The Ladder Construction of Prüfer Modules

Claus Michael Ringel

Dedicated to María Inés Platzeck on the occasion of her 60th birthday

Abstract Let be a ring (associative, with 1). A non-zero module is said to be a Prüfer module provided there exists a surjective, locally nilpotent endomorphism with kernel of finite length. The aim of this note is to construct Prüfer modules starting from a pair of module homomorphisms , where is injective and its cokernel is of finite length. For the ring of integers, one can construct in this way the ordinary Prüfer groups considered in abelian group theory. Our interest lies in the case that is an artin algebra.

1. The construction.

Let be a ring (associative, with 1). The modules to be considered will usually be left -modules. Our main interest will be the case where is an artin algebra, however the basic construction should be of interest for any ring . In fact, the standard examples of what we call Prüfer modules are the Prüfer groups in abelian group theory, thus -modules. Here is the definition of a Prüfer module: it is a non-zero module which has a surjective, locally nilpotent endomorphism with kernel of finite length. If is the kernel of , we often will write , and we will denote the kernel of by . Observe the slight ambiguity: given a Prüfer module , not only but also all non-trivial powers of and maybe many other endomorphisms will have the required properties (surjectivity, local nilpotency, finite length kernel).

The content of the paper is as follows. In the first section we show that any pair of module homomorphisms , where is injective with non-zero cokernel of finite length, gives rise to a Prüfer module. Section 2 provides some examples and section 3 outlines the relationship between Prüfer modules and various sorts of self-extensions of finite length modules. The final sections 4 and 5 deal with degenerations in the sense of Riedtmann-Zwara: we will show that this degeneration theory is intimately connected to the existence of Prüfer modules with some splitting property, and we will exhibit an extension of a recent result by Bautista and Pérez. Our interest in the questions considered here was stimulated by a series of lectures by Sverre Smalø [S] at the Mar del Plata conference, March 2006, and we are indebted to him as well as to M.C.R.Butler and G.Zwara for helpful comments.

For the relevance of Prüfer modules when dealing with artin algebras of infinite representation type, we refer to a forthcoming paper [R5]. The appendix to section 3.3 provides some indications in this direction.

1.1. The basic frame. A pair of exact sequences

yields a module and a pair of exact sequences

by forming the induced exact sequence of using the map :

Recall that a commutative square

is said to be exact provided it is both a pushout and a pullback, thus if and only if the sequence

is exact. Note that our basic setting provides an exact square

Next, we will use that the composition of exact squares is exact:

(E1) The composition of two exact squares

yields an exact square

1.2. The ladder. Using induction, we obtain in this way modules and pairs of exact sequences

for all

We may combine the pushout diagrams constructed inductively and obtain the following ladder of commutative squares:

We form the inductive limit (along the maps ).

Since all the squares commute, the maps induce a map which we denote by :

We also may consider the factor modules and . The map maps into , thus it induces a map

Claim. The map is an isomorphism. Namely, the commutative diagrams

can be rewritten as

with an isomorphism The map is a map from a filtered module with factors (where ) to a filtered module with factors (again with ), and the maps are just those induced on the factors.

It follows: The composition of maps

with the projection map is an epimorphism with kernel It is easy to see that is locally nilpotent, namely we have for all .

Summary. (a) The maps yield a map

with kernel and cokernel .

(b) This map induces an isomorphism . Composing the inverse of this isomorphism with the canonical projection , we obtain an endomorphism

If the cokernel of is non-zero and of finite length, then is a Prüfer module with respect to with basis ; in this case, we call (or better the pair ) the Prüfer module defined by the pair or by the ladder . Prüfer modules which are obtained in this way will be said to be of ladder type.

If necessary, we will use the following notation: , for all and for the Prüfer module. Since is a Prüfer module with basis the cokernel of , we will sometimes write or even

Remark: Using a terminology introduced for string algebras [R3], we also could say: is expanding, is contracting.

Lemma. Assume that with . Then is generated by .

Proof: For , the module is a factor module of thus by induction, is generated by .

1.3. The chessboard. Assume now that both maps are monomorphisms. Then we get the following arrangement of commutative squares:

Note that there are both horizontally as well as vertically ladders: the horizontal ladders yield (and its endomorphism ); the vertical ladders yield (and its endomorphism ).

2. Examples.

(1) The classical example: Let be the ring of integers, and its regular representation. Module homomorphisms are given by the multiplication with some integer , thus we denote such a map just by . Let and . If is odd, then is the ordinary Prüfer group for the prime , and (the subring of generated by ). If is even, then is an elementary abelian 2-group.

(2) Let be the Kronecker algebra over some field . Let be simple projective, indecomposable projective of length 3 and a non-zero map with cokernel (one of the indecomposable modules of length 2). The module is the Prüfer module for if and only if otherwise it is a direct sum of copies of .

(3) Trivial cases: First, let be a split monomorphism. Then the Prüfer module with respect to any map is just the countable sum of copies of . Second, let be an arbitrary monomorphism, let be an endomorphism. Then is the countable sum of copies of .

(4) Assume that there exists a split monomorphism , say and . Then

is a Riedtmann-Zwara sequence as discussed in section 4, thus is a degeneration of .

Remark: Not all Prüfer modules are of ladder type. Consider the generalized Kronecker algebra with countably many arrows starting at the vertex and ending in the vertex . Define a representation as follows: Let be a vector space with a countable basis and let be defined by provided and otherwise. Let be the endomorphism of of , respectively, which sends to and to for Then is a Prüfer module (with respect to , but also with respect to any power of ). Obviously, is a faithful -module. Assume that for some maps with of finite length. Then is generated by , according to Lemma 1.2. However is of finite length and no finite length -module is faithful.

3. Ladder extensions.

3.1. The definition. Let be a non-zero module of finite length. A self-extension is said to be a ladder extension provided there is a commutative diagram with exact rows

such that factors through , say for some

This means that we have a commutative diagram with exact rows of the following kind (here ):

Thus, in order to construct all the ladder extensions of , we may start with an arbitrary epimorphism form its kernel and consider any homomorphism .

According to section 1 we know: Ladder extensions build up to form Prüfer modules.

Lemma. Let be a commutative ring and a -algebra. Then for any

Proof: We deal with the exact sequence induced by or , respectively. But since

Also, any central automorphism of yields isomorphic extensions and . This shows that the extension only depends on the -subspace

Remark. Not all self-extensions are ladder extensions. For example: A non-zero self-extension of a simple module over an artinian ring is never a ladder extension!

Proof: Construct the corresponding ladder, thus the corresponding Prüfer module . The module would be a (serial) module of Loewy length , with arbitrary. But the Loewy length of any module over the artinian ring is bounded by the Loewy length of thus cannot exist.

Example. Here is a further example of a self-extension which is not a ladder extension. Consider the following quiver

with one loop at the vertex b, and one arrow from to . We consider the representations of with the relation The universal covering of has many subquivers of the form

and we consider some representations of ; we present here the corresponding dimension vectors.

There is an obvious exact sequence

Under the covering functor, the representations and are identified, thus we obtain a self-extension. One easily checks that this self-extension is not a ladder extension.

Proposition. Let be an indecomposable module with Auslander-Reiten translate isomorphic to . Assume that there is a simple submodule of with Then the Auslander-Reiten sequence ending (and starting) in is a ladder extension.

Proof. Let be the Auslander-Reiten sequence. Denote by the inclusion map. Since the map factors through , there is a commutative diagram with exact rows of the following form:

Now form the induced exact sequence:

Since the induced sequence splits, thus we obtain a map with It follows that

We do not know whether one can delete the assumption about the existence of .

3.2. Standard self-extensions.

Let be an -module, say with an exact sequence , where denotes a projective cover of . We know that

Note that

(Proof: , thus take and form Since is surjective and is projective, there is with Thus is in the image of )

Thus we can consider

as a subgroup of We call the elements of the standard self-extensions.

Proposition. Standard self-extensions are ladder extensions.

Proof. Here is the usual diagram in which way a map yields a self-extension of

The standard extensions are those where the map factors through , say with

3.3. Modules of projective dimension 1.

Proposition. If the projective dimension of is at most 1, then any self-extension of is standard, thus a ladder extension.

Proof: Consider a module with a projective presentation Any self-extension of is given by a diagram of the following kind:

Since is projective and surjective, there is a map such that The self-extension is given just by

Corollary. If is a hereditary ring, any self-extension is standard, thus a ladder extension.

Example of a ladder extension which is not standard. Consider the quiver

such that Consider the indecomposable length 2 module annihilated by . Then the kernel of is We may visualize this as follows:

There is a ladder extension of , given by the non-trivial map , but this map does not factor through , since is one-dimensional, generated by . Note that factors through , where is the kernel of .

Appendix. Here, we want to indicate that the Corollary can be used in order to obtain a conceptual proof of the second Brauer-Thrall conjecture for hereditary artin algebras.

Assume that there is no generic module. We show: Any indecomposable module is a brick without self-extensions. Assume that there is an indecomposable module which is not a brick or which does have self-extensions. If is not a brick, then the brick paper [R2] shows that there are bricks with self-extensions. Thus, we see that there always is a brick with self-extensions. Take any non-zero self-extension of . According to 3.2, such a self-extension is standard, thus a ladder extension, thus we obtain a corresponding Prüfer module . The process of simplification [R1] shows that all the modules are indecomposable. Thus is not of finite type and therefore there exists a generic module [R5].

But if any indecomposable module is a brick without self-extensions, the quadratic form is weakly positive. Ovsienko asserts that then there are only finitely many positive roots, thus the algebra is of bounded representation type and therefore of finite representation type.

3.4. Warning. A Prüfer module is not necessarily determined by , even if it is of ladder type.

As an example take the generalized Kronecker quiver with vertices and three arrows . and let be the two-dimensional indecomposable representation annihilated by and . Consider a projective cover , let be its kernel, say with inclusion map

0 → ΩH PH H → 0

(*)

The ladders to be considered are given by the various maps such that the image of is not contained in (otherwise, the induced self-extension of will split). In order to specify a self-extension of , we require that is annihilated say by .

We will consider several copies of . If is a generator, let us denote thus, is a basis of .

We start with generated by and consider the exact sequence as displayed above. We see that is a basis of

Now, let us consider two maps , here we denote the generator of by The first map is given by and The second map is defined by and .

Note that , thus and actually this is precisely the self-extension of annihilated by

An easy calculation shows that (and even ) is annihilated by , whereas is faithful. The following displays may be helpful; always, we exhibit the modules:

First the display for the homomorphism .

Now the corresponding display for the homomorphism .

4. Degenerations.

Definition: Let be finite length modules. Call a degeneration of provided there is an exact sequence of the form with of finite length. (such a sequence will be called a Riedtmann-Zwara sequence). The map is called a corresponding steering map. (Note that in case we deal with modules over a finite dimensional -algebra and is an algebraically closed field, then this notion of degeneration coincides with the usual one, as Zwara [Z2] has shown.)

The proof of the following result is essentially due to Zwara, he used this argument in order to show that is a degeneration of if and only if there is an exact sequence (a co-Riedtmann-Zwara sequence) with of finite length.

Proposition. Let be -modules of finite length. The following conditions are equivalent:

(1) is a degeneration of .

(2) There is a Prüfer module and some natural number such that for all

(3) There is a Prüfer module and some natural number such that .

Here is the recipe how to obtain a Prüfer module starting from a degeneration: If is a degeneration of , say with steering module , then there exists a monomorphism with cokernel . The Prüfer module we are looking for is

Proof of the implication (3) (1). Assume that there is a Prüfer module such that We get the following two exact sequences

(1)

in the first, the map is given by applying , in the second the map is given by applying In both sequences, we can replace by Thus we obtain as first sequence a new Riedtmann-Zwara sequence, and as second sequence a dual Riedtmann-Zwara sequence:

(2)

note that both use the same steering module, namely Thus:

Remark. We see: The module is a degeneration of if and only if there exists a module and an exact sequence .

Proof of the proposition. We need further properties of exact squares:

(E2) For any map , and any module , the following diagram is exact:

(E3) Let

be exact. Then is split mono.

(E4) Assume that we have the following exact square

and that is a split monomorphism, then the sequence

splits.

Proofs. (E2) is obvious. (E3): Since is injective, is injective. Let be the cokernel of . We obtain the map by forming the induced exact sequence of , using the zero map . But such an induced exact sequence splits. (E4) Assume that Then

There is the following lemma (again, see Zwara [Z1]):

Lemma (Existence of nilpotent steering maps.) If there is an exact sequence then there is an exact sequence such that the map is nilpotent.

Proof: We can decompose such that the given map maps into , into and such that the induced map belongs to the radical of the category, whereas the induced map is an isomorphism. We obtain the following pair of exact squares

(the left square is exact according to (E2)). The composition of the squares is the desired exact square (note that is isomorphic to ).

Assume that a monomorphism with cokernel and is given. Consider also the canonical embedding and form the ladder for this pair of monomorphisms . The modules are just the modules we are looking for: As we know, there is a Prüfer module with being the kernel of

We construct the maps explicitly as follows:

and

using the recipe (E2). Thus we obtain the following sequence of exact squares:

In particular, we have

Note that the composition is of the form for some .

We also have the following sequence of exact squares:

where the vertical maps are of the form

The composition of these exact squares yields an exact square

Here we may insert the following observation: This sequence shows that the module is a degeneration of the module .

Since the composition is of the form and it follows that is a split monomorphism, see (E3).

Also, we can consider the following two exact squares, with (the upper square is exact, according to (E2)):

The vertical composition on the left is , thus, as we have shown, a split monomorphism. This shows that the exact sequence corresponding to the composed square splits (E4): This yields

Cancelation of gives the desired isomorphism:

Remark to the proof. Given the Riedtmann-Zwara sequence

we have considered the following pair of monomorphisms

The corresponding Prüfer modules are and , respectively. And As we know, we can assume that is nilpotent. Then all the linear combinations

with are also split monomorphisms (with retraction , where ).

Corollary. Assume that is a degeneration of . Then there exists a Prüfer module such that is isomorphic to for some natural number .

5. Application: The theorem of Bautista-Pérez.

Here we assume that we deal with an artin algebra , and all the modules are -modules of finite length.

Proposition. Let be a module with and assume there is given an exact sequence . Then the cokernel of any monomorphism is a degeneration of .

Corollary (Bautista-Pérez). Let be modules, and let and be cokernels of monomorphisms Assume that both and Then the modules and are isomorphic.

Both assertions are well-known in case is an algebraically closed field: in this case, the conclusion of the proposition just asserts that is a degeneration of in the sense of algebraic geometry. The main point here is to deal with the general case when is an arbitrary artin algebra. The corollary stated above (under the additional assumptions that is projective and that are contained in the radical of ) is due to Bautista and Pérez [BP] and this result was presented by Smalø with a new proof [S] at Mar del Plata.

We need the following well-known lemma.

Lemma. Let be a module with Let be a sequence of inclusions of modules with for all Then there is a natural number such that is a split monomorphism for all

Let us use it in order to finish the proof of the proposition. Let and the given monomorphism with cokernel . Let be an additional monomorphism, say with cokernel . Thus we are in the setting of section 1. We apply the Lemma to the chain of inclusions

and see that there is such that splits. This shows that is isomorphic to But we also have the exact sequence

Replacing by , we see that we get an exact sequence of the form

(a Riedtmann-Zwara sequence), as asserted.

Proof of the Corollary. It is well-known that the existence of exact sequences

implies that the modules and are isomorphic [Z1]. But in our case we just have to change one line in the proof of the proposition in order to get the required isomorphism. Thus, assume that both and . Choose such that both the inclusion maps

split. Then is isomorphic both to and to , thus it follows from the Krull-Remak-Schmidt theorem that and are isomorphic.

Remark. Assume that are monomorphisms with cokernels and , respectively, and that and Then splits if and only if splits.

Proof: According to the corollary, we can assume . Assume that splits, thus is isomorphic to . Look at the exact sequence . If it does not split, then , but is isomorphic to

As we have mentioned, the lemma is well-known; an equivalent assertion was used for example by Roiter in his proof of the first Brauer-Thrall conjecture, a corresponding proof can be found in [R4]. We include here a slightly different proof:

Applying the functor to the short exact sequence we obtain the exact sequence

Since the latter term is zero, we see that we have a sequence of surjective maps

being induced by the inclusion maps The maps between the -groups are -linear. Since is a -module of finite length, the sequence of surjective maps must stabilize: there is some such that the inclusion induces an isomorphism

for all Now we consider also some -terms: the exactness of

shows that the connecting homomorphism is zero, and thus that the map (induced by the projection map ) is surjective. But this means that there is a map with , thus is a split epimorphism and therefore the inclusion map is a split monomorphism.

Remark. In general, there is no actual bound on the number . However, in case of dealing with the chain of inclusions

such a bound exists, namely the length of as a -module, or, even better, the length of as an -module, where

Proof: Look at the surjective maps

being induced by the maps (and these maps are not only -linear, but even -linear). Assume that is bijective, for some . As we have seen above, this implies that the sequence

0 → Un Un+1 → W → 0

(*)

splits. Now the map is obtained from as the induced exact sequence using the map . With also any induced exact sequence will split. Thus is a split monomorphism (and will be bijective, again). Thus, as soon as we get a bijection for some , then also all the following maps with are bijective.

Example. Consider the -quiver with subspace orientation:

and let be its path algebra over some field . We denote the indecomposable -modules by the corresponding dimension vectors. Let

Note that a map with cokernel exists only in case the base-field has at least 3 elements; of course, there is always a map with cokernel

We have and it turns out that the module is the following:

The pushout diagram involving the modules (twice) and is constructed as follows: denote by monomorphisms which factor through the indecomposable projective modules , respectively. We can assume that , so that a mesh relation is satisfied. Denote the 3 summands of by , with non-zero maps such that There is the following commutative square, for any we are interested when :

(the only calculation which has to be done concerns the third entries: ). Note that (as well as ) does not split.

But now we deal with a module such that This implies that is isomorphic to . Thus the next pushout construction yields an exact sequence of the form

Acknowledgement. The author is indebted to Dieter Vossieck for a careful reading of the final version of the paper.

References

[BP]    Bautista, R. and Pérez, E.: On modules and complexes without self-extensions. Communications in Algebra 34 (2006), 3139-3152.        [ Links ]

[R1]    Ringel, C.M.: Representations of k-species and bimodules. J.Algebra 41 (1976), 269-302.        [ Links ]

[R2]    Ringel, C.M.: Bricks in hereditary length categories. Resultate der Math. 6 (1983), 64-70.        [ Links ]

[R3]    Ringel, C.M.: Some algebraically compact modules I. In: Abelian Groups and Modules (ed. A. Facchini and C. Menini). Kluwer (1995), 419-439.        [ Links ]

[R4]    Ringel, C.M.: The Gabriel-Roiter measure. Bull. Sci. math. 129 (2005). 726-748.        [ Links ]

[R5]    Ringel, C.M.: Prüfer modules which are not of finite type. In preparation.        [ Links ]

[S]    Smalø, S.: Lectures on Algebras. Mar del Plata, Argentina March 2006. Revista Unión Matemática Argentina 48-2 (2007), 21-45.        [ Links ]

[Z1]    Zwara, G.: A degeneration-like order for modules. Arch. Math. 71 (1998), 437-444        [ Links ]

[Z2]    Zwara, G.: Degenerations of finite-dimensional modules are given by extensions. Compositio Mathematica 121 (2001), 205-218.        [ Links ]

 Claus Michael Ringel
Fakultät für Mathematik, Universität Bielefeld,
POBox 100 131,
D-33 501 Bielefeld, Germany
ringel@math.uni-bielefeld.de

Recibido: 5 de marzo de 2007
Aceptado: 16 de abril de 2007