Download PURE–INJECTIVE AND FINITE LENGTH MODULES OVER

PURE–INJECTIVE AND FINITE LENGTH MODULES
OVER CERTAIN RINGS OF DIFFERENTIAL
POLYNOMIALS
GENNADI PUNINSKI
Abstract. Let F be a field of characteristic zero with a derivation
′
′
such that F ⊂ F , the field of constants of F is algebraically closed,
and R = F [Y,′ ] the ring of differential polynomials over F . Assuming
that the category of finite length modules over R admits almost split
sequences, we classify the finite length and pure-injective modules over
R. In particular, every indecomposable finite length module over R is
uniserial with isomorphic composition factors and the Cantor–Bendixson
rank of the Ziegler spectrum over R is equal to 2. For instance, if k is an
algebraically closed field of characteristic zero, F = k((x)) the Laurent
series field with x′ = 1 or x′ = x, then the ring R = F [Y,′ ], which is the
ring of differential operators over k((x)), is of this kind.
1. Introduction
Let F be a field of characteristic zero with a derivation ′ and R = F [Y,′ ] a
differential polynomial ring, i.e. the ring of left polynomials r = α0 + α1 Y +
· · · + αn Y n , αi ∈ F with Y · α = αY + α′ for α ∈ F . Since α · Y = Y α − α′ ,
hence every r ∈ R can be written as a right polynomial by Y as well.
It is well known (see [5, Ex. 7.31.1]), that R posseses of a left and right
division algorithm, hence R is a left and right principal ideal domain. If
the derivation is nontrivial, then by [5, Thm. 7.28] R is simple, hence by
[2, Ch. 8, Cor. to Thm. 1.1], every finite length module over R can be
represented as R/aR, 0 ̸= a ∈ R. Simple modules over R are of the form
R/aR for an irreducible 0 ̸= a ∈ R, and there is a well known criterion for
the isomorphism of simples (see [2, Ch. 3, Thm. 3.2]).
However, the classification problem even for finite length modules over
R could be wild. For instance, let k be a field of characteristic zero and
F = k(x) the field of rational functions over k with the usual derivation.
Then F [Y,′ ] = B1 (k) is a partial quotient ring of the first Weyl algebra
A1 (k). The technique of the paper by Klinger and Levy [8] could be applied
to B1 (k) showing that for every finite dimensional module over k⟨x, y⟩ there
1
is a finite length module over B1 (k) with the same endomorphism ring. By
making use of this construction Prest and Puninski have proved [12] that
the Cantor–Bendixson rank of the Ziegler spectrum ZgR over R = B1 (k) is
undefined and there is no hope to classify even indecomposable pure-injective
modules over B1 (k). So, this case can be considered as “wild”.
For a contrast, let F be a universal differential field. In this case there is
only one indecomposable finite length module — the trivial module R/Y R ∼
=
F (see [5, Thm. 7.40]). It has been remarked in [12] that there is only 3
isomorphism types of indecomposable pure-injective modules over R : 1)
R/Y R ; 2) the injective envelope E(RR ) and 3) an indecomposable direct
summand of the pure-injective envelope PE(RR ) (all these summands are
isomorphic). For instance, the CB-rank of ZgR is equal to 1 and we are
dealing with “finite type” case.
Let F be a field of characteristic zero with a derivation ′ such that F ′ ⊂ F
and whose field of constants k = {α ∈ F | α′ = 0} is algebraically closed.
We also suppose that the category of finite length modules over R = F [Y,′ ]
admits almost split sequences. Then we classify indecomposable finite length
modules over R showing that each such a module is homogeneous uniserial,
i.e. has the unique composition series with isomorphic composition factors.
Thus, every indecomposable finite length module over R is uniquely determined by its (simple) socle and its length. Moreover, there is no nontrivial
extensions and morphisms between such modules with different socles, so
the whole category of finite length modules over R decomposes into a direct
sum of uniserial categories of global dimension 1. The last assertion is also
in accordance with Dlab and Ringel [3, L. 4.2], since we shall check that
the k-dimensions of Ext(M, N ) and Hom(M, N ) for arbitrary finite length
modules over R are finite and coincide.
However the category of finite length modules over R seems to be similar
to that over integers, we could not find a classical localization (i.e. Ore set)
which corresponds to the given simple module.
So, intuitively that is the case of “tame” representation type. Confirming
this point of view we will give the complete classification of indecomposable
pure-injective modules over R. In fact we will show that this description
looks like very similar to the classification of pure-injective (algebraically
compact) abelian groups: beyond the pure–injective envelopes of indecomposable finite length modules, for every simple module there is a corresponding “prüfer” and “adic” pure-injective module, and also E(RR ) is an
2
analog of rationals. In particular, similarly to abelian groups, the Cantor–
Bendixson rank of the Ziegler spectrum over R is equal to 2.
Precisely we prove that the isolated points in ZgR coincide with the pureinjective hulls of indecomposable finite length modules over R and the points
of CB rank 1 are indecomposable injective torsion modules and dual to them
torsionfree modules. The unique point of CB-rank 2 is the module E(RR ).
Van den Essen and Levelt [4] in particular have described simple modules
over R = F [Y,′ ] for F = k((x)) (there are infinite many simples in every
finite F -dimension). Moreover, Zimmermann [14] proved that the category
of finite length modules over R admits almost split sequences.
The main technical ingredient for our proves is the existence of almost
split sequences in the category of finite length modules over certain differential polynomial rings as it has been shown by Zimmermann [14], [15].
We will calculate these sequences explicitly and then use them for parallel
classification of finite length and pure-injective modules by standard arguments of model theory of modules. However we extracted out whenever it
was possible the special terminology, the spirit of Model Theory of Modules
is dominating throughout. For instance, we prove that every finite length
module over R has pseudo (Lascar) U-rank 1.
Among others we calculate that the Ext-graph for simple modules over
R is a disjunct union of loops. Note that the structure of this graph over
a given k-algebra is extremely important for estimating the possibility to
classify indecomposable finite length modules over it. For example by Prest
and Puninski [12] the presence of “five finger” configuration in the Ext-graph
of simples over a k-algebra R demolishes any attempt to classify finite length
modules over R, since we can interpret in this category the theory of finite
length modules over arbitrary finite dimensional k-algebra.
The general conjecture which we will formulate is that the CB-rank of the
Ziegler spectrum over a differential polynomial ring could be 1, 2 or infinite
according to “finite type”, “tame” and “wild” case and depends on the
model theoretic properties of the trivial module. Note that this trichotomy
is known for hereditary finite dimensional algebras (see [11]).
Note that for the case F = k((x)), the ring R = F [Y,′ ] coincides with
the ring of differential operators on k((x)), and the structure of simples over
this ring plays the crucial role in the classification of simples over ring of
differential operators O[∂] (O = k[[x]]) in van den Essen and Levelt [4]. Also
the categories of finite length modules over a rings of differential polynomials
3
appeared naturally along the investigation of finite type representations of
k-species in Dlab and Ringel [3].
2. Basic notions
A ring R is called a principal ideal domain if R does not contain a zero
divisor and every right (left) ideal of R can be written in the form rR (Rr)
for some r ∈ R. A ring R is called hereditary if every nonzero right (left)
ideal of R is a projective R-module. A nonzero element r ∈ R is called
irreducible if r = st for s, t ∈ R implies that either s or t is invertible.
A module M is uniserial if the lattice of submodules of M is a chain; and
a ring R is uniserial if RR and
RR
are uniserial modules. A commutative
uniserial domain is called a commutative valuation domain. For a module
M by E(M ) (PE(M )) will be denoted its injective (pure-injective) envelope.
For an element m of a module M let ann(m)(R) be a right ideal {r ∈ R |
mr = 0}.
f
g
A nonsplit short exact sequence 0 → A → B → C → 0 of modules
is called AR -sequence if 1) every morphism A → M , which is not a split
monomorphism, factors through f ; 2) every morphism B → C, which is
not a split epimorphism, factors through g. The basic properties of AR sequences can be found in [1].
For the background in model theory of modules, the reader is referred to
[10]. Some more discussion about the Ziegler spectrum and the isolateness
notion in there can be found in [12]. Additionally, let R be a k-algebra over
a field k and M be an R-module. Then M is called pdeudo strongly minimal
if for every pp-formula φ , a k-space φ(M ) is either finite dimensional or
φ(M ) = M . M has pseudo U-rank 1, if for every pp-formula φ, a k-space
φ(M ) is either finite dimensional or cofinite dimensional.
3. Tensors
We shall consider the rings of differential polynomials R = F [Y,′ ] preserving all the above restrictions. So F is a field of characteristic zero with
a derivation ′ , whose field of constants k = {α ∈ F | α′ = 0} is algebraically closed. We shall also suppose that F ′ ⊂ F , and the category
of finite length modules over R admits AR -sequences. By Zimmermann
[14], [15] for F = k((x)) with a derivation given by x′ = 1 or x′ = x, R
is of this sort. Moreover, by [15, Cor. 18] there is ω ∈ F \ F ′ such that
F = F ′ ⊕ k · ω, in particular dim(F/F ′ ) = 1. In fact, we can take ω = 1
4
in any case. Indeed, consider a new indeterminant Y1 = Y ω −1 and anew
derivation d(α) = α′ · ω −1 . Then
αY1 = αY ω −1 = Y α ω −1 − α′ ω −1 = Y1 α − d(α) .
Thus, R[Y,′ ] = R[Y1 ,′ ] and F = d(F ) ⊕ k · 1.
According to Zimmermann [15] for (finite length) modules M, N over R
define (M ⊗F N )R as a vector space M ⊗F N with a right R-module structure
given by (m ⊗ n) · Y = mY ⊗ n + m ⊗ nY .
Lemma 3.1. For every a ∈ R
(m ⊗ n) · a =
∞
∑
mY i ⊗ n ·
i=0
a(i)
,
i!
where ′ is the usual derivative by Y and we set a(0) = a, 0! = 1. In particular,
if mY = 0, then
(m ⊗ n) · a = m ⊗ na ,
and if mY 2 = 0, then
(m ⊗ n) · a = m ⊗ na + mY ⊗ na′ .
Proof. We may assume a = Y p and use the easy induction on p.
Let F be the trivial module R/Y R, where the right R-module structure
is given by α · Y = Y α − α′ ≡ −α′ for α ∈ F . Then there is a canonical
isomorphism M ∼
= F ⊗ M , where m → 1 ⊗ m. In view of [15, Cor. 17, 18]
there is the AR -sequence
f
g
0→F ∼
= R/Y R → R/Y 2 R → R/Y R ∼
= F → 0,
(1)
where f is the left multiplicaton by Y and g the canonical projection. Moreover, every indecomposable finite length module M is the source of AR sequence
f
g
0→F ⊗M ∼
= M → R/Y 2 R ⊗ M → M ∼
= F ⊗ M → 0,
(2)
where f and g are induced from (1), for instance f (m) = Y ⊗ m.
Corollary 3.2. The isolated points in ZgR are exactly PE(M ), where M is
an indecomposable finite length module.
5
Proof. Since for every indecomposable finite length M , there is an AR sequence starting from M , every module PE(M ) is isolated by a minimal
pair in ZgR by [12, Prop. 3.7]. For converse by [12, L. 3.6] it suffices to
prove that there is no injective simple module over R. But this is also
follows from the existence of a nonsplit sequence with a source in the given
simple module.
For 0 ̸= a ∈ R let Mk (a) be a finitely presented module over R with k + 1
generators x0 , . . . , xk and relations
s
∑
i=0
xi
a(s−i)
= 0,
(s − i)!
0 ≤ s ≤ k,
where we put a(0) = a and 0! = 1. In particular, M0 = R/aR with respect
to identification x0 = 1. The following is almost evident.
Remark 3.3. Every element m ∈ Mk can be uniquely repesented in the
∑
form ki=0 xi bi , where deg(bi ) < deg(a) for every i. Moreover, Mk ⊆ Mk+1
with respect to identification xi → xi , i = 0, . . . , k.
Corollary 3.4. Mk /x0 R ∼
= Mk−1 via the morphism xi → xi−1 , i = 1, . . . , k.
Proof. Mk /x0 R has generators w0 = x1 , . . . , wk−1 = xk . We prove that
s
∑
wj
j=0
a(s−j)
=0
(s − j)!
for
0≤s≤k−1
Since s + 1 ≤ k we have
s+1
∑
xi
i=0
a(s+1−i)
=0
(s + 1 − i)!
which is the same as
x0
∑
∑
a(s+1)
a(s−(i−1))
a(s+1)
a(s−j)
+
= x0
+
xi
wj
(s + 1)!
(s − (i − 1))!
(s + 1)!
(s − j)!
s+1
s
i=1
j=0
Since the first term is in x0 R, we obtain the desired.
Suppose that there is another relation
k−1
∑
wj bj = 0
j=0
where one may assume that deg(bj ) < deg(a). Therefore in Mk
k
∑
xl bl−1 = x0 b
l=1
where deg(bj ) < deg(a), hence bj = b = 0 for all j by Remark 3.3.
6
Our next aim is to calculate the AR -sequence starting from Mk .
In the next definition we formally put x−1 = xk+1 = 0. Let yi = Y ⊗ xi ,
zi = 1⊗xi for 0 ≤ i ≤ k be elements of the module R/Y 2 R⊗Mk . Extending
this we put z−1 = yk+1 = 0.
Let ui = −zi−1 + (k + 1 − i)yi , where i = 0, . . . , k + 1. In particular
u0 = (k + 1)y0 , u1 = −z0 + ky1 and uk+1 = −zk .
Lemma 3.5. The submodule of R/Y 2 R ⊗ Mk generated by u0 , . . . , uk+1 is
finitely presented with the following relations:
s
∑
i=0
ui
a(s−i)
= 0,
(s − i)!
0 ≤ s ≤ k + 1.
Therefore this submodule is isomorphic to Mk+1 with respect to identification
u i = xi .
Proof. For s = 0 we get (see Lemma 3.1)
u0 a = (k + 1) y0 a = (k + 1) (Y ⊗ x0 )a = (k + 1) Y ⊗ x0 a = 0 .
Let us consider the case 1 ≤ s ≤ k + 1.
s
∑
i=0
∑
∑
a(s−i)
a(s−j)
a(s−i)
=−
(1 ⊗ xi−1 )
+
Y ⊗ (k + 1 − j) xj
=
ui
(s − i)!
(s − i)!
(s − j)!
= −1⊗
s
∑
xi−1
i=1
s
s
i=1
j=0
∑
∑
a(s−i)
a(s+1−i)
a(s−j)
−Y ⊗
xi−1
+Y ⊗
(k+1−j) xj
(s − i)!
(s − i)!
(s − j)!
s
s
i=1
j=0
The first member of the sum is
−1 ⊗
s−1
∑
l=0
xl
a(s−l−1)
(s − l − 1)!
which is zero in view of relations in Mk since s − 1 ≤ k. The remaining part
of the sum is
[
s−1
∑
Y ⊗
xl −
l=0
=Y ⊗
s−1
∑
l=0
]
a(s−l)
a(s−l)
+ (k + 1 − l)
+ Y ⊗ (k + 1 − s) xs a =
(s − l − 1)!
(s − l)!
xl
a(s−l)
[−(s − l) + k + 1 − l] + Y ⊗ (k + 1 − s) xs a =
(s − l)!
= Y ⊗ (k + 1 − s)
s
∑
l=0
xl
a(s−l)
.
(s − l)!
If s ≤ k the sum is zero in view of relations in Mk . Otherwise s = k + 1 and
the sum is zero because k + 1 − s = 0.
7
Let us prove that there is no other relations between ui -th. Suppose that
∑k+1
i=0 ui bi = 0 for bi ∈ R. In view of above proved relations we may assume
deg(bi ) < deg(a) for every i. So we get
k+1
∑
u i bi = −
i=0
k+1
∑
(1 ⊗ xi−1 ) bi +
i=1
= −1 ⊗
k+1
∑
k
∑
((k + 1 − j) Y ⊗ xj ) bj =
j=0
xi−1 bi − Y ⊗
k+1
∑
i=1
xi−1 b′i
+ Y ⊗ (k + 1 − j)
i=1
Taking into consideration the member 1 ⊗ − we get
∑k+1
i=1
k
∑
xj bj
j=0
xi−1 bi = 0 in Mk ,
hence by Remark 3.3, b1 = · · · = bk+1 = 0 and the original sum is reduced
to u0 b0 = 0. Since u0 b0 = (k + 1)Y ⊗ x0 b0 therefore x0 b0 = 0 in Mk and
b0 = 0.
Let us define now vi = zi + (i + 1)yi+1 , i = 0, . . . , k − 1.
Lemma 3.6. The submodule of R/Y 2 R ⊗ Mk generated by v0 , . . . , vk−1 is
finitely presented with the following relations:
s
∑
vi
i=0
a(s−i)
= 0,
(s − i)!
0 ≤ s ≤ k − 1.
Therefore this module is isomorphic to Mk−1 with respect to identification
vi = xi .
Proof. For s = 0 we have
v0 a = z0 a + y1 a = (1 ⊗ x0 )a + (Y ⊗ x1 ) a =
= 1 ⊗ x0 a + Y ⊗ x0 a′ + Y ⊗ x1 a = 1 ⊗ x0 a + Y ⊗ (x0 a′ + x1 a) .
Since x0 a = 0 and x0 a′ + x1 a = 0 in Mk , we obtain the desired.
Suppose that 1 ≤ s ≤ k − 1. Then
s
∑
∑
∑
a(s−i)
a(s−i)
a(s−j)
=
+
=
(1 ⊗ xi )
(Y ⊗ (j + 1) xj+1 )
(s − i)!
(s − i)!
(s − j)!
s
vi
i=0
=1⊗
s
i=0
s
∑
i=0
j=0
a(s−i)
xi
+Y ⊗
(s − i)!
s
∑
i=0
∑
a(s+1−i)
a(s−j)
xi
+Y ⊗
(j + 1) xj+1
(s − i)!
(s − j)!
s
j=0
The first term in the sum is zero in view of the relations in Mk . The
remaining part is
[
]
s
∑
a(s+1)
a(s+1−l)
a(s+1−l)
Y ⊗ x0
+Y ⊗
xl
+l
+ Y ⊗ (s + 1) xs+1 a =
s!
(s − l)!
(s + 1 − l)!
l=1
∑
a(s+1)
a(s+1−l)
= (s+1)Y ⊗x0
+Y ⊗
xl
[s+1−l+l]+(s+1)Y ⊗xs+1 a =
(s + 1)!
(s + 1 − l)!
s
l=1
8
= (s + 1)Y ⊗
s+1
∑
xl
l=0
a(s+1−l)
(s + 1 − l)!
which is zero since s + 1 ≤ k.
Prove that there is no other relations between vi -th. Let
∑k−1
i=0
vi ci = 0
and we may assume that deg(ci ) < deg(a) for every i. Then
k−1
∑
vi ci =
i=0
=1⊗
k−1
∑
k−1
∑
(1 ⊗ xi + (i + 1)Y ⊗ xi+1 ) ci =
i=0
xi ci + Y ⊗
i=0
k−1
∑
xi c′i + Y ⊗
i=0
k−1
∑
(j + 1) xj+1 cj .
j=0
Considering the member 1 ⊗ − we have
∑k−1
i=0
xi ci = 0 in Mk , hence c0 =
· · · = ck−1 = 0.
We shall identify in the next Lemma the submodule of R/Y 2 R ⊗ Mk
generated by u0 , . . . , uk+1 with Mk+1 and that generated by v0 , . . . , vk−1
with Mk−1 .
Lemma 3.7. R/Y 2 R ⊗ Mk = Mk−1 ⊕ Mk+1 .
Proof. Since yi a = Y ⊗ xi a,
zi b − yi b′ = 1 ⊗ xi b + Y ⊗ xi b′ − Y ⊗ xi b′ = 1 ⊗ xi b ,
and x0 , . . . , xk generate Mk , hence y0 , . . . , yk ; z0 , . . . , zk generate R/Y 2 R ⊗
Mk .
We prove that u0 , . . . , uk+1 , v0 , . . . , vk−1 generate R/Y 2 R ⊗ Mk . For
i = 0, . . . , k − 1 we have ui+1 = −zi + (k − i)yi+1 , vi = zi + (i + 1)yi+1
hence ui+1 + vi = (k + 1)yi+1 and yi+1 =
ui+1 +vi
k+1 .
Therefore y1 , . . . , yk
are contained in the submodule generated by ui -th and vj -th. Since zi =
vi − (i + 1)yi+1 the same is true for z0 , . . . , zk−1 . The rest follows from the
equalities u0 = (k + 1)y0 and uk+1 = −zk .
Thus, it remains only to prove that Mk+1 ∩ Mk−1 = 0. Let
∑k−1
j=0
∑k+1
i=0
u i bi =
vj cj for some bj , cj ∈ R, where we may assume that deg(bi ), deg(cj ) <
deg(a). The direct calculation yields
−1 ⊗
k+1
∑
xi−1 bi + Y ⊗
i=1
=1⊗
k−1
∑
l=0
xl cl + Y ⊗
k
∑
xj [(k + 1 − j) bj − b′j+1 ] =
j=0
x0 c′0
+Y ⊗
k−1
∑
t=1
9
xt [c′t + tct−1 ] + Y ⊗ kxk ck−1 .
Considering the summands of the form 1 ⊗ − we obtain −
∑k−1
l=0
∑k+1
i=1
xi−1 bi =
xl cl in Mk , which in view of relations in Mk implies −b1 = c0 , . . . ,
−bk = ck−1 and bk+1 = 0. Taking into account the members Y ⊗ − we have
(k + 1) b0 − b′1 = c′0 ,
(k + 1 − i) bi − b′i+1 = c′i + ici−1 ,
i = 1, . . . , k − 1
and bk − b′k+1 = kck−1 . Therefore
(k + 1 − i) bi = i ci−1 = −i bi ,
hence bi = 0 for i = 1, . . . , k − 1. Similarly the first and the last equalities
yield b0 = 0 and bk = 0.
Proposition 3.8. Suppose that Mk is an indecomposable module. Then the
almost split sequence starting from Mk is
g
f
0 → Mk → Mk−1 ⊕ Mk+1 → Mk → 0 ,
where f (xi ) =
vi−1 +ui
k+1
for i = 0, . . . , k and g(vi ) = xi , i = 0, . . . , k ; g(uj ) =
−xj−1 , j = 0, . . . , k + 1 (see Corollary 3.4 and Remark 3.3).
Proof. Since Mk is indecomposable, the middle term in the almost split
sequence starting from Mk is R/Y 2 R ⊗ Mk = Mk−1 ⊕ Mk+1 and
f (xi ) = Y ⊗ xi = yi =
Since
ui + vi−1
,
k+1
i = 0, . . . , k .
(
)
vi−1 + ui
xi−1 − xi−1
gf (xi ) = g
=
= 0,
k+1
k+1
hence the image of f is contained in the kernel of g. Let us check the inverse
inclusion. Suppose that


k+1
k−1
∑
∑
u i bi +
vj cj  = 0
g
i=0
j=0
But this is equal to
−
k+1
∑
xi−1 bi +
i=1
k−1
∑
xj cj =
j=0
k−1
∑
xl (−bl+1 + cl ) − xk bk+1 .
l=0
Therefore bk+1 = 0 and bl+1 = cl for l = 0, . . . , k − 1. Thus
k+1
∑
i=0
u i bi +
k−1
∑
j=0
vj cj =
k
∑
i=0
u i bi +
k−1
∑
i=0
vj bj+1 =
k
∑
(ul + vl−1 )bl + u0 b0
l=1
The former summand is in the image of f so as the latter.
10
4. Corollaries
The following is an immediate consequence of Proposition 3.8.
Corollary 4.1. Let M = R/aR, 0 ̸= a ∈ R be an indecomposable module.
Then the module N = R/Y 2 R ⊗ M is finitely presented with generators
x0 = Y ⊗ 1 (the image of 1 in N ), x1 = −1 ⊗ 1 and relations x0 a = 0,
x0 a′ + x1 a = 0. Moreover, M is included in the AR -sequence (2).
For instance, if a = Y , then N = R/Y 2 R. Note that we could not
quarantee that N is an indecomposable module. Of course that is true if
M is simple, since N is a nonsplit extension of a simple by simple, hence
uniserial. Indeed, it can be easily calculated, using the Proposition 3.8 that
for M = R/Y 2 R we have N = R/Y R ⊕ R/Y 3 R.
Corollary 4.2. Let R/aR be an indecomposable module. Then a′ ∈
/ Ra+aR.
Proof. Suppose by way a contradiction that a′ ∈ Ra+aR, hence a′ = sa+at
for s, t ∈ R. By Corollary 4.1, N = R/Y 2 R ⊗ R/aR is generated by x0 , x1
with relations x0 a = 0, x0 a′ + x1 a = 0. Consider the map from N to R/aR
given by x0 → x0 , x1 → −x0 s. Since
x0 a′ + x1 a → x0 a′ − x0 sa = x0 (a′ − sa) = x0 at = 0 ,
this map generates the homomorphism f : N → R/aR which splits the
inclusion R/aR ⊆ N , a contradiction.
Lemma 4.3. Let R/aR, 0 ̸= a ∈ R be an indecomposable module, b, c ∈ R
and c ∈
/ aR + Rb. Then there are d, e, g, h ∈ R such that deg(e) < deg(b),
deg(h) < deg(a), e, h ̸= 0 and
a′ + da = ag + ch ,
ea = bh
(3)
Proof. Let b acts by the right multiplication on the module R/aR, hence c
is not in the image of this map. Consider the module M with generators
y0 , y1 and relations y0 a = 0 , y0 c + y1 b = 0. Let us check that the map
R/aR → M , where 1 → y0 is not a split monomorphism. Otherwise there
is g : M → R/aR such that g(y0 ) = 1 and g(y1 ) = d. Then g(y0 c + y1 b) =
c + db = 0 in R/aR, hence c + db ∈ aR, a contradiction.
Since R/aR → N = R/Y 2 R ⊗ R/aR is the left AR -morphism, hence
there is a homomorphism h : N → M such that the corresponding triangle
is commutative. By Corollary 4.1, N has generators x0 , x1 and relations
11
x0 a = 0, x0 a′ + x1 a = 0. Clearly h(x0 ) = y0 and let h(x1 ) = y0 d + y1 e.
Then
h(x0 a′ + x1 a) = y0 a′ + (y0 d + y1 e) a = y0 (a′ + da) + y1 ea
which is zero in M . Hence the last element is a linear combination of y0 a = 0
(with a coefficient g) and y0 c + y1 b = 0 (with a coefficient h) which is the
same as (3). Clearly we can suppose deg(h) < deg(a). If h = 0 then
a′ = ag − da ∈ aR + Ra, a contradiction. Otherwise e ̸= 0 and deg(e) <
deg(b).
Corollary 4.4. Let R/aR, 0 ̸= a ∈ R be an indecomposable module and
b ∈ R. Then either b acts by right multiplication as an epimorphism of
R/aR or the action on R/Ra by the left multiplication by b has a kernel.
Proof. If c is not in the image of right multiplication by b, then ea = bh by
Lemma 4.3 and h ̸= 0 in R/Ra.
Lemma 4.5. Let R/aR, R/bR be simple modules. Then exactly one of the
following holds true:
1) R/bR ∼
̸ R/aR and b acts by right multiplication as an isomorphism
=
of R/aR ;
2) R/bR ∼
= R/aR, b has a kernel acting by right multiplication on R/aR
and the image of this map is of k-codimension 1.
Proof. Suppose that R/bR ∼
̸ R/aR. If b has a kernel acting by right multi=
plication on R/aR, then ub = av for some u ∈
/ aR, v ∈ R and left multiplication by u defines the nonzero homomorphism R/bR → R/aR, a contradiction. Let c be not in the image of this map. By Lemma 4.3 (see (3)) we
also obtain the isomorphism R/aR → R/bR by the left multiplication by e.
Suppose that R/bR ∼
= R/aR and this isomorphism is given by left multiplication by u. Thus ub = av for u ∈
/ aR, v ∈ R hence u is in the kernel
of this map. For remaining it is enough to prove that Ext(R/aR, R/aR) is
one-dimensional over k. Since a′ ∈
/ aR + Ra by Lemma 4.2, hence this dimension is at least one. Let c be not in the image of the right multiplication
by a. By Lemma 4.3 we obtain
a′ + da = ag + ch ,
ea = ah ,
hence e induces the endomorphism of the simple module R/aR. Since by [15,
L. 8] End(R/aR) is finite dimensional over k and k is algebraically closed,
hence End(R/aR) = k. Therefore 0 ̸= e = h = α ∈ k and c ∈ a′ k + aR + Ra
which is the desired.
12
Corollary 4.6. Ext-graph for simple modules over R is a disjunct union of
loops.
Example 4.7. Let F = k((x)) with the derivation x′ = x and R = R[Y,′ ].
Then the Ext-graph for simple modules over R consists of infinitely many
loops.
1
1
◦
R/aR
...
◦
R/bR
Proof. By [15] and Lemma 4.5 it is enough to prove that there are infinitely
many simple modules over R. But for α1 , α2 , . . . being different in k/Z,
Mi = R/(Y −αi )R are non-isomorphic simple one-dimensional modules.
Lemma 4.8. Let 0 ̸= a ∈ R be irreducible. Then Hom(R/aR, Mk (a)) =
x0 · k is one–dimensional over k. Moreover, if 0 ̸= b ∈ R is irreducible and
R/bR ∼
̸ R/aR, then Hom(R/bR, Mk (a)) = 0.
=
Proof. Let m =
∑k
i=0 xi bi
∈ Mk be such that ma = 0, where deg(bi ) <
deg(a) for every i. In the free module we have
( s
)
k
k
∑
∑
∑
a(s−m)
xm
xi bi a =
gs
(s − m)!
i=0
s=0
m=0
For the coefficient by xk we obtain bk a = agk hence bk = gk = α ∈ k =
End(R/aR). For xk−1 it follows that bk−1 a = a′ gk +agk−1 , hence by Lemma
4.2, gk = bk = 0, m ∈ Mk−1 and we could follow by induction.
If b ∈ R is irreducible then similarly bk b = agk implies bk = gk = 0.
Proposition 4.9. If 0 ̸= a ∈ R is irreducible, then Mk (a) is a uniserial
module with the isomorphic composition factors.
Proof. By Lemma 4.8 it follows that Soc(Mk ) = x0 R is simple. Moreover,
by Remark 3.3, Mk / Soc(Mk ) = Mk /x0 R = Mk−1 , hence the result follows
by induction.
Lemma 4.10. Let M = R/aR, a ∈ R be a simple module of F -dimension
one. Then M = M a ⊕ ann(a)(M ) as a k-space. In particular M a = M a2 .
Proof. By Lemma 4.2 we have a′ ∈
/ M a. Since deg(a′ ) = 0, therefore 1 =
x0 ∈
/ M a and M a is of codimension 1 in the k-space M . Moreover by
Lemma 4.8, ann(a)(M ) = 1 · k. Thus M = M a ⊕ ann(a)(M ). The last
equality yields M a = M a2 .
13
We will show that this is not true even for F -dimension 2.
Example 4.11. Let F = k((x)) with a derivation x′ = x and a = Y 2 +
1/x. Then a is irreducible, and for the simple module M = R/aR we have
ann(a)(M ) ⊆ M a, in particular M a2 ⊂ M a.
Proof. Letting a being reducible, we obtain Y 2 + 1/x = (Y + α)(Y + β)
for α, β ∈ F . Then α + β = 0, hence β = −α and 1/x = −α′ − α2 . If
deg(α) ≥ 0, then 1/x is not reduced in the last equality. For deg(α) ≤ −1
we can not reduce the leading coefficient of α2 .
So, M = R/aR is a simple module, which is 2-dimensional over F . It can
be easily calculated that the action on M by right multiplication by a is
Y α + β → Y (α′′ − 2β ′ ) + (2α′ /x − α/x + β ′′ ) .
Clearly M a consists of Y γ + δ ; γ, δ ∈ F , where the constant term of γ is
zero. In particular, 1 ∈ M a, hence ann(a)(M ) ⊆ M a and M a2 ⊂ M a.
Question 4.12. Is Fitting’s lemma true for simple modules over R, i.e.
does a k-decomposition M = ann(an )(M ) ⊕ M an take place for some n?
The proof of the following claim consists of some matrix calculations.
Lemma 4.13. Let 0 ̸= a ∈ R be irreducible. Then End(Mk (a)) is isomorphic to the ring








α0 α1 . . .
0
..
.
α0 . . .
.. . .
.
.
0
0
...
αk
αk−1
..
.



,


α0
αi ∈ k. In particular, End(Mk ) is an artinian commutative valuation ring.
Proof. Induction by k. For k = 0 we have End(M0 ) = k which is the
desired. For general case if f ∈ End(Mk ) then taking into consideration
the composition series for Mk , we get f (Mi ) ⊆ Mi for every i. Therefore f (Mk−1 ) ⊆ Mk−1 and f induces the endomorphism of the module
Mk /x0 R ∼
= Mk−1 . Thus by the induction assumption we may assume that
f is given by the following equality
 α0 α1 ... αk−2 β
r  
a a′ ...
0 α0 ... αk−3 αk−2 αk−1
 . . .

..
..   0. a. ....
 .. .. . . ...
.
.  ·  .. .. . .
 0 0 ... α
α1
α2
0
0
0
0 ...
0 ...
0
0
α0
0
0 0 ...
0 0 ...
α1
α0
14
a(k−1) a(k)
a(k−2) a(k−1)
..
.
a
0

.. 
=
.′
a
a

a a′ ... a(k−1) a(k)
0 a ... a(k−2) a(k−1)

=  .. .. . .
. . .
0 0 ...
0 0 ...
..
.
..
.′
a
0
a
a
  α0 α1 ...
0 α0 ...
 
.. .. . .
·
 0. 0. ....
0
0
0 ...
0 ...
αk−2 β
s 
αk−3 αk−2 αk−1
..
.
α0
0
0
..
.
α1
α0
0
.. 
. 
,
α
2
α1
α0
where αi , β ∈ k and r, s ∈ R.
Counting on the place 0 × k we obtain
α0 a(k) + α1 a(k−1) + · · · + βa′ + ra = as + a′ αk−1 + · · · + a(k−1) α1 + a(k) α0
or
(β − αk−1 ) a′ = −ra + as
Since a′ ∈
/ aR + Ra by Lemma 4.2, hence β = αk−1 and r = s = αk ∈ k. Zimmermann [15, L. 8] proved that Hom(M, N ) is a finite dimensional
vector space over k for all finite length modules M, N over an arbitrary
differential polynomial ring. From the other hand by [9, Thm. 4.1] for any
simple modules M, N over B1 (k) , Ext(M, N ) is infinite dimensional over k.
Corollary 4.14. Let M , N be finite length modules over R. Then Ext(M, N )
is finite dimensional over k.
Proof. Considering extensions it suffices to prove the claim for simple modules M and N . But then the conclusion follows by Lemma 4.5.
Corollary 4.15. Let M be a finite length module over R and φ a ppformula. Then φ(M ) is either finite dimensional or cofinite dimensional
vector space over k. Thus M has pseudo U-rank 1.
Proof. Since R is an Ore domain similarly to [7, Prop. 1.5] either s | x →
φ(x) for some 0 ̸= s ∈ R and then M s ⊆ φ(M ) is cofinite dimensional; or
φ(x) → xt = 0 for 0 ̸= t ∈ R and φ(M ) ⊆ Hom(R/tR, M ) which is finite
dimensional over k.
5. Classification
For k ≥ 0 let us define the sequence of pp-formulae
( k s
)
∧ ∑
a(s−i)
φk (x0 ) = ∃ x1 , . . . , xk
xi
=0 .
(s − i)!
s=0 i=0
Clearly φ0 ≥ φ1 ≥ . . . .
Lemma 5.1. φk > φk+1 for every k, and φk > ψ for a pp-formula ψ implies
φk+1 ≥ ψ.
15
Proof. The element x0 is a free realization of φk in Mk . Let us consider the
AR -sequence (see Proposition 3.8)
f
0 → Mk → Mk−1 ⊕ Mk+1 → Mk → 0 .
The image of x0 with respect to f is Y ⊗ x0 = y0 =
1
k+1
u0 which is a
free realization of φk+1 in Mk+1 . Since this sequence does not split, hence
φk > φk+1 . Suppose that φk > ψ and (N, n) is a free realization of ψ. Then
the map x0 → n can be extended to the homomorphism h : Mk → N . Since
this map is not a split monomorphism hence it factors through f . In view
of Mk+1 |= φk+1 (u0 ) we obtain N |= φk+1 (n), hence ψ → φk+1 .
Proposition 5.2. Every indecomposable finite length module M over R is
isomorphic to Mk (a) for some k and irreducible a ∈ R. For instance, every indecomposable finite length module over R is uniserial with isomorphic
composition factors.
Proof. We choose 0 ̸= m ∈ Soc(M ) and denote p = ppM (m). Then ma = 0
for some irreducible a ∈ R, hence φ0 ∈ p. Suppose that φ1 ∈
/ p. Then by
Lemma 5.1 the inclusion mR = R/aR ⊆ M is pure hence split and M ∼
=
R/aR. Similarly if φk ∈ p and φk+1 ∈
/ p then M ∼
= Mk . Otherwise φk ∈ p for
every k, thus there is a homomorphism fk : Mk → M such that fk (x0 ) = m.
Since Mk is uniserial with Soc(Mk ) = x0 R, fk should be a monomorphism,
hence lg(M ) ≥ lg(Mk ) = k + 1 for every k, a contradiction.
Remark 5.3. Let M, N be finite length modules over R. Then the k-vector
spaces Hom(M, N ) and Ext(M, N ) have the same finite dimension.
Proof. It follows easily by making use of the standard Hom − Ext sequence.
Let us consider the series of inclusions M0 ⊂ M1 ⊂ M2 ⊂ . . . via the
identification xi → xi and denote M∞ = ∪∞
i=0 Mi .
Lemma 5.4. Mk ⊆ Mk+1 a .
Proof. For k = 0 we have M0 = R/aR with x0 = 1 and M1 is generated by
x0 , x1 with relations x0 a′ + x1 a = 0. Since M0 a is of codimension 1 in M0
and a′ ∈
/ M0 a by Lemma 4.2, hence M0 = M0 a + a′ · k and a′ = x1 a ∈ M1 a.
∼ Mk−1 ,
For k > 0 we have M0 = Soc(Mk ) ⊆ Mk+1 a and Mk /M0 =
Mk+1 /M0 ∼
= Mk , hence we could follow by induction.
16
Lemma 5.5. If 0 ̸= a ∈ R is irreducible, then M∞ (a) is an uniserial
injective module which is pseudo strongly minimal and End(M∞ (a)) is a
noetherian commutative valuation domain

α0 α1 α2

 0 α0 α1


 0 0 α0

..
..
..
.
.
.

...

... 


... 

..
.
where αi ∈ k.
Proof. We prove that Ext(R/bR, M∞ ) = 0 for every 0 ̸= b ∈ R where one
may assume that b is irreducible. If R/bR ̸∼
= R/aR then Ext(R/bR, Mk ) = 0,
hence the desired holds. Otherwise it may be supposed that b = a and the
result follows by Lemma 5.4.
Thus M∞ is a divisible module over a principal ideal domain, hence M∞
is injective by Baer’s criteria.
Now for proving that Hom(R/bR, M∞ ) is finite dimensional over k, we
may suppose that b is irreducible. Then Hom(R/bR, M∞ ) = Hom(R/bR, M0 )
which is finite dimensional over k by [15, L. 8]. Thus, M∞ is pseudo strongly
minimal. In particular, Hom(R/aR, M∞ ) = x0 · k, hence Soc(M∞ ) = M0
∼ M∞ . So M∞ is an uniserial module.
and M∞ /M0 =
Every endomorphism f of M∞ induces the homomorphisms fk : Mk →
Mk so the description of End(M∞ ) follows from Lemma 4.13.
Corollary 5.6. The interval (x = 0; φ0 ] in the lattice of pp-formulae over
R is a chain φ0 > φ1 > . . . .
Proof. By Lemma 5.1 it suffices to prove that ψ < φk for every k yields ψ
is degenerated. Let (N, n) be a free realization of ψ and we may assume
that N is torsion indecomposable, hence of finite length. Since N has a
local endomorphism ring, PE(N ) is also indecomposable. Because ppN (n)
contains φk for every k and PE(N ) is pure-injective, there is a homomorphism f : M∞ → PE(N ) such that f (x0 ) = n. Since M∞ is uniserial with
Soc(M∞ ) = x0 R this map is a monomorphism. Thus, M∞ is a direct summand of PE(N ), hence PE(N ) = M∞ in view of indecomposability. So
PE(N ) is injective so as N as a pure submodule of an injective module over
a noetherian ring. But that contradicts to the existence of AR -sequence
starting from N .
17
Theorem 5.7. Let M be a pure-injective indecomposable module over R.
Then exactly one of the following holds true:
1) M ∼
= PE(N ) where N is an indecomposable module of finite length or
2) M ∼
= M∞ hence injective, or dually
3) M is isomorphic to an indecomposable summand of PE(RR ) hence
torsonfree, or
4) M ∼
= E(RR ).
Proof. Suppose that M has a torsion, hence there is an irreducible 0 ̸= a ∈ R
and 0 ̸= m ∈ M such that ma = 0. So for p = ppM (m) we have φ0 ∈ p. If
φ1 ∈
/ p then since the pair (φ0 /φ1 ) is minimal, hence M is isolated by this
pair and M ∼
/ p we obtain
= PE(R/aR). Similarly for φk ∈ p and φk+1 ∈
M∼
= PE(Mk ). Otherwise as in the proof of Lemma 5.6 we have M ∼
= M∞ .
Notice that in this case M realizes the minimal pair (xa = 0/x = 0) in the
theory Tinj of injective modules.
So we can suppose that M is torsionfree. If M is not injective then
¬ a | m for some m ∈ M and irreducible a ∈ R. By elementary duality (see
[6]) M realizes in the theory Ttf of torsionfree modules the minimal pair
(x = x/a | x). This pair opens on the element 1 in RR and the module
PE(RR ) is the pure-injective envelope of a direct sum of indecomposable
pure-injective modules. So M is isomorphic to a direct summand of PE(R)
which is defined by this pair uniquely.
Proposition 5.8. The Cantor–Bendixson rank of ZgR is equal to 2. The
isolated points in ZgR are precisely the modules 1) from the Theorem 5.7.
The points of CB-rank 1 are precisely the modules of type 2) and 3), and
the unique point of CB rank 2 is the module E(RR ). In particular, there is
no super-decomposable pure-injective module over R.
Proof. From above it is enough to prove that E(RR ) is not isolated even on
level one. Otherwise let the pair φ/ψ isolates E(R) on this level. By the
dichotomy [7] we get s | x → φ and ψ → xt = 0 for some 0 ̸= s, t ∈ R.
Clearly we could find the simple module N = R/rR such that 1 · t ̸= 0 in
N . Then E(N ) ∈ (φ/ψ), a contradiction.
We end up by a general conjecture, concerning the behavior of the category of finite length modules over a differential polynomial ring.
Conjecture 5.9. Let R = F [Y,′ ] be a differential polynomial ring. Then
exactly one of the following holds true:
18
1) CB-rang of ZgR is 1, equivalently trivial module F is pseudo strongly
minimal (finite representation type);
2) CB-rank of ZgR is 2, equivalently F has pseudo U-rank 1 (tame representation type);
3) CB-rank of ZgR is ∞ with all usual badness (wild representation type).
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Moscow State Social University, 107150, Moscow, Losinoostrovskaya, 24,
Russia
19