ) is an algebraic ﬁeld extension of K . Since the centre of [ U ( L ) K /P is closed and [ U ( L ) K /P is complete, it follows that F is complete, so it must in fact be aﬁnite extension of K .We will assume for now that F = K , so applying Lemma 5.5, we see that g has a re-ducing quadruple ( x, y, z, g ′ ). So let L ′ := g ′ ∩ L , then since z / ∈ P , it is clear that z + P ∈ Z ( [ U ( L ) K /P ) = K is not a zero divisor, so using Theorem 4.2, we see that P iscontrolled by L ′ , i.e. P = [ U ( L ) K ( P ∩ \ U ( L ′ ) K ).Let Q := P ∩ \ U ( L ′ ) K , then Q is a semiprime ideal of \ U ( L ′ ) K , so since all locally closedprime ideals in \ U ( L ′ ) K are Dixmier annihilators by induction, it follows from Proposition5.4 that all semiprime ideals arise as an intersection of Dixmier annihilators, i.e. thereexist ﬁnite extensions F j /K , µ j ∈ ( L ′ ) ∗ F j , as j ranges over some indexing set X , and33 = T j ∈ X I ( µ j ) F j .Since z / ∈ Q and Z ( [ U ( L ) K /P ) = K , there exists 0 = β ∈ K such that z − β ∈ Q .Therefore z − β ∈ I ( µ j ) F j for each j . Since β = 0, this means that z / ∈ I ( µ j ) F j , i.e. µ j ( z ) = 0.Now, it is clear that ( x ⊗ , y ⊗ , z ⊗ , g ′ ⊗ K F j ) is a reducing quadruple for g ⊗ K F j ,so applying Lemma ?? gives that if b is a polarisation of g ′ ⊗ K F j at µ j and λ j is anextension of µ j to g ⊗ K F j , then b is a polarisation of g ⊗ K F j at λ j .Therefore, \ D ( λ j ) F j ∼ = [ U ( L ) F j ⊗ \ U ( L ′ ) Fj \ D ( µ j ) F j , so by Lemma 4.1, I ( λ j ) F j = Ann \ U ( L ) Fj \ D ( λ j ) F j is the largest two-sided ideal of [ U ( L ) F j contained in [ U ( L ) F j Ann \ U ( L ′ ) F \ D ( µ j ) F j .But P = [ U ( L ) K Q ⊆ [ U ( L ) K I ( µ j ) F j , and by Proposition 5.6, [ U ( L ) F j = [ U ( L ) K ⊗ K F j ,hence P ⊗ K F j ⊆ [ U ( L ) F j I ( µ j ) F j ⊆ [ U ( L ) F j Ann \ U ( L ′ ) Fj \ D ( µ j ) F j .Thus P ⊗ K F j ⊆ Ann \ U ( L ) Fj \ D ( λ j ) F j and P ⊆ Ann \ U ( L ) K \ D ( λ j ) F j = I ( λ j ) F j .Furthermore, given r ∈ T j ∈ X I ( λ j ) F j , we have that r = P i ≥ x i r i for some r i ∈ \ U ( L ′ ) K byLemma 2.1, with r i → i → ∞ . Then since each I ( λ j ) F j is a prime ideal of [ U ( L ) K ,and z / ∈ I ( λ j ) F j , it follows from Theorem 4.2 that each r i lies in I ( λ j ) F j for every j .This means that r i \ D ( λ j ) F j = 0 for all i, j , so r i \ D ( µ j ) F j = 0 and thus r i ∈ T j ∈ X I ( µ j ) F j = Q for every i . Therefore r ∈ [ U ( L ) K Q = P . Since our choice of r was arbitrary, it followsthat: P = T j ∈ X I ( λ j ) F j .Since P is locally closed and each I ( λ j ) F j is a prime ideal of [ U ( L ) K containing P , itfollows that P = I ( λ j ) F j for some j ∈ X as we require.Finally, take P to be a general locally closed prime ideal. Then F = Z ( [ U ( L ) K /P ) is aﬁnite extension of K , so let g := g ⊗ K F , L := L ⊗ O O F . Then dim F g = dim K g = n , L is a Lie lattice in g , and by Proposition 5.6, there exists a surjection of F -algebras \ U ( L ) F = [ U ( L ) K ⊗ K F ։ [ U ( L ) K /P whose kernel contains P ⊗ K F . Let J be this kernel.Then J is a locally closed prime ideal of \ U ( L ) F and \ U ( L ) F /J ∼ = [ U ( L ) K /P . But Z ( [ U ( L ) F /J ) ∼ = Z ( [ U ( L ) K /P ) = F so it follows from the above discussion that J =Ann \ U ( L ) F [ D ( λ ) F ′ for some ﬁnite extension F ′ /F and some linear form λ of g ⊗ F F ′ suchthat λ ( L ) ⊆ O F ′ .It is clear that J ∩ [ U ( L ) K = P , and hence P = Ann \ U ( L ) K [ D ( λ ) F ′ = I ( λ ) F ′ as required. Corollary 5.9.

Let L be a Lie lattice in g nilpotent. Then given a prime ideal P of [ U ( L ) K , P arises as an intersection of Dixmier annihilators. roof. This is immediate from Theorem 5.8 and Proposition 5.4.

In this section, we will prove our main result, Theorem A, which allows us to describe allweakly rational ideals in [ U ( L ) K in terms of Dixmier annihilators. In section 2, we saw how the adjoint algebraic group G of g acts on g ∗ via the coadjointaction. More generally, if σ is any Lie automorphism of g , then for any λ in g ∗ , we cansimilarly deﬁne σ · λ : g → K, u λ ( σ − ( u )). Note that if L is a Lie lattice in g then σ ( L ) also is.More generally, let us suppose that L , L are Lie lattices in g and σ : L → L is an O -linear Lie isomorphism. Then σ extends to a K -linear isomorphism σ : \ U ( L ) K → \ U ( L ) K of aﬃnoid enveloping algebras.Given a linear form λ ∈ Hom O ( L , O ), σ · λ ∈ Hom O ( L , O ), and if B is a polarisationat λ then B = σ B is a polarisation at σ · λ . Lemma 6.1.

Let I ( λ ) := Ann \ U ( L ) K [ D ( λ ) B E \ U ( L ) K , I ( σ · λ ) := Ann \ U ( L ) K \ D ( σ · λ ) B E \ U ( L ) K . Then σ ( I ( λ )) = I ( σ · λ ) Proof.

By deﬁnition, [ D ( λ ) B = \ U ( L ) K ⊗ \ U ( B ) K K λ and \ D ( σ · λ ) B = \ U ( L ) K ⊗ \ U ( B ) K K σ · λ = \ U ( σ L ) K ⊗ \ U ( σ B ) K K σ · λ . So consider the map Θ : [ D ( λ ) B → \ D ( σ · λ ) B , x ⊗ v σ ( x ) ⊗ v .We will show that Θ is a K -linear isomorphism such that Θ( xm ) = σ ( x )Θ( m ) for all x ∈ [ U ( L ) K , m ∈ [ D ( λ ) B . It will follow from this that x [ D ( λ ) B = 0 if and only if σ ( x ) \ D ( σ · λ ) B = 0, and hence σ ( I ( λ )) = I ( σ · λ ) as required.It is clear that Θ is K -linear, and that it has an inverse deﬁned by x ⊗ v σ − ( x ) ⊗ v ,hence it is an isomorphism of vector spaces.Finally, Θ( x ( y ⊗ v )) = Θ( xy ⊗ v ) = σ ( xy ) ⊗ v = σ ( x )( σ ( y ) ⊗ v ) = σ ( x )Θ( y ⊗ v ).This result becomes particularly useful when comparing Dixmier annihilators, particu-larly using the following lemma. Lemma 6.2.

Suppose λ, µ ∈ Hom O ( L , O ) such that I ( λ ) ∩ U ( g ) = I ( µ ) ∩ U ( g ) , e.g. if I ( λ ) ⊆ I ( µ ) . Then there exists g ∈ G ( K ) such that µ = g · λ .Proof. Since the classical Dixmier module D ( λ ) is dense in [ D ( λ ), it follows that r [ D ( λ ) = 0if and only if rD ( λ ) = 0, and hence I ( λ ) ∩ U ( g ) = Ann U ( g ) D ( λ ). Therefore, if I ( λ ) ∩ U ( g ) = I ( µ ) ∩ U ( g ) then Ann U ( g ) D ( λ ) = Ann U ( g ) D ( µ ), and using [7, Proposition 6.2.3], it followsthat λ and µ lie in the same coadjoint orbit as required.35lso, if I ( λ ) ⊆ I ( µ ) then Ann U ( g ) D ( λ ) ⊆ Ann U ( g ) D ( µ ). But Ann U ( g ) D ( λ ) is a weaklyrational ideal of U ( g ), and hence it is maximal by [7, Proposition 4.7.4], therefore I ( λ ) ∩ U ( g ) = Ann U ( g ) D ( λ ) = Ann U ( g ) D ( µ ) = I ( µ ) ∩ U ( g ). Now, let λ, µ : g → F be linear forms such that λ ( L ) , µ ( L ) ⊆ O F for some ﬁnite extension F/K . We want to compare the Dixmier annihilators I ( λ ) and I ( µ ) in [ U ( L ) K in the casewhere λ and µ lie in the same coadjoint orbit, i.e. µ = g · λ for some g ∈ G ( F ). Explicitly, g = exp(ad( u )) for some u ∈ g ⊗ K F .Our ﬁrst results ensure that it is suﬃcient to consider the case where F = K . Proposition 6.3.

Let

F/K be a ﬁnite extension, and let λ : g → K be K -linear. Thenthere exists a polarisation b of g at λ such that b ⊗ K F is a polarisation for g ⊗ K F atthe extension λ F : g ⊗ K F → F .Proof. Using induction on dim( g ). If dim( g ) = 1 then it is obvious, because g and g ⊗ F are the only polarisations. So suppose the result holds whenever dim( g ) < n .If λ ( a ) = 0 for some non-zero ideal a of g , then using induction we may choose a polari-sation a ⊆ b such that ba ⊗ K F = b ⊗ K F a ⊗ K F is a polarisation for g ⊗ K F a ⊗ K F at λ F . Hence b ⊗ K F is a polarisation for g ⊗ K F at λ F .So from now on, we may assume that λ ( a ) = 0 for all non-zero ideals a of g . Then itfollows from Proposition 2.7 that g has a reducing quadruple ( x, y, z, g ′ ), and λ ( z ) = 0.Clearly ( x ⊗ , y ⊗ , z ⊗ , g ′ ⊗ K F ) is a reducing quadruple for g ⊗ K F .Let b be a polarisation for g ′ at λ | g ′ such that b ⊗ K F is a polarisation for g ′ ⊗ K F . Thenusing Lemma 2.7 we see that b is a polarisation for g at λ , and b ⊗ K F is a polarisationfor g ⊗ K F at λ F as required. Corollary 6.4.

Let

F/K be a ﬁnite extension, and let λ : g → F be K -linear such that λ ( L ) ⊆ O F . Then for any ﬁnite extension L/F , I ( λ ) F = I ( λ ) L .Proof. This is immediate from Proposition 6.3 and Theorem 4.4.So, from now on, we will assume that λ and µ take values in K , and µ = exp(ad( u )) · λ for some u ∈ g . Let σ := exp(ad( u )) ∈ G ( K ), and ﬁx a natural number N ∈ N such that u ∈ p − N L . Also let c be the nilpotency class of g , i.e. c is minimal such that ad( g ) c = 0.Since σ is a Lie automorphism of g , it follows that σ L is an O -Lie lattice in g , hencethere exists a natural number n ∈ N such that p n L ⊆ σ L and p n σ L ⊆ L . Lemma 6.5.

For any n ≥ cN + v p ( c !) , p n L ⊆ σ L and p n σ L ⊆ L .Proof.

Since σ = exp(ad( u )), where u = p − N v for some v ∈ L , it follows that for all w ∈ L : σ ( w ) = w + p − N [ v, w ] + 12 p − N [ v, [ v, w ]] + · · · + 1 c ! p − cN ( ad ( v )) c ( w ) (7)36ut for each 0 ≤ i ≤ c , (ad( v )) i ( w ) ∈ L , v p ( i ! p − iN ) = − iN − v p ( i !) ≥ − cN − v p ( c !) ≥ − n ,so i ! p − iN (ad( v )) i ( w ) ∈ p − n L . Hence σ L ⊆ p − n L , and p n σ L ⊆ L .Also, σ is an isomorphism, and σ − = exp(ad( − u )), with − u ∈ p − N L . So since σ − : σ L → L is a Lie-isomorphism, it follows from the above discussion that p n L ⊆ σ L .It is clear that since σ : L → σ L is a continuous isomorphism of O -Lie lattices, it extendsto a continuous isomorphism σ : [ U ( L ) K → \ U ( σ L ) K of K -algebras. Moreover, for any n ∈ N , σ induces an isomorphism σ : \ U ( p n L ) K → \ U ( p n σ L ) K , and thus using Lemma6.5, for n ≥ cN + v p ( c !), there is an injective K -algebra homomorphism σ : \ U ( p n L ) K → [ U ( L ) K . Proposition 6.6.

Given n ∈ N such that n ≥ cN + v p ( c !) , if I is a two-sided ideal of [ U ( L ) K , then σ : \ U ( p n L ) K → [ U ( L ) K maps I ∩ \ U ( p n L ) K into I .Proof. Consider the sequence of continuous O -linear maps σ i := P ≤ j ≤ i j ! (ad( u )) j : \ U ( p n L ) K → [ U ( L ) K . Clearly each of these sends I ∩ \ U ( p n L ) K into I .We will show that σ i converges pointwise to σ as i → ∞ , and since all ideals in [ U ( L ) K are closed, it will follow that σ ( I ∩ \ U ( p n L ) K ) ⊆ I .Let δ := ad( u ), and let v be the p -adic ﬁltration on [ U ( L ) K induced from [ U ( L ). Then forall u ∈ L , v ( δ ( u )) ≥ v ( u ) − N .Since δ is a derivation, a standard inductive argument shows that for all a , · · · , a r ∈ [ U ( L ) K : X ≤ j ≤ i j ! δ j ( a a · · · a r ) = X ≤ j ≤ i X j + ··· + j r = j Y ≤ m ≤ r j m ! δ j m ( a j ) !! (8)So, if x ∈ [ U ( L ) K , then x = P ( r,u ) λ u u · · · u r , where the sum is taken over all r ≥ u = u · · · u r for u i ∈ L , and v p ( λ u ) − nr → ∞ as r → ∞ . Therefore, ﬁxing t ∈ N , wehave:( σ − σ t )( x ) = X ( r,u ) λ u ( σ − σ t )( u · · · u r ) = X ( r,u ) λ u X j>t X j + ··· + j r = j Y ≤ m ≤ r j m ! δ j m ( u m ) !!! (9)For each r ≥

0, let A r := { α ∈ [ c ] r : | α | > t } , where [ c ] = { , · · · , c − } . Note that δ c ( u ) = 0 for all u ∈ L .Then ( σ − σ t )( x ) = P ( r,u ) λ u Ç P α ∈ A r Q ≤ m ≤ r α m ! δ α m ( u m ) å , and since A r is ﬁnite, P α ∈ A r Q ≤ m ≤ r α m ! δ α m ( u m )is a ﬁnite sum. 37ince α m < c for all α ∈ A r , we have that v p ( α m !) ≤ v p ( c !). Also, since v ( δ ( u )) ≥ v ( u ) − N ,it follows that v ( δ α m ( u )) ≥ v ( u ) − α m N . Therefore v ( α m ! δ α m ( u m )) ≥ v ( u m ) − α m N − v p ( c !)for all m ≤ r .Thus for each pair ( r, u ), v ( P α ∈ A r Q ≤ m ≤ r α m ! δ α m ( u m )) ≥ v ( u )+ · · · + v ( u r ) − ( | α | N + v p ( c !) r ) ≥− ( | α | N + rv p ( c !)) ≥ − r ( cN + v p ( c !)), the last inequality follows since | α | ≤ rc .Therefore, v Ç λ u Ç P α ∈ A r Q ≤ m ≤ r α m ! δ α m ( u m ) åå ≥ v p ( λ u ) − r ( cN + v p ( c !)) ≥ v p ( λ u ) − nr → ∞ as r → ∞ .Moreover, for r ≤ tc , A r = ∅ , so we have:( σ − σ t )( x ) = X ( r,u ) ,r> tc λ u X α ∈ A r Y ≤ m ≤ r α m ! δ α m ( u m ) ! (10)Therefore, v (( σ − σ t )( x )) ≥ inf { v p ( λ u ) − nr : u = u · · · u r with r > tc } , and this tends toinﬁnity as t → ∞ . Hence ( σ − σ t )( x ) → t → ∞ .So σ ( x ) = lim t →∞ σ t ( x ), so if x ∈ I then σ ( x ) ∈ I as required.Now, let b be a polarisation for g at λ , and let b ′ be a polarisation for g at µ . Since µ = σ · λ , it follows that σ b is also a polarisation for g at µ . Also, it is clear that σ b ∩ σ L = σ ( b ∩ L ), so let B := b ∩ L and B ′ := b ′ ∩ L .Consider the Dixmier modules [ D ( λ ) B := [ U ( L ) K ⊗ \ U ( B ) K K λ , [ D ( µ ) σ B := \ U ( σ L ) K ⊗ \ U ( σ B ) K K µ , [ D ( µ ) B ′ := [ U ( L ) K ⊗ \ U ( B ′ ) K K µ .Then [ D ( λ ) B and [ D ( µ ) B ′ are [ U ( L ) K -modules, topological completions of the U ( g )-modules D ( λ ) b and D ( µ ) b ′ respectively, while [ D ( µ ) σ B is a \ U ( σ L ) K -module, a topological comple-tion of D ( µ ) σ b .Let I ( µ ) := Ann \ U ( L ) K [ D ( µ ) B ′ E [ U ( L ) K , and let I ′ ( µ ) := Ann \ U ( σ L ) K [ D ( µ ) σ B E \ U ( σ L ) K . Lemma 6.7. I ′ ( µ ) = σ ( I ( λ )) , and given n ∈ N such that p n L ⊆ σ L , I ( µ ) ∩ \ U ( p n L ) K = I ′ ( µ ) ∩ \ U ( p n L ) K .Proof. Since σ : L → σ L is a Lie isomorphism, it follows from Lemma 6.1 that I ′ ( µ ) = σ ( I ( λ )).Let C := σ b ∩ p n L , then the \ U ( p n L ) K -aﬃnoid Dixmier module [ D ( µ ) C := \ U ( p n L ) K ⊗ [ U ( C ) K K µ embeds densely into [ D ( µ ) σ B . So if x ∈ \ U ( p n L ) K then x [ D ( µ ) C = 0 if and only if x [ D ( µ ) σ B = 0. 38ut since σ b and b ′ are polarisations of g at µ with σ b ∩ p n L = C and b ′ ∩ p n L = p n B ′ , wecan apply Theorem 4.4 to get that Ann \ U ( p n L ) K [ D ( µ ) C = Ann \ U ( p n L ) K [ D ( µ ) p n B ′ . Therefore,given x ∈ \ U ( p n L ) K : x [ D ( µ ) σ B = 0 ⇐⇒ x [ D ( µ ) C = 0 ⇐⇒ [ D ( µ ) p n B ′ = 0 ⇐⇒ x [ D ( µ ) B ′ = 0Therefore I ( µ ) ∩ \ U ( p n L ) K = I ′ ( µ ) ∩ \ U ( p n L ) K as required.Now we can prove the main result of this section, which allows us to compare Dixmierannihilators for λ, µ ∈ g ∗ in the same coadjoint orbit: Theorem 6.8.

Let g be a nilpotent K -Lie algebra, with nilpotency class c , and let L bean O -Lie lattice in g . Let λ, µ : g → K be K -linear maps such that λ ( L ) ⊆ O K , andsuppose that µ = exp(ad( u )) · λ for some u ∈ p − N ( L ⊗ O O K ) . Then given n ∈ N suchthat n ≥ N c + v p ( c !) , I ( λ ) ∩ \ U ( p n L ) K = I ( µ ) ∩ \ U ( p n L ) K .Proof. Firstly, note that u lies in p − N ( L ⊗ O O F ) for some ﬁnite extension F of K , andwe may assume further that λ and µ take values in F , possibly after extending F .Using Corollary 6.4, we see that I ( λ ) = I ( λ ) F , i.e. I ( λ ) = Ann \ U ( L ) K [ D ( λ ) F , and similarly I ( µ ) = Ann \ U ( L ) K [ D ( µ ) F , so we may safely assume that F = K . So λ and µ are K -linearforms of g , u ∈ p − N L , and setting σ := exp(ad( u )), since n ≥ N c + v ( c !) we see usingLemma 6.5 that p n L ⊆ σ L and p n σ L ⊆ L .We will prove that I ( µ ) ∩ \ U ( p n L ) K ⊆ I ( λ ), and after replacing σ by σ − = exp(ad( − u )),it will follow that I ( λ ) ∩ \ U ( p n L ) K ⊆ I ( µ ) as required.By Lemma 6.7, we see that I ( µ ) ∩ \ U ( p n L ) K = σ ( I ( λ )) ∩ \ U ( p n L ) K ⊆ σ ( I ( λ )) ∩ \ U ( p n σ L ) K = σ ( I ( λ ) ∩ \ U ( p n L ) K ).But since I ( λ ) is a two-sided ideal of [ U ( L ) K , it follows from Proposition 6.6 that σ ( I ( λ ) ∩ \ U ( p n L ) K ) ⊆ I ( λ ), and hence I ( µ ) ∩ \ U ( p n L ) K ⊆ I ( λ ) as required. Now, let P be a weakly rational ideal of [ U ( L ) K . Using Theorem 5.8 and Proposition 5.4,we see that P arises an an intersection of Dixmier annihilators: P = T j ∈ X I ( λ j )for some λ j : g → K K -linear, such that λ j ( L ) ⊆ O K for each j .Also, P ∩ U ( g ) is a weakly rational ideal of U ( g ) and hence it is maximal by [7, Proposition4.7.4]. So since P ∩ U ( g ) ⊆ I ( λ j ) ∩ U ( g ) for all j , it follows that I ( λ j ) ∩ U ( g ) = I ( λ k ) ∩ U ( g )for all j, k ∈ X .Therefore, using Lemma 6.2, this means that for every j, k ∈ X , there exists a j,k ∈ G ( K ) such that a j,k · λ j = λ k , i.e. all λ j lie in the same coadjoint orbit.39 roposition 6.9. Let λ : g → K be a K -linear map such that λ ( L ) ⊆ O K . Then thereexists an integer N ≥ such that for any linear form µ : g → K in the G -coadjoint orbitof λ with µ ( L ) ⊆ O K , µ = exp(ad( u )) · λ for some u ∈ p − N ( L ⊗ O O K ) .Proof. Let Y be the coadjoint orbit in g ∗ = Hom K ( g , K ) containing λ , and recall fromLemma 2.14 that there exists an aﬃne algebraic subgroup S of G such that G ∼ = S × Y as varieties, where the natural morphism G → Y, g g · λ is just the projection to thesecond factor. Consider the following sequence in the category of K -varieties deﬁned over K : ad( g ) → exp G ∼ = S × Y → Y. (11)Let U be the set of all µ ∈ Y such that µ ( L ) ⊆ O K . Then U is an aﬃnoid subdomainof Y (when Y is considered as a rigid variety) isomorphic to Sp ‘ S ( g ), where ‘ S ( g ) is the π -adic completion of the symmetric algebra S ( g ) with respect to the lattice S ( L ). Sinceexp is an isomorphism, we may take the inverse image V := exp − (1 × U ) of 1 × U , whichwill be an aﬃnoid subdomain of ad( g ).But ad( g ) ∼ = g /Z ( g ) is a union of open discs containing p − n ( L /Z ( L )) for n ∈ N . So since V is aﬃnoid, it follows that V is contained in p − N ( L /Z ( L ) ⊗ O O K ) for some N ∈ N .Therefore, for any µ ∈ U , we can choose u ∈ p − N ( L ⊗ O O K ) such that µ is the imageof ad( u ) under the composition ad( g ) → G → Y , i.e. µ = exp(ad( u )) · λ as required.Now we can ﬁnally prove our main theorem: Proof of Theorem A.

Let P be a weakly rational ideal of [ U ( L ) K , then since P is prime,we see using Corollary 5.9 that P = T j ∈ X I ( λ j ) for linear forms λ j all lying in the samecoadjoint orbit.Since λ j ( L ) ⊆ O K for each j , it follows from Proposition 6.9 that we can choose N ∈ N , u j,k ∈ p − N ( L ⊗ O O K ) for each j, k ∈ X such that λ k = exp(ad( u j,k )) · λ j .Therefore, let c be the nilpotency class of g , and choose n ∈ N with n ≥ N c + 2 v ( c !).Then using Theorem 6.8, we see that I ( λ j ) ∩ \ U ( p n L ) K = I ( λ k ) ∩ \ U ( p n L ) K for each j, k ∈ X .Therefore, P ∩ \ U ( p n L ) K = T j ∈ X I ( λ j ) ∩ \ U ( p n L ) K = I ( λ j ) ∩ \ U ( p n L ) K for any j ∈ X . Hence P ∩ \ U ( p n L ) K = Ann \ U ( p n L ) K \ D ( λ j ) is a Dixmier annihilator as required.We suspect that we can always take n = 0 in the statement of Theorem A, but we donot have a proof of this. As outlined in the introduction, our aim is to prove the deformed Dixmier-Moeglin equiv-alence for [ U ( L ) K , and using Proposition 5.3, we see that this just means proving that if40 is a weakly rational ideal of [ U ( L ) K then P ∩ \ U ( π n L ) K is locally closed in \ U ( π n L ) K for n suﬃciently high. In fact, in the case where L is nilpotent, we suspect moreover that P ∩ \ U ( π n L ) K is maximal.So now that we have established that weakly rational ideals become Dixmier anni-hilators after passing to a suitably high deformation \ U ( π n L ) K , it remains to prove thatDixmier annihilator ideals I ( λ ) are maximal. In this section, we will establish this insome important cases, and in particular prove Theorem B. Fix

F/K a ﬁnite, Galois extension, with G := Gal ( F/K ), let R be a K -algebra and let S := R ⊗ K F . There is a natural action of G on S by automorphisms via σ · ( r ⊗ α ) = r ⊗ σ ( α ), and clearly R is invariant under this action.We will make a further assumption on S , namely that if α , · · · , α n ∈ F are linearlyindependent over K and r ⊗ α + · · · + r n ⊗ α n = 0 in S then r i = 0 for all i . UsingLemma 2.1, it is easily veriﬁed that this property is satisﬁed if we take R to be the aﬃnoidenveloping algebra [ U ( L ) K . Proposition 7.1.

Let I be a G -invariant ideal of S , and let J := I ∩ R . Then I = J ⊗ K F .In fact, if r ⊗ α + · · · + r n ⊗ α n ∈ I with α , · · · , α n linearly independent over K , then r , · · · , r n ∈ J .Proof. For each s ∈ S , let l ( s ) be the minimal number of simple tensors required to sumto s , so l ( s ) = 0 if and only if s = 0. If s ∈ I and l ( s ) = 1 then s = r ⊗ α is a simpletensor, with α = 0, thus r = α − s ∈ I ∩ R = J . Hence s ∈ J ⊗ K F .So, we will assume, for induction, that for all s ∈ I with l ( s ) < n , s ∈ J ⊗ K F ,and moreover that if s = r ⊗ α + · · · + r m ⊗ α m with m < n and α , · · · , α m linearlyindependent over K , then r i ∈ J for all i .Now, suppose s ∈ I with l ( s ) = n >

1. Then choose a subﬁeld F ′ ⊆ F with K ⊆ F ′ such that F ′ is minimal with respect to the property that there exist r , · · · , r n ∈ R , α , · · · , α n ∈ F ′ such that s = r ⊗ α + · · · + r n ⊗ α n . Since l ( s ) = n , it follows immedi-ately that α , · · · , α n are linearly independent over K , and note that since α − s ∈ I and l ( s ) = l ( α − s ), we may assume that α = 1.If F ′ = K then s ∈ R so l ( s ) = 1, a contradiction. Hence F ′ = K , so since F/K is aGalois extension, there exists σ ∈ G such that the set F ′′ := { α ∈ F ′ : σ ( α ) = α } 6 = F ′ .Clearly F ′′ is a subﬁeld of F ′ with K ⊆ F ′′ .Since α = 1 ∈ F ′′ , choose m ≥ t , · · · , t n ∈ R , β , · · · , β n ∈ F ′ with β , · · · , β m ∈ F ′′ and s = t ⊗ β + · · · + t n ⊗ β n . Clearly m < n byminimality of F ′ and the fact that F ′′ ( F ′ .Now, since I is G -invariant, σ ( s ) − s ∈ I . But σ ( s ) − s = t ⊗ ( σ ( β ) − β ) + · · · + t n ⊗ ( σ ( β n ) − β n ) = t m +1 ⊗ ( σ ( β m +1 ) − β m +1 ) + · · · + t n ⊗ ( σ ( β n ) − β n ).If t i ∈ J for all i > m then t ⊗ β + · · · + t m ⊗ β m ∈ I , and hence since m < n and β , · · · , β m are linearly independent over K , t i ∈ J for all i by induction, and hence s ∈ J ⊗ K F . 41herefore, we will assume for contradiction that t i / ∈ J for some i > m , so since σ ( s ) − s ∈ I , it follows from induction that σ ( β m +1 ) − β m +1 , · · · , σ ( β n ) − β n are linearlydependent over K .So, there exist γ m +1 , · · · , γ n ∈ K , not all zero, such that σ ( γ m +1 β m +1 + · · · + γ n β n ) = γ m +1 β m +1 + · · · + γ n β n . So let β ′ := γ m +1 β m +1 + · · · + γ n β n ∈ F ′ , and clearly β ′ ∈ F ′′ bythe deﬁnition of F ′′ . We may assume without loss of generality that γ m +1 = 0. Thus: s = t ⊗ β + · · · + t n ⊗ β n = t ⊗ β + · · · + t m ⊗ β m + γ − m +1 t m +1 ⊗ ( β ′ − γ m +2 β m +2 − · · · − γ n β n ) + t m +2 ⊗ β m +2 + · · · + t n ⊗ β n = t ⊗ β + · · · + t m ⊗ β m + γ − m +1 t m +1 ⊗ β ′ + ( t m +2 − γ − m +1 γ m +2 t m +1 ) ⊗ β m +2 + · · · +( t n − γ − m +1 γ n t n ) ⊗ β n .But since β , · · · , β m , β ′ ∈ F ′′ , this contradicts the maximality of m . Therefore s ∈ J ⊗ K F .Finally, to complete the induction, suppose that s = r ⊗ α + · · · + r n ⊗ α n ∈ I with α , · · · , α n linearly independent over K . Then l ( s ) ≤ n so we have proved that s ∈ J ⊗ K F ,so there exist t , · · · , t m ∈ J , β , · · · , β m ∈ F such that s = t ⊗ β + · · · + t m ⊗ β m .So, extend { α , · · · , α n } to a basis { α , · · · , α d } of F/K , and set r n +1 = · · · = r d = 0.Then for each i = 1 · · · , m , β i = P ≤ j ≤ d γ i,j α j for some γ i,j ∈ K , and: s = P ≤ i ≤ m t i ⊗ β i = P ≤ i ≤ m P ≤ j ≤ d γ i,j t i ⊗ α j = P ≤ j ≤ d Ç P ≤ i ≤ m γ i,j t i å ⊗ α j .But we know that s = P ≤ j ≤ d r j ⊗ α j , and hence P ≤ j ≤ d Ç r j − P ≤ i ≤ m γ i,j t i å ⊗ α j = 0, and thisimplies that r j = P ≤ i ≤ m γ i,j t i ∈ J for each j as required. Corollary 7.2.

If we assume S has ﬁnite left and right Krull dimension, and P be isprime ideal of S , then { σ ( P ) : σ ∈ G } form a complete set of minimal prime ideals above ( P ∩ R ) ⊗ K F .Proof. Let Q = P ∩ R . Then for every σ ∈ G , σ ( Q ) = Q , and hence Q ⊆ σ ( P ). So,let I := ∩ σ ∈ G σ ( P ), then I is a G -invariant two-sided ideal of S , and I ∩ R = Q , thus I = Q ⊗ K F by Proposition 7.1.If P ′ is a prime ideal of S with I ⊆ P ′ ⊆ P , then Q σ ∈ G σ ( P ) ⊆ I ⊆ P ′ , which implies that σ ( P ) ⊆ P ′ ⊆ P for some σ ∈ G , so P ⊆ σ − ( P ).But S has ﬁnite Krull dimension, so it follows from [16, Lemma 6.4.5] that S has ﬁniteclassical Krull dimension, and hence there exists n ≥ P = P ( P ( · · · ( P n of S . But P ⊆ σ − ( P ) ( σ − ( P ) ( · · · ( σ − ( P n ) is also a chain of prime ideals, and thus P = σ − ( P ) and σ ( P ) = P .42ut since σ ( P ) ⊆ P ′ ⊆ P , this implies that P = P ′ , and hence P is minimal primeabove I as required. The same argument shows that σ ( P ) is minimal prime above I forevery σ ∈ G .In particular, we could take R = [ U ( L ) K , in which case S = [ U ( L ) F has ﬁnite Krull di-mension as required. Therefore, we can apply this result in the case where P is a Dixmierannihilator.So, let λ : g → F be a linear form such that λ ( L ) ⊆ O F , and P := Ann \ U ( L ) F [ D ( λ ) F , thenfor every σ ∈ G , σ ( P ) = Ann \ U ( L ) F \ D ( σ · λ ) F by Lemma 6.1. Proposition 7.3.

Let λ, µ : g → F be linear forms such that λ ( L ) , µ ( L ) ⊆ O F . Let P = I ( λ ) E [ U ( L ) K , Q = I ( µ ) E [ U ( L ) K .Then if P ⊆ Q , there exists σ ∈ G such that if P σ := Ann \ U ( L ) F \ D ( σ · λ ) F and Q :=Ann \ U ( L ) F [ D ( µ ) F then P σ ⊆ Q .Proof. Firstly, let P = Ann \ U ( L ) F [ D ( λ ) F . Then since P , Q are primitive ideals ofAnn \ U ( L ) F [ D ( λ ) F by Theorem 5.8 and Theorem 4.4, it follows from Corollary 7.2 thatthe set { σ ( P ) : σ ∈ G } consists of all minimal prime ideals above ( P ∩ [ U ( L ) K ) ⊗ K F = P ⊗ K F , and also that Q is minimal prime above Q ⊗ K F .Therefore, if P ⊆ Q then P ⊗ K F ⊆ Q ⊗ K F ⊆ Q , and hence there exists σ ∈ G suchthat σ ( P ) ⊆ Q . But using Lemma 6.1 we see that σ ( P ) = Ann \ U ( L ) F \ D ( σ · λ ) F = P σ asrequired. Now, ﬁx linear forms λ, µ : g → K such that λ ( L ) , µ ( L ) ⊆ O , and suppose that I ( λ ) ⊆ I ( µ ). Let us assume further that L is powerful, i.e. [ L , L ] ⊆ p L .Let a be an abelian ideal of g , and recall from section 3.2 that we deﬁne a ⊥ = { u ∈ g : λ ([ u, a ]) = 0 } . Also, using Proposition 2.4, we may ﬁx a polarisation b of g at λ such that a ⊆ b , and construct the aﬃnoid Dixmier module [ D ( λ ) b with respect to this polarisation.Note that since λ ([ a , b ]) = 0, b ⊆ a ⊥ .Suppose a ⊥ has codimension s in g , and ﬁx a basis { u , · · · , u s } for L / ( a ⊥ ∩ L ). Then if A := a ∩ L , it follows from Corollary 3.4 and Proposition 3.5 that the action [ U ( A ) K → End K [ D ( λ ) has image contained in K h ∂ , · · · , ∂ s i , where ∂ i = ddu i , and this image contains ∂ , · · · , ∂ s . Proposition 7.4.

The quotient \ U ( A ) K I ( λ ) ∩ \ U ( A ) K has Krull dimension s .Proof. We will prove that the image R of [ U ( A ) K under the action ρ : [ U ( A ) K → End K [ D ( λ ) b has Krull dimension s . Since the Dixmier annihilator I ( λ ) does not de-pend on the choice of polarisation by Theorem 4.4, the result will follow.43ote that since A is abelian, [ U ( A ) K is isomorphic to a commutative Tate-power seriesring, and hence R is a commutative aﬃnoid algebra. Also, since ∂ , · · · , ∂ s ∈ ρ ( [ U ( A ) K ),it follows that there exist m ≥ π m ∂ , · · · , π m ∂ s ∈ ρ ( [ U ( A )), and hence everypower series in K h π m ∂ , · · · , π m ∂ s i lies in ρ ( [ U ( A ) K ) = R .Therefore we have inclusions of commutative aﬃnoid algebras, K h π m ∂ , · · · , π m ∂ s i ֒ −→ R ֒ −→ K h ∂ , · · · , ∂ s i , which gives rise to a chain of open embeddings of the associatedaﬃnoid spectra: Sp K h ∂ , · · · , ∂ s i ֒ −→ Sp R ֒ −→ Sp K h π m ∂ , · · · , π m ∂ s i .Now, in [9] the notion of the analytic dimension dim X of a rigid variety X is deﬁned,and it is proved to be equal to the supremum of the Krull dimensions of every aﬃnoidalgebra A such that Sp A is an aﬃnoid subdomain of X . In particular, if Sp A is anaﬃnoid subdomain of Sp B in the sense of Deﬁnition 2.1, then K.dim( A ) ≤ K.dim( B ).Therefore, since the Tate algebras K h ∂ , · · · , ∂ s i and K h π m ∂ , · · · , π m ∂ s i both havedimension s , it remains to prove that the embeddings Sp R → Sp K h π m ∂ , · · · , π m ∂ s i and Sp K h ∂ , · · · , ∂ s i → Sp R deﬁne aﬃnoid subdomains.For convenience, set D := Sp K h ∂ , · · · , ∂ s i and D := Sp K h π m ∂ , · · · , π m ∂ s i . Then D can be realised as the unit disc in s -dimensional rigid K -space, while D is a larger disccontaining D , so explicitly D = { ( x , · · · , x s ) ∈ D : | x i | ≤ i } .But since Sp R contains D , we could instead write D = { ( x , · · · , x s ) ∈ Sp R : | x i | ≤ i } , and this is a Weierstrass subdomain of Sp R in the sense of [5, Deﬁnition 3.3.7],and hence D is an aﬃnoid subdomain of Sp R by [5, Proposition 3.3.11]. ThereforeK.dim( R ) ≥ dim D = s .Now, choose a basis { v , · · · , v r } for A , so that [ U ( A ) K ∼ = K h v , · · · , v r i . Using Propo-sition 3.5 we see that for each i , ρ ( v i ) = f i ( ∂ , · · · , ∂ s ) for some polynomial f i . So, let T := K h π m ∂ , · · · , π m s ∂ s i , then since ρ ( [ U ( A ) K ) contains T , it follows that the aﬃnoidalgebra B := T h ζ , · · · , ζ r i / ( ζ i − f i ( ∂ , · · · , ∂ s )) surjects onto R = ρ ( [ U ( A ) K ), where each a ∈ T is sent to a , and each ζ i is sent to ρ ( v i ). Therefore, K.dim( R ) ≤ K.dim( B ).But clearly there is a map T → B , inducing a map of aﬃnoid varieties Sp B → Sp T ,and the proof of [5, Proposition 3.3.11] shows that this corresponds to the embedding ofthe Weierstrass subdomain { x ∈ Sp T : | f i ( x ) | ≤ i } , into Sp T , and hence Sp B is an aﬃnoid subdomain of Sp T = D by [5, Proposition 3.3.11], and hence K.dim( B ) ≤ dim( D ) = s .Therefore s ≤ K.dim( R ) ≤ K.dim( B ) ≤ s , forcing equality, so K.dim( R ) = s asrequired. Corollary 7.5.

Suppose that L is a powerful Lie lattice in g , λ, µ : g → K are linearforms with λ ( L ) , µ ( L ) ⊆ O , and suppose that I ( λ ) ⊆ I ( µ ) . Then for any abelian ideal a of g , if A := a ∩ L then I ( λ ) ∩ [ U ( A ) K = I ( µ ) ∩ [ U ( A ) K .Proof. Let a ⊥ = { u ∈ g : λ ([ u, a ]) = 0 } , and let a ⊥ = { u ∈ g : µ ([ u, a ]) = 0 } . Thenif s i is the codimension of a ⊥ i in g , for i = 1 ,

2, then it follows from Proposition 7.4that [ U ( A ) K /I ( λ ) ∩ [ U ( A ) K has Krull dimension s and [ U ( A ) K /I ( µ ) ∩ [ U ( A ) K has Krulldimension s . 44ince I ( λ ) ⊆ I ( µ ), clearly there is a surjection [ U ( A ) K /I ( λ ) ∩ [ U ( A ) K → [ U ( A ) K /I ( µ ) ∩ [ U ( A ) K , and by Corollary 3.4 both these algebras are commutative domains. Therefore s ≤ s , with equality if and only if I ( λ ) ∩ [ U ( A ) K = I ( µ ) ∩ [ U ( A ) K , so we will prove that s = s .Since I ( λ ) ⊆ I ( µ ), it follows from Lemma 6.2 that there exists g ∈ G ( K ) such that µ = g · λ . We will prove that the image of a ⊥ under g is a ⊥ , and hence a ⊥ and a ⊥ musthave the same dimension, and hence the same codimension in g as required.Firstly, if u ∈ a ⊥ , then λ ([ u, g ]) = 0, so given v ∈ g , µ ([ g ( u ) , v ]) = ( g · λ )([ g ( u ) , g ( g − ( v ))]) = λ ( g − g [ u, g − ( v )]) = λ ([ u, g − ( v )]) = 0, and hence µ ([ g ( u ) , g ]) = 0 and g ( u ) ∈ a ⊥ . Thus g ( a ⊥ ) ⊆ a ⊥ Conversely, since λ = g − · µ , the same argument shows that g − ( a ⊥ ) ⊆ a ⊥ , and hence g ( a ⊥ ) = a ⊥ as required. A particularly useful technique, which we hope should ultimately help us complete theproof of the deformed Dixmier-Moeglin equivalence in full generality, is to look for asuitable generating set for Dixmier annihilators.Speciﬁcally, recall that an ideal I of [ U ( L ) K is controlled by a closed ideal A of L if I = [ U ( L ) K ( I ∩ [ U ( A ) K ), i.e. I is generated as a left ideal by a subset of [ U ( A ) K . If A = a ∩ L for some ideal a of g , we may also say that I is controlled by a for convenience. Lemma 7.6.

Let I be a two-sided ideal of [ U ( L ) K , and let a be an ideal of g : • If I is controlled by a , then I is controlled by any ideal a ′ of g with a ⊆ a ′ . • If I = ∩ i ∈ X J i for some two-sided ideals J i of [ U ( L ) K , and each J i is controlled by a ,then I is controlled by a . • If F/K is a ﬁnite extension, and J is a two-sided ideal of [ U ( L ) F such that J ∩ [ U ( L ) K = I , then if J is controlled by a ⊗ K F , I is controlled by a .Proof. For the ﬁrst statement, it is clear that if I is controlled by A = a ∩L ⊆ a ′ ∩L = A ′ ,then I = [ U ( L ) K ( I ∩ [ U ( A ) K ) ⊆ [ U ( L ) K ( I ∩ \ U ( A ′ ) K ) ⊆ I , forcing equality.Now, let A = a ∩ L , and let { u , · · · , u d } be a basis for L such that { u r +1 , · · · , u d } isa basis for A for some r ≥

0. Using Lemma 2.1, we see that every element s ∈ [ U ( L ) K can be written as s = P α ∈ N d λ α u α · · · u α d d for some unique λ α ∈ K with λ α → α → ∞ . For each β ∈ N r , set s β = P γ ∈ N d − r λ ( β,γ ) u γ r +1 r +1 · · · u γ d d ∈ [ U ( A ) K , and it follows that s = P β ∈ N r u β · · · u β r r s β .We will ﬁrst prove that I is controlled by A if and only if for every s ∈ I , s β ∈ I forall β ∈ N r . One direction is clear since if I = [ U ( L ) K ( I ∩ [ U ( A ) K ) then every element of I is a sum of elements of the form su with u ∈ I ∩ [ U ( A ) K , so su = P β ∈ N r u β · · · u β r r s β u and45 β u ∈ I as required.Conversely, if s β ∈ I for all s ∈ I then u β · · · u β r r s β ∈ [ U ( L ) K ( I ∩ [ U ( A ) K ) for each β , sosince [ U ( L ) K ( I ∩ [ U ( A ) K ) is closed in [ U ( L ) K , it follows that s ∈ [ U ( L ) K ( I ∩ [ U ( A ) K ) andhence I is controlled by A .Now, if I = ∩ i ∈ X J i for some two-sided ideals J i of [ U ( L ) K , then given s = P β ∈ N r u β · · · u β r r s β ∈ I , we know that s ∈ J i = [ U ( L ) K ( J i ∩ [ U ( A ) K ) for every i , and hence s β ∈ J i for every i , β , i.e. s β ∈ ∩ i ∈ X J i = I for all β , and hence I is controlled by A .For the ﬁnal statement, it is clear that { u r +1 , · · · , u d } is an F -basis for A ⊗ O O F . Soif J is controlled by A ⊗ O O F then given s ∈ I ⊆ J , s β ∈ J for every β . Therefore s β ∈ J ∩ [ U ( L ) K = I and hence I is controlled by A as required.Now, given a linear form λ : g → K , recall that we deﬁne g λ := { u ∈ g : λ ([ u, g ]) = 0 } . Lemma 7.7.

Let σ be a Lie automorphism of g , and let σ · λ be the linear form deﬁnedby ( σ · λ )( u ) = λ ( σ − ( u )) . Then g σ · λ = σ ( g λ ) .Proof. Firstly, u ∈ g σ · λ if and only if σ · λ ([ u, g ]) = λ ( σ − ([ u, g ])) = 0. But since σ isa Lie automorphism, this is true if and only if λ ([ σ − ( u ) , g ]) = 0, i.e. σ − ( u ) ∈ g λ and u ∈ σ ( g λ ) as required. Proposition 7.8.

Suppose that the subalgebra g λ is an ideal in g . Then P = I ( λ ) iscontrolled by g λ .Proof. We proceed by induction on n := dim K g . If n = 1 then g λ = g and the statementis obvious, so the base case holds.For the inductive step, if there exists a non-zero ideal t of g such that λ ( t ) = 0 then clearly t ⊆ g λ . Let g := g / t , let λ be the linear form of g induced by λ , and let L := L / t ∩ L .Then g λ = g λ / t is an ideal in g , so using induction, the Dixmier annihilator I ( λ ) of \ U ( L ) K is controlled by g λ = g λ / t .But using Lemma 3.1 we see that \ D ( λ ) ∼ = [ D ( λ ) as [ U ( L ) K -modules, and thus I ( λ ) = I ( λ ) / t [ U ( L ) K and it follows immediately that I ( λ ) is controlled by g λ .Therefore, we may assume that λ ( t ) = 0 for all non-zero ideals t of g . Note that since g λ is an ideal in g , it follows that t = g λ ∩ ker( λ ) is an ideal in g , and clearly λ ( t ) = 0, so t = 0. Thus λ is injective when restricted to g λ , and hence g λ is one-dimensional over K .Write g λ = Kz for some z ∈ Z ( g ) with λ ( z ) = 0, and since Z ( g ) ⊆ g λ , it follows that Z ( g ) = Kz . Naturally we may assume that z ∈ L .Using Proposition 2.7, g has a reducing quadruple ( x, y, z, g ′ ), with y, z ∈ L , and we canchoose a polarisation b of g at λ with b ⊆ g ′ . Since b is also a polarisation of g ′ at µ := λ | g ′ , it follows from Lemma 2.3 that dim K b = (dim K g + dim K g λ ) = (dim K g ′ +dim K g ′ µ ). But dim K g ′ = dim K g −

1, so this means that dim K g ′ µ = dim K g λ + 1 = 2.But clearly y, z ∈ g ′ µ , and hence g ′ µ = Kz ⊕ Ky , and clearly this is an ideal in g , andtherefore in g ′ . 46ow, using Theorem 4.2, we see that I ( λ ) is controlled by L ′ = g ′ ∩ L , so let Q := I ( λ ) ∩ \ U ( L ′ ) K – a semiprime ideal of \ U ( L ′ ) K such that I ( λ ) = Q [ U ( L ) K = [ U ( L ) K Q . Wewill prove that Q is controlled by Kz , and it will follow that I ( λ ) is controlled by Kz .Using Corollary 5.9, we know that Q = ∩ i ∈ I Ann \ U ( L ′ ) K \ D ( µ i ) F i for some ﬁnite extensions F i /K , linear forms µ i : g ′ → F such that µ i ( L ′ ) ⊆ O F . Thus I ( λ ) = [ U ( L ) K Q ⊆ [ U ( L ) K Ann \ U ( L ′ ) K \ D ( µ i ) F i for each i .Setting λ i as any extension of µ i to L , it follows that \ D ( λ i ) F i = [ U ( L ) F i ⊗ \ U ( L ′ ) Fi \ D ( µ i ) F i .Since I ( λ ) ⊗ K F i is a two-sided ideal of [ U ( L ) F i contained in [ U ( L ) F i Ann \ U ( L ′ ) Fi \ D ( µ i ) F i , itfollows from Lemma 4.1 that I ( λ ) ⊆ Ann \ U ( L ) K \ D ( λ i ) F i = I ( λ i ).Using Lemma 6.2, this means that λ and λ i lie in the same G -coadjoint orbit for all i ,i.e. λ i = g i · λ for some g i ∈ G ( F )So, ﬁxing i , let F = F i for convenience, and let µ = λ | g ′ , then µ extends to an F -linearform on g ′ F = g ′ ⊗ K F , and g ′ µF = ( g ′ µ ) ⊗ K F . Let σ be the Lie automorphism of g ′ F given by the restriction of g i to g ′ F , then µ i = σ · µ , so it follows from Lemma 7.7 that g ′ µ i F = σ ( g ′ µF ) = g ′ µ ⊗ F since g ′ µ is an ideal in g , and hence is preserved by σ .It follows from induction that Ann \ U ( L ′ ) Fi \ D ( µ i ) is controlled by g ′ µ ⊗ K F i , and henceAnn \ U ( L ′ ) K \ D ( µ i ) is controlled by g ′ µ by Lemma 7.6. Since this is true for all i , it alsofollows from Lemma 7.6 that Q = ∩ i ∈ I Ann \ U ( L ′ ) K \ D ( µ i ) F i is controlled by g ′ µ = Kz ⊕ Ky .Finally, consider the subalgebra h = Span K { x, y, z } of g , let H := h ∩ L , A := g ′ µ ∩ L = O z ⊕ O y , and let I := ( Q ∩ [ U ( A ) K ) \ U ( H ) K . Then I is a proper two-sided ideal of \ U ( H ) K containing z − λ ( z ), thus I = ( z − λ ( z )) \ U ( H ) K by Lemma 4.3.Therefore, Q ∩ [ U ( A ) K = I ∩ [ U ( A ) K = ( z − λ ( z )) [ U ( A ) K , and hence Q = ( Q ∩ \ U ( A ′ ) K ) \ U ( L ′ ) K = ( z − λ ( z )) \ U ( L ′ ) K , i.e. Q is controlled by O z So since g λ = Kz , it follows that Q is controlled by g λ , and thus I ( λ ) is controlled by g λ as required. Proposition 7.9.

Suppose that a is any ideal of g containing g λ . Then I ( λ ) is controlledby a .Proof. This proof is very similar to the proof of Proposition 7.8, but uses it as a keystep. Again, we proceed by induction on n := dim K g . If n = 1 then g λ = g = a and thestatement is obvious, so the base case holds.For the inductive step, if there exists a non-zero ideal t of g such that λ ( t ) = 0 thenclearly t ⊆ g λ ⊆ a . Let g := g / t , let a := a / t , let λ be the linear form of g inducedby λ , and let L = L / t ∩ L . Then g λ = g λ / t , and a is an ideal in g containing g µ . So,using induction, the Dixmier annihilator I ( λ ) of \ U ( L ) K is controlled by a .But using Lemma 3.1, [ D ( λ ) ∼ = [ D ( λ ) as [ U ( L ) K -modules, and thus I ( λ ) = I ( λ ) / t [ U ( L ) K and it follows immediately that I ( λ ) is controlled by a .47herefore, we may assume that λ ( t ) = 0 for all non-zero ideals t of g , and it follows that Z ( g ) = Kz for some z ∈ Z ( L ) such that λ ( z ) = 0. Furthermore, we may assume that a is the ideal of g generated by g λ , i.e the subspace g λ + [ g , g λ ] + [ g , [ g , g λ ]] + · · · , since if I ( λ ) is controlled by this ideal, then it will be controlled by any larger ideal by Lemma 7.6.If g λ = a then g λ is an ideal of g , and the result follows from Proposition 7.8. So we mayassume that g λ = a , and hence a is not central in g . Therefore, since g is nilpotent, thereexists y ∈ a ∩ L with y / ∈ Z ( g ) such that [ y, g ] ⊆ Z ( g ) = Kz . So setting g ′ := ker(ad( y )),it follows that g = g ′ ⊕ Kx and ( x, y, z, g ′ ) is a reducing quadruple.Moreover, λ ([ y, g λ ]) = 0 and since [ y, g ] ⊆ Kz and λ ( z ) = 0, this means that[ y, g λ ] = 0. Also, [ y, [ g , g ]] = 0, so [ y, [ g , [ g [ · · · , [ g , g λ ] · · · ]] = 0, and hence [ y, a ] = 0by the construction of a , and thus a ⊆ g ′ .Now, using Theorem 4.2, we see that I ( λ ) is controlled by L ′ = g ′ ∩ L , so let Q := I ( λ ) ∩ \ U ( L ′ ) K – a semiprime ideal of \ U ( L ′ ) K such that I ( λ ) = Q [ U ( L ) K = [ U ( L ) K Q . Wewill prove that Q is controlled by a , and it will follow that I ( λ ) is controlled by a .Using Proposition 2.7, we can choose a polarisation b of g at λ with b ⊆ g ′ . Since b isalso a polarisation of g ′ at µ = λ | g ′ , it follows from Lemma 2.3 that dim K b = (dim K g +dim K g λ ) = (dim K g ′ + dim K g ′ µ ). But dim K g ′ = dim K g −

1, so this means that dim K g ′ µ =dim K g λ + 1.So since g ′ µ contains g λ and y , and since λ ([ x, y ]) = 0, y / ∈ g λ . Therefore g λ ⊕ Ky ⊆ g ′ µ and dim K g λ ⊕ Ky = dim K g ′ µ , so g ′ µ = g λ ⊕ Ky ⊆ a .Using Corollary 5.9, we know that Q = ∩ i ∈ I Ann \ U ( L ′ ) K \ D ( µ i ) F i for some ﬁnite extensions F i /K , linear forms µ i : g ′ → F such that µ i ( L ′ ) ⊆ O F . Thus I ( λ ) = [ U ( L ) K Q ⊆ [ U ( L ) K Ann \ U ( L ′ ) K \ D ( µ i ) F i for each i .Setting λ i as any extension of µ i to L , it follows that \ D ( λ i ) F i = [ U ( L ) F i ⊗ \ U ( L ′ ) Fi \ D ( µ i ) F i .Since I ( λ ) ⊗ K F i is a two-sided ideal of [ U ( L ) F i contained in [ U ( L ) F i Ann \ U ( L ′ ) Fi \ D ( µ i ) F i , itfollows from Lemma 4.1 that I ( λ ) ⊆ Ann \ U ( L ) K \ D ( λ i ) F i = I ( λ i ).Using Lemma 6.2, this means that λ and λ i lie in the same G -coadjoint orbit for all i ,i.e. λ i = g i · λ for some g i ∈ G ( F )So, ﬁxing i , let F = F i for convenience, and let µ = λ | g ′ , then µ extends to an F -linearform on g ′ F = g ′ ⊗ K F , and g ′ µF = ( g ′ µ ) ⊗ K F . Let σ be the Lie automorphism of g ′ F given by the restriction of g i to g ′ F , then µ i = σ · µ , so it follows from Lemma 7.7 that g ′ µ i F = σ ( g ′ µF ) ⊆ σ ( a ⊗ F ) ⊆ a ⊗ F .Therefore, using induction, Ann \ U ( L ) Fi \ D ( µ i ) F i is controlled by a ⊗ F i , and henceAnn \ U ( L ) K \ D ( µ i ) F i is controlled by a using Lemma 7.6. Since this holds for all i and Q = ∩ i ∈ I Ann \ U ( L ′ ) K \ D ( µ i ) F i , it also follows from Lemma 7.6 that Q is controlled by a , andhence I ( λ ) is controlled by a as required. 48 .4 Special Linear forms Deﬁnition 7.1.

We call a linear form λ : g → K special if the ideal a of g generated by g λ satisﬁes λ ([ a , a ]) = 0 . This condition is not universal, but as we will see, it is very useful. Using Proposition7.9, we see that if λ is special then P = I ( λ ) is controlled by an ideal a of g satisfying λ ([ a , a ]) = 0. We will now study the Dixmier annihilators associated to special linearforms. Lemma 7.10.

Let

F/K be a ﬁnite extension, and let λ : g → F be a linear form suchthat λ ( L ) ⊆ O F . Then there exists a unique ideal t of g , maximal with respect to theproperty that λ ( t ) = 0 , and in fact t = I ( λ ) ∩ g .Proof. Firstly, it is clear that if t , t E g with λ ( t ) = λ ( t ) = 0, then λ ( t + t ) = 0, so itfollows that there exists a unique ideal t which is maximal with respect to the propertythat λ ( t ) = 0.Moreover, by Lemma 2.3, t ⊗ F ⊆ b for any polarisation b of g F at λ , hence t [ D ( λ ) F = t [ U ( L ) F ⊗ \ U ( B ) F F λ = [ U ( L ) F t ⊗ \ U ( B ) F F λ = [ U ( L ) F ⊗ \ U ( B ) F t F λ = 0, so t ⊆ I ( λ ) ∩ g .Conversely, let t ′ := I ( λ ) ∩ g , then t ′ [ D ( λ ) F = 0, so t ′ ⊗ F λ = 0, which implies that t ′ ⊆ b and λ ( t ′ ) = 0 as required. Theorem 7.11.

Assume that L is a powerful Lie lattice, and λ : g → K is a speciallinear form such that λ ( L ) ⊆ O . Then the Dixmier annihilator I ( λ ) is a maximal idealof [ U ( L ) K .Proof. Let t = I ( λ ) ∩ g , and let L = L / L ∩ t , then to prove that I ( λ ) is maximal in [ U ( L ) K , we need only prove that I ( λ ) / t [ U ( L ) K is maximal in \ U ( L ) K = [ U ( L ) K / t [ U ( L ) K .But λ ( t ) = 0 by Lemma 7.10, so t ⊆ g λ and λ induces a special linear form λ of g / t , and it follows from Lemma 3.1 that [ D ( λ ) ∼ = \ D ( λ ) as [ U ( L ) K -modules, and thusAnn \ U ( L ) K \ D ( λ ) = I ( λ ) / t [ U ( L ) K .Therefore, after quotienting out by t , we may assume that I ( λ ) ∩ g = 0, and hence λ ( a ) = 0 for all non-zero ideals a of g .Now, since λ is special, we may choose an ideal a of g with g λ ⊆ a and λ ([ a , a ]) = 0. But[ a , a ] is an ideal of g , and hence [ a , a ] = 0, i.e. a is an abelian ideal of g . So suppose that Q is a maximal ideal of [ U ( L ) K with I ( λ ) ⊆ Q , we will prove that I ( λ ) = Q , and hence Q is maximal.Since Q is maximal, it is locally closed, so by Theorem 5.8, Q = I ( µ ) for some ﬁniteextension F/K and some linear form µ : g → F with µ ( L ) ⊆ O F , and after extending F if necessary, we may assume that F/K is a Galois extension.Using Proposition 7.3, we see that there exists σ ∈ Gal ( F/K ) such that Ann \ U ( L ) F \ D ( σ · λ ) F ⊆ Ann \ U ( L ) F [ D ( µ ), and note that Ann \ U ( L ) F \ D ( σ · λ ) F ∩ [ U ( L ) K = I ( λ ) by Lemma 6.1.Also, note that g σ · λF = σ ( g λF ) by Lemma 7.7. But since λ takes values in K , g λF =( g λ ) ⊗ K F , so it follows that g σ · λF = σ (( g λ ) ⊗ K F ) is contained in the abelian ideal49 ( a ⊗ K F ) = a ⊗ K F . Therefore σ · λ is a special linear form of g ⊗ K F .So let us ﬁrst suppose that F = K , and hence σ = 1, then since I ( λ ) ⊆ I ( µ ), we know that µ = g · λ for some g ∈ G ( K ), and hence g µ = g ( g λ ) by Lemma 7.7. Hence g µ ⊆ g ( a ) = a and hence µ is also a special linear form, and it follows from Proposition 7.9 that I ( λ )and I ( µ ) are both controlled by a , i.e. if A = a ∩ L then I ( λ ) = [ U ( L ) K ( I ( λ ) ∩ [ U ( A ) K )and I ( µ ) = [ U ( L ) K ( I ( µ ) ∩ [ U ( A ) K ).But since a is abelian, we see using Corollary 7.5 that I ( λ ) ∩ [ U ( A ) K = I ( µ ) ∩ [ U ( A ) K ,and hence I ( µ ) = [ U ( L ) K ( I ( µ ) ∩ [ U ( A ) K ) = [ U ( L ) K ( I ( λ ) ∩ [ U ( A ) K ) = I ( λ ) as required.More generally, since Ann \ U ( L ) F \ D ( σ · λ ) F ⊆ Ann \ U ( L ) F [ D ( µ ) and σ · λ is special, it followsfrom this argument that Ann \ U ( L ) F \ D ( σ · λ ) F = Ann \ U ( L ) F [ D ( µ ). Thus I ( λ ) = Ann \ U ( L ) F \ D ( σ · λ ) F ∩ [ U ( L ) K = Ann \ U ( L ) F [ D ( µ ) F ∩ [ U ( L ) K = I ( µ ) = Q , and hence I ( λ ) = Q is maximal as re-quired. Corollary 7.12.

Let

F/K be a ﬁnite, Galois extension, L a powerful Lie lattice, λ : g → F a linear form such that λ ( L ) ⊆ O F and the extension of λ to g ⊗ K F is special. Then I ( λ ) is a maximal ideal of [ U ( L ) K .Proof. Using Theorem 7.11, we see that P = Ann \ U ( L ) F [ D ( λ ) is a maximal ideal of [ U ( L ) F .So let us suppose that Q is a maximal ideal of [ U ( L ) K with I ( λ ) ⊆ Q .Choose a prime ideal Q ′ of [ U ( L ) F which is minimal prime above Q ⊗ K F . Then since I ( λ ) ⊗ K F ⊆ Q ⊗ K F ⊆ Q ′ , it follows that Q ′ contains a minimal prime above I ( λ ) ⊗ K F .But using Corollary 7.2, this means that σ ( P ) ⊆ Q ′ for some σ ∈ Gal ( F/K ), and since P is maximal, σ ( P ) is maximal, which implies that Q ′ = σ ( P ) is minimal prime above I ( λ ) ⊗ K F .Therefore, Q ′ ∩ [ U ( L ) K = σ ( P ) ∩ [ U ( L ) K = I ( λ ), but since Q ⊆ Q ′ ∩ [ U ( L ) K , it followsthat Q = I ( λ ) as required. We are now ready to prove Theorem B, which now requires only the following result:

Lemma 7.13.

Assume g is nilpotent and metabelian (i.e. [ g , g ] is abelian). Then anylinear form λ : g → K is special.Proof. We want to prove that for any linear form λ : g → K , g λ is contained in an ideal a of g such that λ ([ a , a ]) = 0. Firstly, since [ g , g ] is an abelian ideal of g , using Lemma2.4 we see that there exists a polarisation b of g at λ such that [ g , g ] ⊆ b .But since b is a subalgebra of g containing [ g , g ], it follows that b is an ideal of g .Also, g λ ⊆ b by Lemma 2.3, and of course λ ([ b , b ]) = 0. Therefore λ is special. Proof of Theorem B.

As per Deﬁnition 1.2, we want to prove that for all prime ideals P of [ U ( L ) K , there exist n suﬃciently high such that P ∩ \ U ( π n L ) K is weakly rational if andonly if it is primitive if and only if it is locally closed.50e already know using Proposition 5.3 that if that if P ∩ \ U ( π n L ) K is locally closed, thenit is primitive and weakly rational, and also note if P ∩ \ U ( π N L ) K is weakly rational forany N , then P ∩ \ U ( π n L ) K is weakly rational for all n ≥ N .So we can assume that there exists such an N ≥

0, and we will prove that P ∩ \ U ( π n L ) K is locally closed, and in fact maximal, for suﬃciently high n ≥ N .Since P ∩ \ U ( π N L ) K is weakly rational, it follows from Theorem A that for all suﬃcientlyhigh n ≥ N , there exists a ﬁnite extension F/K and a linear form λ : g → F such that λ ( π n L ) ⊆ O F and P ∩ \ U ( π n L ) K = Ann \ U ( π n L ) K [ D ( λ ) F . After extending F if necessary,we may assume that F/K is Galois. Since g is metabelian, it is clear that g ⊗ K F is alsometabelian, so it follows from Lemma 7.13 that λ is special.Choose n such that π n L is a powerful Lie lattice, i.e. such that π n ∈ p O , which is truefor all suﬃciently high n . Then using Corollary 7.12 we see that Ann \ U ( π n L ) K [ D ( λ ) F = P ∩ \ U ( π n L ) K is a maximal ideal of \ U ( π n L ) K as required.Finally, it follows from Corollary 3.4 that \ U ( π n L ) K / Ann \ U ( π n L ) K [ D ( λ ) F is a domain, andhence \ U ( π n L ) K /P ∩ \ U ( π n L ) K is a simple domain. References [1] K. Ardakov; S. Wadsley, Ù D -modules on Rigid Analytic Spaces I J. Reine Angew.Math. , 221-275 (2019).[2] K. Ardakov; S. Wadsley, On irreducible representations of compact p -adic analyticgroups, Annals of Mathematics , 453-557 (2013).[3] J. Bell, On the importance of being primitive Revista Colombiana de Matem´aticas , 87-112 (2019).[4] A. Borel, Linear Algebraic Groups,

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