# Primitive ideals in rational, nilpotent Iwasawa algebras

aa r X i v : . [ m a t h . R T ] F e b Primitive ideals in rational, nilpotent Iwasawa algebras

Adam JonesFebruary 9, 2021

Abstract

Given a p -adic ﬁeld K and a nilpotent uniform pro- p group G , we prove that allprimitive ideals in the K -rational Iwasawa algebra KG are maximal, and can bereduced to a particular standard form. Setting L as the associated Z p -Lie algebraof G , our approach is to study the action of KG on a Dixmier module [ D ( λ ) overthe aﬃnoid envelope [ U ( L ) K , and to prove that all primitive ideals can be reducedto annihilators of modules of this form. Contents p -valued Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Prime ideals in KG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.5 Completions of KG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.6 Dixmier modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Primitive Ideals 38 KG to KG p n . . . . . . . . . . . . . . . . . . . . . . . . 395.3 Extension from KG p n to KG . . . . . . . . . . . . . . . . . . . . . . . . 41 Fix p > K \ Q p be a ﬁnite extension with ring of integers O , uniformiser π , residue ﬁeld k . Let G be a compact p -adic Lie group, and recall that we deﬁne the completed groupalgebra of G over O as: O G := lim ←− O [ G/N ] (1)where the limit is taken over all open normal subgroups N of G . Continuous, O -linearrepresentations of G are closely related to O G -modules.This paper is part of an ongoing project to classify the prime ideal structure of O G ,towards which much progress has been made in [4], [5], [3], [15] and [14]. In the samevein as those works, we aim to prove that all prime ideals in O G can be reduced to aparticular standard form. Speciﬁcally, recall the following deﬁnition [14, Deﬁnition 1.1]: Deﬁnition 1.1.

We say that a prime ideal P of O G is standard if there exists a closed,normal subgroup H of G such that: • G := GH is torsionfree. • H − ⊆ P . • The image of P in O G is centrally generated.We say that P is virtually standard if P ∩ O U is a ﬁnite intersection of standard primeideals of O U for some open normal subgroup U of G . The essence of this deﬁnition is that P is standard when it can be constructed using only augmentation ideals of the form ( H − O G , for H a closed subgroup of G , and centrallygenerated ideals, i.e. the obvious prime ideals.In our case, we will assume further that G is a uniform pro- p group in the sense of [11,Deﬁnition 4.1]. This is a safe reduction since all compact p -adic Lie groups have an open,uniform normal subgroup.Let us recall the main conjecture within the study of two-sided ideals in non-commutativeIwasawa algebras, ﬁrst proposed in [1, Question N], and stated in [14, Conjecture 1.1]: Conjecture 1.2.

Let G be a solvable, uniform pro- p group, and let P be a prime idealin O G . Then P is virtually standard, and moreover if p ∈ P then P is standard. ote: There is a version of this conjecture for non-solvable groups, which requires us toexclude the case where O G/P is a ﬁnitely generated O -module, but this will not concernus here.When the prime ideal P contains p , we can reduce to studying the mod- p Iwasawa algebra : kG := O G ( π ) = lim ←− k [ G/N ].We know that Conjecture 1.2 holds for all prime ideals P of kG whenever G is nilpotentby [5, Theorem A], and also when G is abelian-by-procyclic by [15, Theorem 1.4].In the case where the prime ideal P does not contain p , however, the picture is very diﬀer-ent. Deﬁne the rational Iwasawa algebra or Iwasawa algebra of continuous distributions as KG := O G ⊗ O K .This is a Noetherian, topological K -algebra, and the prime ideals of O G not containing p are in bijection with prime ideals in KG , via the map P P ⊗ O K . We aim to provethe analogue of Conjecture 1.2 for prime ideals in KG . Note:

1. This conjecture is trivially true for G abelian.2. The requirement that prime ideals in KG are only virtually standard is necessary,since they are not all standard. For example, if G = Z p × Z p , and K contains a p ’th rootof unity ζ , and we let P be the kernel of the map KG → K Z p , ( r, s ) ζ r s . Then P is aprime ideal of KG and P is not standard.In this paper, we will prove a version of Conjecture 1.2 for KG , in the case where G isnilpotent. There is an alternative way of describing standard prime ideals in O G and KG , and thusformulating Conjecture 1.2, which will be of more practical use:Firstly, for any two-sided ideal I of O G , recall from [5, Deﬁnition 5.2] that we deﬁne I † := { g ∈ G : g − ∈ I } , a closed, normal subgroup of G , and we say that I is faithful if I † = 1, i.e. if the natural map G → (cid:0) O GI (cid:1) × , g g + I is injective.Then setting G I := GI , the kernel of the natural surjection O G → O G I is the augmen-tation ideal ( I † − O G , and the image of I under this surjection is a faithful ideal of O G I . Note: If I is prime and p ∈ I , it follows from [5, Lemma 5.2] then G I is torsionfree, butthis need not be true if p / ∈ I . Roughly speaking, this is why we can only generally assertthat prime ideals not containing p are virtually standard , and not standard.If P is a faithful, prime ideal of O G , then to prove that P is standard, we see usingDeﬁnition 1.1 that it is only required to prove that P is centrally generated. Using [6,Corollary A], we know that Z ( O G ) = O Z ( G ), so P is centrally generated precisely when3 = ( P ∩ O Z ( G )) O G .More generallly, if I is a right ideal of O G and H is a closed subgroup of G , we say that H controls I if I = ( I ∩ O H ) O G , i.e. I is generated as a right ideal by a subset of O H .Deﬁne the controller subgroup of I by I χ := T { U ≤ o G : U controls I } , and it followsfrom [7, Theorem A] that a closed subgroup H of G controls an ideal I E RG if and onlyif I χ ⊆ H , so in particular I χ controls I .If I is a two-sided ideal, then I χ is a closed, normal subgroup of G by [5, Lemma5.3(a)], and to prove that I is centrally generated, all that is required is to prove that I is controlled by Z ( G ), i.e. I χ ⊆ Z ( G ).So, to summarise, given a prime ideal P of O G , to prove that P is standard, we needonly to prove that the quotient G P = GP † is torsionfree, and that the image of P underthe surjection O G → O G P is controlled by Z ( G P ). Therefore, we deduce the followingalternative formulation for Conjecture 1.2: Alternative Formulation:

Let G be a solvable, uniform pro- p group. We conjecturethat every faithful prime ideal of O G is controlled by Z ( G ). When studying Iwasawa algebras, rather than studying general prime ideals, we may beinterested speciﬁcally in classifying primitive ideals , i.e. the annihilators of simple O G -modules.However, since G is a pro- p group, the Iwasawa algebra O G has a unique maximal leftideal m = ( G − , π ), which is in fact two-sided. Therefore m is the only primitive idealin O G . The rational Iwasawa algebra KG , on the other hand, has many simple modulesand primitive ideals. Theorem A.

Let G be a nilpotent, uniform pro- p group. Then every primitive ideal of KG is maximal and virtually standard. Moreover, every faithful, primitive ideal of KG is standard. As explained above, we see that to prove Theorem A, it suﬃces to show that all faithful,primitive ideals in KG are controlled by Z ( G ).Now, recall from [18, III 2.1.2] the deﬁnition of a p -valuation ω : G → R ∪ {∞} . We willrecap the key properties of p -valuations in section 2, but for now, just recall that if G isuniform, then G carries a complete p -valuation given by ω ( g ) := sup { n ∈ N : g ∈ G p n +1 } ,so this concept gives rise to a larger class of torsionfree compact p -adic Lie groups which,in particular, contains the class of all closed subgroups of uniform groups.If we assume that ( G, ω ) is a complete, nilpotent p -valued group of ﬁnite rank, then itfollows from [5, Theorem A] that all faithful prime ideals in the mod- p Iwasawa algebra kG are controlled by Z ( G ). One might think that these techniques could be generalisedto the characteristic 0 case to prove the same result. Unfortunately, the author showed in[14] that these techniques fail in characteristic 0, and they can only be used to establisha much weaker control theorem for primitive ideals ([14, Theorem 1.2]).4owever, in this paper, we will adapt the argument given in [14] with some newtechniques, and prove the following much stronger control theorem for general primeideals: Theorem B.

Let G be a nilpotent, complete p -valued group of ﬁnite rank. Then thereexists an abelian normal subgroup A of G such that A controls every faithful prime idealin KG . Of course, if we could show that this subgroup A is central, then Theorem A would followimmediately, and would remain true for prime ideals as opposed to just primitive ideals.But unfortunately, this need not always be the case.For example, if G = H ⋊ Z p for H abelian and ( G, H ) Z ( G ), then the subgroup A given by Theorem B is H , which is not central. We will prove Theorem B in section 3.Theorem B is the strongest result we have obtained to date concerning general prime ide-als in KG , but all subsequent results require the additional assumption that our primeideals are primitive .The key idea is that we want to deﬁne a class of KG -representations M whose annihilatorideals completely describe the primitive ideal structure of KG . Using [24, Theorem 5.2],we have a dense, faithfully ﬂat embedding fo KG into the locally analytic distributionalgebra D ( G, K ) as deﬁned in [25, Deﬁnition 2.1, Proposition 2.3], so it makes sense torestrict to the class of coadmissible D ( G, K )-modules, which naturally have the structureof KG -modules. However, since D ( G, K ) is non-noetherian, this may present diﬃculties,so instead we restrict our attention to larger, Noetherian completions of KG :Returning to the case where G is uniform, let L G = log( G ) be the Z p -Lie algebra of G asdeﬁned in [11, Theorem 4.30], and set L := p L G . Recall from from [16, Deﬁnition 1.2]that we deﬁne the aﬃnoid enveloping algebra of L with coeﬃcients in K to be: [ U ( L ) K := Ç lim ←− n ∈ N U ( L ) /π n U ( L ) å ⊗ O K (2)This is a Noetherian, Banach K -algebra, and recall from [2, Theorem 10.4] that thereexists a continuous, dense embedding of K -algebras: KG ֒ −→ [ U ( L ) K , g exp(log( g )) . (3)Unlike the embedding KG → D ( G, K ), this map is not faithfully ﬂat, but we can stilluse it to study the representation theory of KG via the representation theory of L .In section 2, we will recall from a previous work [16] how we deﬁne the Dixmier module [ D ( λ ) of [ U ( L ) K , corresponding to a linear form λ ∈ Hom Z p ( L , O ). It follows from [16,Theorem A] that using the annihilators of these modules, we can completely describe theprimitive ideal structure of [ U ( L ) K .So now, we are interested in the restricted action of KG on [ D ( λ ), and the key resultwe need in the proof of Theorem A is the following: Theorem C.

Let G be a nilpotent, uniform pro- p group such that L is powerful, let F/K be a ﬁnite extension, and let λ ∈ Hom Z p ( L , O F ) such that λ | Z ( L ) is injective. Then P :=Ann KG [ D ( λ ) F is controlled by Z ( G ) . ote: To say that L is powerful just means that [ L , L ] ⊆ p L .We will prove this result in section 4. The key idea is that we know that the annihilator P := Ann KG [ D ( λ ) is controlled by an abelian normal subgroup A of G by Theorem B,so we consider the action of KA on [ D ( λ ), and prove that the kernel of this action iscontrolled by Z ( G ).In section 5, we will apply [16, Theorem A], to prove that it suﬃces to know thatDixmier annihilators are controlled by Z ( G ) to establish the same result for all primitiveideals, and Theorem A will follow immediately from Theorem C. Acknowledgments:

I am very grateful to Konstantin Ardakov for many helpful com-ments. I would also like to thank EPSRC and the Heilbronn Institute for MathematicalResearch for funding me in my research.

Notation:

For g, h ∈ G , we denote the group commutator by ( g, h ) := ghg − h − .Moreover, we will write H E ic G to mean that H is a closed, isolated normal subgroup of G , i.e. GH is torsionfree. Let us ﬁrst recap some basic notions of ring ﬁltrations and valuations. Throughout, let R be any ring. Deﬁnition 2.1. A ﬁltration on R is a map w : R → Z ∪ {∞} such that w (0) = ∞ andfor all r, s ∈ R : • w ( r + s ) ≥ min { w ( r ) , w ( s ) } . • w ( rs ) ≥ w ( r ) + w ( s ) We say that w is separated if w ( r ) = ∞ if and only if r = 0 , and w is a valuation if w ( rs ) = w ( r ) + w ( s ) for all r, s ∈ R . For each n ∈ Z , we deﬁne F n R := { r ∈ R : w ( r ) ≥ n } , and deﬁne the associated gradedring gr w R to be ⊕ n ∈ Z F n RF n +1 R with multiplication ( r + F n +1 R )( s + F m +1 R ) = rs + F n + m +1 R .Note that w is a valuation if and only if gr w R is a domain.If r ∈ R and w ( r ) = n then we denote gr( r ) := r + F n +1 R ∈ gr R .Recall from [20, Ch. II Deﬁnition 2.2.1] that a ﬁltration w : R → Z ∪ {∞} is Zariskian if the

Rees ring ˜ R := ⊕ n ∈ Z F n R is Noetherian, and F R ⊆ J ( F R ). We will not use thisdeﬁnition very often, but we will usually always assume that our ﬁltrations are Zariskian.Note that if w is Zariskian, then it is separated and both R and gr w R are Noetherian,since they arise as quotients of the Rees ring.6 xample:

1. If R carries a ﬁltration w , then the matrix ring M n ( R ) carries a ﬁltration w n ( A ) = min { w ( a i,j ) : i, j = 1 , · · · , n } – the standard matrix ﬁltration .2. If I is a two-sided ideal of R and R carries a ﬁltration w , then the quotient ring RI carries the quotient ﬁltration given by w ( r + I ) = sup { w ( r + y ) : y ∈ I } . Note that gr wRI = gr w R gr I , and if w is Zariskian then w is Zariskian.Now, recall the following deﬁnition ([15, Deﬁnition 3.1]) Deﬁnition 2.2.

Let Q be a simple artinian ring, and let v : Q → Z ∪ {∞} be a ﬁltration.We say that v is a non-commutative valuation if the completion b Q of Q with respect to v is isomorphic to a matrix ring M n ( Q ( D )) , where: • Q ( D ) is the ring of quotients of some non-commutative DVR D with uniformiser ν , • the extension of v to b Q is given by the standard matrix ﬁltration corresponding tothe ν -adic ﬁltration on Q ( D ) . Note that if v is a non-commutative valuation on Q , then for all z ∈ Z ( Q ), q ∈ Q , v ( qz ) = v ( q ) + v ( z ), a property which will be very useful to us in section 3.The following construction allows us to deﬁne a non-commutative valuation on the ar-tinian ring of quotients Q ( R ) of a Zariskian ﬁltered ring R . This construction was derivedin [5, Section 3], and we state it fully since we will need it for some proofs in section 3. Construction 2.3.

Let R be a prime, Noetherian ring with a Zariskian ﬁltration w suchthat gr w R is commutative and the graded ideal ( gr w R ) ≥ is non-nilpotent. Then for eachminimal prime ideal q of gr w R , we can construct a non-commutative valuation on Q ( R ) using the following data: • S := { r ∈ R : gr( r ) / ∈ q } – an Ore set in R such that S − R = Q ( R ) • w ′ – a Zariskian ﬁltration on Q ( R ) such that w ′ ( r ) ≥ w ( r ) for all r ∈ R , and w ′ ( s − r ) = w ′ ( r ) − w ( s ) for all s ∈ S . The associated graded gr w ′ Q ( R ) is thehomogeneous localisation of gr w R at q . • Q ′ – the completion of Q ( R ) with respect to w ′ , an artinian ring. • U – the positive part of Q ′ , a Noetherian ring. • z – a regular, normal element of J ( U ) such that z n U = F nw ′ ( z ) Q ′ for all n ∈ Z . • v z,U – the z -adic ﬁltration on Q ′ , topologically equivalent to w ′ . • b Q – a simple quotient of Q ′ . • V – the image of U in b Q . • z – the image of z in V . • v z,V – the z -adic ﬁltration on b Q . B – a maximal order in b Q , equivalent to V , satisfying B ⊆ z − r V for some r ≥ ,isomorphic to M n ( D ) for some non-commutative DVR D . • v z, B – the z -adic ﬁltration on B . • v q – the J ( B ) -adic ﬁltration on b Q , topologically equivalent to v z, B .Then v = v q deﬁnes a non-commutative valuation on Q ( R ) , whose completion is b Q , andthe natural map R → Q ( R ) is continuous. Moreover, if w ( x ) ≥ then v ( x ) ≥ . Given a ring R and a group H , recall from [22] that a crossed product of R with H , denoted R ∗ H , is a ring extension R ⊆ S , free as a left R -module with basis { h : h ∈ H } ⊆ S × in bijection with H such that for each g, h ∈ H : • gR = Rg and • gRhR = ghR .Furthermore, given a sequence H , · · · , H r of groups, we denote an iterated crossed prod-uct R ∗ H ∗ H ∗· · ·∗ H r inductively to mean a crossed product of R ∗ H ∗· · ·∗ H r − with H r .Let us recap some properties of crossed products that we will use throughout. Lemma 2.4.

Let R be a Noetherian Q -algebra, F a ﬁnite group. Then if P is a primeideal of a crossed product S = R ∗ F , then: i . P ∩ R is semiprime in R . ii . J := ( P ∩ R ) · S is semiprime in S , and P is a minimal prime above J . iii . S/J = (

P/P ∩ R ) ∗ F .Proof. We will prove that P ∩ R is an F -prime ideal, i.e. it is F -invariant and for any F -invariant ideals A, B of R , if AB ⊆ P ∩ R then A ⊆ P ∩ R or B ⊆ P ∩ R .Having established this, part i follows from the fact that all minimal primes above P ∩ R form a single F -orbit by [22, Lemma 14.2( ii )], part iii is obvious since J = ⊕ g ∈ F ( P ∩ R )¯ g ,and part ii is part iii together with [21, Proposition 10.5.8] and [22, Theorem 4.4].So, suppose A, B E R are F -invariant, i.e. for all g ∈ F , ¯ gA = A ¯ g and ¯ gB = B ¯ g , andsuppose that AB ⊆ P ∩ R . Then AS, BS are two-sided ideals of S , and ( AS )( BS ) ⊆ P .So since P is prime, we can assume without loss of generality that AS ⊆ P .So since AS = ⊕ g ∈ F A ¯ g , it follows that A ⊆ P ∩ R , and hence P ∩ R is F -prime asrequired. Lemma 2.5.

Let R be a Noetherian ring, F a ﬁnite group. Then if P is a primitive idealof a crossed product R ∗ F , then P ∩ R is semiprimitive. roof. Let S = R ∗ F , then since P is primitive, P = Ann S M for some irreducible S -module M . Since F is ﬁnite, M is ﬁnitely generated over R , so since R is Noetherian,we can choose a maximal R -submodule U of M .For each g ∈ F , gU g − is a maximal R -submodule of M , so set M g := M/gU g − ,an irreducible R -module, and let Q g := Ann R M g , a primitive ideal of R . Clearly if r ∈ P ∩ R = Ann R M then rN g = 0 for all g ∈ F , so P ∩ R ⊆ ∩ g ∈ F Q g .Also, ∩ g ∈ F gU g − is an S -submodule, so by simplicity of M , ∩ g ∈ F gU g − = 0. So if r ∈ ∩ g ∈ F Q g then rM g = 0 for all g , so rM ⊆ gU g − for all g , i.e. rM ⊆ ∩ g ∈ F gU g − = 0 andhence r ∈ Ann R M = P ∩ R . Hence: P ∩ R = ∩ g ∈ F Q g Hence P ∩ R is semiprimitive as required. p -valued Groups Let G be a group. Recall from [18, III 2.1.2] that we deﬁne a p -valuation on G to be amap ω : G → R ∪ {∞} such that for all g, h ∈ G : • ω ( g ) = ∞ if and only if g = 1. • ω ( g − h ) ≥ min { ω ( g ) , ω ( h ) } . • ω (( g, h )) ≥ ω ( g ) + ω ( h ). • ω ( g p ) = ω ( g ) + 1. • ω ( g ) > p − .Note that if ( G, ω ) is a p -valued group then G is torsionfree, and carries a topology de-ﬁned by the metric d ( g, h ) := c − ω ( g − h ) for some c >

1. We will always assume that G is complete with respect to this topology, in which case we can deﬁne p -adic exponentia-tion in G , i.e. for all g ∈ G , α ∈ Z p , if α = lim n →∞ α i for α i ∈ Z , we deﬁne g α := lim n →∞ g α i ∈ G .Given d ∈ N , we say that G has ﬁnite rank d if there exists a subset g := { g , · · · , g d } ⊆ G such that for every g ∈ G , there exists a unique α ∈ Z dp such that g = g α := g α · · · g α d d ,and ω ( g ) := min { v p ( α i ) + ω ( g i ) : i = 1 , · · · , d } . We call such a subset g an ordered basis for ( G, ω ). Example: If G is uniform, and ω ( g ) := sup { n ∈ N : g ∈ G p n +1 } , then ( G, ω ) is acomplete p -valued group, and any minimal topological generating set for G is an orderedbasis for ( G, ω ). Deﬁnition 2.6.

We say that a p -valuation ω : G → R ∪ {∞} is abelian if: • There exists n ∈ N such that ω ( G ) ⊆ n Z . • For all g, h ∈ G , ω (( g, h )) > ω ( g ) + ω ( h ) . G, ω ) is any integer valued p -valued group of ﬁnite rank (e.g.a uniform group), then we can choose c > ω c ( g ) := ω ( g ) − c is an abelian p -valuation on G .Now, suppose that ( G, ω ) is a complete p -valued group of rank d , with ordered basis g = { g , · · · , g d } then the Iwasawa algebra O G is isomorphic to the power series ring O [[ b , · · · , b d ]] as an O -module (and as a ring if G is abelian), where each variable b i corresponds with g i − ω is an abelian p -valuation, taking values in n Z for some n ∈ N , then recallfrom [14, Section 2.2] that we can deﬁne a ﬁltration w on O G via: w ( P α ∈ N d λ α b α · · · b α d d ) = inf { v π ( λ α ) + P i ≤ d enα i ω ( g i ) : α ∈ N d } ,where e is the ramiﬁcation index of K/ Q p .We call w the Lazard ﬁltration on O G . Using [24, Theorem 4.5], we see that gr w O G ∼ = k [ t, t , · · · , t d ], where k is the residue ﬁeld of K , t = gr( π ) and t i = gr( b i ),and hence is commutative. Note that for any g ∈ G , w ( g − ≥ enω ( g ), with equality if g = g i for some i .Furthermore, since O G is complete with respect to w and gr w O G is Noetherian, it followsfrom [20, Ch. II Theorem 2.1.2] that w is a Zariskian ﬁltration. Hence for any two-sidedideal I of O G , the quotient ﬁltration w on O G/I is Zariskian.In particular, if I is a prime ideal then we can use Construction 2.3 to deﬁne a non-commutative valuation v on the Goldie ring of quotients Q ( O G/I ) such that the naturalmap τ : ( O G, w ) → ( Q ( O G/I ) , v ) is continuous. K G

Fixing (

G, ω ) a complete p -valued group of ﬁnite rank, we will now examine some basicproperties of prime ideals in KG . First of all, the following lemma allows us to simplifythe statement of Theorem C to remove reference to the ﬁnite extension F/K : Lemma 2.7.

Let

F/K be a ﬁnite extension, and let I ′ a right ideal of F G . Setting I := I ′ ∩ KG , we have that if I ′ is controlled by U ≤ c G then I is controlled by U .Proof. We will ﬁrst suppose that U is open in G . Then given r ∈ I , choose a complete setof coset representatives { g , · · · , g r } for U in G , then r = P ≤ i ≤ r r i g i for some r i ∈ KU ⊆ F U .So since I = I ′ ∩ KG and I ′ is controlled by U , it follows that r i ∈ I ′ ∩ F U ∩ KG = I ′ ∩ KU = I for each i , and hence I is controlled by U .So, let I χ be the controller subgroup of I , i.e. the intersection of all open subgroups of G controlling I . So since this includes all open subgroups of G controlling I ′ , we have that I χ ⊆ I ′ χ , hence any closed subgroup controlling I ′ also controls I .Now, recall that a two-sided ideal P of a ring R is completely prime if the quotient RP isa domain. The following result is the characteristic 0 analogue of [5, Theorem 8.6], andit uses a similar argument. 10 heorem 2.8. Let P be a prime ideal of KZ ( G ) . Then P KG is a completely primeideal of KG , and if P is faithful then P KG is faithful.Proof.

Let Z := Z ( G ). We will prove that if P is a prime ideal of O Z with p / ∈ P then P O G is completely prime, and it is faithful if P is faithful. The result for the rationalIwasawa algebras follows immediately.Let Q be the ﬁeld of fractions of O Z/P . If we let w be the Lazard ﬁltration on O Z ,then since w is a Zariskian ﬁltration and the associated graded is a commutative, inﬁ-nite dimensional k -algebra, it follows from Construction 2.3 that there exists a valuation v ′ on Q such that the natural map τ : O G → Q is continuous, and if w ( x ) ≥ v ′ ( τ ( x )) ≥ v ′ ( τ ( z − z ∈ Z then v ′ ( τ ( z − n ) = 0 for all n since v ′ isa valuation, which is a contradiction since ( z − n converges to zero in O G , and hence in Q by continuity of τ . Therefore v ′ ( τ ( z − > z ∈ Z ( G ), and after choosing anordered basis { z , · · · , z n } for Z and an integer M such that M v ′ ( τ ( z i − ≥ w ( z i − i , then we obtain an equivalent valuation v := M v ′ on Q such that v ( τ ( x )) ≥ w ( x )for all x ∈ O Z .Recall that if we ﬁx an ordered basis { g , · · · , g e } for GZ , then every element of O G hasthe form P α ∈ N e µ α c α for some µ α ∈ O Z where c i = g i −

1. Deﬁne a map u : O G → Z ∪ {∞} via: u : O G → Z ∪ {∞} , X α ∈ N e µ α c α inf { v ( τ ( µ α )) + w ( c α ) : α ∈ N e } . (4)Since v is a separated valuation, it is clear that u ( P α ∈ N e µ α c α ) = ∞ if and only if µ α ∈ P for all α , i.e. if and only if P α ∈ N e µ α c α ∈ P O G . Therefore u − ( ∞ ) = P O G . So followingthe proof of [5, Theorem 8.6], we will prove that u is a valuation on O G , from which itwill follow that P O G = u − ( ∞ ) is a completely prime ideal.Firstly, it is clear from the deﬁnition that u ( r + s ) ≥ min { u ( r ) , u ( s ) } , u ( µ ) = v ( τ ( µ )) and u ( µr ) = u ( µ ) + u ( r ) for all r, s ∈ O G , µ ∈ O Z . It is also clear that if r , r , · · · ∈ O G with r i → i → ∞ then u ( r + r + · · · ) ≥ inf { u ( r i ) : i ≥ } , therefore to prove that u is a ﬁltration it remains to prove that u ( c α c β ) ≥ u ( c α ) + u (( c ) β ) for all α, β ∈ N r .Write c α c β = P γ ∈ N e λ α,βγ c γ , then by the deﬁnition of the Lazard ﬁltration, w ( P γ ∈ N e λ α,βγ c γ ) =inf { w ( λ α,βγ ) + w ( c γ ) : γ ∈ N d } . So since u ( x ) ≥ w ( τ ( x )) for all x ∈ O Z , we have: u ( c α c β ) = inf { v ( τ ( λ α,βγ )) + w ( c γ ) : γ ∈ N e } ≥ inf { w ( λ α,βγ ) + w ( c γ ) : γ ∈ N e } = w ( c α ) + w ( c β ) = u ( c α ) + u ( c β ).So u is a ﬁltration on O G , and to verify that it is a valuation, we will show thatthe associated graded gr u O G is a domain. First note that the deﬁnition of u gives riseto a natural inclusion of graded rings gr v O Z/P → gr u O G , and this gives rise to anisomorphism of graded rings gr v ( O Z/P )[ Y , · · · , Y e ] → gr u O G where Y i is sent to gr( c i ).11herefore gr u O G is a domain and u is a valuation as required.Finally, if P is faithful, then suppose g ∈ G and g − ∈ P O G . Then write g = zg α · · · g α e e for some z ∈ Z , α i ∈ Z p , and it follows that: h − z −

1) + ( z − P = γ ∈ N e (cid:0) αγ (cid:1) c α + P = γ ∈ N e (cid:0) αγ (cid:1) c α .Therefore, we see that z − ∈ P and hence z = 1 since P is faithful. It also followsthat for each 0 = γ ∈ N e , (cid:0) αγ (cid:1) ∈ P , and hence (cid:0) αγ (cid:1) = 0 since P ∩ O = 0. This is onlypossible if α = ( α , · · · , α e ) = 0, and hence h = zg α · · · g α e e = 1 and P O G is faithful aswe require.In particular, it follows from this result that standard prime ideals in KG are completelyprime. K G

For the rest of this section, ﬁx G a uniform pro- p group, let L := p log( G ) be its Z p -Liealgebra of G , and let g = L ⊗ Z p Q p .To reiterate, we aim to study the action of KG on certain [ U ( L ) K -modules using thedense embedding KG → [ U ( L ) K . However, this embedding is not faithfully ﬂat, sorepresentation theoretic information is lost when passing from KG to [ U ( L ) K .Perhaps a better choice for a completion of KG would be the distribution algebra D ( G, K ) of G with coeﬃcients in K in the sense of [24]. In this case, the natural denseembedding KG → D ( G, K ) is faithfully ﬂat by [24, Theorem 4.11], but unfortunately D ( G, K ) is not Noetherian, so it would be diﬃcult in practice to extract general ring-theoretic information from D ( G, K ).However, for each n ≥

0, consider the crossed products D p n = D p n ( G ) := \ U ( p n L ) K ∗ GG pn as deﬁned in [2, Proposition 10.6], which arise as a Banach completions of KG with respectto the extension of the dense embedding KG p n → \ U ( p n L ) K to KG = KG p n ∗ GG pn . Thesealgebras give rise to an inverse system: KG → D ( G, K ) → · · · D p → D p → D p → D = [ U ( L ) K .i.e. D ( G, K ) = lim ←− n →∞ D p n , so since D ( G, K ) is faithfully ﬂat over KG , we want to ap-proximate D ( G, K ) using the Noetherian Banach algebras D p n , and thus limit how muchinformation we lose. Indeed, using [2, Proposition 10.6(e), Corollary 10.11], we see thatfor all KG -modules M , D p n ⊗ KG M = 0 for all suﬃciently high n . Lemma 2.9.

Let A be a free abelian pro- p group of rank d , A := p log( A ) . Then AA p = C × · · · × C d where each C i = h c i i = h g i A p i is a cyclic group of order p , and D p = D p ( A ) is an iterated crossed product: D p = \ U ( p A ) K ∗ C ∗ · · · ∗ C d . here for each i = 1 , · · · , d c ir = c ri for ≤ r < p , and c ip = g pi .Proof. Firstly, it is clear that since A = Z dp that AA p = Z dp ( p Z p ) d = ( Z p p Z p ) d = C × · · · × C d asrequired.For the second statement, it suﬃces to prove that KA = KA p ∗ C ∗ · · · ∗ C d , and thatthis decomposition satisﬁes the same properties, since it will be preserved after passingto the completion.Choose a Z p -basis { g , · · · , g d } for A , and we may assume that C i = h c i i where c i = g i A p .Then every element r ∈ KA has the form P α ∈ [ p − d r α g α · · · g α d d for some r α ∈ KA p , and r is sent to P α ∈ [ p − d r α c α · · · c α d d under the isomorphism KA → KA p ∗ AA p .So, since g i , g j and g i g j are sent to c i , c j and c i c j respectively, it follows that c i c j = c i · c j for each i, j . Hence KA = KA p ∗ C ∗ · · · ∗ C d .Finally, for 0 ≤ r < p , g ri is sent to c ri , and hence c ir = c ri , and g pi ∈ KA p is sent to g pi , so c ip = g pi as required. Recall from [16, Deﬁnition 2.3] the following deﬁnition:

Deﬁnition 2.10.

Let λ : g → K be a Q p -linear form such that λ ( L ) ⊆ O (i.e. λ ∈ Hom Z p ( L , O ) ). • A polarisation of g K = g ⊗ Q p K at λ is a solvable subalgebra b of g K such that forany subspace b ⊆ V ⊆ g K , λ ([ V, V ]) = 0 if and only if V = b . • Given a polarisation b of g K at λ , let B := b ∩ ( L ⊗ Z p O ) , and let K λ be the one-dimensional b -module induced by λ . Deﬁne the aﬃnoid Dixmier module of [ U ( L ) K induced by λ to be [ D ( λ ) = [ D ( λ ) b := [ U ( L ) K ⊗ \ U ( B ) K K λ Note:

If it is unclear what the base ﬁeld K is, we may sometimes write [ D ( λ ) K .So, ﬁxing λ ∈ Hom Z p ( L , O ), let b be a polarisarion of g at λ , and we see that KG acts on [ D ( λ ) b via the embedding KG → [ U ( L ) K . Set P := Ann KG [ D ( λ ), and using [16, Theorem4.4], we see that this does not depend on the choice of polarisation. Deﬁnition 2.11.

Deﬁne the λ -scalar ideal of g to be a λ , the largest ideal of g suchthat λ ( a ) = 0 . Also, set A λ := a λ ∩ L , and deﬁne the λ -scalar subgroup of G , A λ :=exp( p A ) E ic G . Note:

For any choice of polarisation b of g ⊗ Q p K at λ , it follows from [16, Lemma 2.3]that a λ ⊆ b . Lemma 2.12.

Let A λ be the λ -scalar subgroup of G . Then if P = Ann KG [ D ( λ ) , then A λ = P † = { g ∈ G : g − ∈ P } . In particular, P is faithful if and only if the restrictionof λ to Z ( g ) is injective. roof. Firstly, since a λ ⊆ b , we see that a λ [ D ( λ ) = a λ [ U ( L ) K ⊗ \ U ( B ) K K λ = [ U ( L ) K a λ ⊗ \ U ( B ) K K λ = 0.So since A λ − ⊆ a λ [ U ( L ) K , it is clear that A λ − ⊆ P , i.e. A λ ⊆ P † .Now, since T = P † is a closed, normal subgroup of G , T := p log( T ) is an ideal of L ,and it contains p log( A λ ) = A λ . Also, since ( T − [ D ( λ ) = 0, it follows that T [ D ( λ ) = 0.This is only possible if T ⊆ B and λ ( T ) = 0.Setting t := T ⊗ Z p Q p , t is an ideal of g , a λ ⊆ t and λ ( t ) = 0. So by the deﬁnition of a λ ,this means that a λ = t .So, for any u ∈ T , there exists i ∈ N such that π i u ∈ A λ = a λ ∩ L , and this means that u ∈ A λ . So A λ = T , and it follows immediately that A λ = T .Finally, since G is nilpotent, L is nilpotent, and thus if A λ = 0, then it must have non-trivial intersection with Z ( g ). So since P is faithful if and only if A λ = 1 (i.e. if and onlyif a λ = 0), and any subspace of Z ( g ) is an ideal of g , it follows that P is faithful preciselywhen nothing in Z ( g ) is sent to zero under λ , i.e. λ | Z ( g ) is injective.This lemma is useful to know, because it implies that for any Dixmier annihilator P , P † is a closed, isolated normal subgroup of G , and hence we can replace G by G P = GP † ,which is still a nilpotent, uniform group, and P = P ( P † − KG becomes a faithful Dixmierannihilator.Note that this lemma explains why we need the assumption that λ | Z ( g ) is injective in thestatement of Theorem C, since it is generally untrue that non-faithful prime ideals in KG are controlled by Z ( G ). In this section, we will study general prime ideals within the rational Iwasawa algebra KG of a p -valuable group G . This is of course equivalent to studying prime ideals in O G that do not contain p .The methods we use are inspired by those used in [5] to prove that faithful prime idealsin the mod- p Iwasawa algebra kG are standard, namely the study of Mahler expansionsof G -automorphisms . Unfortunately, these methods do not work in characteristic 0, asdemonstrated in [14, Section 3.3], and the best result they can be used to obtain is aweak control theorem for faithful primitive ideals ([14, Theorem 1.2]).The methods we will employ in this section involve considering logarithms of G -automorphisms, which in many ways is the characteristic 0 version of their Mahler ex-pansions. And while these methods are not yet suﬃcient to prove standardness for allprime ideals, we can employ them together with techniques from [5, Section 7] to adaptthe argument used in [14, Section 3.4] and ultimately reprove the weak control theorem[14, Theorem 1.2] for all faithful prime ideals, and not just primitives. This will culminatein the proof of Theorem B. 14 .1 Bounded Ring Automorphisms Let R be a ring carrying a complete Zariskian ﬁltration w . Recall from [5] that a function f : R → R is bounded if inf { w ( f ( r )) − w ( r ) : r ∈ R } > −∞ , in which case we deﬁne the degree of f to be the number deg w ( f ) := inf { w ( f ( r )) − w ( r ) : r ∈ R } .If we set B ( R ) as the space of bounded, additive maps f : R → R , then B ( R ) is a ringwith pointwise addition and composition as multiplication, and deg w deﬁnes a completeseparated ﬁltration on B ( R ). Lemma 3.1. If f : R → R is an additive map such that deg w ( f ) > , and I is a two-sided ideal of R such that f ( I ) ⊆ I . Then if w is the quotient ﬁltration on R/I , and f : R/I → R/I is the map induced from f , then deg w ( f ) > .Proof. Let µ := deg w ( f ) >

0. Then given r ∈ R , w ( f ( r )) − w ( r ) ≥ µ , we want to provethat w ( f ( r + I )) − w ( r + I ) > w ( r + I ) = sup { w ( r + y ) : y ∈ I } , so let us suppose for contradiction that w ( r + I ) ≥ w ( f ( r + I )), and hence there exists y ∈ I such that w ( r + y ) ≥ w ( f ( r ) + u )for all u ∈ I . In particular, since f ( I ) ⊆ I , w ( r + y ) ≥ w ( f ( r ) + f ( y )) = w ( f ( r + y )) ≥ w ( r + y ) + µ > w ( r + y ) – contradiction.Therefore w ( f ( r + I )) > w ( r + I ) for all r ∈ R , so since w is integer valued, it followsthat deg w ( f ) ≥ > R is a Z p -algebra, with p = 0 and w ( p ) >

0. The following lemmawill be useful to us several times in this section:

Lemma 3.2.

Given m ∈ N , a, b ∈ R such that w ( a ) = 0 , w ( b ) ≥ , and a and b commute: • v p (cid:0) p m k (cid:1) = m − v p ( k ) for all < k < p m . • min { m − v p ( k ) + ( p m − k ) w ( b ) : 0 < k < p m } → ∞ as m → ∞ • w (( a + b ) p m − a p m ) ≥ min { p m w ( b ) , m − v p ( k ) + ( p m − k ) w ( b ) : 0 < k < p m } , andthus ( a + b ) p m − a p m → as m → ∞ .Proof. Firstly, if k = a + a p + · · · + a t p t for some 0 ≤ a i < p , we deﬁne s ( k ) := a + a + · · · + a t . Then using [18, III 1.1.2.5] we see that v p ( k !) = k − s ( k ) p − . Therefore, v p (cid:0) p m k (cid:1) = v p Ä p m ! k !( p m − k )! ä = p m − s ( p m ) − k + s ( k ) − ( p m − k )+ s ( p m − k ) p − = s ( k )+ s ( p m − k )+1 p − .But since k < p m , we may assume that t = m −

1, i.e. k = a + a p + · · · + a m − p m − .And since k = 0, let i be maximal such that a m − i = 0, so k = a m − i p m − i + · · · + a m − p m − and hence v p ( k ) = m − i .Now, p m = ( p − p m − i + ( p − p m − i +1 + · · · + ( p − p m − + p m − i , and thus p m − k =( p − a m − i ) p m − i + ( p − a m − i +1 − p m − i +1 + · · · + ( p − a m − − p m − and we deduce that s ( p m − k ) = ip − s ( k ) − ( i −

1) = i ( p − − s ( k ) + 1.Therefore, v p (cid:0) p m k (cid:1) = s ( k )+ s ( p m − k ) − p − = i ( p − p − = i = m − v p ( k ) as required.15o prove the second statement, we just need to prove that for any k , m − v p ( k ) + ( p m − k ) w ( b ) → ∞ as m → ∞ :If v p ( k ) ≤ m then m − v p ( k ) + ( p m − k ) w ( b ) ≥ m → ∞ .If v p ( k ) > m then k = p v p ( k ) y with v p ( y ) = 0, so m − v p ( k )+( p m − k ) w ( b ) ≥ p v p ( k ) ( p m − v p ( k ) − y ) w ( b ) ≥ p m → ∞ as required.Finally, ( a + b ) p m − a p m = P ≤ k

Lemma 3.3.

The sequence ϕ n − converges to in B ( R ) as v p ( n ) → ∞ .Proof. We just need to prove that deg w ( ϕ n − → ∞ as v p ( n ) → ∞ . Since deg w ( ϕ n − > n , it suﬃces to prove that deg w ( ϕ p m − → ∞ as m → ∞ .Now, ϕ p m − ϕ −

1) + 1) p m − P ≤ k

1, it follows from Lemma 3.2 that if k > w ( (cid:0) p m k (cid:1) ( ϕ − p m − k ) ≥ w (cid:0) p m k (cid:1) +( p m − k ) deg w ( ϕ − ≥ m − v p ( k )+( p m − k ) deg w ( ϕ − →∞ as m → ∞ , and clearly if k = 0 then deg w ( (cid:0) p m k (cid:1) ( ϕ − p m − k ) ≥ p m deg w ( ϕ − → ∞ as required.Now, let us suppose that R is a prime, Noetherian ring, gr w R is commutative, and thatthe positively graded ideal (gr w R ) ≥ is not nilpotent. Then the simple, artinian ring Q ( R ) carries a non-commutative valuation v , which we can describe using Construction2.3. Clearly any automorphism ϕ of R extends to Q ( R ).The following theorem allows us to pass from ( R, w ) to ( Q ( R ) , v ) without diﬃculty. Theproof is similar to that of an analogous result in a characteristic p setting, namely theproof of [15, Proposition 3.5]. Theorem 3.4. If ϕ ∈ Aut ( R ) and deg w ( ϕ − > , then there exists n ∈ N with deg v ( ϕ n − > .Proof. Using Construction 2.3, we see that we have a sequence of ﬁltrations w ′ , v z,U , v z,V , v z, B , v on Q ( R ), and our strategy is to prove that ϕ n − n withrespect to each of these in turn. Let us ﬁrst consider w ′ .Since deg w ( ϕ − >

0, i.e. w ( ϕ ( r ) − r ) > w ( r ) for all r ∈ R , it follows that the inducedgraded automorphism ϕ : gr w R → gr w R, r + F n R ϕ ( r ) + F n R is just the identity.Therefore, since gr w ′ Q ( R ) = (gr R ) q , it follows that the induced morphism ϕ : gr w ′ Q ( R ) → gr w ′ Q ( R ) is also the identity, and hence deg w ′ ( ϕ − ≥ n ∈ N such that deg w ′ ( ϕ n − ≥ w ′ ( z ). But since z n U = F nw ′ ( z ) Q ( R ), it follows that if r ∈ z n U then w ′ (( ϕ n − r )) ≥ w ′ ( r ) + w ′ ( z ) ≥ nw ′ ( z ) + w ′ ( z ) = ( n + 1) w ′ ( z ), and hence ( ϕ n − z n U ) ⊆ F ( n +1) w ′ ( z ) Q ( R ) = z n +1 U , andhence deg v z,U ( ϕ n − ≥ b Q is a simple quotient of the completion Q ′ of Q ( R ) with respect to v z,U , i.e. a quotient of Q ′ by a maximal ideal m . But since Q ′ is artinian, all maximalideals are minimal prime ideals, and hence there are only ﬁnitely many of them. Since ϕ n is continuous, ϕ n ( m ) is also a minimal prime ideal of Q ′ , i.e. ϕ n permutes the setof minimal prime ideals of Q ′ . Since this set is ﬁnite, all permutations have ﬁnite order,so there exists m such that ϕ mn ( m ) = m .Therefore, there exists n such that ϕ n induces an automorphism ϕ n : b Q → b Q , andsince Q ( R ) is simple, the composition Q ( R ) ֒ −→ Q ′ ։ b Q must be injective. Therefore,we can think of ϕ n as a extension of ϕ n to b Q , so sometimes we may just call it ϕ n forconvenience.Now, if r ∈ z n V , then r is the image of z n s in b Q , for some s ∈ U . So since deg v z,U ( ϕ n − ≥

1, it follows that ( ϕ n − z n s ) ∈ z n +1 U , and hence ( ϕ n − r ) ∈ z n +1 V . Thereforedeg v z,V ( ϕ n − ≥ B is a maximal order in b Q , equivalent to V , and B ⊆ z − r V . So let I := { x ∈ V : Bx ⊆ V } , then I is a two-sided ideal of V with z r V ⊆ I . Since deg v z,V ( ϕ n − >

0, itfollows from Lemma 3.3 that we can choose m ∈ N such that ( ϕ mn − V ) ⊆ z r V , andhence there exists n ∈ N such that ( ϕ n − V ) ⊆ I .In particular, ϕ n ( I ) ⊆ I , and it follows from Noetherianity of V that ϕ n ( I ) = I ,and hence I = ϕ − n ( I ).Therefore, given x ∈ ϕ n ( B ), x = ϕ n ( b ) for some b ∈ B , and given c ∈ I , xc = ϕ n ( b ) c = ϕ n ( bϕ − n ( c )), and since ϕ − n ( c ) ∈ I , it follows that bϕ − n ( c ) ∈ V , and hence xc ∈ V .So setting O ( I ) = { b ∈ B : bI ⊆ V } , it follows that O ( I ) contains B and ϕ n ( B ). But O ( I ) is an order in b Q , equivalent to V , by [21, Lemma 5.1.12], so since B and ϕ n ( B ) aremaximal orders, this means that O ( I ) = B = ϕ n ( B ), and hence ϕ n is an automorphismof B .Also, we can choose n such that deg v z,V ( ϕ n − ≥ r + 1, and hence ( ϕ n − z n B ) ⊆ ( ϕ n − z n − r V ) ⊆ z n +1 v ⊆ z n +1 B . Therefore deg z, B ( ϕ n − ≥ B = M n ( D ) for some non-commutative DVR D , so let ν be a uniformiser in D ,and it follows that all two-sided ideals of B have the form ν n B . Since v is the ν -adicﬁltration on b Q , and v is topologically equivalent to v z, B , we know that there exists k ∈ N such that z r B ⊆ ν B . By Lemma 3.3, we can choose n ≥ n such that deg v z, B ( ϕ n − ≥ k ,i.e. ( ϕ n − B ) ⊆ z k B ⊆ µ B .So, if we assume for induction that ( ϕ n − ν i B ) ⊆ ν i +1 B for all i < m , then ( ϕ n − ν m ) = ϕ n ( ν m ) − ν m = ( ϕ n − ν m − ) ϕ n ( ν ) + ν m − ( ϕ n − ν ). But ( ϕ n − ν ) ∈ ν B ,( ϕ n − ν m − ) ∈ ν m B and ϕ n ( ν ) ∈ ν B , therefore ( ϕ n − ν m ) ∈ ν m +1 B .17t follows that ( ϕ n − ν m B ) ⊆ ν m +1 B for all m , and hence deg v ( ϕ n − ≥ Now, let (

G, ω ) be a complete, p -valued group of ﬁnite rank. We may assume that ω isan abelian p -valuation as deﬁned in Deﬁnition 2.6, and we let w be the correspondingLazard ﬁltration on O G – a complete Zariskian ﬁltration.Fix an ordered basis g = { g , · · · , g d } for ( G, ω ), so that O G is isomorphic to the spaceof power series O [[ b , · · · , b d ]] as an O -module, where b i = g i −

1. Recall that for any g ∈ G , w ( g − ≥ enω ( g ), with equality if g = g i for some i .Now, recall the following deﬁnition ([5, Deﬁnition 4.5]): Deﬁnition 3.5.

An automorphism ϕ ∈ Aut ( G ) is bounded if inf { ω ( ϕ ( g ) g − ) − ω ( g ) : g ∈ G } > p − , and we deﬁne the degree of ϕ to be the number deg ω ( ϕ ) = inf { ω ( ϕ ( g ) g − ) − ω ( g ) : g ∈ G } .Let Aut ω ( G ) be the group of bounded automorphisms of G . Lemma 3.6. If ϕ ∈ Aut ω ( G ) then ϕ extends to a continuous, O -linear automorphism of O G such that deg w ( ϕ − > .Proof. Clearly ϕ extends to an O -linear automorphism of O G , so we need only provethat deg w ( ϕ − >

0, and it will follow that ϕ is continuous and extends to O G .Firstly, for each i = 1 , · · · , d , ( ϕ − g i −

1) = ϕ ( g i ) − g i = ( ϕ ( g i ) g − i − g i . Butsince ϕ ∈ Aut ω ( G ), we know that ω ( ϕ ( g i ) g − i ) > ω ( g ) + p − , so w ( ϕ ( g i ) g − i − ≥ enω ( ϕ ( g i ) g − i ) ≥ enω ( g i ) = w ( g i − w (( ϕ − g i − − w ( g i − > a, b ∈ O [ G ]:( ϕ − ab ) = ϕ ( a ) ϕ ( b ) − ab = ( ϕ ( a ) − a ) ϕ ( b ) + a ( ϕ ( b ) − b ) = ( ϕ − a ) ϕ ( b ) + a ( ϕ − b ).If we assume that w (( ϕ − a )) − w ( a ) > w (( ϕ − b )) − w ( b ) >

0, then it followsthat w ( ϕ ( b )) = w ( b ) and w (( ϕ ( a ) − a ) ϕ ( b ) + a ( ϕ ( b ) − b )) > min { w ( a ) + w ( ϕ ( b ) , w ( a ) + w ( b ) } = w ( a ) + w ( b ) = w ( ab ), therefore w (( ϕ − ab )) − w ( ab ) > w (( ϕ − g i − − w ( g i − > i = 1 , · · · , d , it follows that for every α ∈ N d , λ ∈ O , w ( λ ( ϕ − g − α · · · ( g d − α d )) − w ( λ ( g − α · · · ( g d − α d ) > O [ G ] is generated as an additive group by the monomials λb α · · · b α d d ,it follows that w ( ϕ ( s )) = w ( s ) for all s ∈ O [ G ].Now, given r ∈ O [ G ] with w ( r ) = t , it follows from the deﬁnition of w that r = P α ∈ A λ α b α · · · b α d d + s , where A := { α ∈ N d : P ≤ i ≤ d α i enω ( g i ) = t } and w ( s ) > t . Since w ( ϕ ( s )) = w ( s ), it is clear that w (( ϕ − s )) ≥ w ( s ) > t , so to prove that w (( ϕ − r )) >t , it remains to show that w (( ϕ − P α ∈ A λ α b α · · · b α d d )) > t .But w ( λ α b α · · · b α d d ) = t for all α ∈ A , so we have seen that w (( ϕ − λ α b α · · · b α d d )) > t foreach α , and it follows immediately that w (( ϕ − P α ∈ A λ α b α · · · b α d d )) > t as required.18ow, ﬁx a prime ideal P of O G such that p / ∈ P , and ﬁx an automorphism ϕ ∈ Aut ω ( G )such that P is invariant under the extension of ϕ to O G , i.e. ϕ ( P ) = P . Hence ϕ inducesan automorphism ϕ of O GP . Theorem 3.7.

There exists a non-commutative valuation v on Q ( O G/P ) such that thenatural map τ : ( O G, w ) → ( Q ( O G/P ) , v ) is continuous, and there exists n ∈ N suchthat deg v ( ϕ n − ≥ v ( p ) .Proof. Let w be the quotient ﬁltration on O GP , so that gr w O GP ∼ = gr w O G gr w P . So since w isZariskian, it follows that gr w O GP is commutative and Noetherian, and since O GP is com-plete with respect to w , it follows from [20, Theorem 2.1.2] that w is Zariskian.Moreover, since p / ∈ P and w ( p ) >

0, it follows that w ( p n ) = w ( p n ) for all n ∈ N , becauseif w ( p n ) > w ( p n ) then p n + r ∈ P for some r ∈ O G with w ( r ) > w ( p n ), and hence1 + p − n r ∈ P ⊗ O K with w ( p − n r ) >

0, but 1 + p − n r is a unit in KG – a contradiction.Therefore, it follows that gr( p ) lies in the positively graded piece of the associated gradedgr O GP and is non-nilpotent. Hence (gr w O GP ) ≥ is non-nilpotent.Therefore, using Construction 2.3, we can deﬁne a non-commutative valuation v on Q ( O G/P ) such that the inclusion ( O G/P, w ) → ( Q ( O G/P ) , v ) is continuous. Sincethe surjection ( O G, w ) → ( O G/P, w ) is continuous, it follows that the composition τ isalso continuous.Now, since deg w ( ϕ − > ϕ − P ) ⊆ P , it follows from Lemma3.1 that deg w ( ϕ − >

0. Using Theorem 3.4, it follows that there exists n ∈ N suchthat deg v ( ϕ n − > Given a complete p -valued group ( G, ω ) of ﬁnite rank, and a faithful prime ideal P of O G such that p / ∈ P , we now want to take steps towards proving a control theorem for P . Again, assume that there is an automorphism ϕ ∈ Aut ω ( G ) with ϕ = 1 such that ϕ ( P ) = P , and let ϕ be the automorphism of O G/P induced from ϕ .We will now assume further that there is a closed, central subgroup A of G such that: • ϕ ( g ) g − ∈ A for all g ∈ G . • For all a ∈ A , ϕ ( a ) = a .It is straightforward to show that these properties are satisﬁed for ϕ n for all n ∈ N .Now, using Theorem 3.7, we ﬁx a non-commutative valuation v on Q = Q ( O G/P ) suchthat the natural map τ : ( O G, w ) → ( Q, v ) is continuous, and a natural number n ∈ N such that deg v ( ϕ n − ≥ v ( p ). After replacing ϕ by ϕ n if necessary, we may assume that n = 1.Let b Q be the completion of Q with respect to v , and clearly ϕ extends continuouslyto b Q . 19lso, since A is central in G , P ∩ O A is a prime ideal of O A . Therefore, the ﬁeld offractions of O AP ∩O A is contained in b Q . So, let F be the closure of this ﬁeld of fractions in b Q , then F is a central subﬁeld of b Q carrying a complete valuation v F = v | F . Deﬁnition 3.8.

Deﬁne the logarithm of ϕ to be the derivation of b Q deﬁned by thelogarithm series at ϕ . Speciﬁcally: log( ϕ ) = X k ≥ ( − k +1 k ( ϕ − k (5) Note:

1. This deﬁnition never makes sense if p ∈ P , because in this case if p | k then k = 0 in b Q .2. This deﬁnition makes sense when p / ∈ P because deg v ( ϕ − ≥ v ( p ) so deg v ( ϕ − k ≥ kv ( p ) for all k and deg v ( ( − k +1 k ( ϕ − k ) ≥ kv ( p ) − v ( k ) = kv ( p ) − v p ( k ) v ( p ) → k → ∞ . So since B ( b Q ) is complete with respect to deg v by [5, Lemma 2.4], it followsthat the logarithm series must converge to an element of B ( b Q ), and since ϕ is an auto-morphism, this must be a derivation.3. See the proof of [23, Theorem 4] for details of why log( ϕ ) is a derivation of b Q . Proposition 3.9.

Fix g ∈ G , then: i . deg(log( ϕ )) ≥ . ii . The series log( ϕ ( g ) g − ) := P k ≥ − k +1 k τ ( ϕ ( g ) g − − k converges in F . iii . log( ϕ )( g ) = log( ϕ ( g ) g − ) τ ( g ) . iv . log( ϕ )( F ) = 0 v . log( ϕ ) is a continuous, F -linear derivation.Proof. i . Since deg v ( ϕ ) ≥ v ( p ), it follows that deg( ( − k +1 k ( ϕ − k ) ≥ kv ( p ) − v ( k ) =( k − v p ( k )) v ( p ) ≥

1, and hence deg v (log( ϕ )) ≥ ii – iii . For any g ∈ G , ( ϕ − g ) = τ ( ϕ ( g ) − g ) = τ ( ϕ ( g ) g − − τ ( g ), and if wesuppose for induction that ( ϕ − k ( g ) = τ ( ϕ ( g ) g − − k τ ( g ), then ( ϕ − k +1 ( g ) =( ϕ − τ ( ϕ ( g ) g − − k τ ( g )) = ϕ ( τ ( ϕ ( g ) g − − k ) ϕ ( τ ( g )) − τ ( ϕ ( g ) g − − k τ ( g ).But since ϕ ( g ) g − ∈ A , it follows from our assumption that ϕ ( τ ( ϕ ( g ) g − − k ) = τ ( ϕ ( g ) g − − k , and hence ( ϕ − k +1 ( g ) = τ ( ϕ ( g ) g − − k ϕ ( τ ( g )) − τ ( ϕ ( g ) g − − k τ ( g ) = τ ( ϕ ( g ) g − − k +1 τ ( g ).Therefore, log( ϕ )( g ) = P k ≥ − k +1 k τ ( ϕ ( g ) g − − k τ ( g ), so multiplying on the right by τ ( g ) − gives that log( ϕ ( g ) g − ) = P k ≥ − k +1 k τ ( ϕ ( g ) g − − k converges to log( ϕ )( g ) τ ( g ) − ,and since τ ( ϕ ( g ) g − − ∈ τ ( O A ) ⊆ F , it follows that log( ϕ ( g ) g − ) ∈ F . iv . Again, since ϕ ( a ) = a for all a ∈ A , it follows that ( ϕ − τ ( a )) = 0, and hencelog( ϕ )( a ) = 0. So since log( ϕ ) is O -linear and continuous, it follows that log( ϕ )( s ) = 0for all s ∈ τ ( O A ) = O A/ O A ∩ P . 20oreover, since F is the completion of the ﬁeld of fractions of τ ( O A ), and log( ϕ ) isa derivation, it follows that log( ϕ )( s ) = 0 for all s ∈ F as required. v . We know that log( ϕ ) is a continuous derivation, and since log( ϕ )( F ) = 0 it followsthat it is F -linear. Remark:

Without our assumptions that ϕ ( g ) g − ∈ A for all g , and ϕ is trivial whenrestricted to A , this proposition fails.Fix λ := inf { v (log( ϕ )( ϕ ( g ) g − )) : g ∈ G } , and let U := { g ∈ G : v (log( ϕ ( g ) g − )) > λ } .The following lemma depends on the assumption that P is faithful: Lemma 3.10. ≤ λ < ∞ , and U is a proper, open subgroup of G containing G p and ( G, G ) .Proof. For convenience, set ψ ( g ) := ϕ ( g ) g − . Then since ϕ ( g ) g − ∈ Z ( G ) for all g ∈ G ,it follows that ψ : G → A is a group homomorphism.Using Proposition 3.9, we see that deg( ϕ ) ≥

1, and that log( ϕ )( g ) = log( ψ ( g )) g forall g ∈ G . So since v ( g ) = 0, it follows that v (log( ψ ( g ))) = v (log( ϕ )( g )) ≥ deg( ϕ ) ≥ g ∈ G , and hence λ ≥ λ = ∞ then v (log( ψ ( g ))) = ∞ and hence log( ψ ( g )) = 0 for all g ∈ G . But the functionon b Q deﬁned by the logarithm series is injective by [11, Corollary 6.25( ii )], and hence τ ( ψ ( g ) −

1) = 0 for all g ∈ G , i.e. ψ ( g ) − ∈ P . But P is faithful, so this means that ψ ( g ) = ϕ ( g ) g − = 1 for all g ∈ G , and hence ϕ is trivial – contradicting our assumption.Therefore 1 ≤ λ < ∞ , and clearly λ is an integer, so there exists g ∈ G such that v (log( ψ ( g ))) = λ , and hence U ( G .Furthermore, given g ∈ G , v (log( ψ ( g p ))) = v (log( ψ ( g ) p )) = v ( p log( ψ ( g )) = v (log( ψ ( g )))+ v ( p ) ≥ λ + v ( p ) > λ , and hence G p ⊆ U .Also, if g, h ∈ U then v (log( ψ ( gh ))) = v (log( ψ ( g ) ψ ( h ))) = v (log( ψ ( g )) + log( ψ ( h ))) ≥ min { v (log( ψ ( g ))) , v (log( ψ ( h ))) } > λ ,so since GG p is a ﬁnite group and UG p is closed under the group operation, it follows that U is an open subgroup of G .Finally, if g, h ∈ G , v (log( ψ ( g, h ))) = v (log(( ψ ( g ) , ψ ( h )))) = v (log(1)) = ∞ > λ , so( G, G ) ⊆ U as required.So, using [5, Lemma 4.2], choose an ordered basis { g , · · · , g d } for ( G, ω ) such that { g p , · · · , g pr , g r +1 , · · · , g d } is an ordered basis for U for some 1 ≤ r ≤ d . Thus v (log( ϕ ( g i ) g − i )) = λ for i = 1 , · · · , r and v (log( ϕ ( g i ) g − i )) > λ for all i > r .Now, set a := log( ϕ ( g ) g − ). Then a ∈ F by Proposition 3.9, and v ( a ) = λ < ∞ so a = 0, thus a is a unit in F . So for each i = 1 , · · · , d , set z i := a − log( ϕ ( g i ) g − i ) ∈ F .21ince F is central, it follows from the deﬁnition of a non-commutative valuation that v ( z i ) = v (log( ϕ ( g i ) g − i )) − v ( a ) for each i , so v ( z ) = · · · = v ( z r ) = 0, and v ( z i ) > i > r .From now on, set δ := a − log( ϕ ) : b Q → b Q , which is an F -linear derivation of b Q , andusing Proposition 3.9 we see that for all g ∈ G , δ ( g ) = a − log( ϕ )( g ) = a − log( ϕ ( g ) g − ) g .But if g = g α := g α · · · g α d d for some α ∈ Z dp thenlog( ϕ ( g ) g − ) = log(( ϕ ( g ) g − ) α · · · ( ϕ ( g d ) g − d ) α d )) = α log( ϕ ( g ) g − ) + · · · + α d log( ϕ ( g d ) g − d ).Therefore, δ ( g α ) = ( α z + · · · + α d z d ) g α for all α ∈ Z dp .Furthermore, since δ is F -linear, it follows that δ n ( g α ) = ( α z + · · · α d z d ) n g α for all n ∈ N . We will now show how we can study convergence of δ p m to prove a control theorem for P . Notation:

For any α ∈ Z p , denote by α ′ the unique integer in { , · · · , p − } such that α ≡ α ′ (mod p ). Also, let V = { z ∈ F : v ( q ) ≥ } , V + = { z ∈ F : v ( q ) > } , so that VV + is a ﬁeld extension of F p . Lemma 3.11.

For all m ∈ N , z p m , · · · , z p m r are F p -linearly independent modulo V + .Proof. Let us suppose that there exist integers α , · · · , α r ∈ { , , · · · , p − } such that α z p m + · · · + α r z p m r ∈ V + , i.e. v ( α z p m + · · · + α r z p m r ) > α p m i ≡ α i (mod p ) by Fermat’s Little theorem, so 0 ≡ α z p m + · · · + α r z p m r ≡ ( α z + · · · + α r z r ) p m (mod V + ), which implies that α z + · · · + α r z r ∈ V + .But z i = a − log( ϕ ( g i ) g − i ), so α i z i = a − log(( ϕ ( g i ) g − i ) α i ). But since g ϕ ( g ) g − deﬁnes a group homomorphism, it follows that α i z i = a − log( ϕ ( g α i i ) g − α i i ), and hence α z + · · · + α r z r = a − log( ϕ ( g α ) g − α · · · ϕ ( g α r r ) g − α r r ) = a − log( ϕ ( g α ) g − α ) ∈ V + .Therefore, it follows that v (log( ϕ ( g α ) g − α )) > v ( a ) = λ , and hence g α ∈ U by thedeﬁnition of U . But we know that { g p , · · · , g pr , g r +1 , · · · , g d } is an ordered basis for U ,which means that p | α i , i.e. α i = 0, for all i as required. Lemma 3.12.

For any m ∈ N , α ∈ Z dp , δ p m ( g α ) = ( α ′ z p m + · · · + α ′ r z p m r ) g α + z ( m,α ) g α for some z ( m,α ) ∈ V + .Proof. Firstly, we know that δ p m ( g α ) = ( α z + · · · + α d z d ) p m g α , and we know that v ( α r +1 z r +1 + · · · + α d z d ) >

0. Also, using Lemma 3.11, we see that v ( α z + · · · + α r z r ) = 0if p ∤ α i for some i . Therefore:( α z + · · · + α d z d ) p m ≡ ( α z + · · · + α r z r ) p m ≡ α p m z p m + · · · + α p m r z p m r (mod V + ).22ince α p m i ≡ α i ≡ α ′ i (mod p ) for all m , it follows that α p m z p m + · · · + α p m r z p m r ≡ α ′ z p m + · · · + α ′ r z p m r (mod V + ).Therefore, ( α z + · · · + α d z d ) p m = α ′ z p m + · · · + α ′ r z p m r + z ( m,α ) for some z ( m,α ) ∈ V + ,and hence δ p m ( g α ) = ( α ′ z p m + · · · + α ′ r z p m r ) g α + z ( m,α ) g α as required.Now, deﬁne T := z z · · · z r z p z p · · · z pr . . · · · .. . · · · .. . · · · .z p r − z p r − · · · z p r − r . Then using Lemma 3.12, we see that: à δ ( g α ) ...δ p r − ( g α ) í = T à α ′ g α ...α ′ r g α í + à z (0 ,α ) g α ...z ( r − ,α ) g α í . (6)But T is a matrix of Vandermonde type, in the sense of [4, Section 1.1], and the entriesof T all lie in V and are F p -linearly independent modulo V + by Lemma 3.11, so it followsfrom [4, Lemma 1.1] that det( T ) has value zero. Therefore T is invertible, and its inversehas value 0.Deﬁne u : b Q r → b Q, ( s , · · · , s r ) T s + · · · + s r , and deﬁne: ι : b Q → b Q, s uT − à δ ( τ ( s )) ...δ p r − ( τ ( s )) í (7)Note that ι is continuous and F -linear. Also for any α ∈ Z dp , using (6) we see that: ι ( g α ) = uT − à δ ( g α ) ...δ p r − ( g α ) í = u à α ′ g α ...α ′ r g α í + uT − à z (0 ,α ) g α ...z ( r − ,α ) g α í .But the entries of T − all have value at least 0, so we deduce that ι ( g α ) = ( α ′ + · · · + α ′ r ) g α + z ( α ) g α for some z ( α ) ∈ V + .For each m ∈ N , deﬁne ι m : b Q → b Q, q ι p m ( p − ( q ), which is also continuous and F -linear. Also, for each k ≥

0, deﬁne b Q k := { q ∈ Q : v ( q ) ≥ k } . Proposition 3.13.

Given m ∈ N : • The composition ι m τ : O G → b Q is continuous and O -linear. • There exists k m ∈ N such that k m → ∞ as m → ∞ , and for all α ∈ Z dp , m ( g α ) ≡ ® g α if v p ( α + · · · + α r ) = 00 if v p ( α + · · · + α r ) > mod b Q k m ) .Moreover, choose a sequence of integers m < m < · · · such that k m < k m < · · · , thenfor every s ∈ O G , i ∈ N , ( ι m i − ι m i +1 )( τ ( s )) ∈ b Q k mi .Proof. The ﬁrst statement is obvious, since τ and ι m are both continuous and O -linear.Now, we know that ι is F -linear, and for any α ∈ Z dp , ι ( g α ) = ( α ′ + · · · + α ′ r + z ( α ) ) g α for some z ( α ) ∈ F with v ( z ( α ) ) ≥

1. Therefore, ι m ( g α ) = ι p m ( p − ( g α ) = ( α ′ + · · · + α ′ r + z ( α ) ) p m ( p − g α .But if v p ( α ′ + · · · + α ′ r ) = 0 then ( α ′ + · · · + α ′ r ) p − ≡ p ), so ( α ′ + · · · + α ′ r + z ( α ) ) p − =1 + y ( α ) for some y ( α ) ∈ F with v ( y ( α ) ) ≥

1, and ι m ( g α ) = (1 + y ( α ) ) p m g α .On the other hand, if v p ( α ′ + · · · + α ′ r ) > v ( ι m ( g α )) ≥ p m ( p − γ := inf { v ( y ( α ) ) : α ∈ Z dp , v ( α + · · · + α r ) = 0 } ≥

1, and for each m ∈ N , deﬁne t m := min { p m γ, m − v p ( k ) + ( p m − k ) γ : 0 < k < p m } . Then since γ = v ( y ( α ) ) for some α ∈ Z dp , it follows from Lemma 3.2 that t m → ∞ as m → ∞ .Furthermore, also using Lemma 3.2, we see that: v ((1 + y ( α ) ) p m − ≥ min { p m v ( y ( α ) ) , m − v p ( k ) + ( p m − k ) v ( y ( α ) ) : 0 < k < p m } ≥ t m .Hence v ( ι m ( g α ) − g α ) ≥ t m for all m whenever v ( α + · · · + α r ) = 0.So, let k m := min { p m ( p − , t m } → ∞ as m → ∞ , then v ( ι m ( g α )) ≥ k m if v ( α + · · · + α r ) >

0, and v ( ι m ( g α ) − g α ) ≥ k m if v p ( α + · · · + α r ) = 0 as required.In particular, if k m < k m < · · · , then for every g ∈ G , ι m i ( g ) ≡ ι m i +1 ( g ) (mod b Q k mi )for all i , i.e. ( ι m i − ι m i +1 )( g ) ∈ b Q k mi . So since ι m is O -linear for every m , it follows that( ι m i − ι m i +1 )( τ ( s )) ∈ b Q k mi for every s ∈ O G , and since ( ι m i − ι m i +1 ) τ is continuous, thismeans that ( ι m i − ι m i +1 )( τ ( s )) ∈ b Q k mi for every s ∈ O G as required.Using this proposition, it follows that there that there exists a continuous, O -linear map ι : O G → b Q such that ι ( P ) = 0, ι ( s ) ≡ ι m ( τ ( s )) (mod b Q k mi ) for every i , and for every α ∈ Z dp : ι ( g α ) = ® g α if v p ( α + · · · + α r ) = 00 if v p ( α + · · · + α r ) > U contains ( G, G ) and G p by Lemma 3.10, the quotient GU has the structureof an F p -vector space, with basis { g U, · · · , g r U } . Therefore, the map χ : GU → Z p Z , g α U α + · · · + α r + p Z is a non-zero F p -linear map, so ker( χ ) = VU for some proper open subgroup V of G , and it follows that for all g ∈ G : ι ( g ) = ® g if g / ∈ V g ∈ V (9)24ow we are ready to prove a control theorem. Firstly, let C ∞ ( G, O ) be the space oflocally constant functions f : G → O , and recall from [7, Proposition 2.5 and Lemma 2.9]that there is a natural action ρ : C ∞ ( G, O ) → End O O G such that for any open subgroup U of G , if f ∈ C ∞ ( G, O ) is constant on the cosets of U then the action of f on O G canbe described explicitly:If x ∈ O G and x = P g ∈ C x g g , where C is a set of coset representatives for U in G and x c ∈ O U , then ρ ( f )( x ) = P g ∈ C f ( g ) x c g In particular, deﬁne f : G → O , g ® g / ∈ V g ∈ V . Then clearly f ∈ C ∞ ( G, O ) isconstant on the cosets of V , so ρ ( f )( g ) = ® g if g / ∈ V g ∈ V .Therefore, it follows from (9) that ι ( g ) = τ ρ ( f )( g ) for all g ∈ G , so since ι , τ and ρ ( g ) arecontinuous and O -linear, it follows that ι = τ ρ ( f ). Therefore, since ι ( P ) = 0, it followsthat ρ ( f )( P ) ⊆ P . Proposition 3.14. P is controlled by V .Proof. This is now identical to the proof of [14, Theorem 1.4]. Firstly, suppose that C = { x , · · · , x t } is a complete set of coset representatives for V in G , then for all r ∈ O G , r = P i ≤ t r i x i for some r i ∈ O V .Suppose we can choose C such that if r ∈ P then r ∈ P ∩ O V . Then since rx − x i ∈ P for all i = 1 · · · , t and rx − x i has x component r i , it follows that r i ∈ P ∩ O V for each i , and hence P is controlled by V .It remains to prove that we can choose such a set C of coset representatives such thatif P i ≤ t r i x i ∈ P , then at least one of the r i lies in P ∩ O V .Since G p ⊆ V , it follows that V has ordered basis { g p , · · · , g ps , g s +1 , · · · , g d } and thus C = { g b · · · g b r s : 0 ≤ b i < p } is a complete set of coset representatives for V in G .So for each b ∈ [ p − s , let g b = g b · · · g b r r (here [ p −

1] = { , , · · · , p − } ).Then if r = P b ∈ [ p − s r b g b ∈ P , then ρ ( f )( r ) = P b ∈ [ p − s f ( g b ) r b g b , and since ρ ( f )( P ) ⊆ P thisalso lies in P . But f ( g b ) = 1 if b = 0, and f ( g ) = 0 hence ρ ( f )( r ) = P b ∈ [ p − s \{ } r b g b ∈ P .Therefore, r g = r − ρ ( f )( r ) ∈ P , and thus r ∈ P ∩ O U as required.So, since prime ideals in O G not containing P correspond bijectively with prime idealsin KG , altogether we have now prove the following theorem: Theorem 3.15.

Let ( G, ω ) be a complete, p -valued group of ﬁnite rank, and let P be afaithful, prime ideal of KG . Also, let ϕ ∈ Aut ω ( G ) be an automorphism of G , and let A be a closed, central subgroup of G such that: • ϕ = 1 . ϕ ( P ) = P . • ϕ ( g ) g − ∈ A for all g ∈ G . • ϕ ( a ) = a for all a ∈ A .Then P is controlled by a proper, open subgroup of G . This result is, in essence, the characteristic 0 version of [5, Theorem B]. This resultwas suﬃcient to fully prove Conjecture 1.2 in characteristic p for G nilpotent in [5], butunfortunately our additional assumption that ϕ ( g ) g − is ﬁxed by ϕ for all g ∈ G restrictsthe usefulness of this result, which is why, as we will see, we cannot assume that thesubgroup A described in the statement of Theorem B is central. Now we are ready to prove Theorem B, and the remainder of the argument is similar tothe proof of [14, Theorem 1.2], as given in [14, Section 3.5].Firstly, recall that a prime ideal P of KG is non-splitting if for any closed subgroup H of G that controls P , P ∩ KH is a prime ideal of KH . Furthermore, a right ideal I of KG is virtually non-splitting if I = P KG for some non-splitting prime ideal P of KU ,where U is some open subgroup of G .The following theorem, analogous to [5, Theorem 5.8] and partially proved in [14,Theorem 4.8], essentially proves that establishing a control theorem for virtually non-splitting ideals is suﬃcient to establish it for all primes. Theorem 3.16.

Let ( G, ω ) be a complete p -valued group of ﬁnite rank, let A be a closedsubgroup of G , and suppose that all faithful, virtually non-splitting right ideals of KG arecontrolled by A . Then all faithful, prime ideals of KG are controlled by A .Proof. Let P be a faithful, prime ideal of KG , and let P = I ∩ · · · ∩ I m be an essentialdecomposition for P in the sense of [21, Deﬁnition 5.6], with each I j virtually prime, and I , · · · , I m forming a single G -orbit.Setting m = 1, I = P , it is clear that such a decomposition exists, so we will assume that m is maximal such that a decomposition of this form exists. We know that m is ﬁnitebecause KG/P has ﬁnite uniform dimension in the sense of [21]. So, by [14, Proposition4.4], each I j is a virtually non-splitting right ideal of KG . Furthermore, since P is faithful,it follows from [14, Lemma 4.2] that each I j is faithful.Therefore, by assumption, I j is controlled by A , so I j = ( I j ∩ KA ) KG for each j . Sosince P = I ∩ · · · ∩ I r , we have that( P ∩ KA ) KG = (( I ∩ KA ) ∩ · · · ∩ ( I r ∩ KA )) KG = ( I ∩ KA ) KG ∩ · · · ∩ ( I r ∩ KA ) KG = I ∩ · · · ∩ I r = P by [14, Lemma 4.1( i )].Thus P is controlled by A as required.Now, in [14], we deﬁned the closed, isolated normal subgroup C G ( Z ( G )) of G to be26 G ( Z ( G )) := { g ∈ G : if ( h, G ) ⊆ Z ( G ) then ( g, h ) = 1 } .The main result in that paper ([14, Theorem 1.2]) was that all faithful, primitive idealsin KG are controlled by C G ( Z ( G )). With the results proved in this section, we can nowgeneralise this result to all prime ideals. Theorem 3.17.

Let ( G, ω ) be a complete p -valued group of ﬁnite rank, let P be a faithfulprime ideal of KG . Then P is controlled by C G ( Z ( G )) .Proof. First, suppose that P is non-splitting, and let H := P χ be the controller subgroupof P :Then Q := P ∩ KH is a faithful, prime ideal of KH by the deﬁnition of non-splitting,and since H is the smallest subgroup of G controlling P by [7, Theorem A], Q is notcontrolled by any proper subgroup of H . Also, note that H is a normal subgroup of G by the proof of [5, Lemma 5.2], so for any g ∈ G , ( g, H ) ⊆ H .Now, choose g ∈ G such that ( g, G ) ⊆ Z ( G ), and let A := Z ( G ) ∩ H . Let ϕ be theautomorphism of H induced by conjugation by g . Then clearly ϕ ( Q ) = Q , and for all h ∈ H , ϕ ( h ) h − = ( g, h ) ∈ Z ( G ) ∩ H = A . Moreover, since A is central in G , it followsthat ϕ ( a ) = a for all a ∈ A . So applying Theorem 3.15 gives that if ϕ = 1, then Q iscontrolled by a proper subgroup of H – contradiction.Therefore ϕ = 1, i.e. g centralises H .Therefore, if g ∈ G and ( g, G ) ⊆ Z ( G ), then ( g, H ) = 1, and hence H is contained in C G ( Z ( G )) as required. Thus P is controlled by C G ( Z ( G )).Now suppose that I E r KG is a faithful and virtually non-splitting right ideal of KG .Then I = P KG for some open subgroup U of G , and some faithful, non-splitting prime P of KU .We have proved that P is controlled by C U ( Z ( U )), and C U ( Z ( U )) = C G ( Z ( G )) ∩ U by[14, Lemma 4.9], and hence I is controlled by C G ( Z ( G )).So, using Theorem 3.16, it follows that every faithful, prime ideal of KG is controlledby C G ( Z ( G )) as required.Now we can ﬁnally complete the proof of Theorem B. So suppose that ( G, ω ) is a nilpo-tent, p -valued group of ﬁnite rank. Proof of Theorem B.

Consider the following sequence of subgroups, A = G and for each i ≥ A i +1 = C A i ( Z ( A i )). Since G is nilpotent, each A i must also be nilpotent.Therefore, if A i is non-abelian then there must exist g ∈ A i with g / ∈ Z ( A i ) such that( g, A i ) ⊆ Z ( A i ) (i.e. g ∈ Z ( A i )). But by deﬁnition, if h ∈ A i +1 = C A i ( Z ( A i )) then( g, h ) = 1, so since g is not central, this means that A i +1 = A i .Moreover, if A i is abelian, then clearly A i +1 = A i , and hence A j = A i for all j > i . Butsince A i +1 is a closed, isolated normal subgroup of A i , the chain G = A ⊇ A ⊇ A ⊇ · · · i ≥ A j = A i for all j ≥ i , and hence A i is abelian. Let A := A i , and we will prove that all faithful, prime ideals in KG arecontrolled by A .Let P be a faithful, non-splitting prime ideal of KG , and let us suppose for contradictionthat P is not controlled by A . But trivially, P is controlled by G = A , so let 0 ≤ j < i be maximal such that P is controlled by A j .Since P is non-splitting and A j is a closed subgroup of G , Q := P ∩ KA j is a faithful primeideal of KA j , so it follows from Theorem 3.17 that Q is controlled by C A j ( Z ( A j )) = A j +1 ,and hence P = QKG = ( Q ∩ KA j +1 ) KA j KG = ( P ∩ KA j +1 ) KG is controlled by A j +1 – contradiction.Therefore, every faithful, non-splitting prime ideal of KG is controlled by A .Now suppose that I E r KG is a faithful and virtually non-splitting right ideal of KG .Then I = P KG for some faithful, non-splitting prime P of KU , where U is some opensubgroup of G .Again, let B = U and for j ≥ B j +1 := C B j ( Z ( B j )). Then using [14, Lemma4.9], B j ⊆ A j for each j , and hence B i is abelian. So since P is a faithful, non-splittingprime ideal of KU , it follows that P is controlled by B i ⊆ A , and hence I = P KG =( P ∩ KB i ) KU KG ⊆ ( I ∩ KA ) KG is controlled by A . So using Theorem 3.16, it followsthat every faithful, prime ideal of KG is controlled by A . In this section, we will study the action of the rational Iwasawa algebra KG on the aﬃ-noid Dixmier module [ D ( λ ), and ultimately prove Theorem C.Throughout, ﬁx G a nilpotent, uniform pro- p group, L = p log( G ), and g = L ⊗ Z p Q p .We will assume further that L is powerful , i.e. [ L , L ] ⊆ p L . Let λ : g → K be a Q p -linear map such that λ ( L ) ⊆ O and λ | Z ( g ) is injective. Let b be apolarisation of g K := g ⊗ Q p K at λ , and let B := b ∩ L K , where L K := L ⊗ Z p O .Fix P := Ann KG [ D ( λ ), which does not depend on the choice of polarisation by [16,Theorem 4.5]. Lemma 4.1. P is a faithful, completely prime ideal of KG .Proof. Since λ | Z ( g ) is injective, it follows from Lemma 2.12 that P is a faithful ideal of KG .Furthermore, since L K is powerful, it follows from [16, Corollary 3.4] that if I :=Ann \ U ( L ) K [ D ( λ ) then \ U ( L ) K I is a domain. So since P = I ∩ KG , this means that KGP is adomain, and hence P is completely prime ideal as required.28herefore, using Theorem B, it follows that P is controlled by an abelian subgroup A of G . Let A := p log( A ), and let a := A ⊗ Z p Q p . Then a is an abelian ideal of g , sosince P is independent of the choice of polarisation, we may assume that a ⊆ b . In otherwords, we may assume that the subgroup A acts by scalars on the submodule K λ of [ D ( λ ) = [ U ( L ) K ⊗ \ U ( B ) K K λ .Our approach will be to study the action of KA on [ D ( λ ), and prove that the kernel ofthis action is centrally generated.Let { x , · · · , x r } be an O -basis for L K / B . Then using [16, Lemma 3.3], we see that [ D ( λ )is isomorphic as a K -vector space to K h x , · · · , x r i . Lemma 4.2.

There exists an O -basis { x , · · · , x r } for L K / B such that if we let ∂ i := ddx i ∈ End K [ D ( λ ) , then each u ∈ a acts on [ D ( λ ) by a polynomial f u ∈ K [ ∂ , · · · , ∂ s ] forsome s ≤ r where f u (0) = λ ( u ) . Moreover, if u ∈ A then f u ∈ O [ ∂ , · · · , ∂ s ] .Furthermore, s = 0 if and only if A is central, and for each i = 1 , · · · , s , ∂ i lies in theimage of U ( a ) under the action.Proof. Firstly, let a ⊥ = { u ∈ g K : λ ([ u, a ]) = 0 } , and let s := dim K ga ⊥ . Then ﬁx abasis { u , · · · , u r } for L K / B such that { u s +1 , · · · , u r } is a basis for ( a ⊥ ∩ L K ) / B . Thenit follows from [16, Proposition 3.5] that each u ∈ a acts on [ D ( λ ) by a polynomial f u ∈ K [ ∂ , · · · , ∂ s ], and that ∂ , · · · , ∂ s lie in the image of U ( a ) under this action.Furthermore, we see using [16, Proposition 3.3] that f u = P α ∈ N s α ! · · · α s ! λ (ad( u s ) α r · · · ad( u ) α ( u )) ∂ α · · · ∂ α s s ,so clearly the constant term is λ ( u ). Moreover, if u ∈ A then since L is powerful,ad( u r ) α r · · · ad( u ) α ( u ) ∈ p α + ··· + α r L , and hence λ (ad( u r ) α r · · · ad( u ) α ( u )) ∈ p α + ··· + α r O for all α ∈ N r .So since v p ( α i !) ≤ α i for each i , it follows that α ! · · · α r ! λ (ad( u r ) α r · · · ad( u ) α ( u )) ∈O as required.Finally, if s = 0 then a acts by scalars on [ D ( λ ), and hence [ a , g ] ⊆ P and λ ([ a , g ]) = 0.So since P is faithful, it follows from Lemma 2.12 that [ g , a ] = 0, and hence A is central.Conversely, if a is central then clearly [ U ( A ) K acts by scalars on [ D ( λ ), so since ∂ , · · · , ∂ s lie in the image of this action, and do not act by scalars, it follows that s = 0. Note:

It follows from this lemma that the image of [ U ( A ) K in End K [ D ( λ ) is contained in K h ∂ · · · , ∂ s i . We will now prove some technical results using techniques from rigid geometry. A de-tailed introduction to the theory of rigid geometry and its applications can be found in[8] and [9], whose results we will often use in this section.29irst, recall from [8, Deﬁnition 3.1.1] that an aﬃnoid algebra over K is a quotient of theTate algebra K h t , · · · , t r i for some r ∈ N . It follows from [8, Proposition 3.1.5] that anyaﬃnoid algebra R carries a complete, separated ﬁltration w R . Lemma 4.3.

Let φ : K h u , · · · , u d i → R be a map of aﬃnoid algebras, and let a , · · · , a r ∈ R lie in the image of φ . Then there exists m ∈ N such that the image of φ inside R con-tains the aﬃnoid K -subalgebra topologically generated by π m a , · · · , π m a r .Proof. For each i , we know that a i = φ ( r i ) for some r i ∈ K h u , · · · , u d i , so choose m ∈ N such that w inf ( π m r i ) ≥ w R ( π m a i ) ≥ i .Then there exist K -algebra maps Θ : K h X , · · · , X r i → K h u , · · · , u d i and Θ : K h X , · · · , X r i → R sending X i to π m r i and π m a i respectively, and it is clear that Θ = φ Θ . Therefore,the image of φ must contain the image of Θ , which is precisely the aﬃnoid K -algebratopologically generated by π m a , · · · , π m a r as required.Now, let K be the algebraic closure of K . Recall that for each ǫ ∈ R , we deﬁne the d -dimensional disc of radius ǫ to be the space D dǫ := { α ∈ K d : v π ( α i ) ≥ ǫ for each i } This is an aﬃnoid space, in the sense of [8, Deﬁnition 3.1.1], isomorphic to Sp K h u , · · · , u d i .Thus all discs are isomorphic, regardless of the radius.Moreover, the Tate algebra K h u , · · · , u d i can be realised as the space of analytic func-tions on D d , i.e. the set of all power series in u , · · · , u d converging on the unit disc, whilefor each n ∈ N , the subalgebra K h π n u , · · · , π n u d i is precisely those functions whichconverge on D d − n .Following [19, 5.1.2], for each non-constant polynomial g ( t ) := b + b t + · · · + b n t n ∈ K [ t ]with b ∈ O , deﬁne χ ( g ) := max ≤ j ≤ n − v π ( b j ) j . Lemma 4.4.

Let g ( t ) ∈ K [ t ] be a polynomial with g (0) ∈ O , and let β ∈ K with v π ( β ) > . Then χ ( βg ) < χ ( g ) .It follows that if f ( t ) , · · · , f d ( t ) ∈ K [ t ] are polynomials with f i (0) ∈ O for each i , and v ( β ) > then setting µ i := max ≤ j ≤ d χ ( β i f j ) for each i ≥ , we have that µ > µ > µ > · · · .Proof. Suppose g ( t ) = b + b t + · · · + b n t n , with b ∈ O , b n = 0. Then by deﬁnition; χ ( βg ) = max ≤ j ≤ n − v π ( βb j ) j = max ≤ j ≤ n − v π ( b j ) j − v π ( β ) j So since v π ( β ) >

0, this maximum is strictly less than max ≤ j ≤ n − v π ( b j ) j = χ ( g ).To prove the second statement, it suﬃces to prove that µ > µ and apply induction. Sosuppose µ = χ ( βf i ) and µ = χ ( f j ), then we have that µ = χ ( βf i ) < χ ( f i ) ≤ χ ( f j ) = µ . 30ecall from [19, Theorem 5.1.2] that if we assume b = 0, then the set X ( g ) := { α ∈ K : v π ( g ( α )) ≥ } is an aﬃnoid subdomain of A ,anK := K , whose G -connected component about 0 is thedisc D χ ( g ) = { α ∈ K : v π ( α ) ≥ χ ( g ) } .Furthermore, if b = 0, it is clear that χ ( g ) = χ (1 + g ), and that X ( g ) = { α ∈ K : v π ( g ( α )) ≥ } = { α ∈ K : v π (1 + g ( α )) ≥ } = X (1 + g ), so we reach the sameconclusion. Lemma 4.5.

Suppose that K contains a ( p − ’st root of p . Then given polynomials f , · · · , f r ∈ O [ t ] , there exists α ∈ K , k ∈ { , · · · , r } such that v p ( f i ( α )) ≥ − for all i and v p ( f k ( α )) < p − − .Proof. If ω ∈ K and ω p − = p then v p ( ω ) = p − . So for each j ≥ Y j := { α ∈ K : v π ( ω j f i ( α )) ≥ i } ,and set µ j := max i =1 , ··· ,d χ ( ω j f i ). Then using [19, Theorem 5.1.2] we see that Y j is an aﬃnoidsubdomain of A an and the G -connected component of Y j about zero is the closed disc D µ j .We want to ﬁnd α ∈ K such that v p ( f i ( α )) ≥ − i , i.e. v p ( pf i ( α )) ≥

0, andsince ω p − = p , this just means that α ∈ Y p − . So it remains to ﬁnd an element α inthe connected component D µ p − of Y p − such that v p ( f k ( α )) < p − − k , i.e. v π ( f k ( α )) < v π ( p )( p − − j ≥

0, ﬁx i j = 1 , · · · , d such that χ ( ω j f i j ) = µ j . Using Lemma 4.4 we seethat µ > µ > µ > · · · , and we know that for each j , the G -connected component of X ( ω j − f i j − ) about zero is D µ j − .In particular, since µ j − > µ j we have that D µ j − ( D µ j , so since D µ j is G -connected,this means that D µ j X ( ω j − f i j − ). So for each j , we may choose α j ∈ D µ j \ X ( ω j − f i j − ).But X ( ω j − f i j − ) = { α ∈ K : v π ( f i j − ( α )) ≥ − ( j − v π ( ω ) } , so this means that v π ( f i j − ( α j )) < − v π ( ω )( j − v p ( ω ) = p − so v π ( ω ) = v π ( p ) p − , thus v π ( f i j − ( α j )) < − v p ( π ) j − p − = v p ( π )( p − − jp − )So, ﬁnally, choose j = p −

1, and let k := i j − . Then α j ∈ D µ p − ⊆ Y p − and v π ( f k ( α j ))

1) as required.

Corollary 4.6.

Suppose that K contains a ( p − ’st root of p . Then given polynomials f , · · · , f r ∈ O [ t , · · · , t m ] , there exists α ∈ K m , k ∈ { , · · · , r } such that v p ( f i ( α )) ≥ − for all i and v p ( f k ( α )) < p − − .Proof. If m = 1, this is precisely Lemma 4.5, so assume that m > α , · · · , α m − ) ∈ O m − . For each i = 1 , · · · , r , let g i ( t ) := f i ( α , · · · , α m − , t ) ∈ O [ t ].Then using Lemma 4.5, there exists α m ∈ K , k ∈ { , · · · , r } such that − ≤ v p ( g i ( α m )) for all i and v p ( g k ( α m )) ≤ p − −

1. So let α = ( α , · · · , α m ) ∈ K m , andit follows that − ≤ v p ( f i ( α )) for all i , and v p ( f k ( α )) ≤ p − − .3 Almost-polynomial maps Now we will further explore the action of the abelian lattice A on the Dixmier module [ D ( λ ) ∼ = K h x · · · , x s i . Deﬁnition 4.7.

A map φ : K h u , · · · , u d i → K h t , · · · , t r i of K -algebras is called an almost-polynomial map if • φ ( u i ) ∈ O [ t , · · · , t r ] for each i , • t , · · · , t r are contained in the image of φ . Using Lemma 4.3, we see that if φ is an almost-polynomial map then there exist m ∈ N such that im( φ ) contains K h π m t , · · · , π m t r i . Example:

1. Since A is abelian, we know that [ U ( A ) K ∼ = K h u , · · · , u d i by [16, Lemma2.1], and using Lemma 4.2 we see that the action ρ : [ U ( A ) K → End K [ D ( λ ) has imagecontained in K h ∂ , · · · , ∂ s i , and the map K h u , · · · , u d i → K h ∂ , · · · , ∂ s i is an almost-polynomial map.2. If φ : K h u , · · · , u d i → K h t , · · · , t r i is an almost-polynomial map then so is therestriction K h pu , · · · , pu d i → K h t , · · · , t r i . Moreover, if F/K is a ﬁnite extension, thenthe scalar extension φ F : F h u , · · · , u d i → F h t , · · · , t r i is also an almost polynomialmap. Lemma 4.8.

Let φ : K h u , · · · , u d i → K h t , · · · , t r i be an almost-polynomial map, andlet f i := φ ( u i ) ∈ K [ t , · · · , t r ] for each i . Then setting Y := { α ∈ K r : v π ( f i ( α )) ≥ foreach i } , we have that: i . Y is an aﬃnoid subdomain of A r,anK . ii . The image of φ is contained in the set of all functions in K h t , · · · , t r i convergingon Y .Proof. Set A := im( φ ), then t , · · · , t r ∈ A by Deﬁnition 4.7. Since K h t , · · · , t r i isaﬃnoid, it follows from Lemma 4.3 that there exists m ∈ N such that A contains T = K h π m t , · · · , π m t r i , we may of course choose m to be arbitrarily large.If we set B := T h ζ , · · · , ζ d i / ( ζ i − f i ( t , · · · , t r ) : i = 1 , · · · , d ), then there is a naturalsurjection from B to A , identical on T , which sends t i to f i ( t ). This gives rise to a closedembedding of aﬃnoid varieties Sp A ֒ −→ Sp B . i . Since each f i is a polynomial, it is clear that there exists N > α ∈ K r and v π ( α ) < − N then v π ( f i ( α )) < i . So by choosing m > N we may assumethat Y = { α ∈ D r − m : v π ( f i ( α )) ≥ } .Hence using [8, Lemma 3.3.10( i )] and the proof of [8, Proposition 3.3.11], we see that Y = Sp B and that Y is an aﬃnoid subdomain of A r,anK .32 i . Notice that K h u , · · · , u d i is precisely the set of functions converging on the open unitdisc D d , so it follows that the image of K h u , · · · , u d i under φ is contained in the set offunctions converging on { α ∈ K r : ( f ( α ) , · · · , f d ( α )) ∈ D d } = { α ∈ K r : v π ( f i ( α )) ≥ i } = Y as required.The following result will be essential later when proving a control theorem. Proposition 4.9.

Let φ : K h u , · · · , u d i → K h t , · · · , t r i be an almost-polynomial map,and let f i := φ ( u i ) ∈ K [ t , · · · , t r ] for each i . Then there exists k ∈ { , · · · , d } such that exp( pf k ( t )) does not lie in φ ( K h pu , · · · , pu d i ) .Proof. We may assume that K contains a p − p . If we prove the result inthis case, then it follows generally, since if K ′ := K ( p − √ p ) and we can ﬁnd k such thatexp( pf k ) does not lie in the image of K ′ h pu , · · · , pu d i under the scalar extension of φ ,then it will also not lie in the image of K h pu , · · · , pu d i under φ .Let Y := { α ∈ K r : v π ( pf i ( α )) ≥ i } . Then using Lemma 4.8 we see that Y isan aﬃnoid subdomain of A ,anK , and that φ ( K h pu , · · · , pu d i ) is contained in the set ofall functions in K h t , · · · , t r i converging on Y . So it remains to prove that for some k ,exp( pf k ) does not converge on Y , and thus cannot lie in the image of K h pu , · · · , pu d i .Using [17, Example 0.4.1], the disc of convergence for exp is { λ ∈ K : v p ( λ ) > p − } , so itremains only to ﬁnd α ∈ Y such that v p ( pf k ( α )) ≤ p − for some k , i.e. v p ( f k ( α )) ≤ p − − f , · · · , f d ∈ O [ t , · · · , t r ], so using Corollary 4.6 we know that there exists α ∈ K r such that v p ( f i ( α )) ≥ − i and v p ( f k ( α )) < p − − k . Hence v π ( pf i ( α )) ≥ i , and hence α ∈ Y , and v p ( pf k ( α )) ≤ p − .Now we will explore more closely the image of the Tate algebra under an almost polyno-mial map. Theorem 4.10.

Suppose that φ : K h u , · · · , u d i → K h t , · · · , t r i is an almost-polynomialmap, and let S := φ ( K h u , · · · , u d i ) . Then S is an integrally closed domain of Krulldimension r .Proof. First, we will prove that S has Krull dimension r , and this is very similar to theproof of [16, Proposition 7.5]:Since t , · · · , t r ∈ S , it follows from Lemma 4.3 that S contains K h π m t , · · · , π m t r i forsome m ≥

0. Therefore we have inclusions of commutative aﬃnoid algebras, K h π m t , · · · , π m t r i ֒ −→ S ֒ −→ K h t , · · · , t r i , which gives rise to a chain of open embeddings of the associated aﬃ-noid spectra: Sp K h t , · · · , t r i ֒ −→ Sp S ֒ −→ Sp K h π m t , · · · , π m t r i .The notion of the analytic dimension dim X of a rigid variety X is deﬁned in [13], whereit is proved to be equal to the supremum of the Krull dimensions of every aﬃnoid algebra R such that Sp R is an aﬃnoid subdomain of X . In particular, if Sp B is an aﬃnoidsubdomain of Sp A in the sense of [8, Deﬁnition 3.3.9], then K.dim( B ) ≤ K.dim( A ).Therefore, since the Tate algebras K h t , · · · , t s i and K h π m t , · · · , π m t r i both havedimension r , it remains to prove that the embeddings Sp S → Sp K h π m t , · · · , π m t r i and33p K h t , · · · , t r i → Sp S deﬁne aﬃnoid subdomains.For convenience, set D := Sp K h t , · · · , t r i and D := Sp K h π m t , · · · , π m t r i . Then D can be realised as the unit disc in r -dimensional rigid K -space, while D is a deformeddisc containing D , so ﬁxing coordinates, D = { ( x , · · · , x r ) ∈ D : v π ( x i ) ≥ i } .But since Sp S contains D , we could instead write D = { ( x , · · · , x r ) ∈ Sp S : v π ( x i ) ≥ i } , and this is a Weierstrass subdomain of Sp S in the sense of [8, Deﬁnition3.3.7], and hence D is an aﬃnoid subdomain of Sp S by [8, Proposition 3.3.11]. ThereforeK.dim( S ) ≥ dim( D ) = r .Now, set T := K h π m t , · · · , π m t r i , and let f i ( t , · · · , t r ) = φ ( u i ) ∈ T for each i . Deﬁne B := T h ζ , · · · , ζ d i / ( ζ i − f i ( t , · · · , t r ) : i = 1 , · · · , d )then B is an aﬃnoid algebra which naturally surjects onto S = φ ( K h u , · · · , u d i ), whereeach a ∈ T is sent to a , and each ζ i is sent to φ ( u i ). Therefore, K.dim( S ) ≤ 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 [8, Proposition 3.3.11] shows that this corresponds to the embeddingof the Weierstrass subdomain Y = { x ∈ Sp T : v π ( f i ( x )) ≥ i } into Sp T , andhence Sp B is an aﬃnoid subdomain of Sp T = D by [8, Proposition 3.3.11], and henceK.dim( B ) ≤ dim( D ) = r .Therefore r ≤ K.dim( S ) ≤ K.dim( B ) ≤ r , forcing equality, so K.dim( S ) = r as re-quired. Moreover, this implies that S and B have the same Krull dimension, and hence S is a quotient of B by a minimal prime ideal.To prove that S is integrally closed, we will prove that the aﬃnoid variety Y = Sp B is normal , i.e. at every point p ∈ Y , the ring of germs of aﬃnoid functions O Y,p (as deﬁnedin [8, Deﬁnition 4.1]) is reduced and integrally closed. Using this, it will follow from [9,Proposition 7.3.8] that the localisation B q of B at every prime ideal q of B is reducedand integrally closed.Since S is a minimal prime quotient of B , it will follow that S is an integrally closeddomain as required.Using [10, Theorem 5.1.3], we see that since the aﬃne variety A rK is smooth, and hencenormal, its analytiﬁcation A r,anK is also normal. So since Y = { x ∈ A r,anK : v π ( f i ( x )) ≥ i } is an aﬃnoid subdomain of A r,anK by Lemma 4.8(ii), it follows that O A r,anK ,p = O Y,p for every point p ∈ Y . Hence O Y,p is reduced and integrally closed as required.

In this subsection, we will prove a control theorem for kernels of almost-polynomial maps.Throughout, we will assume that K contains a p ’th root of unity ζ .Fix A a free abelian pro- p group of rank d , let A := p L A be the associated Z p -Lie algebraof A , and let φ : [ U ( A ) K → K h t , · · · , t s i be an almost-polynomial map.Consider the crossed product D p = D p ( A ) = \ U ( p A ) K ∗ AA p deﬁned in Section 2.5. Thisis a Banach completion of KA with respect to the extension of the dense embedding34 : KA p → \ U ( p A ) K to KA , and there is a natural map τ : D p → [ U ( A ) K . Deﬁne φ ′ : D p → K h t , · · · , t s i and φ A : KA → K h t , · · · , t s i making the following diagramcommute: D p [ U ( A ) K KA K h t , · · · , t s i τ φ ′ φι φ A From now on, set I = ker( φ ′ ), and let Q := ker( φ A ) = I ∩ KA , and deﬁne: U := { a ∈ A : φ ( a ) ∈ φ ( \ U ( p A ) K ) } . Proposition 4.11. U is a proper open subgroup of A containing A p .Proof. Since φ is a ring homomorphism, it is clear that for all a, b ∈ U , ab ∈ U , andsince KA p is a subalgebra of \ U ( p A ) K , it is clear that A p ⊆ U . Therefore, since AA p is aﬁnite group, and UA p is closed under multiplication, it follows that U is a subgroup of A containing A p , and hence it it open.Finally, since φ is an almost polynomial map, it follows from Proposition 4.9 that thereexists u ∈ A such that exp( pφ ( u )) = φ (exp( pu )) does not lie in the image of \ U ( p A ) K under φ . But a := exp( pu ) ∈ A and hence a / ∈ U . Therefore U is a proper subgroup of G .Using this proposition, we can ﬁx a Z p -basis { a , · · · , a d } for A such that { a , · · · , a r , a pr +1 , · · · , a pd } is a Z p -basis for U , so a , · · · , a r ∈ U and a r +1 , · · · , a d / ∈ U .Since A is a free abelian pro- p group, we have that AA p is a direct product of d copiesof the cyclic group of order p , where the i ’th copy is generated by the image of a i in AA p .Setting c i := a i A p , it follows from Lemma 2.9 that: D p = \ U ( p A ) K ∗ h c i ∗ · · · ∗ h c d i (10)where c it = c ti for 0 ≤ t < p and c ip = a pi .From now on, let S := φ ( \ U ( p A ) K ) ⊆ K h t , · · · , t s i , and let B := \ U ( p A ) K ∗h c i∗· · ·∗h c r i ≤ D p . Then since a , · · · , a r lie in U , the image of B under φ is S by the deﬁnition of U .Furthermore, since KU = KA p ∗ UA p = KA p ∗ h c i ∗ · · · ∗ h c r i , it is clear that KU ⊆ B .Let J := I ∩ B E B be the kernel of the restriction of φ ′ to B , and let I ′ := J D p – anideal of D p contained in I . Lemma 4.12. I is a prime ideal of D p , minimal prime above I ′ .Proof. Since D p /I ∼ = im( φ ′ ) ≤ K h t , · · · , t s i , it is clear that I is a prime ideal of D p .Since D p is a crossed product of B with a ﬁnite group, it follows from Lemma 2.4( ii ) that I is minimal prime above I ′ = ( I ∩ B ) D p . 35ow, we will need the following small result from Galois theory [27]: Lemma 4.13.

Let F be a ﬁeld of characteristic 0, containing a p ’th root of unity ζ . Let r ∈ F , and suppose that r has no p ’th root in F . Choose a p ’th root α ∈ F of r , and let F ′ := F ( α ) . Then if β ∈ F ′ and β p ∈ F then β = cα m for some c ∈ F , ≤ m < p .Proof. Since F ′ is the splitting ﬁeld for the polynomial x p − r over F , it is clear that F ′ is a Galois extension of F . So since [ F ′ : F ] = p this means that Gal ( F ′ /F ) has order p .In fact, if we consider the element σ ∈ Gal ( F ′ /F ) sending α to ζ α , then Gal ( F ′ /F )is cyclic of order p , generated by σ .The result is clear if β ∈ F , so assume β / ∈ F and β p ∈ F . Then β is a root of thepolynomial x p − β p ∈ F [ x ], and hence σ ( β ) is also a root. Therefore σ ( β ) = ζ m β for some0 ≤ m < p , so σ ( α − m β ) = ζ − m α − m ζ m β = α − m β .But since σ generates Gal ( F ′ /F ), it follows that α − m β is ﬁxed by the Galois group, sosince F ′ /F is a Galois extension, this means that c := α − m β ∈ F , and hence β = cα m asrequired.For clarity, we will introduce/recall the following data: • I = ker( φ ′ ) E D p . • Q = I ∩ KA E KA . • U = { a ∈ A : φ ( a ) ∈ φ ( \ U ( p A ) K ) } = h a , · · · , a r , a pr +1 , · · · , a pd i . • B = \ U ( p A ) K ∗ h c i ∗ · · · ∗ h c r i ≤ D p . • S = φ ( \ U ( p A ) K ) = φ ′ ( B ) ≤ K h t , · · · , t s i . • J = I ∩ B E B . • I ′ = J D p E D p . • R := D p /I ′ Proposition 4.14. R is a domain.Proof. Since D p = B ∗ h c r +1 i ∗ · · · ∗ h c d i and I ′ = J D p , it follows from Lemma 2.4( iii )that R ∼ = S ∗ h c r +1 i ∗ · · · ∗ h c d i , where ¯ c pi = φ ( a pi ) for each i . So using [22, Theorem 4.4]we see that R is reduced. Therefore, we may consider the usual semisiple artinian ringof quotients Q ( R ) of R , which has the form: Q ( R ) = Q ( S ) ∗ h c r +1 i ∗ · · · ∗ h c d i ,where Q ( S ) is the ﬁeld of fractions of S . Note that since S = φ ( \ U ( p H ) K ) is the image ofa Tate algebra under an almost-polynomial map, it follows from Theorem 4.10 that S isan integrally closed domain. It remains to prove that Q ( R ) is a ﬁeld.Let T := Q ( S ), and for each i = 1 , · · · , d − r , deﬁne T i := T i − ∗ h c r + i i , so that T d − r = Q ( R ). 36learly T is a ﬁeld, so we will use induction to show that T i is a ﬁeld for each i , so inparticular, Q ( R ) is a ﬁeld. So assume that for some j > T , · · · , T j − are all ﬁelds:Then since T j = T j − ∗ h c r + j i where ¯ c pr + j = φ ( a pr + j ) ∈ S , it follows that T j = T j − [ x ] / ( x p − φ ( a pr + j ))So we only need to show that the polynomial x p − φ ( a pr + j ) ∈ T j − [ x ] is irreducible overthe ﬁeld T j − .Since K contains a p ’th root of unity, we see using standard Galois theory that this justmeans we need to show that this polynomial has no root in T j − , i.e. that there is no b ∈ T j − such that b p = φ ( a pr + j ).Let us suppose for contradiction that b p = φ ( a pr + j ) for some b ∈ T j − = T j − ∗ h c r + j − i .Then since φ ( a pr + j ) ∈ S ⊆ T j − and T j − is a ﬁeld containing K , it follows from Lemma4.13 that b = b ¯ c k r + j − for some b ∈ T j − , 0 ≤ k < p .Therefore, b p = φ (( a r + j a − k r + j − ) p ) ∈ S , so applying a second induction, for each i >

0, wecan ﬁnd integers 0 ≤ k , · · · , k i − < p and b i ∈ T j − i such that b pi = φ (( a r + j a − k r + j − a − k r + j − · · · a − k i − r + j − i +1 ) p ) ∈ S . Taking i = j we have that b j ∈ T = Q ( S ) and b pj ∈ S . So since S is integrally closed,it follows that b j ∈ S ⊆ K h t , · · · , t s i .Now, ( b j φ ( a − r + j a k r + j − · · · a k j − r +1 )) p = 1, so it follows that there is a p ’th root of unity ζ ∈ K such that: ζ b j = φ ( a r + j a − k r + j − a − k r + j − · · · a − k j − r +1 ).Therefore, since b j ∈ S , this means that φ ( a r + j a − k r + j − a − k r + j − · · · a − k j − r +1 ) ∈ S = φ ( \ U ( p A ) K ),or in other words a r + j a − k r + j − a − k r + j − · · · a − k j − r +1 ∈ U by the deﬁnition of U .This is the required contradiction since { a , · · · , a r , a pr +1 , · · · , a pd } is a Z p -basis for U ,and each k i is less than p . Theorem 4.15.

Let φ : [ U ( A ) K → K h t , · · · , t s i be an almost-polynomial map. Then thekernel Q of the restriction of this map to KA is controlled by U .Proof. If I = ker( φ ′ ) E D p , then using Proposition 4.14 we see that R = D p / ( I ∩ B ) D p is a domain. But we know that I is minimal prime above ( I ∩ B ) D p by Lemma 4.12, soit follows that I = ( I ∩ B ) D p .So, if r ∈ Q = I ∩ KA then since KA = KU ∗ AU , r = P a ∈ A//U s a a for some s a ∈ KU ⊆ B .So since r ∈ I = ( I ∩ B ) D p it follows that s a ∈ I ∩ B ∩ KU = Q ∩ KU for each a , and hence r ∈ ( Q ∩ KU ) KA . Since our choice of r was arbitrary, this means that Q = ( Q ∩ KU ) KA ,i.e. Q is controlled by U . 37 .5 Control Theorem for Dixmier annihilators Now we can ﬁnally conclude our proof of Theorem C. Again, G is a nilpotent, uniformpro- p group, whose Z p -Lie algebra L = p log( G ) is powerful.Fix a linear form λ : L → O such that the restriction of λ to Z ( g ) is injective, then P :=Ann KG [ D ( λ ) is a faithful prime ideal of KG by Lemma 4.1, and KGP is a domain.

Proof of Theorem C

Firstly, for any ﬁnite extension

F/K , if we let I = Ann F G \ D ( λ F ),where λ F is the scalar extension of λ , then clearly I ∩ KG = P . So if we prove that I is controlled by Z ( G ), then it will follow from Lemma 2.7 that P is controlled by Z ( G ).Therefore, we may pass to ﬁeld extensions of K without issue. In particular we mayassume that F = K contains a p ’th root of unity.Since P is a faithful, prime ideal of KG , it follows from Theorem B that P is controlledby an abelian subgroup of G . So let A = P χ be the controller subgroup of P , then A is an abelian normal subgroup of G , so if we let A := p log( A ) then A is an abelianideal of L . We want to prove that A is central in G , or equivalently that A is central in L .Using Lemma 4.2, we see that the image of [ U ( A ) K in End K [ D ( λ ) is contained in a Tatealgebra K h ∂ , · · · , ∂ s i such that if s > [ U ( A ) K → K h ∂ , · · · , ∂ s i is analmost polynomial map. Moreover, s = 0 if and only if A is central, so let us assumethat s > U of G such that the kernel Q of the restriction of φ to KA is controlled by U . But clearly Q = P ∩ KA , so this is a contradiction since A is the controller subgroup of P . The aim of this section is to prove our main result Theorem A. The essence of ourargument is to compare general primitive ideals in KG to Dixmier annihilators. Fix G a uniform, nilpotent pro- p group, and as usual let L := p L G , and g := L ⊗ Z p Q p . Deﬁnition 5.1.

Given a prime ideal P of KG , we say that P is weakly rational if Z ( KG/P ) is a ﬁnite ﬁeld extension of K . It follows from [12, Theorem 1.1(1)] that any primitive ideal of KG is weakly rational. Lemma 5.2.

Let P ⊆ Q be weakly rational ideals of KG , and suppose that P is faithful.Then Q is faithful.Proof. Let F = Z ( KG/P ), F = Z ( KG/Q ), then F , F are ﬁnite ﬁeld extensions of K ,and clearly the natural surjection KG/P ։ KG/Q reduces to a ﬁeld extension F ֒ −→ F .38ince Q † = { g ∈ G : g − ∈ Q } is a normal subgroup of G , using nilpotence of G we seethat if Q † = 1, then there exists z ∈ Q † ∩ Z ( G ) with z = 1. Thus z + P, P ∈ F ⊆ F ,and z + Q = 1 + Q , which implies that z + P = 1 + P and hence z − ∈ P . So since P is faithful, z = 1 – contradiction.Therefore, Q † = 1, and hence Q is faithful.Now, recall from Section 2.5 the deﬁnition of the Banach completions D p n = \ U ( p n L ) K ∗ GG pn of KG for each n ∈ N . Proposition 5.3.

Let P be a primitive ideal of KG , then for all suﬃciently high n ∈ N ,there exists a primitive ideal Q n of D p n = \ U ( p n L ) K ∗ GG pn such that Q n ∩ KG = P .Proof. Since P is primitive, P = Ann KG M for some irreducible KG -module M . Us-ing [2, Proposition 10.6(e), Corollary 10.11], we see that for n suﬃciently high, c M := D p n ⊗ KG M = 0.Since M is irreducible and c M = 0, the natural map M → c M , m ⊗ m is injective. Andsince D p n is a Banach completion of KG with respect to some ﬁltration w , it follows that c M is a completion of M = KGm with respect to the ﬁltration v ( rm ) = sup { w n ( r + y ) : y ∈ KG and ym = 0 } .Therefore, if r ∈ P , i.e. rM = 0, then taking limits shows that r c M = 0, so P ⊆ Ann KG c M = (Ann D pn c M ) ∩ KG .Now, since D p n is Noetherian and c M is a ﬁnitely generated D p n -module, we can choosea maximal submodule U ≤ c M , and let M ′ := c M /U – an irreducible D p n -module.Since M is irreducible, the composition M ֒ −→ c M ։ M ′ is either injective or zero. If itis zero then M ⊆ U , and hence c M ⊆ U and M ′ = 0. This contradiction implies that thecomposition is injective.Finally, let Q n = Ann D pn M ′ , then Q n is a primitive ideal of D p n , and P ⊆ Ann KG c M ⊆ Ann KG M ′ = Q n ∩ KG . Also, if r ∈ Q n ∩ KG then rM ′ = 0, so since M ⊆ M ′ , rM = 0and r ∈ P . Thus P = Q n ∩ KG as required. K G to K G p n Now we begin to explore how we can relate primitive ideals in Iwasawa algebras to Dixmierannihilators:

Proposition 5.4.

Given a primitive ideal P of KG , there exists m ∈ N with m ≥ ,ﬁnite extensions F , · · · , F r /K and Q p -linear maps λ i : g → F i with λ i ( p m L ) ⊆ O F i foreach i = 1 , · · · , r , such that: P ∩ KG p m = Ann KG pm \ D ( λ ) F ∩ · · · ∩ Ann KG pm \ D ( λ r ) F r Proof.

Using Proposition 5.3, if P is primitive, then for any suﬃciently high n ≥ Q of D p n = \ U ( p n L ) K ∗ GG pn such that Q ∩ KG = P , and hence Q ∩ KG p n = P ∩ KG p n . 39et I = Q ∩ \ U ( p n L ) K , then using Lemma 2.5 we see that I is a semiprimitive ideal of \ U ( p n L ) K , so choose primitive ideals J , J , · · · , J r of \ U ( p n L ) K such that I = J ∩ J ∩· · · ∩ J r .Since each J i is primitive, it follows from [16, Theorem A] that there exists m ≥ n such that for each i , J i ∩ \ U ( p m L ) K = Ann \ U ( p m L ) K \ D ( λ i ) F i for F i /K a ﬁnite extension, λ i : g → F i Q p -linear with λ i ( p m L ) ⊆ O F i . Thus: P ∩ KG p m = Q ∩ KG p m = I ∩ KG p m = ( J ∩ \ U ( p m L ) K ) ∩ · · · ∩ ( J r ∩ \ U ( p m L ) K ) ∩ KG p m is an intersection of Dixmier annihilators as required.Now, we want to show that all faithful, primitive ideals of KG are centrally generated,which we know is true for Dixmier annihilators by Theorem C. Proposition 5.4 allows usto compare general primitive ideals to Dixmier annihilators, and the following result usesthis to prove a reduced version of Theorem A: Theorem 5.5.

Let G be a nilpotent, uniform pro- p group with centre Z , and let P bea faithful, primitive ideal of KG . Then there exists N ∈ N such that for all n ≥ N , P ∩ KG p n is controlled by Z p n .Proof. Using Proposition 5.4, we see that for some m ≥

1, there are ﬁnite extensions F , · · · , F r and Q p -linear maps λ i : g → F i with λ ( p m L ) ⊆ O F i such that P ∩ KG p m =Ann KG pm \ D ( λ ) F ∩ · · · ∩ Ann KG pm \ D ( λ r ) F r .For each i = 1 , · · · , r , set J i := Ann KG pm \ D ( λ i ) F i for convenience, clearly these are primeideals of KG p m , thus P ∩ KG p m is semiprime and J , · · · , J r are the minimal primes above P ∩ KG p m , hence they are all G -conjugate by the proof of [5, Lemma 5.4(b)]. Note thatfor all n ≥ m , J i ∩ KG p n = Ann KG pn [ D ( λ ) F i for each i .Also, since P is faithful, P ∩ KG p m is faithful, so J † ∩ · · · ∩ J † r = P † = 1. But since J † , · · · , J † r are G -conjugate and G is orbitally sound by [5, Proposition 5.9], this meansthat the subgroup 1 must have ﬁnite index in J † i for each i , which means that they areﬁnite. But G is torsionfree, thus J † i = 1 for all i , i.e. J , · · · , J r are faithful.So since J i = Ann KG pm \ D ( λ i ) F i is faithful, it follows from Lemma 2.12 that λ i is in-jective when restricted to Z ( g ).Now, since m ≥

1, note that for all n ≥ m , p log( G p n ) = p n − log( G ) is a powerfulLie lattice. Therefore, using Theorem C, we see that J i ∩ KG p n = Ann KG pn \ D ( λ i ) F i iscontrolled by Z ( G p n ) for each i , and using [5, Lemma 8.4(a)], Z ( G p n ) = Z ( G ) ∩ G p n = Z p n .Therefore, setting B i,n := J i ∩ KG p n = Ann KG pn \ D ( λ i ), B i,n = ( B i,n ∩ KZ p n ) KG p n foreach i , so using [14, Lemma 4.1(a)]: P ∩ KG p n = B ,n ∩ · · · ∩ B r,n = ( B ,n ∩ KZ p n ) KG p n ∩ · · · ∩ ( B r,n ∩ KZ p n ) KG p n = ( B ,n ∩ · · · ∩ B r,n ∩ KZ p n ) KG p n = ( P ∩ KZ p n ) KG p n Hence P ∩ KG p n is controlled by Z p n as required.40 .3 Extension from K G p n to K G

The results of the previous subsection show that we can establish Theorem A after passingto G p n for suﬃciently high n . We now just need to extend to KG . Lemma 5.6.

Let P be a weakly rational ideal of KG . Then P ∩ KZ ( G ) is a maximalideal of KZ ( G ) .Proof. Since P is prime in KG , Q := P ∩ KZ ( G ) is prime in KZ ( G ). So setting F := Z ( KG/P ), it is clear that KZ ( G ) /Q ֒ −→ F . So since KZ ( G ) /Q is a domaincontaining K , and F is a ﬁnite extension of K , it follows that KZ ( G ) /Q is a ﬁeld, andhence Q is maximal. Proposition 5.7.

Let G be a nilpotent, uniform pro- p group, and let P ⊆ P be faithful,primitive ideals of KG . Then there exists n ∈ N such that P ∩ KG p n = P ∩ KG p n . Itfollows that if P is a faithful, primitive ideal of KG then P is maximal.Proof. Using Theorem 5.5, we see that there exist N , N ∈ N such that for all n ≥ N , n ≥ N , P i ∩ KG p ni is controlled by Z ( G ) p ni for each i . So choose n ≥ max { N , N } and we have that P ∩ KG p n , P ∩ KG p n are controlled by Z ( G ) p n .Since P is primitive, it is weakly rational, so using Lemma 5.6 we see that P ∩ KZ ( G )is a maximal ideal of KZ ( G ). So since P ∩ KZ ( G ) ⊆ P ∩ KZ ( G ), we have that P ∩ KZ ( G ) = P ∩ KZ ( G ), and hence P ∩ KZ ( G ) p n = P ∩ KZ ( G ) p n . Therefore: P ∩ KG p n = ( P ∩ KZ ( G ) p n ) KG p n = ( P ∩ KZ ( G ) p n ) KG p n = P ∩ KG p n .Finally, given a faithful, primitive ideal P of KG , let Q be a maximal ideal of KG containing P . Since P and Q are primitive, they are weakly rational, so since P isfaithful, Q is faithful by Lemma 5.2. Thus, by the above, there exists n ∈ N such that P ∩ KG p n = Q ∩ KG p n is controlled by Z ( G ) p n .But P ∩ KZ ( G ) is prime in KZ ( G ), so P ∩ KG p n = ( P ∩ KZ ( G ) p n ) KG p n is prime in KG p n by Theorem 2.8. So since P ∩ KG p n = Q ∩ KG p n , it follows from [22, Theorem16.6( iii )] that P = Q , and hence P is maximal. Theorem 5.8.

Let G be a nilpotent, uniform pro- p group. Then all faithful, primitiveideals of KG are controlled by Z ( G ) .Proof. Let P be a faithful, primitive ideal of KG , and let Z = Z ( G ). We want to provethat P is controlled by Z .Using Theorem 5.5, we know that there exists n ∈ N such that P ∩ KG p n is controlled by Z p n , and hence is prime in KG p n by Theorem 2.8. So let J := ( P ∩ KG p n ) KG , then us-ing Lemma 2.4 we see that J is a semiprime ideal of KG , and P is minimal prime above J .Let Q := P ∩ KZ , then Q is prime in KZ , so QKG is prime in KG by Theorem 2.8.And since P ∩ KG p n = ( P ∩ KZ p n ) KG p n , we have that: J = ( P ∩ KG p n ) KG = ( P ∩ KZ p n ) KG ⊆ QKG .But clearly

QKG ⊆ P , so since QKG is prime and P is minimal prime above J , it followsthat P = QKG = ( P ∩ KZ ) KG , and hence P is controlled by Z as required.41ow we can ﬁnally prove our main result. First, we just need a small Lemma: Lemma 5.9.

Let G be a uniform pro- p group, let N be a closed, normal subgroup of G .Then there exists an open, uniform normal subgroup U of G such that N ∩ U is a closed,isolated normal subgroup of U .Proof. Recall from [28, Deﬁnition 1.6] the deﬁnition of the isolater i G ( N ) of N in G ,and recall from [28, Proposition 1.7, Lemma 1.8] that i G ( N ) is a closed, isolated normalsubgroup of G , and N is open in i G ( N ).Therefore, there exists n ∈ N such that if g ∈ i G ( N ) then g p n ∈ N . So if g = h p n ∈ U := G p n and g p = h p n +1 ∈ N ⊆ i G ( N ), then h ∈ i G ( N ), so g = h p n ∈ N . Hence N ∩ U isisolated in U as required. Proof of Theorem A.

Let P be a primitive ideal of KG , and we want to prove that P isvirtually standard, i.e. that P ∩ KU is a ﬁnite intersection of standard ideals for someopen, normal subgroup U of G .Firstly, if P is faithful, then it follows from Theorem 5.8 that P is controlled by Z ( G ),and hence is standard, and using Proposition 5.7 we see that P is maximal. So we canassume that P is not faithful.Let N := P † = { g ∈ G : g − ∈ P } . Then N is a closed, normal subgroup of G , so byLemma 5.9, there exists an open, uniform normal subgroup U of G such that N ∩ U isisolated in U . Let Q := P ∩ KU , then Q is a semiprimitive ideal in KU by Lemma 2.5,and Q † = N ∩ U is a closed, isolated normal subgroup of U .Let U := UQ † , and let Q := Q ( Q † − KU . Then U is a nilpotent, uniform pro- p groupand Q is a faithful semiprimitive ideal of KU . Therefore, it follows that Q is a ﬁniteintersection of faithful, primitive ideals in KU . Since all faithful primitives in KU aremaximal and standard, this means that Q is a ﬁnite intersection of maximal standardideals.Therefore, since Q is a homomorphic image of Q , this means that Q is a ﬁnite intersec-tion of maximal, standard ideals, and it follows from Deﬁnition 1.1 that P is a virtuallystandard prime ideal of KG . Therefore, it remains to show that P is maximal.Using Lemma 2.4( ii ), we see that P is minimal prime above the semiprime ideal ( P ∩ KU ) KG . So since P ∩ KU is semimaximal in KU , it follows from [22, Theorem 16.6( iii )]that P is maximal in KG as required. References [1] K. Ardakov; K. A. Brown, Ring-theoretic properties of Iwasawa algebras: a survey.

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