Some model theory over hereditary noetherian
domains
Mike Prest
Department of Mathematics
Manchester University
Manchester, M13 9PL, United Kingdom
Gennadi Puninski ∗
Moscow State Social University
Losinoostrovskaya 24
107150 Moscow
Russia
Abstract
Questions in the model theory of modules over hereditary noetherian domains are investigated with particular attention being paid to
differential polynomial rings and to generalized Weyl algebras. We
prove that there exists no isolated point in the Ziegler spectrum over
a simple hereditary generalized Weyl algebra A of the sort considered
in [2] over a field k with char(k) = 0 (the first Weyl algebra A1 (k) is
such) and the category of finite length modules over A does not have
any almost split sequence. We show that the theory of all modules over
a wide class of generalized Weyl algebras and related rings interprets
the word problem for groups and in the case that the field is countable
there exists a superdecomposable pure–injective module over A. This
class includes, for example, the universal enveloping algebra U sl2 (k).
1
Preliminaries
A main theme in the model theory of modules over a ring R is the investigation of the pure-injective indecomposable modules over R, the set
∗
This paper was written during the visit of the second author to the University of
Manchester supported by EPSRC grant GR/L68827. He would like to thank the University for the kind hospitality.
1
of (isomorphism types of) which may be given a quasi–compact topology
forming the Ziegler spectrum ZgR of R. There are various natural questions about these objects. First of all one can try to describe the points of
the Ziegler spectrum, that is to classify the pure–injective indecomposable
modules over a ring R in the sense of producing a list of invariants for the
isomorphism classes of these modules, for instance in terms of R itself.
A second aim is to understand the structure of the topological space
ZgR , for instance its complexity, where a natural measure of complexity
of ZgR is its Cantor–Bendixson rank (CB-rank). One may even ask the
coarser question: does the CB-rank of ZgR exist? This turns out [34] to be
connected with the question of the existence of a superdecomposable (i.e.
without indecomposable direct summands) pure–injective module over R
and even this question seems to be very difficult. The information about
the structure of ZgR is very often involved in the question of decidibility of
the theory of modules over R, so this is an additional motivation for trying to
describe ZgR . Even at the first level of complexity of ZgR , that concerning
the isolated points, the isolated finitely presented ones (rather, the pure–
injective envelopes of finitely presented modules with a local endomorphism
ring) correspond to modules with a minimal left almost split map and these
are of great algebraic importance.
Let us describe some background. A classical object of investigation in
the model theory of modules is the theory of abelian groups. Kaplansky [19]
gave the complete list of pure–injective (=algebraically compact) indecomposable abelian groups and proved that every pure–injective abelian group
is the pure–injective envelope of a direct sum of indecomposable ones so, in
our terminology, there exists no superdecomposable pure–injective abelian
group. It can be calculated using this that the CB-rank of ZgZZ is equal to 2
(a result essentially due to Garavaglia [12] - see [25] 10.28). The decidability
of the theory of abelian groups can be deduced from this description, but it
was before the model theory of modules was born when Szmeliew [33] proved
this beautiful result. The model theory of abelian groups and of modules
over a commutative Dedekind domain was investigated by Eklof and Fisher
[7] and the situation here is almost as good as for abelian groups. For decidability one should also require that a commutative Dedekind domain be
“effectively given” (in particular it should be countable) before decidability
of the theory of modules can be addressed.
Remarkable progress was made recently in the model theory of modules over a serial ring where a complete set of invariants for indecomposable pure–injective modules was given by Eklof and Herzog [8]. However a
2
superdecomposable pure–injective module already exists over any commutative valuation domain without Krull dimension (see Puninski [28], Salce
[32]) and so the CB-rank of ZgR is not in general defined for these rings. It
has been calculated by Puninski [29] that the CB-rank of ZgR over a commutative valuation domain of Krull dimension α is equal to 2α, in particular
every even (but no odd) value occurs. Answers to questions concerning decidibility of the theory of modules over such a ring are far from being clear
even in the case that the ring has finite Krull dimension.
Over a von Neumann regular ring every pure–injective module is injective
so we are dealing here with the theory of injective modules (see Goodearl
[13]) and superdecomposable injective modules were involved in this theory
very naturally by Goodearl and Boyle [14]. However the precise calculation
of the CB-rank of ZgR was made only for commutative von Neumann regular
ringa R by Garavaglia (in fact in terms of elementary Krull dimension, see
[25, Ch. 16]). It turned out that the CB-rank of ZgR is equal to a rank
of B(R) (the Boolean algebra of idempotents of R) which is defined by
transfinite factorization by the ideal generated by atoms. So for an atomless
Boolean algebra B the CB-rank of ZgR does not exist and, also, every ordinal
occurs as the CB-rank of ZgR for some Boolean algebra R. For the case of
arbitrary von Neumann regular rings there is no corresponding criterion
known for the existence of a superdecomposable pure-injective.
Recent progress in the representation theory of finite–dimensional algebras has increasingly involved pure–injective modules and there has been
some progress on the above questions. The reader may consult [25, Chapter
17] to see how questions about the existence of a superdecomposable pureinjective module, decidability of the theory of modules and tameness of an
algebra are conjecturally connected. It has been proved (see [27], [31] and
references therein) that over an arbitrary tame hereditary finite dimensional
algebra R the CB-rank of ZgR is equal to 2 and in the approach of [27] the
model theory of modules over some noncommutative Dedekind prime rings
is very much involved.
The most understandable class of hereditary noetherian prime rings
(hnp-rings) is the class of so–called bounded hnp-rings where the model
theory of modules, as does the whole theory of modules, looks very similar
to the theory of abelian groups. This can be found in Marubayashi [22] and
(in more model–theoretic form) in Prest [27]. The first results in the model
theory of modules for the unbounded case were obtained by Puninski, Prest
and Rothmaler [30], but at this time only the first steps in this area were
taken.
3
In this paper we obtain some new information about the structure of ZgR
over certain noetherian domains with particular attention being paid to the
generalized Weyl algebras and to differential polynomial rings. If we consider
a differential polynomial ring R over a universal field with derivation ([11])
then it is a V-ring, meaning that all simple modules are injective. We show
that the model theory of modules over this ring is very easy: a complete
classification of indecomposable pure–injective modules is possible and the
CB-rank of ZgR is equal to 1. For instance, every finite length point in ZgR
is isolated and this is also true for the ring of differential polynomials over
the field of Laurent series (over a field of characteristic zero) as was proved
through ingenious calculations by Zimmermann [35].
We prove that for a simple hereditary generalized Weyl algebra A in the
sense of [2] over a field k with char(k) = 0 (including the first Weyl algebra
A1 (k)) the situation is completely the opposite: there is no isolated point
in ZgA and there is no almost split sequence in the category of finite length
modules over A. In particular the CB-rank of ZgA is undefined and the
classification problem over these algebras seems to be hopeless: for example
we show that for a countable field there are 2ω points in ZgA and even
a superdecomposable pure-injective module exists. The situation for the
algebra B1 , which is a differential polynomial ring, is similar.
This is done using a construction of Klingler and Levy [20] which they
used for proving wildness of the category of finite length modules over A1 .
Supporting this point of view we prove that the theory of modules over
a wide class of generalised Weyl algebras interprets the word problem for
groups. This is proved by interpreting the theory of k⟨X, Y ⟩-modules into
the theory of modules over such an algebra. For instance this result can
be applied to A1 (k), B1 (k) and, in consequence, to the universal enveloping
algebra U sl2 (k) provided char(k) = 0, showing that the theory of modules
over these algebras is extremely complicated.
Unfortunately the question of the existence of a superdecomposable pureinjective module over A1 for an uncountable field remains open, just as in
the case of wild finite–dimensional algebras.
2
Basic notions
We need some notions from the model theory of modules - we recall these
here (see, for instance [26], for more detail). So suppose that M is any
module over the ring R. The pp-definable subgroups of M are the subgroups
4
which are projections of solution sets of systems of R-linear equations. That
is, let H be any finite (m-by-n say) matrix with entries from R, let annM H =
{a ∈ M m : aH = 0} and consider the image of this group under projection
to first coordinate from M m to M . This image is a typical pp-definable
subgroup of M and, if we let φ(x) denote the corresponding formula (in
the usual language for R-modules) namely ∃x2 , . . . , xm (x, x2 , . . . , xm )H = 0
then it is denoted by φ(M ). Here φ is a typical pp formula (with the one free
variable x). For an element a ∈ M the pp-type of a in M is the collection
ppM (a) of all pp formulas φ(x) such that a ∈ φ(M ). There is an obvious
pre-ordering on pp formulas, namely ψ ≤ φ iff for all right R-modules M we
have ψ(M ) ≤ φ(M ). There is also a map φ 7→ Dφ between pp formulas for
right R-modules and pp formulas for left R-modules which induces a duality
between the corresponding partial orders: that is D reverses the ordering
and D2 is equivalent to the identity.
A morphism A −→ B between right modules is pure if for every left
module L we have that the induced map A⊗L −→ B⊗L is monic. A module
N is pure-injective (also called algebraically compact) if it is injective over
pure embeddings. Every module M is a pure submodule of its pure-injective
envelope PE(M ) which is unique to isomorphism over M . The set ZgR of
isomorphism types of indecomposable pure-injective right R-modules may
be topologised [34] by specifying, as a basis of open sets, the sets of the form
(φ/ψ) = {N ∈ ZgR : φ(N ) > ψ(N )} where φ > ψ are pp formulas. This
quasicompact space is called the (right) Ziegler spectrum of R. Elementary
duality D above extends (see [15]) to the right and left Ziegler spectra of a
ring, allowing one to define, at least for certain points N of ZgR , the dual
point DN in the left Ziegler spectrum. If M is a module and φ > ψ are
pp formulas such that φ(M ) > ψ(M ) and there is no pp formula θ with
φ(M ) > θ(M ) > ψ(M ) then we say that φ/ψ is an M -minimal pair. By a
minimal pair we mean a pair of pp formulas which is M -minimal for every
module M , equivalently for every module N ∈ ZgR . If φ/ψ is a minimal
pair then (φ/ψ) is a singleton {N } say with N thus isolated. We say that a
pair φ > ψ of pp formulas opens in the module M if φ(M ) > ψ(M ).
Recall that a module M is totally transcendental (a concept from model
theory) iff it is Σ-pure-injective (that is, iff any direct product of copies of M
is pure-injective) iff M has the descending chain condition on pp-definable
subgroups (e.g. see [25] and references therein).
We need the following fact, which may be extracted from [16] or [25, Section 11.3]. Let M be a finitely presented module with local endomorphism
ring. Then PE(M ) is indecomposable (hence a point of the Ziegler spec5
trum); moreover, if M ′ also is finitely presented with local endomorphism
ring and if PE(M ) ∼
= PE(M ′ ) then M ∼
= M ′.
By a ring R we mean an associative ring with unit and all modules
in the sequel (if the contrary is not stated) will be right unital modules
MR , therefore we write endomorphisms of a module on the left. A ring is
right artinian (right noetherian) if it has the descending (ascending) chain
condition on right ideals. The ring R is noetherian if R is right noetherian
and left noetherian. A ring R is prime if aRb ̸= 0 for every 0 ̸= a, b ∈ R and
R is a domain if ab ̸= 0 for every 0 ̸= a, b ∈ R. A ring R is right hereditary
if every right ideal is projective as a right module over R and hereditary if
R is right and left hereditary.
A hereditary noetherian prime ring is referred to as an hnp-ring. We
use the term hnp-domain instead of hereditary noetherian domain. An hnpring R is said to be right bounded if every essential right ideal of R contains
a nonzero two–sided ideal and is said to be bounded if it is left and right
bounded. There is a dichotomy for hnp-rings: by Lenagan [21] every hnpring is either bounded or primitive (has a simple faithful module) but not
both. An hnp-ring without idempotent ideals is called a (noncommutative)
Dedekind prime ring — we say a Dedekind domain if it is a domain. Of course
commutative Dedekind domains are of this sort as are maximal orders over
them. A nontrivial example is given by the first Weyl algebra A1 (k) =
k⟨X, Y | Y X − XY = 1⟩ over a field k of characteristic zero (see below
for more examples). An hnp-ring has enough invertible ideals if every nonzero ideal contains an invertible ideal. Every bounded hnp-ring and every
Dedekind prime ring has enough invertible ideals.
The ring R is said to be a right V-ring if every simple right module over R
is injective. For example a commutative ring is a V -ring iff it is von Neumann
regular. A ring R is an right RD-ring [30] if every finitely presented right
module over R is a direct summand of a direct sum of modules of the form
R/ri R, ri ∈ R. This property is left/right symmetric (see, e.g. [30]) and a
commutative ring is RD iff it is Prüfer i.e. has distributive lattice of ideals.
Fact 2.1 [30]Let R be an RD-ring. Then every pp-formula φ(x) over R
is equivalent to a finite conjunction of pp-formulas of the form ai | xbi ,
ai , bi ∈ R and also is equivalent to a finite sum of dual pp-formulas of the
form ∃ y (yci = x ∧ ydi = 0), ci , di ∈ R. Every hereditary noetherian prime
ring with enough invertible ideals is RD.
We say that the module MR has Krull dimension (see [24] for the definition of this dimension) if the lattice of submodules of M does not contain a
6
dense chain, in other words if every interval in this lattice contains a simple
(that is, two-point) subinterval.
For a module M over a ring R and m ∈ M , r ∈ R define annM (r) = {m ∈
M | mr = 0}, annR (m) = {r ∈ R | mr = 0}. Correspondingly annM (J) for
J ⊂ R means {m ∈ M | mJ = 0} and Ann(M ) = {r ∈ R | M r = 0}. For
a morphism f : M → N between modules we define ker(f ) = {m ∈ M |
f (m) = 0} and coker(f ) = N/f (M ).
3
Ziegler spectrum
Lemma 3.1 [23, p. 321]Let R be an arbitrary domain and let 0 ̸= r ∈ R.
Then for any module MR there are canonical isomorphisms (of modules over
the centre of R) Hom(R/rR, M ) ∼
= annM (r) and Ext(R/rR, M ) ∼
= M/M r.
Proof. The isomorphism Hom(R/rR, M ) ∼
= annM (r) is clear. Consider
the short exact sequence 0 → R → R → R/rR → 0 where the map R →
R is given by left multiplication by r and where this is a monomorphism
because R is a domain. Applying the functor Hom(−, M ) we obtain the
exact sequence:
0 → Hom(R/rR, M ) → Hom(R, M ) → Hom(R, M ) → Ext(R/rR, M ) → 0
(the last zero is since Ext(R, M ) = 0). Since Hom(R, M ) ∼
= M , this leads
to the exact sequence
0 → annM (r) → M → M → Ext(R/rR, M ) → 0
where the map M → M is given by right multiplication by r, as required.
2
Corollary 3.2 Let R be a domain, 0 ̸= a, b ∈ R. Then Ext(R/aR, R/bR) ̸=
0 iff Ra + bR ̸= R iff Ext(R/Rb, R/Ra) ̸= 0.
Proof. By Lemma 3.1 Ext(R/aR, R/bR) is isomorphic to the cokernel of the
action on R/bR given by right multiplication by a. Thus Ext(R/aR, R/bR) ̸=
0 iff this action is not onto iff Ra + bR ̸= R. The rest follows by symmetry.
2
Lemma 3.3 Let M = R/I be a simple injective right module over a right
noetherian ring R. Then M is isolated in ZgR by the minimal pair (xI =
0/x = 0).
7
Proof. Let r1 , . . . , rn ∈ R be generators for I and φ(x) be the pp-formula
∧i xri = 0 (that is xI = 0). Then M is isolated by the pp-pair (φ(x)/x = 0).
Indeed let N be a pure-injective indecomposable module over R and suppose
that N |= φ(n) for some 0 ̸= n ∈ N . Then nR ∼
= M , hence nR is a direct
summand of N , thus, by indecomposability, nR = N ∼
= M. 2
Lemma 3.4 Suppose that for a ring R there are two sets Φ, Ψ of ppformulas such that every pp-formula may be written as a finite sum of elements of Φ and also as a finite conjunction of elements of Ψ. Then a basis
of ZgR is given by the sets of the form (φ/φ ∧ ψ) with φ ∈ Φ and ψ ∈ Ψ.
Proof. One easily checks that (
(φ/ ∧j ψj ) = ∪j (φ/φ ∧ ψj ). 2
By Fact 2.1 we obtain
∑
i φi /ψ)
= ∪i (φi /φi ∧ ψ) and then that
Corollary 3.5 ZgR over any RD-ring R has a basis of open sets consisting
of sets of the form (φ/ψ) where φ = φ(x) is ∃ y (x = yr ∧ ys = 0) for some
r, s ∈ R and ψ is φ ∧ u | xt for some u, t ∈ R. In particular this is true for
every hereditary noetherian prime ring with enough invertible ideals.
Lemma 3.6 Let R be a hereditary noetherian domain with enough invertible
ideals. Then the following conditions are equivalent:
1) the set of pure-injective hulls of finite length points is dense in ZgR ;
2) there are no simple injective left R-modules.
Proof.
1) ⇒ 2). Otherwise let M = R/J be a simple injective left module
and choose a set of generators r1 , . . . , rn for the left ideal J. By Lemma
3.3 the left pair (Jx = 0/x = 0) is minimal. By elementary duality the
∑
pair (x = x/ i ri | x) is minimal and since R ̸= J this pair opens in RR .
Thus the unique pure–injective indecomposable module in which this pair
opens is a direct summand of PE(R) and hence is a torsionfree module, a
contradiction.
2) ⇒ 1). Let (φ/ψ) be a nonempty basic open set: we must find a
finitely generated torsion module M in which (φ/ψ) opens (for then M is
a direct sum of indecomposable finitely generated torsion modules, hence
by [9] finite length modules and (φ/ψ) must open in at least one of these
and hence in the (indecomposable) pure-injective hull of this module). By
Corollary 3.5 we may suppose that φ is of the form ∃ y (x = yr ∧ ys = 0),
r, s ∈ R and that ψ is of the form φ(x) ∧ u | xt, u, t ∈ R. If s ̸= 0 then R/sR
8
is a cyclic torsion module and the pair (R/sR, r) is a free realization of φ.
Thus R/sR |= ¬ u | rt as desired.
Otherwise s = 0, hence (since the pair (R, r) is a free realization of
φ) rt ∈
/ Ru in particular rt ̸= 0, hence rt ∈
/ I for some nonzero right
ideal I (otherwise rtR is a simple module, hence R is an artinian ring, a
contradiction). If u = 0 then (φ/ψ) opens in R/I. Suppose u ̸= 0. Since
R/Ru is not an injective left module and R is an RD-domain, by arguments
of [30] the pp-type ppR/Ru (rt) does not contain the pp-formula a | x for
some 0 ̸= a ∈ R. Thus rt ∈
/ aR + Ru, hence ¬ u | rt is true in R/aR. 2
The following proposition relates isolation in the Ziegler spectrum to the
existence of left almost split morphisms.
Proposition 3.7 Let R be any ring. Suppose that M is a finitely presented
module with End(M ) local (hence with PE(M ) an indecomposable module).
Then M is isolated by a minimal pair iff there is f : M → N , which is
not a split monomorphism, with N finitely presented and such that every
morphism g : M → M ′ with M ′ in M od − R which is not a pure embedding
factors through f . In this case f is either a non-split monomorphism or an
epimorphism with a simple essential kernel: the latter case occurs exactly
when M is absolutely pure.
Proof. ⇒. Suppose that PE(M ) is isolated by a minimal pair. Let c̄
be a finite generating tuple for M . Then by [34, 8.10] tpM (c̄) contains a
minimal pair (φ/ψ) (that is, φ ∈ ppM (c̄), ψ ∈
/ ppM (c̄) and φ/ψ is a minimal
pair). Since M is finitely presented and hence ppM (c̄) is finitely generated,
without loss of generality c̄ in M is a free realization of φ. Let b̄ in N be a
free realization of ψ. Then there is f : M → N taking c̄ to b̄ which is not a
split monomorphism (since it is strictly pp-type-increasing).
If g : M → M ′ is not a pure embedding then ppM ′ (gc̄) > φ and hence
([25, 9.26]) ppM ′ (gc̄) ≥ ψ so there is a factorization of g through f , as required. If it is not monic then, immediately from the factorization property,
the kernel of f must be simple. The only case in which f is not monic
is, by the factorisation property of f , that in which every monomorphism
g : M → M ′ is pure.
⇐. Let f : M → N be a morphism as given, let c̄ generate M , say
ppM (c̄) ↔ φ and let ppN (f c̄) ↔ ψ. Since f is not a split monomorphism and
hence not pure (M , N being finitely presented) we have φ > ψ. Certainly
M ∈ (φ/ψ). Suppose that there were ξ with φ > ξ > ψ. Let ā in M ′ be
a free realization of ξ and let g : M → M ′ be a morphism taking c̄ to ā.
9
Since g factors through f we obtain ψ ≥ ξ, a contradiction. Thus (φ/ψ) is
a minimal pair, as required. 2
In particular, for M a finitely presented module with local endomorphism
ring, M is isolated by a minimal pair iff there is a left almost split map (see
[1]) with domain M in the category of finitely presented modules.
Proposition 3.8 Let R be a hereditary noetherian prime ring. Suppose that
M is an indecomposable module of finite length and suppose that there are
infinitely many simple modules S such that Ext(S, M ) ̸= 0. Then PE(M )
is not isolated by a minimal pair in ZgR .
Proof. If PE(M ) is isolated then take f : M → N as in the proposition
3.7 and let C = coker(f ). Since M is not absolutely pure (for Ext(S, M ) ̸=
0) for some simple module S and R is noetherian), C ̸= 0. Since N is
finitely presented, N = N ′ ⊕ N ′′ with N ′ torsion and N ′′ torsionfree. Since
f (M ) ⊆ N ′ , we may suppose that N is torsion, so both N and C are
of finite length. Let S be a simple module with Ext1R (S, M ) ̸= 0 — say
0 → M → X → S → 0 is a nonsplit sequence. Then we have the following
commutative diagram:
0 −→ M
↓
0 −→ M
f
−→ N −→ C −→ 0
↓h
↓ h′
g
−→ X −→ S −→ 0
with h given by Proposition 3.7 (since the extension is non–split, g is not
pure) and with the induced map h′ being nonzero since f is not split. But
then there are infinitely many simple modules S with Hom(C, S) ̸= 0 —
contradicting that C is of finite length. 2
Corollary 3.9 Let R be a hereditary noetherian prime ring such that for
every simple module T there are infinitely many simple modules S with
Ext(S, T ) ̸= 0. Then no point PE(M ) with M of finite length is isolated
by a minimal pair.
Proof. Let M be an indecomposable finite length module over R. By
Proposition 3.8 it is enough to prove that Ext(S, M ) ̸= 0 for infinitely many
simple modules S — we use induction on the length of M . Choose a simple submodule T of M . Then M is included in the short exact sequence
0 → T → M → M ′ → 0. By induction there are infinitely many S such
10
that Ext(S, M ′ ) ̸= 0. For such S we have, since R is hereditary, the exact
sequence
0 → Hom(S, T ) → Hom(S, M ) → Hom(S, M ′ ) →
→ Ext(S, T ) → Ext(S, M ) → Ext(S, M ′ ) → 0
as desired.
2
Lemma 3.10 Let R be a hereditary noetherian domain and let MR be an
indecomposable module of finite length such that Hom(R/aR, M ) has Krull
dimension over End(M ) for every 0 ̸= a ∈ R. If PE(M ) is an isolated point
in ZgR then it is isolated by a minimal pair.
Proof. Suppose PE(M ) is isolated by a pair (φ/ψ): so M |= φ(m) ∧
¬ ψ(m) for some m ∈ M . Since M is torsion ma = 0 for some 0 ̸= a ∈ R.
Clearly the pair (φ ∧ xa = 0/ψ ∧ xa = 0) also isolates PE(M ). Since
Hom(R/aR, M ) = annM (a) has Krull dimension over End(M ), and hence
there is an M -minimal pair between φ and ψ, we can suppose that (φ/ψ)
is itself an M -minimal pair. We prove that (φ/ψ) is a minimal pair in the
largest theory of modules over R. Otherwise φ > ξ > ψ for some pp-formula
ξ. Since (φ/ψ) isolates M , both (φ/ξ) and (ξ/ψ) must open in M , which is
a contradiction. 2
Theorem 3.11 Let R be a hereditary noetherian domain with enough invertible ideals and without injective simple left modules such that for every
simple module M the following holds:
i) there are infinitely many simple modules S such that Ext(S, M ) ̸= 0;
ii) Hom(R/aR, M ) has Krull dimension over End(M ) for 0 ̸= a ∈ R.
Then there are no isolated points in ZgR .
Proof. Suppose M is an isolated point in ZgR . By Lemma 3.6 we can
take M = PE(N ) for some N of finite length. Since N is torsion and (ii)
holds, by Lemma 3.10 PE(M ) is isolated by a minimal pair, contradicting
Corollary 3.9. 2
4
Generalized Weyl algebras
Let k be a field and let σ be an automorphism of the polynomial ring k[H].
The k-algebra A = k⟨X, Y | Y X = a(H), XY = σ(a)⟩, where b(H)X =
11
Xσ −1 (b), b(H)Y = Y σ(b) is a generalised Weyl algebra (GWA). Let B =
k(H)[X, X −1 , σ] be the skew Laurent polynomial ring which is the localization of A at k[H]\0. It follows from [2] that a GWA A is a noetherian domain
of Krull dimension 1 and B is a principal ideal domain. An arbitary element
u of a GWA A is of the form u = b−m (H)Y m + · · · + b0 (H) + · · · + bn (H)X n
where bi (H) ∈ k[H], alternatively a similar form with coefficients written
on the right. We define deg(u) = n + m + 1 (this is called “length” in [4]),
for instance deg(0) = 0, deg(2) = 1 and deg(Y + X) = 3.
The automorphism σ induces an action on the set of (irreducible) polynomials over k[H] hence on the set of maximal ideals over k[H]. In the
case that σ(H) = H − 1 the orbits of this action look like ZZ, “linear” in
the terminology of [4]. We define an equivalence relation on the set of irreducible polynomials over k[H] by p ∼ q if p = σ k (q) for some k ∈ ZZ. The
orbit is said to be degenerate if it contains an irreducible polynomial p such
that p | a(H) and nondegenerate otherwise. A degenerate orbit contains
polynomials p1 , p2 , . . . , pn+1 such that σ ki (pi ) = pi+1 , for some ki > 0 and
such that the orbit is split into finitely many intervals (−∞, p1 ], (p1 , p2 ], . . . ,
(pn , pn+1 ], (pn+1 , +∞) each of which corresponds to a simple module.
By [4] every simple module over A is either k[H]-torsion or k[H]-torsionfree.
The following is the complete description of the simple k[H]-torsion modules
over a GWA A where char(k) = 0, σ(H) = H − 1. Since we are working
from the opposite side to [4], this description is dual to [4].
1) For a nondegenerate orbit with irreducible polynomial p: the module
M = A/pA.
In particular Mλ = A/(H − λ)A is a simple module for λ ∈ k not a root
of a(H).
For a degenerate orbit containing irreducible polynomials p1 , . . . , pn+1 ,
where pi | a(H):
2) A/(Y, p1 )A corresponding to (−∞, p1 ];
3) A/(Y, pi+1 , X ki )A corresponding to (pi , pi+1 ];
4) A/(X, σ(pn+1 ))A corresponding to (pn+1 , +∞).
Modules of type 3) are exactly the finite-dimensional simple modules.
For c(H) ∈ k[H] define [c] = {p(H) | p(H) is irreducible and σ k (p) | c(H)
for some k ∈ ZZ}. For c(H), b(H) ∈ k[H] define c < b if there is no
irreducible p(H) such that p | b and σ k (p) | c for k ∈ ZZ, k ≥ 0. An element
p ∈ A, p = a0 + · · · + an X n is called l-normal if a0 (H) < an (H), a(H).
Every k[H]-torsion module M can be decomposed into a direct sum of
cyclic modules k[H]/pl k[H] for certain irreducible polynomials p and all p’s
appearing in this way form the set we denote by Sup(M ).
12
If M is a k[H]-torsionfree simple module, then there exists p ∈ A, of the
form p = a0 + · · · + an X n that is irreducible in B, l-normal and M = Mp is
at one end of a short exact sequence:
0 → A ∩ pB/pA → A/pA → A/A ∩ pB = Mp → 0
(∗)
where the module Vp = A ∩ pB/pA is k[H]-torsion and Sup(Vp ) ⊆ [a0 ] ∩
([an ∪ a]) ([4, Lem. 5.7, Thm. 5.8]).
The scheme of the proof of the following result was communicated to the
second author by V. Bavula (see also [3]).
Theorem 4.1 Let A be a GWA, char(k) = 0 with σ(H) = H − 1. Then
for every simple module M over A which is not finite-dimensional there are
infinitely many simple modules S such that Ext(S, M ) ̸= 0 and there are
infinitely many simple modules T such that Ext(M, T ) ̸= 0.
Proof.
Case 1. M is k[H]-torsionfree.
Then M = Mp can be included in the short exact sequence (∗). We
prove that Ext(Mλ , M ) ̸= 0 and Ext(M, Mλ ) ̸= 0 for every λ ∈ k such that
λ + n is not a root of ai (H), a(H) for every n ∈ ZZ. Because Mλ ∼
= Mµ iff
λ − µ ∈ ZZ and there are infinitely many orbits but finitely many roots we
can find infinitely many nonisomorphic Mλ .
We prove that Ext(Mλ , M ) ̸= 0. Apply Hom(Mλ , −) to the sequence
(∗):
0 → Hom(Mλ , Vp ) → Hom(Mλ , A/pA) → Hom(Mλ , M ) →
→ Ext(Mλ , Vp ) → Ext(Mλ , A/pA) → Ext(Mλ , M ) → . . .
Since Sup(Mλ ) ∩ Sup(Vp ) = ∅, hence Ext(Mλ , Vp ) = 0 it is enough to
prove that Ext(Mλ , A/pA) ̸= 0 hence, by Lemma 3.2, Ext(A/Ap, A/A(H −
λ)) ̸= 0.
Consider the action on A/A(H − λ) given by left multiplication by p =
a0 + · · · + X n an (here it does not matter whether p is l-normal hence we
can choose the side for coefficients freely). Every element in A/A(H − λ)
is represented by an element with canonical form u = Y m b−m + · · · + X l bl
where bi ∈ k, hence pu = Y m a0 (λ − m)b−m + · · · + X n+l an (λ + l)bl . Thus
deg(pu) = m + l + n + 1 = deg(u) + n ≥ 1 + 1 = 2 so, for instance, the image
of X is not in the image of this map.
We prove that Ext(M, Mλ ) ̸= 0. Applying Hom(−, Mλ ) to (∗) we obtain
13
0 → Hom(M, Mλ ) → Hom(A/pA, Mλ ) → Hom(Vp , Mλ ) →
→ Ext(M, Mλ ) → Ext(A/pA, Mλ ) → Ext(Vp , Mλ ) → . . .
Since Vp and Mλ are torsion k[H]-modules and Sup(Vp ) ∩ Sup(Mλ ) = ∅,
we obtain Hom(Vp , Mλ ) = Ext(Vp , Mλ ) = 0 and Ext(M, Mλ ) ∼
= Ext(A/pA, Mλ ).
Thus this case is symmetric to that just proved.
Case 2. M = A/pA is k[H]-torsion and p belongs to a non-degenerate
orbit.
Consider the collection of modules Nα = A/(α + X)A, 0 ̸= α ∈ k.
We prove that Nα is a simple module for α ̸= 0. Clearly α + X is an
irreducible polynomial in B which is l-normal and, because [a0 ] = [α] = ∅,
the corresponding module Vp is zero. We have X ≡ −α and (α + X)Y =
αY + a(H − 1) hence Y ≡ −α−1 a(H − 1) in Nα . Thus Nα ∼
= k[H] with
module structure given by b(H)X = Xb(H +1) = −αb(H +1) and b(H)Y =
Y b(H − 1) = −α−1 a(H − 1)b(H − 1).
We prove that Nα ∼
̸ Nβ for α ̸= β, α, β ̸= 0. Let the morphism
=
Nα → Nβ , given by left multiplication by b(H). Then b(H)(α + X) =
b(H)α − βb(H + 1). Considering the leading coefficient bk of b(H) we obtain
bk α = bk β hence bk = 0 and b(H) = 0.
We prove that Ext(Nα , M ) ̸= 0 for α = 1, 2, . . . equivalently Ext(A/Ap, A/A(α+
X)) ̸= 0. The action on A/A(α + X) given by left multiplication by p is
simply the action on k[H] by the same multiplication, hence is not onto.
For Ext(M, Nα ) ̸= 0 for α = 1, . . . we obtain the same action on k[H] by
right multiplication on p.
Case 3. M = A/(Y, p)A where p | a(H), hence pb(H) = a(H).
We check that Ext(Nα , M ) ̸= 0 for α = 1, 2, . . . so consider the action
on M given by right multiplication by (α + X). A canonical form for a
representative of u ∈ M is u = b0 (H)+. . .+bn (H)X n where deg(bi ) < deg(p)
for every i. Thus u(α+X) = b0 (H)α+(b1 (H)α+b0 (H))X +· · ·+(bn (H)α+
bn−1 (H))X n + bn (H)X n+1 (here deg(bi (H)α + bi−1 (H)) < deg(p), hence the
form is still canonical). Therefore deg(u(α + X)) = n + 2 = deg(u) + 1 ≥
1 + 1 = 2 hence X is not in the image of this action.
We prove that Ext(M, Nα ) ̸= 0 for α = 1, 2, . . .. Let I = ann(Y, p) =
{(u, v) ∈ R ⊕ R | Y u + pv = 0} — a submodule of the right module R2 . We
prove that I is generated by the columns (X, −b(H)) and (p(H − 1), −Y ).
Since Y X = a(H) = pb(H) and Y p(H − 1) = p(H)Y these pairs are in I.
We prove that all other relations are generated by these. Because we have X
and p(H − 1) on the first coordinate we need prove only that there exists no
14
nontrivial relation of the form Y · (b−m (H)Y m + · · · + b0 (H)) = pr, deg(bi ) <
deg(p). The left-hand-side is t = b−m (H + 1)Y m+1 + · · · + b0 (H + 1)Y and
such an element lies in pA iff p | bi (H + 1) for each i hence bi = 0 since p is
irreducible.
Thus by applying Hom(−, Nα ) to the exact sequence 0 → (Y, p)A →
A → M → 0 and considering the above presentation of (Y, p)A we obtain
Ext(M, Nα ) = S/T where S = {(m, n) ∈ Nα ⊕ Nα | mX − nb(H) = 0,
mp(H − 1) − nY = 0} and T = {(m, n) ∈ Nα ⊕ Nα | m = kY , n = kp
for some k ∈ Nα }. Put l = (−b(H − 1), α). Computing in Nα we have
−b(H −1)X = −Xb(H) = αb(H) and −b(H −1)p(H −1) = −a(H −1) = αY
because αY = (α + X)Y − a(H − 1) ≡ −a(H − 1) in Nα . So it remains to
prove that l ∈
/ T . Otherwise (−b(H − 1), α) = (kY, kp) for some k ∈ Nα .
Thus α ∈ Nα p, hence 1 ∈ Nα p . Therefore 1 is in the image of the action on
Nα given by right multiplication by p and this action is multiplication by p
in k[H] which is not onto, a contradiction.
Case 4. M = A/(X, σ(p))A. This case is dual to the previous and the
required collection of modules is A/(Y + β)A, β = 1, 2, . . .. 2
For the case σ(H) = H − 1, char(k) = 0 by [2, Cor. 3.2] A is simple iff
there is no finite–dimensional simple module over A iff every degenerate orbit
contains only one irreducible polynomial p such that p | a(H). Similarly,
[2], such a GWA A is hereditary iff A is simple and p2 does not divide a(H)
for every irreducible polynomial p.
Corollary 4.2 Let A be a GWA with char(k) = 0 and σ(H) = H − 1.
Then:
1) no infinite-dimensional simple module is the source of a left almost
split morphism in the category of finite length modules over A;
2) if A is simple then no simple module is the source of a left almost
split morphism in the category of finite length modules over A;
3) if A is simple and hereditary there is no isolated point in ZgA . In
particular the category of finite length modules over A does not have any
almost split sequence.
Proof. By Theorem 4.1 for every simple module M which is not finite– dimensional there are infinitely many simple modules S such that Ext(S, M ) ̸=
0. By Proposition 3.7 and the proof of Proposition 3.8 the points 1) and 2)
follow.
For case 3) since A is a Dedekind domain and by [2, Thm. 6], Ext(M, N )
is finite-dimensional over k for any finite length modules M, N . As (see
15
below) there is no finitely generated injective left module over A we can
apply Theorem 3.11. 2
Corollary 4.3 Let A be a simple hereditary GWA with char(k) = 0 and
σ(H) = H − 1. Then no finitely generated module over A is injective.
Proof. Suppose M is a finitely generated injective module over A. Since A,
hence M , is noetherian, M has a maximal proper submodule and hence a
simple factor module N which is injective by hereditarity. Now by Theorem
4.1 Ext(S, N ) ̸= 0 for some simple S, a contradiction. 2
Let A1 (k) = k⟨X, Y | Y X − XY = 1⟩ be the first Weyl algebra for
k a field of characteristic zero. So A1 is a generalized Weyl algebra with
a(H) = H and σ(H) = H − 1. B1 (K) is the corresponding localization of
A with respect to the Ore set k[H] \ 0. Then (see [23]) A1 is a hereditary
simple noetherian domain and B is a simple left and right principal ideal
domain. Thus both A1 and B1 are Dedekind domains.
Fact 4.4 [23, Thm. 4.1, 5.2, 5.3]Let R be either A1 or B1 and let I, J be
nonzero right ideals of R. Then
1) Hom(R/I, R/J) is a finite dimensional vector space over k.
2) Ext(R/I, R/J) is a finite dimensional vector space over k for R = A1
and infinite dimensional over k for R = B1 .
Theorem 4.5 Let k be a field with char(k) = 0. Then there is no isolated
point in the Ziegler spectrum of A1 (k) or B1 (k).
Proof. Since A1 (k) is a simple hereditary GWA we can apply Corollary 4.2.
B is a Dedekind domain without injective simple modules by [23, Cor. 4.2].
Also by [23] and Fact 4.4 (i) and (ii) of Theorem 3.11 holds. Thus the
conclusion of the Theorem gives us the required result. 2
5
Examples
Our first remark in this section is an immediate consequence of Garavaglia’s
unpublished [12, Theorem 1] (see [25, 10.10]).
Remark 5.1 Let R be a left noetherian ring. Then PE(RR ) is the pure–
injective envelope of a direct sum of indecomposable pure–injective modules.
16
Proof. Otherwise there is a superdecomposable pp-type p(x) consistent
with the theory of RR . Choose φ ∈ p− such that the left ideal φ(R) is
maximal among such ideals. By [34] φ is not large in p hence there are
ψ, ξ ∈ p− such that φ → ψ, ξ and ψ + ξ ∈ p. Clearly φ(R) ⊂ ψ(R), a
contradiction. 2
Example 5.2 Let R be a principal ideal domain that is a V-ring and not a
division ring. Then the CB-rank of ZgR is equal to 1. The isolated points in
ZgR are the simple modules and the duals of the simple left modules, these
being the indecomposable direct summands of PE(R). The unique nonisolated point of ZgR is E(R).
Proof. Let M = R/aR be a simple module over R, hence 0 ̸= a is an
irreducible element. Then by Lemma 3.3 M is isolated by the minimal pair
(xa = 0/x = 0). By symmetry and elementary duality the pair (x = x/a | x)
is minimal and hence by [34] defines a unique indecomposable pure–injective
module and in no superdecomposable pure-injective module. Since this pair
opens in RR it follows that this pure-injective module is a direct summand
of PE(R) and, in particular, is a torsionfree module.
Consider an arbitrary pure–injective indecomposable module M over R.
If mr = 0 for some 0 ̸= m ∈ M , 0 ̸= r ∈ R then mR is a module of finite
length, hence contains a simple submodule. Because every simple module is
injective M must be simple. Otherwise M is torsionfree. If M is a divisible
module it is injective and hence M ∼
= E(R). Otherwise there is m ∈ M and
an irreducible element b ∈ R such that m ∈
/ M b. Then the pair (x = x/b | x)
opens in M and so, since this pair is minimal, M is isolated.
Suppose that M = E(R) were isolated by a pair (φ/ψ) where, by Lemma
3.5, we can suppose that φ is ∃ y (x = yr ∧ ys = 0) for some r, s ∈ R and ψ
is φ ∧ u | xt for some u, t ∈ R. Since M is a divisible module we have u = 0,
hence ψ is φ ∧ xt = 0 and t ̸= 0. Since M is torsionfree, s = 0. Therefore φ
is r | x and so the pair (φ/ψ) opens in R and hence, by Remark 5.1, opens
in one of the indecomposable direct summands of PE(R), none of which is
injective, contradicting that (φ/ψ) isolates E(R). 2
Example 5.3 Let R be a differential polynomial ring over a universal field
with derivation. Then ZgR consists of three points: the unique simple module
V , the dual DV of the unique simple left module and E(R). Here V , DV
are isolated points and E(R) is a point of CB-rank 1. The modules V and
E(R) are totally transcendental but DV is not and PE(R) = PE(DV (α) ) for
some α ≥ 2.
17
Proof. By [11] there is a unique simple module V over R. By Example
5.2, V is the unique isolated injective point in ZgR and by duality there is a
unique isolated torsionfree point DV . Thus PE(R) = PE(DV (α) ) and α ≥ 2
since PE(R) is a decomposable module (since R is not local). The module
E(R) cannot be isolated. Since R is a noetherian ring injective modules over
R are totally transcendental. Since R is not left artinian, the module RR is
not totally transcendental, hence DV is not. 2
Notice that for a right V -ring Ext(S, T ) = 0 for arbitrary simple right
modules S, T and hence the Ext-graph between simple modules is degenerate.
Example 5.4 Let k be a field of characteristic zero, let F = k((x)) be the
Laurent series field with the usual derivation and let R = F [y,′ ] be the
differential polynomial ring over F . Then every finite length point in ZgR
is isolated by a minimal pair.
Proof. By Zimmermann [35] the category of finite length modules over R
has almost split sequences. By Proposition 3.7 the pure–injective envelope
of every indecomposable finite length module is isolated by a minimal pair.
2
In the following example we consider a GWA that is not simple and has
global dimension 2 by [2, Thm. 5].
Example 5.5 Let A be a GWA with char(k) = 0, σ(H) = H − 1 and
a(H) = H(H − 1). Then for every simple module M over A, Ext(S, M ) ̸= 0
for infinitely many simple modules S. In particular, no simple module over
A is the domain of a left almost split morphism in the category of finite
length modules over A.
Proof. By Theorem 4.1 and the classification of simple modules we need
consider only the case where M = A/(X, Y, H − 1), the unique simple finitedimensional module over A. Note that dimk (M ) = 1. Let pα = (H+1)+αX,
0 ̸= α ∈ k. Since pα is l-normal we have the short exact sequence
0 → Vα = Vpα → A/pα A → Sα = A/A ∩ pα B → 0
where Sα is simple. We prove that Ext(Sα , M ) ̸= 0 for every α ̸= 0.
Applying the functor Hom(−, M ) we obtain the exact sequence
0 → Hom(Sα , M ) → Hom(A/pα A, M ) → Hom(Vα , M ) →
18
→ Ext(Sα , M ) → Ext(A/pα A, M ) → Ext(Vα , M ) → . . .
Since pα is l-normal by [2, L. 3.7], pα acts by right multiplication as a
monomorphism on every simple k[H]-torsion module. Since M is a finite
dimensional module, pα acts as an isomorphism of M over k. In particular
Ext(A/pα A, M ) = 0. Clearly M ∼
= k with the corresponding A-module
structure given by β(H − 1) = βX = βY = 0, β ∈ k. If the homomorphism
f : A/pα A → M is induced by left multiplication by an element of A with
image β ∈ M then β(H + 1 + αX) = 2β = 0, hence Hom(A/pα A, M ) = 0
and Ext(Sα , M ) ∼
= Hom(Vα , M ).
In the following and elsewhere we sometimes identify an element of A
with its image in a given cyclic module.
We prove that Vα is a cyclic module with generator mα = Y 2 − 6αY −
2
α (H − 2)(H + 1) and relations mα (H − 1) = mα X = 0, and that Sα =
A/(pα A + mα A) is 2–related.
We show that every element in A/pα A has a unique representative of
the form u = Y m b−m + · · · + Y b−1 + b0 (H), where bi ∈ k for i < 0. Since
X ≡ −α−1 (H +1) and Y H ≡ −α(H −1)(H −2), it follows that every element
in A/pα A can be represented in this form. Suppose that this representation
is not unique, therefore for some 0 ̸= u as above u = pα v, where v =
Y t c−t (H) + · · · + X l cl (H), that is
Y m b−m + · · · + Y b−1 + b0 (H) = (H + 1 + αX)[Y t c−t (H) + · · · + X l cl (H)] .
If m > 0 then b−m equals the leading coefficient of the right side, that is
b−m = (H +1−t)c−t (H), a contradiction. Thus u = b0 (H), hence deg(u) = 1
and deg(pα v) = 1 + deg(v) ≥ 1 + 1 = 2, a contradiction again.
Now calculate in A: pα (Y 2 − Y α(H + 1)) = Y 2 (H − 1) + Y [α(H −
2)(H − 3) − αH(H + 1)] − α2 (H − 1)(H − 2)(H + 1) = mα (H − 1). Thus
mα (H − 1) = pα a for some a ∈ A, hence mα (H − 1) = 0 in A/pα A and
mα = pα a · (H − 1)−1 ∈ pα B. Thus mα ∈ Vα . If mα = 0 in A/pα A
hence mα = pα a′ for a′ ∈ A, then a′ (H − 1) = a, a contradiction since
a = Y 2 − Y α(H + 1).
We prove that mα X = 0 in A/pα A. By direct calculation we obtain the
equality mα X = (Y 2 −6αY −α2 (H −2)(H +1))X = Y H(H −1)−6αH(H −
1) − α2 X(H − 1)(H + 2) = ((H + 1) + αX)(Y (H − 1) − α(H − 1)(H + 2)),
hence mα X ∈ pα A as desired.
Next we show that every element u ∈ Hα = A/(mα A + pα A) has a
canonical representative Y a1 + a0 (H), where a1 ∈ k. Since in Hα we have
the additional relation Y 2 ≡ 6αY + α2 (H − 2)(H + 1), every element u ∈ Hα
19
can be represented in this form. If this representation is not unique, then
for some 0 ̸= u = Y a1 + a0 (H) we obtain u ∈ pα A + mα A that is, in A/pα A
we have u = mα v for some v ∈ A. Since mα (H − 1) = mα X = 0 in A/pα A
we can suppose that v = c−t Y t + · · · + c0 , ci ∈ k. Thus
Y a1 + a0 (H) = [Y 2 − 6αY − α2 (H − 2)(H + 1)] · (c−t Y t + · · · + c0 ) ,
where this will be an identity after transformating the right–hand–side to
canonical form in Aα /pα A. If t ≥ 0, then (after calculations in A/pα A) the
leading term of the right side is Y t+2 c−t , t + 2 ≥ 2, a contradiction (since
there is no such term on the left–hand–side).
We prove that Vα = (mα A + pα A)/pα A. If not then Sα is a proper
homomorphic image of A/(pα A+mα A). Every element u of A/(pα A+mα A)
has a representative in canonical form Y a1 + a0 (H), a1 ∈ k and each of its
proper factors is finite dimensional. (If a1 = 0 this is clear, otherwise set
a1 = 1 and, calculating modulo pα A, we obtain (Y + a0 (H))X = a0 (H)X +
H(H − 1) = −α−1 (H + 1)a0 (H + 1) + H(H − 1) and if this is zero we
obtain (H + 1)a0 (H + 1) = αH(H − 1), a contradiction). Thus Sα is simple
finite-dimensional, hence isomorphic to M . But we have seen above that
Hom(A/pα A, M ) = 0, a contradiction.
Thus Vα = (mα A + pα A)/pα A and Sα = A/(pα , mα )A. We prove that
there are no other relations on Vα . Otherwise, modulo the relations mα (H −
1) = mα X = 0, we can suppose that any relation is of the form mα (Y m +
b−m+1 Y m−1 + . . . + b0 ) = pα (ck (H)Y k + . . . + cl (H)X k ), where bi ∈ k, k ≤ 0.
The leading (i.e., most negatively-indexed) coefficient of the left part is 1
and of the right part is (H + 1)ck (H), a contradiction.
Thus there is f : Vα → M such that f (mα ) = 1.
It remains to prove only that Sα ̸∼
= Sβ for α ̸= β, α, β ̸= 0. Otherwise
(see the canonical form and relations in Sβ ), we obtain a homomorphism
g : Sα → Sβ induced by left multiplication by u = a0 (H)+Y a1 . Hence in Sβ
u(H +1+αX) = a0 (H)(H +1)+Y a1 (H +1)+αXa0 (H +1)+H(H −1)a1 α =
a0 (H)(H + 1) + Y a1 − β(H − 1)(H − 2)a1 − β −1 α · (H + 1)a0 (H + 1) +
H(H − 1)a1 α. Since this element is zero in Sβ we obtain a1 = 0 and hence
a0 (H) = β −1 αa0 (H + 1), so α = β, a contradiction.
2
Conjecture 5.6 Let A be a GWA with char(k) = 0 and σ(H) = H − 1.
Then there are no almost split sequences in the category of finite length
modules over A and there are no isolated points in ZgA .
20
6
Decidability
In the paper [20] Klingler and Levy, starting with a special sextuple of simple
modules over the first Weyl algebra A1 (k) construct a map from the category of modules over the free associative algebra k⟨X, Y ⟩ to the category of
modules over A1 with very nice properties (e.g. this map preserves endomorphism rings). The proof of the following result involves this construction
very heavily.
Theorem 6.1 Let R be a right noetherian k-algebra over a field k which has
a sextuple of simple nonisomorphic modules T, T1 , . . . , T5 such that End(T ) ∼
=
End(Ti ) ∼
= k and Ext(Ti , T ) ̸= 0 for every i. Then the theory of k⟨X, Y ⟩modules is interpretable in the theory of R-modules.
Proof. Let Vi be a nonsplit extension of T by Ti : thus Vi contains a simple
submodule T and Vi /T ∼
= Ti . Clearly the Vi are nonisomorphic uniserial
modules of length 2. Also End(Vi ) ∼
= k since if f ∈ End(Vi ) then since
soc(Vi ) = T we have that f induces the map f |T ∈ End(T ) = k. So we
obtain a homomorphism End(Vi ) → k which is clearly onto. If the kernel
of this map is nonzero, then f |T = 0 for some 0 ̸= f ∈ End(Vi ) hence f
induces a nonzero homomorphism from V /T = Ti to T , a contradiction.
Set Vi = R/Ki and T = R/J.
Claim. Each of annVi (Ki ), annT (J) and annVi (J) is 1-dimensional over k.
Let a = 1 + J ∈ T and let b ∈ annT (J). Then there is an endomorphism
of T taking a to b. By hypothesis this endomorphism is multiplication by
an element of k and so b ∈ ak, as required. The same argument shows
that annVi (Ki ) is 1-dimensional. Finally, let c ∈ annVi (J). Then we have
a morphism T → cR → cR + T /T → Vi /T which is zero iff c ∈ T . So by
hypothesis, c ∈ T , i.e. annVi (J) = annT (J) as required.
Since Ki , J are finitely generated (R is right noetherian) we have that
“xKi = 0” and “xJ = 0” are pp-formulas and, by the Claim, define minimal
pp-definable subgroups of Vi . Since 1 + Ki generates Vi there exists ri such
that ri + Ki ∈ annVi (J) and so if we take θi (x, y) to be the pp formula
(xKi = 0 ∧ yJ = 0 ∧ y = xri ) then θi is a linking formula which defines
a map between annVi (Ki ) and annVi (J) which, by minimality, defines an
isomorphism between them. That is, for every element a ∈ annVi (J) there is
a unique element c ∈ annVi (Ki ) such that θi (a, c) is true in Vi and, moreover,
the (k-linear ) map so defined is an isomorphism.
Now we recall the construction of [20] Section 2. Given a k⟨X, Y ⟩-module
(Λ)
N , choose a basis (nλ )λ∈Λ of N over k. Define the module W (N ) = V1 ⊕
21
(Λ)
. . . ⊕ V5 where for λ ∈ Λ the λth copy of Vi is generated by, say, the
element aiλ and where we may choose aiλ ∈ annVi (Ki ). Also let K(N )
2
5
be the image in W((N ) of the composition
of the morphism T (Λ) → T (Λ)
)
1 1 0 1 1
5
which has matrix
with the natural inclusion of T (Λ)
1 0 1 X Y
(Λ)
into W (N ). Then define F (N ) = W (N )/K(N ). Set Ui = Vi
and let Wi
be image of Ui in F (N ). Let āiλ denote the image of aiλ in F (N ).
Note that annF (N ) (Ki ) = annWi (Ki ). For, if we had a ∈ annF (N ) (Ki ) \
annWi (Ki ) then there would be a nonzero morphism from Vi to F (N )/Wi
which, by [20] proof of 2.7, has no composition factor isomorphic to Ti —
contradiction.
Set Ai = annF (N ) (Ki ) and let Bi = {b ∈ F (N ) | there exists a ∈
F (N ) such that F (N ) |= θi (a, b)}. Note that Bi = Ai ri and that Bi ⊆
annF (N ) (J) = soc(F (N )). Moreover, since annF (N ) (Ki ) = annWi (Ki ) and
by choice of θi , θi defines an isomorphism between Ai and Bi .
Clearly Bi is a pp-definable subgroup of F (N ) and (b̄iλ = āiλ ri )λ∈Λ is a
k-basis of Bi . We will show that there is a pp-definable (in the language of
R-modules) action of k⟨X, Y ⟩ on B1 under which the map given by b̄1λ → nλ
is an isomorphism of k⟨X, Y ⟩-modules.
5
By definition of K(N ) it is the subset of soc(W (N )) = T (Λ) consisting
of all elements of the form
(c + d, c, d, c + dX, c + dY )
(∗)
where c, d are any elements of T (Λ) and where dX, dY are defined in the
∑
obvious way (e. g. if nλ X = µ nµ αµ with αµ ∈ k and if (d, 0, 0, 0, 0) is b1λ
∑
then (0, 0, 0, dX, 0) is µ b4µ αµ ).
Now, given b1 ∈ B1 there are b3 ∈ B3 , b4 ∈ B4 , b5 ∈ B5 such that
b1 + b3 + b4 + b5 = 0 (namely if b1 is the image of (d, 0, 0, 0, 0) ∈ soc(W (N ))
then b3 , b4 and b5 are respectively the images of c3 = (0, 0, d, 0, 0), c4 =
(0, 0, 0, dX, 0), c5 = (0, 0, 0, 0, dY ) in F (N )). We claim that b3 , b4 b5 are
uniquely determined. For if b′3 ∈ B3 , b′4 ∈ B4 , b′5 ∈ B5 are such that
b1 + b′3 + b′4 + b′5 = 0 then choose preimages c′3 ∈ U3 , c′4 ∈ U4 , c′5 ∈ U5 in
W (N ), of these elements. Thus we obtain (c3 − c′3 ) + (c4 − c′4 ) + (c5 − c′5 ) ∈
(U3 ⊕ U4 ⊕ U5 ) ∩ K(N ) and hence it has the form (∗) above. But then, with
the notation as in (∗), we obtain c + d = 0, c = 0 and hence also d = 0.
Thus (c3 − c′3 ) + (c4 − c′4 ) + (c5 − c′5 ) = 0 and hence c3 = c′3 and so b′3 = b3
and similarly b′4 = b4 and b′5 = b5 , as required.
22
Now consider b4 . There are b′′1 ∈ B1 , b′′2 ∈ B2 , b′′5 ∈ B5 such that b′′1 +
b′′2 + b4 + b′′5 = 0 and we claim that these are unique. For otherwise, arguing
as above, we would obtain a non–zero element of (U1 ⊕ U2 ⊕ U5 ) ∩ K(N )
which, with the notation of (∗), would imply d = 0 and c + dX = 0 and
hence also c = 0, a contradiction.
Therefore we obtain a map from B1 to itself, given by taking b1 as above
to b′′1 . Note that this map may be defined by a pp formula ρX (x, y) meaning
that ρX (b1 , b′′1 ) holds and that in F (N ) it is true that for all x ∈ B1 there is
a unique y ∈ B1 such that ρX (x, y) holds. This map takes b1 = b̄1λ via b4 =
b̄4λ X (i.e. the image of b4λ X) to b′′1 = b̄1λ X and hence B1 , equipped with this
map, is an isomorphic copy of the k[X]-module N under the map b1λ → nλ .
Similarly we use the fifth coordinate to pp-define another action on B1 which
replicates the action of Y on N . Thus we obtain pp-formulas φ(x), ρX (x, y),
ρY (x, y) in the language of R-modules such that, on any module M of the
form F (N ), both ρX , ρY define maps from φ(M ) to itself and the resulting
k⟨X, Y ⟩-module is isomorphic to N . Now let σ be a sentence in the language
of R-modules which says that ρX , ρY are total and functional when restricted
to φ(−). Thus in any R-module M satisfying σ we obtain a k⟨X, Y ⟩-module
and any k⟨X, Y ⟩-module can occur in this way. We deduce that the theory
of R-modules interprets the theory of k⟨X, Y ⟩-modules. 2
Now consider the case of a GWA. Notice that if k is algebraically closed
then every automorphism σ : k[H] → k[H] fixing k pointwise is (up to linear
change of variables): σ(H) = H − 1 or σ(H) = qH, q ∈ k, q ̸= 0.
Consider the case σ(H) = H − 1.
Proposition 6.2 Let A be a generalized Weyl algebra over a field k with
char(k) = 0 and σ(H) = H − 1. Then the theory of k⟨X, Y ⟩-modules is
interpretable in the theory of modules over A. In particular the theory of
modules over A interprets the word problem for groups.
Proof. Let T = N1 = A/(1 + X)A, Ti = A/(H − αi )A, i = 1, . . . , 5 where
αi + k is not a root of a(H) for every k ∈ ZZ and αi − αj ∈
/ ZZ for i ̸= j. We
check that the conditions of Proposition 6.1 hold.
Clearly T ∼
= k[H] where the module structure is given by b(H)X =
Xb(H + 1) = (X + 1)b(H + 1) − b(H + 1) = −b(H + 1) and b(H)Y =
Y b(H − 1) = [(X + 1)Y − a(H − 1)] b(H − 1) = −a(H − 1)b(H − 1).
We prove that T is a simple module. Calculating in T , suppose that
b(H) = a0 + · · · + an−1 H n−1 + an H n ∈ T with n ≥ 1 and ai ∈ k. Then
b(H)X = −b(H + 1) ∈ T , hence b(H) − b(H + 1) is in the submodule
23
generated by b(H) and is of smaller degree. We prove that this polynomial
is nonzero. Computing the coefficient of H n−1 we obtain that it is equal to
an−1 − nan − an−1 = −nan ̸= 0.
Next, if the homomorphism f : T → T is induced by left multiplication
by b(H) then, in T , b(H)(1 + X) = b(H) − b(H + 1) = 0, a contradiction if
deg(b) ≥ 1. Thus End(T ) ∼
= k. Also clearly T is k[H] torsionfree.
Since αi + k is not a root of a(H) for k ∈ ZZ the modules Ti are simple
by [4, Cor. 4.1] and nonisomorphic since αi − αj ∈
/ ZZ for i ̸= j. Since Ti is
a k[H]-torsion module, we have T ̸∼
= Ti for every i. The canonical form for
an element u ∈ Ti is u = b−m Y m + · · · + b0 + · · · + bn X n where bi ∈ k. We
prove that End(Ti ) ∼
= k. Let g : Ti → Ti be induced by left multiplication
by u. Thus, computing in T , u(H − αi ) = b−m (H + m − αi )Y m + · · · + (H −
αi )b0 + · · · + bn (H − n − αi )X n ≡ b−m mY m + · · · + 0 + · · · + bn (−n)X n = 0.
Hence bi = 0 for i ̸= 0.
For Ext(Ti , T ) ̸= 0 consider the action on T given by right multiplication
by H − αi . Since b(H) ∈ T is mapped to b(H)(H − αi ) which is of greater
degree, this map is not onto. 2
Let U sl2 (k) denote the universal enveloping algebra of the Lie algebra
sl2 (k) over a field k.
Corollary 6.3 The theories of modules over A1 , B1 , U sl2 (k) with char(k) =
0 all intepret the theory of k⟨X, Y ⟩-modules. Hence all these theories interpret the word problem for groups.
Proof. By [2] the factor algebras Uλ = A/(C − λ)A (C is the Casimir
element) of U sl2 (k) are generalized Weyl algebras with σ(H) = H − 1 and
a(H) = λ − H(H − 1). Also A1 is a generalized Weyl algebra. Thus we can
apply Proposition 6.2.
For B1 [23, Prop. 4.4] the B1 modules Ti = B1 /(X + iY )B1 , i = 0, . . . , 5
are simple and nonisomorphic (otherwise they would be isomorphic as A1 modules, which is not the case) and End(Ti ) ∼
= k. Now apply Fact 4.4 and
Proposition 6.1. 2
Now consider the case σ(H) = qH for q ̸= 0 not a root of unity.
Proposition 6.4 Let A be a generalized Weyl algebra over a field k with
char(k) = 0 and σ(H) = qH where q ̸= 0 is not a root of unity. Then the
theory of k⟨X, Y ⟩-modules is interpretable in the theory of modules over A.
In particular the theory of modules over A interprets the word problem for
groups.
24
Proof. Let T = A/(1 + XH)A, Ti = A/(H − αi )A, i = 1, . . . , 5 where αi + k
is not a root of a(H) for any k ∈ ZZ and αi − αj ∈
/ ZZ for i ̸= j. We check
that the conditions of Proposition 6.1 hold.
The automorphism σ has only one cyclic orbit when it acts on the set
of maximal ideals of k[H] and this orbit is a singleton {H}. Indeed let
b(H) be an irreducible polynomial, deg(b) = l ≥ 1 and σ n (b) = αb for some
α ∈ k. Computing the leading coefficient we obtain α = q ln and then for
the constant term b0 = q ln b0 hence b0 = 0 and H | b(H).
We prove that the module T is simple. Clearly (1 + XH) is irreducible
in B = k[X, X −1 , σ] and l-normal. By [4, Th. 5.13] we need to show that
A = HA + [B ∩ (1 + XH)A] and this is the case since 1 ∈ HA + (1 + XH)A.
Now we obtain the desired conclusion by [4, L. 5.7].
Since XH ≡ −1 and (1 + XH)Y = Y + qHa(qH) so Y ≡ −qHa(qH)
in T it follows that every element u ∈ T has a canonical representative
u = b0 (H) + Xb1 + · · · + X n bn where bi ∈ k for i ≥ 1. For the following
calculations we need the fact that HX k = (HX)X k−1 = XHq −1 X k−1 =
(XH + 1 − 1)q −1 X k−1 ≡ −q −1 X k−1 , k ≥ 1 and X l H = X · X l−1 H =
XHq l−1 X l−1 ≡ −q l−1 X l−1 , l ≥ 1.
We prove that End(T ) ∼
= k. Let f : T → T be induced by left multiplication by u as above. Then u(1 + XH) = [b0 (H) − σ −1 (b0 (H))] + X(b1 −
qb1 ) + · · · + X n (bn − q n bn ) = 0, hence (q is not a root of unity) bi = 0 for
i ≥ 1 and b0 = σ(b0 ). If deg(b0 ) ≥ 1 then for the leading coefficient ck of b0
we obtain ck = q k ck , a contradiction.
The modules Ti are simple and nonisomorphic as in the proof of Proposition 6.2. For End(Ti ) ∼
= k we similarly obtain (see the proof of that Propom
sition) (b−m Y + · · · + b0 + · · · + bn X n )(H − αi ) = b−m (q −m αi − αi )Y m +
· · · (αi − αi )b0 + · · · + bn (q n αi − αi )X n hence bi = 0 for i ̸= 0.
We check that Ext(Ti , T ) ̸= 0. Consider the action of right multiplication
by (H − αi ) on T . We obtain (b0 (H) + Xb1 + · · · + X n−1 bn−1 + X n bn )(H −
αi ) = (see computations above) = [b0 (H)(H − αi ) − b1 ] + X(−αi b1 − qb2 ) +
X n−1 (−αi bn−1 − q n−1 bn ) + X n (−αi bn ). Since αi ̸= 0, supposing that 1 is in
the image of this action, we obtain bn = . . . = b1 = 0 and 1 = b0 (H)(H −αi ),
a contradiction. 2
The quantum Weyl algebra A1 (q), q ̸= 0, 1 (see [4, Sect. 6]) is a generalized Weyl algebra A where a(H) = H and σ(H) = q −1 (H − 1).
Corollary 6.5 The theory of A1 (q)-modules when 0 ̸= q is not a root of
unity and char(k) = 0 interprets the theory of k⟨X, Y ⟩-modules, and hence
interprets the word problem for groups.
25
Proof. Consider the change of variables H ′ = H − (1 − q)−1 . Then a(H ′ ) =
H ′ + (1 − q)−1 and σ(H ′ ) = q −1 H ′ . 2
Let k be a field with a derivation ′ and let R = k[Y,′ ] be the differential
polynomial ring. Then every element r ∈ R can be written in the form
r = r0 +Y r1 +. . .+Y n rn and the commutation law is given by αY = Y α+α′ ,
α ∈ k. By [11] every differential polynomial ring over a field of characteristic
zero is a simple principal ideal domain, hence a Dedekind domain. Put
k ′ = {α ∈ k | α′ = 0}. Clearly k ′ is a subfield of k.
Proposition 6.6 Let R = k[Y,′ ] be a differential polynomial ring with
char(k) = 0 and suppose that there are α1 , . . . , α5 ∈ k such that every differential operator Di = α′ − αi α is mono not onto and all differential operators
Dij = α′ − αij α , αij = αi − αj , for i ̸= j are mono. Then the theory of
k ′ ⟨X, Y ⟩-modules is interpretable in the theory of R-modules. In particular
the theory of modules over R interprets the word problem for groups.
Proof. We prove that the sextuple of modules T = R/Y R, Ti = R/(Y −
αi )R satisfies the conditions of Proposition 6.1. Since Y −αi is an irreducible
polynomial, all modules Ti are simple and the same true for T . Clearly Ti ∼
=
′
′
k with a module structure given by αY = Y α + α = (Y − αi )α + αi α + α =
αi α + α′ and correspondingly αY = α′ for T . We prove that End(Ti ) ∼
= k′ .
Every endomorphism f of Ti is a left multiplication by α ∈ k and because
f (Y −αi ) = 0 we have α(Y −αi ) = α′ = 0 and so α ∈ k ′ . Since the argument
reverses we obtain End(T ) ∼
= k′ .
Suppose that Ti ∼
= Tj for i ̸= j. The isomorphism g : Ti → Tj is induced
by left multiplication by α ∈ k hence, computing in Tj , 0 = α(Y − αi ) =
α′ + αj α − ααi = α′ − α(αi − αj ). Thus if g is nonzero then α ̸= 0 and α
is in the kernel of the differential operator Dij , a contradiction. Similarly
T ∼
̸ Ti since Di is mono.
=
We prove that Ext(Ti , T ) ̸= 0. Right multiplication by Y − αi defines
the map T → T with α(Y − αi ) = α′ − αi α = Di (α). Since this map is not
onto, we obtain the desired conclusion. 2
For example in the case of B1 we have that k = k1 (x) is a field of rational
functions. We can choose αi = i, i = 1, . . . , 5. Since a differential equation
f ′ − nf , 0 ̸= n ∈ ZZ can not be solved in k (there are no exponentials enx
in k) it follows that all operators Di , Dij are mono. They are not epi since
1/x is not in the image of these operators.
26
φ
•
θ1
θ1 + θ2
θ11 •
θ
•
• 21
•
• θ2
•
•
θ22
θ12 •
θ1 ∧ θ2
•
ψ
Figure 1:
7
Superdecomposable pure–injective modules
Let P be a subposet of a modular lattice. We say that P is wide if, given any
two distinct comparable points φ > ψ in P , there exist incomparable points
θ1 , θ2 ∈ P with φ > θ1 , θ2 > ψ and there exist θ11 , θ12 , θ21 , θ22 ∈ P such that
θ1 + θ2 ≥ θ11 > θ1 > θ12 ≥ θ1 ∧ θ2 and θ1 + θ2 ≥ θ21 > θ2 > θ22 ≥ θ1 ∧ θ2
(see Figure 1). Here the join + and the meet ∧ refer to the operations in the
modular lattice (if P is itself sublattice then the definition simplifies since
we may take θi1 = θ1 + θ2 and θi2 = θ1 ∧ θ2 ).
Theorem 7.1 [34, 7.8]Let R be any ring and L denote the lattice of ppformulas in one free variable over R.
1) If there is a superdecomposable pure–injective module over R then L
has a wide subposet.
2) If L is countable (in particular, if R is countable) and L contains a
wide poset then there is a superdecomposable pure–injective module over R.
Let A be any finitely presented module. An A-pointed module is a morphism f : A → M (or pair (M, f (m)) where m is a fixed generating tuple
for A). A (pointed) morphism from f : A → M to g : A → N is a morphism h : M → N such that hf = g. In the case A = R we refer just
27
to pointed modules. We define an equivalence relation on A-pointed modules by: f : A → M is equivalent to g : A → N iff there are morphisms
u : M → N and v : N → M such that uf = g and vg = f . It is immediate
from [25, 2.7, 8.5] that if m is a generating tuple of A then f and g above
are equivalent iff ppM (f (m)) = ppN (g(m)).
The set of equivalence classes of pointed finitely presented modules forms
a poset where f ≥ g iff there is u such that uf = g. It is also immediate
from [25, 2.7, 8.5, 8.15] that this poset is isomorphic to the lattice L.
A wide poset of A-pointed modules is a set of A-pointed modules such
that corresponding set P of equivalence classes is wide.
Proposition 7.2 The following conditions for any ring R are equivalent.
1) There is wide poset of pp-formulas in one free variable over R.
2) There is a wide poset of pointed finitely presented modules.
3) There is a wide poset of A-pointed finitely presented modules for some
nonzero finitely presented module A.
Proof. This follows from the above discussion, together with the observation that A-pointed modules yield Rn -pointed modules and the fact (which
follows directly from [25, 10.8]) that the existence of a wide poset of Rn pointed modules implies the existence of a wide poset of R-pointed modules.
2
Lemma 7.3 Let R be any ring and let C be a full subcategory of the category
of R-modules. Suppose that F is a full embedding from C to M od − S. If
P is a wide poset of A-pointed modules in C, then F (P ) is a wide poset of
F A-pointed modules in M od − S.
Proof. By F (P ) we mean the collection of F A-pointed modules F A → F M ,
where A → M is in P . The result follows directly from the definitions using
the fact that F is full. 2
If k is any field, we let f d − k⟨X, Y ⟩ denote the category of finite–
dimensional k⟨X, Y ⟩-modules (regarded as a full subcategory of M od −
k⟨X, Y ⟩).
Proposition 7.4 There is a wide poset of pointed modules in f d − k⟨X, Y ⟩.
Proof. One may produce such a system explicitly but we may argue indirectly as follows. First, there is a superdecomposable pure–injective module
over k⟨X, Y ⟩ — for instance, the injective hull of k⟨X, Y ⟩ [18]. So by 7.1
28
•
•
Figure 2:
there is a wide poset of pp-formulas in L. Next, recall that the category
of k⟨X, Y ⟩-modules is interpreted in the category of modules over the path
algebra R′ of the quiver shown in Figure 2 — given a k⟨X, Y ⟩-module M
form the representation of this quiver which has M (as a k-vectorspace) at
both vertices and has the arrows being, respectively, the identity map of M ,
multiplication by X, and by Y (with respect to the identification). Therefore there exists a wide subposet of the lattice of pp-formulas for R′ -modules
and hence, by Proposition 7.2, a wide poset of finitely presented (= finite
dimensional) R′ -modules. Next by [6, Thm. 3] there is a full embedding F
of the category f d − R′ into f d − k⟨X, Y ⟩. By Lemma 7.3 F carries the wide
poset of pointed modules in f d − R′ into a wide poset of A-pointed modules
in f d − k⟨X, Y ⟩ where A is some module in f d − k⟨X, Y ⟩. 2
Lemma 7.5 Let R be a right noetherian ring satisfying the conditions of
Section 2.1 [20] (that is, R is a k-algebra over a field k which has a sextuple of
simple nonisomorphic modules T, T1 , . . . , T5 such that End(T ) ∼
= End(Ti ) ∼
=
k and Ext(Ti , T ) ̸= 0 for every i). Let F denote the construction of [20]
and let A be a finite–dimensional k⟨X, Y ⟩-module. Then the image under F
of any wide poset of A-pointed modules in f d − k⟨X, Y ⟩ is a wide poset of
F A-pointed modules in mod − R.
Proof. Let P be a wide poset of A-pointed finite–dimensional k⟨X, Y ⟩modules. For each A → M in P , we have M finite dimensional and hence,
by [20, 2.12], F M is of finite length and hence finitely presented. Thus we
obtain a poset of F A-pointed modules in mod − R. It will be enough to
show that [A → M ] ≥ [A → N ] iff [F A → F M ] ≥ [F A → F N ]. Certainly,
if we have [α : A → M ] ≥ [β : A → N ], say f : M → N with f α = β,
then by the construction of [20, 2.11.4] there is a morphism g in mod − R
with gF (α) = F (β) and hence [F A → F M ] ≥ [F A → F N ]. Conversely
if [F A → F M ] ≥ [F A → F N ] then there is a morphism g in mod − R
with gF (α) = F (β). By [20, 2.12] there is f in f d − k⟨X, Y ⟩ with g = F f .
29
Without loss of generality we may suppose (see the construction, 2.11, of
[20]) that F (f )F (α) = F (f α) and hence by [20, 2.12], that f α = β and
hence that [A → M ] ≥ [A → N ] as required. 2
Corollary 7.6 If R is as in Lemma 7.5 and is countable then there is a superdecomposable pure–injective R-module and there are 2ℵ0 indecomposable
pure–injective modules over R. In particular if the field k is countable then
this is true for A1 (k), B1 (k), A1 (q) and U sl2 (k).
Proof. The existence of a superdecomposable pure-injective is by Proposition 7.2 and Theorem 7.1. Then the existence of 2ℵ0 indecomposable pureinjectives follows by results of Garavaglia and Ziegler (see [25, 10.10, 10.22]).
For the last part apply Corollary 6.3 and Corollary 6.5. 2
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