Let
$\unicode[STIX]{x1D6EC}$
be a connected hereditary artin algebra. The modules considered here are left
$\unicode[STIX]{x1D6EC}$
-modules of finite length,
$\operatorname{mod}\unicode[STIX]{x1D6EC}$
denotes the corresponding category. The subcategories of
$\operatorname{mod}\unicode[STIX]{x1D6EC}$
we deal with are always assumed to be closed under direct sums and direct summands (in particular closed under isomorphisms). In this setting, a subcategory is a torsion class (the class of torsion modules for what is called a torsion pair or a torsion theory) provided it is closed under factor modules and extensions. The torsion classes form a partially ordered set with respect to inclusion, it will be denoted by
$\operatorname{tors}\unicode[STIX]{x1D6EC}$
. This poset clearly is a lattice (even a complete lattice). It is easy to see that a torsion class
${\mathcal{C}}$
in
$\operatorname{mod}\unicode[STIX]{x1D6EC}$
is functorially finite if and only if it has a cover (a cover for
${\mathcal{C}}$
is a module
$C$
such that
${\mathcal{C}}$
is the set of modules generated by
$C$
), we denote by
$\operatorname{f - tors}\unicode[STIX]{x1D6EC}$
the set of functorially finite torsion classes in
$\operatorname{mod}\unicode[STIX]{x1D6EC}$
.
In a recent paper [Reference Iyama, Reiten, Thomas and TodorovIRTT], Iyama et al. have discussed the question whether the poset
$\operatorname{f - tors}\unicode[STIX]{x1D6EC}$
(with the inclusion order) also is a lattice.
Theorem. The poset
$\operatorname{f - tors}\unicode[STIX]{x1D6EC}$
is a lattice if and only if
$\unicode[STIX]{x1D6EC}$
is representation-finite or
$\unicode[STIX]{x1D6EC}$
has precisely two simple modules.
Iyama et al. have shown this in the special case when
$\unicode[STIX]{x1D6EC}$
is a
$k$
-algebra with
$k$
an algebraically closed field (so that
$\unicode[STIX]{x1D6EC}$
is Morita equivalent to the path algebra of a quiver). The aim of this note is to provide a proof in general.
Here is an outline of the essential steps of the proof. Recall that a module is called exceptional provided it is indecomposable and has no self-extensions. A pair of modules
$X,Y$
will be called an
$\operatorname{Ext}$
-pair provided both
$X,Y$
are exceptional,
$\operatorname{Hom}(X,Y)=\operatorname{Hom}(Y,X)=0$
and
$\operatorname{Ext}^{1}(X,Y)\neq 0$
,
$\operatorname{Ext}^{1}(Y,X)\neq 0$
. We follow the strategy of [Reference Iyama, Reiten, Thomas and TodorovIRTT] by establishing the existence of
$\operatorname{Ext}$
-pairs for any connected hereditary artin algebra which is representation-infinite and has at least three simple modules (Proposition 5). On the other hand, we show directly that the set of functorially finite torsion classes which contain a fixed
$\operatorname{Ext}$
-pair has no minimal elements (Proposition 4).
1 Normalization
Let
${\mathcal{X}}$
be a class of modules. We denote by
$\operatorname{add}({\mathcal{X}})$
the modules which are direct summands of direct sums of modules in
${\mathcal{X}}$
. A module
$M$
is generated by
${\mathcal{X}}$
provided
$M$
is a factor module of a module in
$\operatorname{add}({\mathcal{X}})$
, and
$M$
is cogenerated by
${\mathcal{X}}$
provided
$M$
is a submodule of a module in
$\operatorname{add}({\mathcal{X}})$
. The subcategory of all modules generated by
${\mathcal{X}}$
is denoted by
${\mathcal{G}}({\mathcal{X}}).$
In case
${\mathcal{X}}=\{X\}$
or
${\mathcal{X}}=\operatorname{add}X$
, we write
${\mathcal{G}}(X)$
instead of
${\mathcal{G}}({\mathcal{X}})$
, and use the same convention in similar situations. We write
${\mathcal{T}}(X)$
for the smallest torsion class containing the module
$X$
(it is the intersection of all torsion classes containing
$X$
, and it can be constructed as the closure of
$\{X\}$
using factor modules and extensions).
Since
$\unicode[STIX]{x1D6EC}$
is assumed to be hereditary, we write
$\operatorname{Ext}(X,Y)$
instead of
$\operatorname{Ext}^{1}(X,Y)$
.
Following Roiter [Reference RoiterRo], we say that a module
$M$
is normal provided there is no proper direct decomposition
$M=M^{\prime }\oplus M^{\prime \prime }$
such that
$M^{\prime }$
generates
$M^{\prime \prime }$
(this means: if
$M=M^{\prime }\oplus M^{\prime \prime }$
and
$M^{\prime }$
generates
$M^{\prime \prime }$
, then
$M^{\prime \prime }=0$
). Of course, given a module
$M$
, there is a direct decomposition
$M=M^{\prime }\oplus M^{\prime \prime }$
such that
$M^{\prime }$
is normal and
$M^{\prime }$
generates
$M^{\prime \prime }$
and one can show that
$M^{\prime }$
is determined by
$M$
uniquely up to isomorphism, thus we call
$M^{\prime }=\unicode[STIX]{x1D708}(M)$
a normalization of
$M$
. This was shown already by Roiter [Reference RoiterRo], and later by Auslander–Smalø [Reference Auslander and SmaløAS]. It is also a consequence of the following Lemma which will be needed for our further considerations.
Lemma 1.
(a) Let
$(f_{1},\ldots ,f_{t},g)\;:\;X\rightarrow X^{t}\oplus Y$
be an injective map for some natural number
$t$
, with all the maps
$f_{i}$
in the radical of
$\operatorname{End}(X)$
. Then
$X$
is cogenerated by
$Y$
.(b) Let
$(f_{1},\ldots ,f_{t},g)\;:\;X^{t}\oplus Y\rightarrow X$
be a surjective map for some natural number
$t$
, with all the maps
$f_{i}$
in the radical of
$\operatorname{End}(X)$
, then
$Y$
generates
$X$
.
Proof. (a) Assume that the radical
$J$
of
$\operatorname{End}(X)$
satisfies
$J^{m}=0$
. Let
$W$
be the set of all compositions
$w$
of at most
$m-1$
maps of the form
$f_{i}$
with
$1\leqslant i\leqslant t$
(including
$w=1_{X}$
). We claim that
$(gw)_{w\in W}\;:\;X\rightarrow Y^{|W|}$
is injective. Take a nonzero element
$x$
in
$X$
. Then there is
$w\in W$
such that
$w(x)\neq 0$
and
$f_{i}w(x)=0$
for
$1\leqslant i\leqslant t$
. Since
$(f_{1},\ldots ,f_{t},g)$
in injective and
$w(x)\neq 0$
, we have
$(f_{1},\ldots ,f_{t},g)(w(x))\neq 0.$
But
$f_{i}w(x)=0$
for
$1\leqslant i\leqslant t,$
thus
$g(w(x))\neq 0.$
This completes the proof.
(b) This follows by duality. ◻
Corollary [Uniqueness of normalization]. Let
$M$
be a module. Assume that
$M=M_{0}\oplus M_{1}=M_{0}^{\prime }\oplus M_{1}^{\prime }$
such that both
$M_{0}$
and
$M_{0}^{\prime }$
generate
$M$
. Then there is a module
$N$
which is a direct summand of both
$M_{0}$
and
$M_{0}^{\prime }$
which generates
$M$
.
Proof. We may assume that
$M$
is multiplicity free. Write
$M_{0}\simeq N\oplus C,~M_{0}^{\prime }\simeq N\oplus C^{\prime },$
such that
$C,C^{\prime }$
have no indecomposable direct summand in common. Now,
$N\oplus C$
generates
$N\oplus C^{\prime }$
,
$N\oplus C^{\prime }$
generates
$N\oplus C$
, and
$N\oplus C$
generates
$C$
. We see that
$N\oplus C$
generates
$C$
, such that the maps
$C\rightarrow C$
used belong to the radical of
$\operatorname{End}(C)$
(since they factor through
$\operatorname{add}(N\oplus C^{\prime })$
and no indecomposable direct summand of
$C$
belongs to
$\operatorname{add}(N\oplus C^{\prime })$
). Lemma 1 asserts that
$N$
generates
$C$
, thus it generates
$M$
.◻
Proposition 1. If
$T$
has no self-extensions, then
$T$
is a cover for the torsion class
${\mathcal{T}}(T)$
. Conversely, if
${\mathcal{T}}$
is a torsion class with cover
$C$
, then
$\unicode[STIX]{x1D708}(C)$
has no self-extensions.
Proof. For the first assertion, one has to observe that
${\mathcal{G}}(T)$
is closed under extensions, thus equal to
${\mathcal{T}}(T).$
This is a standard result say in tilting theory. Here is the argument: let
$g^{\prime }\;:\;T^{\prime }\rightarrow M^{\prime }$
and
$g^{\prime \prime }\;:\;T^{\prime \prime }\rightarrow M^{\prime \prime }$
be surjective maps with
$T^{\prime },T^{\prime \prime }$
in
$\operatorname{add}T$
. Let
$0\rightarrow M^{\prime }\rightarrow M\rightarrow M^{\prime \prime }\rightarrow 0$
be an exact sequence. The induced exact sequence with respect to
$g^{\prime \prime }$
is of the form
$0\rightarrow M^{\prime }\rightarrow Y_{1}\rightarrow T^{\prime \prime }\rightarrow 0$
with a surjective map
$g_{1}\;:\;Y_{1}\rightarrow M$
. Since
$\unicode[STIX]{x1D6EC}$
is hereditary and
$g^{\prime }$
is surjective, there is an exact sequence
$0\rightarrow T^{\prime }\rightarrow Y_{2}\rightarrow T^{\prime \prime }\rightarrow 0$
with a surjective map
$g_{2}\;:\;Y_{2}\rightarrow Y_{1}.$
Since
$\operatorname{Ext}(T^{\prime \prime },T^{\prime })=0$
, we see that
$Y_{2}$
is isomorphic to
$T^{\prime }\oplus T^{\prime \prime }$
, thus in
$\operatorname{add}T$
. And there is the surjective map
$g_{1}g_{2}\;:\;Y_{2}\rightarrow M.$
For the converse, we may assume that
$C$
is normal and have to show that
$C$
has no self-extension. Let
$C_{1},C_{2}$
be indecomposable direct summands of
$C$
and assume for the contrary that there is a nonsplit exact sequence
$$\begin{eqnarray}0\rightarrow C_{1}\rightarrow M\rightarrow C_{2}\rightarrow 0.\end{eqnarray}$$
Now
$M$
belongs to
${\mathcal{T}}$
, thus it is generated by
$C$
, say there is a surjective map
$C^{\prime }\rightarrow M$
with
$C^{\prime }\in \operatorname{add}C.$
Write
$C^{\prime }=C_{2}^{t}\oplus C^{\prime \prime }$
such that
$C_{2}$
is not a direct summand of
$C^{\prime \prime }$
. Consider the surjective map
$C_{2}^{t}\oplus C^{\prime \prime }\rightarrow M\rightarrow C_{2}$
. Since the last map
$M\rightarrow C_{2}$
is not a split epimorphism, all the maps
$C_{2}\rightarrow C_{2}$
involved belong to the radical of
$\operatorname{End}(C_{2})$
. According to Lemma 1,
$C^{\prime \prime }$
generates
$C_{2}.$
This contradicts the assumption that
$C$
is normal.◻
Remark.
Proposition 1 provides a bijection between the isomorphism classes of normal modules without self-extensions and torsion classes with covers. This is one of the famous Ingalls–Thomas bijections; see for example [Reference Obaid, Nauman, Fakieh and RingelONFR] or also [Reference RingelR3].
We recall that a torsion class is functorially finite if and only if it has a cover. Of course, if
$C$
is a cover of the torsion class
${\mathcal{T}}$
, then
$\unicode[STIX]{x1D707}(C)$
is a minimal cover of
${\mathcal{T}}$
.
Proposition 2. Let
${\mathcal{T}}$
be a nonzero functorially finite torsion class. Then there is an indecomposable module
$U$
in
${\mathcal{T}}$
such that any nonzero map
$V\rightarrow U$
with
$V\in {\mathcal{T}}$
is a split epimorphism.
Proof. Let
$C$
be a minimal cover of
${\mathcal{T}}$
. Since
$C$
has no self-extensions, it is a direct summand of a tilting module. In particular, the quiver of
$\operatorname{End}(C)$
is directed. It follows that
$C$
has an indecomposable direct summand
$U$
such that any nonzero map
$C\rightarrow U$
is a split epimorphism. Assume now that
$V$
belongs to
${\mathcal{T}}$
and
$f\;:\;V\rightarrow U$
is a nonzero map. There is a surjective map
$g\;:\;C^{t}\rightarrow V$
for some
$t$
. Since the composition
$fg\;:\;C^{t}\rightarrow U$
is nonzero, it is split epi, thus also
$f$
is split epi.◻
Remark.
As we have mentioned, normal modules have been considered by Roiter, but actually, he used a slightly deviating name, calling them “normally indecomposable”.
2 Inclusions of functorially finite torsion classes
If
${\mathcal{X}}$
is a class of modules and
$U$
is an indecomposable module, we denote by
${\mathcal{X}}_{U}$
the class of modules in
${\mathcal{X}}$
which have no direct summand isomorphic to
$U$
.
Proposition 3. Assume that
${\mathcal{T}}$
is a torsion class and that
$U$
is an indecomposable module in
${\mathcal{T}}$
. The following assertions are equivalent:
(i) The class
${\mathcal{T}}_{U}$
is a torsion class.(ii) Any nonzero map
$V\rightarrow U$
with
$V\in {\mathcal{T}}$
is split epi.
Proof. (i)
$\;\Longrightarrow \;$
(ii). We assume that
${\mathcal{T}}_{U}$
is a torsion class. Let
$f\;:\;V\rightarrow U$
be a nonzero map with
$V\in {\mathcal{T}}$
. We claim that
$f$
is surjective. Note that
$f(V)$
and
$U/f(V)$
both belong to
${\mathcal{T}}$
, since
${\mathcal{T}}$
is closed under factor modules. If
$f$
is not surjective, then
$f(V)$
is a factor module of
$V$
and a proper nonzero submodule of
$U$
, whereas
$U/f(V)$
is a proper nonzero factor module of
$U$
. It follows that both
$f(V)$
and
$U/f(V)$
belong to
${\mathcal{T}}_{U}$
. Since we assume that
${\mathcal{T}}_{U}$
is a torsion class, it is closed under extensions, and therefore
$U$
belongs to
${\mathcal{T}}_{U}$
, a contradiction.
Write
$V=V^{\prime }\oplus U^{t}$
for some
$t$
with
$V^{\prime }$
in
${\mathcal{T}}_{U}$
. If
$f$
is not split epi, then Lemma 1(b) asserts that
$V^{\prime }$
generates
$U$
. But we assume that
${\mathcal{T}}_{U}$
is a torsion class, thus closed under direct sums and factor modules. Therefore, if
$V^{\prime }$
generates
$U$
, then
$U$
has to belong to
${\mathcal{T}}_{U}$
, again a contradiction. Altogether we have shown that
$f$
is split epi.
(ii)
$\;\Longrightarrow \;$
(i). We assume now that any nonzero map
$V\rightarrow U$
with
$V\in {\mathcal{T}}$
is a split epimorphism, and we have to show that
${\mathcal{T}}_{U}$
is a torsion class. In order to see that
${\mathcal{T}}_{U}$
is closed under factor modules, let
$V$
belong to
${\mathcal{T}}_{U}$
and let
$W$
be a factor module of
$V$
. Assume that
$U$
is a direct summand of
$W$
, thus
$U$
is a factor module of
$V$
. The projection
$p\;:\;V\rightarrow U$
is a nonzero map, thus by assumption
$p$
is a split epimorphism. But this implies that
$U$
is a direct summand of
$V$
, whereas
$V$
belongs to
${\mathcal{T}}_{U}$
. This shows that
$W$
belongs to
${\mathcal{T}}_{U}$
.
In order to show that
${\mathcal{T}}_{U}$
is closed under extensions, consider a module
$M$
with a submodule
$V$
such that both
$V$
and
$M/V$
belong to
${\mathcal{T}}_{U}.$
Since
${\mathcal{T}}$
is closed under extension,
$M$
belongs to
${\mathcal{T}}$
. Assume that
$U$
is a direct summand of
$M$
, say
$M=U\oplus M^{\prime }$
. If
$V\subseteq M^{\prime }$
, then
$M/V=U\oplus M^{\prime }/V$
shows that
$U$
is a direct summand of
$M/V$
in contrast to our assumption that
$M/U$
belongs to
${\mathcal{T}}_{U}$
. Thus
$V\not \subseteq M^{\prime }.$
It follows that
$V$
is not contained in the kernel of the canonical projection
$q\;:\;M\rightarrow M/M^{\prime }\simeq U$
, thus the restriction of
$q$
to
$V$
is a nonzero map
$V\rightarrow U$
. The condition (ii) asserts that this map
$V\rightarrow U$
is split epi, therefore
$V$
does not belong to
${\mathcal{T}}_{U}$
, a contradiction. This shows that
$M$
belongs to
${\mathcal{T}}_{U}$
.◻
Proposition 4. Let
${\mathcal{E}}$
be a class of indecomposable modules with the following property: if
$E$
belongs to
${\mathcal{E}}$
, there is
$E^{\prime }$
in
${\mathcal{E}}$
with
$\operatorname{Ext}(E,E^{\prime })\neq 0.$
Then the set of functorially finite torsion classes
${\mathcal{T}}$
which contain
${\mathcal{E}}$
has no minimal elements.
Proof. Let
${\mathcal{T}}$
be a functorially finite torsion class which contains
${\mathcal{E}}$
. According to Proposition 2, there is an indecomposable module
$U$
in
${\mathcal{T}}$
such that any nonzero map
$V\rightarrow U$
with
$V\in {\mathcal{T}}$
is a split epimorphism. According to Proposition 3, the class
${\mathcal{T}}_{U}$
is a torsion class. Since
${\mathcal{T}}$
is functorially finite, also
${\mathcal{T}}_{U}$
is functorially finite.
We claim that
${\mathcal{E}}$
is contained in
${\mathcal{T}}_{U}.$
Thus, let
$E$
belong to
${\mathcal{E}}$
. Since
$E$
is indecomposable, we have to show that
$E$
is not isomorphic to
$U$
. By assumption, there is
$E^{\prime }$
in
${\mathcal{E}}$
with
$\operatorname{Ext}(E,E^{\prime })\neq 0.$
Thus, there is a nonsplit exact sequence
$0\rightarrow E^{\prime }\rightarrow M\rightarrow E\rightarrow 0$
. Since
$E,E^{\prime }$
both belong to
${\mathcal{E}}\subseteq {\mathcal{T}}$
and
${\mathcal{T}}$
is closed under extensions,
$M$
belongs to
${\mathcal{T}}$
. Since the given map
$M\rightarrow E$
is not split epi, it follows that
$E$
is not isomorphic to
$U$
. Thus
${\mathcal{E}}\subseteq {\mathcal{T}}_{U}$
. Since
${\mathcal{T}}_{U}$
is properly contained in
${\mathcal{T}}$
, we see that
${\mathcal{T}}$
is not minimal in the set of functorially finite torsion classes which contain
${\mathcal{E}}$
.◻
3 Construction of
$\operatorname{Ext}$
-pairs
The aim of this section is to show the following proposition.
Proposition 5. A connected hereditary artin algebra which is representation-infinite and has at least three simple modules has
$\operatorname{Ext}$
-pairs.
Given a finite-dimensional artin algebra
$R$
, we denote by
$Q(R)$
its
$\operatorname{Ext}$
-quiver: its vertices are the isomorphism classes
$[S]$
of the simple
$R$
-modules
$S$
, and given two simple
$R$
-modules
$S,S^{\prime }$
, there is an arrow
$[S]\rightarrow [S^{\prime }]$
provided
$\operatorname{Ext}(S,S^{\prime })\neq 0.$
If
$R$
is hereditary, then clearly
$Q(R)$
is directed. If necessary, we endow
$Q(R)$
with a valuation as follows: Given an arrow
$S\rightarrow S^{\prime }$
, consider
$\operatorname{Ext}(S,S^{\prime })$
as a left
$\operatorname{End}(S)^{\text{op}}$
-module or as a left
$\operatorname{End}(S^{\prime })$
-module and put
$$\begin{eqnarray}v([S],[S^{\prime }])=(\dim _{\operatorname{End}(S)^{\text{op}}}\operatorname{Ext}(S,S^{\prime }))(\dim _{\operatorname{ End}(S^{\prime })}\operatorname{Ext}(S,S^{\prime }))\end{eqnarray}$$
(note that in contrast to [Reference Dlab and RingelDR], we only need the product of the two dimensions, not the pair). Given a vertex
$i$
of
$Q(R)$
, we denote by
$S(i),P(i),I(i)$
a simple, projective or injective module corresponding to the vertex
$i$
, respectively.
The valuation of any arrow can be interpreted as follows (
$\unicode[STIX]{x1D70F}$
is the Auslander–Reiten translation).
Lemma 2. If
$Q(\unicode[STIX]{x1D6EC})=(1\rightarrow 2)$
, then the arrow
$1\rightarrow 2$
has valuation at least
$2$
if and only if
$I(2)$
is not projective if and only if
$P(1)$
is not injective. If the arrow
$1\rightarrow 2$
has valuation at least
$3$
, then
$\unicode[STIX]{x1D70F}S(1)$
is neither projective, nor a neighbor of
$P(1)$
in the Auslander–Reiten quiver, consequently
$\operatorname{Hom}(P(1),\unicode[STIX]{x1D70F}^{2}S(1))\neq 0,$
thus
$\operatorname{Ext}(\unicode[STIX]{x1D70F}S(1),P(1))\neq 0$
.◻
In the proof of Proposition 5, we have to construct some exceptional modules. Two general results will be needed.
Lemma 3. Let
$e$
be an idempotent of the artin algebra
$\unicode[STIX]{x1D6EC}$
and
$\langle e\rangle$
the two-sided ideal generated by
$e$
. Let
$M$
be a
$\unicode[STIX]{x1D6EC}$
-module with
$eM=0.$
Then
$M$
is exceptional as a
$\unicode[STIX]{x1D6EC}$
-module if and only if
$M$
is exceptional when considered as a
$\unicode[STIX]{x1D6EC}/\langle e\rangle$
-module.
Proof. Of course, if
$0\rightarrow M\rightarrow M^{\prime }\rightarrow M\rightarrow 0$
is an exact sequence in
$\operatorname{mod}\unicode[STIX]{x1D6EC}$
, then
$eM^{\prime }=0$
, thus it is an exact sequence in
$\operatorname{mod}\unicode[STIX]{x1D6EC}/\langle e\rangle$
.◻
A
$\unicode[STIX]{x1D6EC}$
-module
$M$
is said to be sincere provided there is no nonzero idempotent
$e\in \unicode[STIX]{x1D6EC}$
with
$eM=0$
.
Lemma 4. Any connected hereditary artin algebra
$\unicode[STIX]{x1D6EC}$
has sincere exceptional modules.
(Let us add that sincere exceptional modules are even faithful, see for example [Reference RingelR2, Corollary 2.3].)
Proof, using induction on the number
$n$
of vertices of
$Q(\unicode[STIX]{x1D6EC})$
. If
$n=1$
, then any simple
$\unicode[STIX]{x1D6EC}$
-module is a sincere exceptional module.
Now assume that
$n\geqslant 2$
. Up to duality, we can assume that there exists a simple injective module
$S$
such that the full subquiver
$Q^{\prime }$
of
$Q(\unicode[STIX]{x1D6EC})$
whose vertices are the isomorphism classes
$[S^{\prime }]$
of the simple modules
$S^{\prime }$
which are not isomorphic to
$S$
is connected. Let
$\unicode[STIX]{x1D6EC}^{\prime }$
be the restriction of
$\unicode[STIX]{x1D6EC}$
to
$Q^{\prime }$
. By induction, there is a sincere exceptional
$\unicode[STIX]{x1D6EC}^{\prime }$
-module
$M^{\prime }$
. We form the universal extension
$M$
of
$M^{\prime }$
by
$S$
, thus there is an exact sequence
$$\begin{eqnarray}0\rightarrow M^{\prime }\rightarrow M\rightarrow S^{t}\rightarrow 0\end{eqnarray}$$
with
$t\geqslant 1$
such that
$S$
is not a direct summand of
$M$
and
$\operatorname{Ext}(S,M)=0.$
It is well-known (and easy to see) that
$M$
is indecomposable and has no self-extensions.◻
The proof of Proposition 5 requires to look at four special cases.
Case 1. The algebra
$\unicode[STIX]{x1D6EC}$
is tame.
We use the structure of the Auslander–Reiten quiver of
$\unicode[STIX]{x1D6EC}$
as presented in [Reference Dlab and RingelDR]. Since we assume that
$\unicode[STIX]{x1D6EC}$
has at least 3 vertices, there is a tube of rank
$r\geqslant 2$
. The simple regular modules in this component form an
$\operatorname{Ext}$
-cycle of cardinality
$r$
, say
$X_{1},\ldots ,X_{r}$
. There is a unique indecomposable module
$Y$
with a filtration
$Y=Y_{0}\supset Y_{1}\supset \cdots \supset Y_{r-1}=0$
such that
$Y_{i-1}/Y_{i}=X_{i}$
for
$1\leqslant i\leqslant r-1.$
Clearly, the pair
$Y,X_{r}$
is an
$\operatorname{Ext}$
-pair.
Case 2. The quiver
$Q=Q(\unicode[STIX]{x1D6EC})$
is not a tree.
Deleting, if necessary, vertices, we may assume that the underlying graph of
$Q$
is a cycle. Let
$w$
be a path from a source
$i$
to a sink
$j$
of smallest length, let
$Q^{\prime }$
be the subquiver of
$Q$
given by the vertices and the arrows which occur in
$w$
. Not every vertex of
$Q$
belongs to
$Q^{\prime }$
, since otherwise
$Q$
is obtained from
$Q^{\prime }$
by adding just arrows, thus by adding a unique arrow, namely an arrow
$i\rightarrow j$
. But then this arrow is also a path from a sink to a source, and it has length
$1$
. By the minimality of
$w$
, we see that also
$w$
has length
$1$
and therefore
$Q$
has just the two vertices
$i,j$
. But then
$Q$
can have only one arrow, thus is a tree. This is a contradiction.
Let
$Q^{\prime \prime }$
be the full subquiver given by all vertices of
$Q$
which do not belong to
$Q^{\prime }$
. Of course,
$Q^{\prime \prime }$
is connected (it is a quiver of type
$\mathbb{A}$
). According to Lemma 4, there is an exceptional module
$X$
with support
$Q^{\prime }$
and an exceptional module
$Y$
with support
$Q^{\prime \prime }$
. Since
$Q^{\prime },Q^{\prime \prime }$
have no vertex in common, we see that
$\operatorname{Hom}(X,Y)=0=\operatorname{Hom}(Y,X)$
.
There is an arrow
$i\rightarrow j^{\prime \prime }$
with
$j^{\prime \prime }$
a vertex of
$Q^{\prime \prime }$
. This arrow shows that
$\operatorname{Ext}(X,Y)\neq 0.$
Similarly, there is an arrow
$i^{\prime \prime }\rightarrow j$
with
$i^{\prime \prime }$
a vertex of
$Q^{\prime \prime }$
. This arrow shows that
$\operatorname{Ext}(Y,X)\neq 0.$
We consider now algebras
$\unicode[STIX]{x1D6EC}$
with
$\operatorname{Ext}$
-quiver
$1\rightarrow 2\rightarrow 3$
. We denote by
$\unicode[STIX]{x1D6EC}^{\prime }$
the restriction of
$\unicode[STIX]{x1D6EC}$
to the subquiver with vertices
$1,2$
, and by
$\unicode[STIX]{x1D6EC}^{\prime \prime }$
the restriction of
$\unicode[STIX]{x1D6EC}$
to the subquiver with vertices
$2,3.$
Given a representation
$M$
, let
$M_{3}$
be the sum of all submodules of
$M$
which are isomorphic to
$S(3),$
then
$M/M_{3}$
is a
$\unicode[STIX]{x1D6EC}^{\prime }$
-module.
Lemma 5. Let
$X,Y$
be
$\unicode[STIX]{x1D6EC}$
-modules. If
$X_{3}=0$
and
$\operatorname{Ext}(Y/Y_{3},X)\neq 0$
, then also
$\operatorname{Ext}(Y,X)\neq 0.$
Proof. The exact sequence
$0\rightarrow Y_{3}\rightarrow Y\rightarrow Y/Y_{3}\rightarrow 0$
yields an exact sequence
$$\begin{eqnarray}\operatorname{Hom}(Y_{3},X)\rightarrow \operatorname{Ext}(Y/Y_{3},X)\rightarrow \operatorname{Ext}(Y,X).\end{eqnarray}$$
The first term is zero, since
$Y_{3}$
is a sum of copies of
$S(3)$
and
$X_{3}=0$
. Thus, the map
$\operatorname{Ext}(Y/Y_{3},X)\rightarrow \operatorname{Ext}(Y,X)$
is injective.
Case 3.
$Q(\unicode[STIX]{x1D6EC})=(1\rightarrow 2\rightarrow 3)$
, and
$v(1,2)\geqslant 2,~v(2,3)\geqslant 2.$
Let
$X=S(2)$
and let
$Y$
be the universal extension of
$X$
using the modules
$S(1)$
and
$S(3)$
(thus, we form the universal extension from above using copies of
$S(1)$
and the universal extension from below using copies of
$S(3)$
). Clearly,
$Y$
is exceptional. Since the socle of
$Y$
consists of copies of
$S(3)$
, we have
$\operatorname{Hom}(S(2),Y)=0.$
Since the top of
$Y$
consists of copies of
$S(1)$
, we have
$\operatorname{Hom}(Y,S(2))=0.$
Since
$v(1,2)\geqslant 2$
, the module
$Y/Y_{3}$
is not a projective
$\unicode[STIX]{x1D6EC}^{\prime }$
-module. As a consequence,
$\operatorname{Ext}(Y/Y_{3},S(2))\neq 0.$
Lemma 5 shows that also
$\operatorname{Ext}(Y,S(2))\neq 0.$
By duality, we similarly see that
$\operatorname{Ext}(S(2),Y)\neq 0.$
Case 4.
$Q(\unicode[STIX]{x1D6EC})=(1\rightarrow 2\rightarrow 3)$
, and
$v(1,2)\geqslant 3,~v(2,3)=1.$
Let
$X=P(1)/P(1)_{3}$
(thus
$X$
is the projective
$\unicode[STIX]{x1D6EC}^{\prime }$
-module with top
$S(1)$
). Let
$Y=\unicode[STIX]{x1D70F}X$
, where
$\unicode[STIX]{x1D70F}=D\operatorname{Tr}$
is the Auslander–Reiten translation in
$\operatorname{mod}\unicode[STIX]{x1D6EC}$
. Of course, both modules
$X,Y$
are exceptional. Since
$Y=\unicode[STIX]{x1D70F}X,$
we know already that
$\operatorname{Ext}(X,Y)\neq 0.$
We claim that
$Y/Y_{3}=\unicode[STIX]{x1D70F}^{\prime }S(1)$
, where
$\unicode[STIX]{x1D70F}^{\prime }$
is the Auslander–Reiten translation of
$\unicode[STIX]{x1D6EC}^{\prime }$
. Since
$P(1)_{3}=S(3)^{a}$
for some
$a\geqslant 1$
, a minimal projective presentation of
$X$
has the form
$$\begin{eqnarray}0\rightarrow S(3)^{a}\rightarrow P(1)\rightarrow X\rightarrow 0,\end{eqnarray}$$
thus the defining exact sequences for
$Y=\unicode[STIX]{x1D70F}X$
is of the form
$$\begin{eqnarray}0\rightarrow Y\rightarrow I(3)^{a}\rightarrow S(1)\rightarrow 0.\end{eqnarray}$$
In order to obtain
$\unicode[STIX]{x1D70F}^{\prime }S(1)$
, we start with a minimal projective presentation
$$\begin{eqnarray}0\rightarrow S(2)^{a}\rightarrow P^{\prime }(1)\rightarrow S(1)\rightarrow 0,\end{eqnarray}$$
where
$P^{\prime }(1)$
is the projective cover of
$S(1)$
as a
$\unicode[STIX]{x1D6EC}^{\prime }$
-module (actually,
$P^{\prime }(1)=X$
). Since
$\unicode[STIX]{x1D708}(2,3)=1$
, the number
$a$
in (*) and (**) is the same. The defining exact sequences for
$Y=\unicode[STIX]{x1D70F}X$
and
$\unicode[STIX]{x1D70F}^{\prime }S(1)$
are part of the following commutative diagram with exact rows and columns:
The left column shows that
$Y/Y_{3}=\unicode[STIX]{x1D70F}^{\prime }S(1)$
.
As we have mentioned in Lemma 2,
$v(1,2)\geqslant 3$
implies that
$\operatorname{Ext}(\unicode[STIX]{x1D70F}^{\prime }S(1),P^{\prime }(1))\neq 0$
. According to Lemma 5, we see that
$\operatorname{Ext}(Y,X)\neq 0.$
Finally, let us show that
$X,Y$
are orthogonal. Since
$Y=\unicode[STIX]{x1D70F}X$
and
$X$
is exceptional, we see that
$\operatorname{Hom}(X,Y)=0.$
On the other hand, any homomorphism
$Y\rightarrow X$
vanishes on
$Y_{3}$
, since
$X$
has no composition factor
$S(3)$
. Now
$Y/Y_{3}$
is indecomposable and not projective as a
$\unicode[STIX]{x1D6EC}^{\prime }$
-module, whereas
$X$
is a projective
$\unicode[STIX]{x1D6EC}^{\prime }$
-module, thus
$\operatorname{Hom}(Y,X)=\operatorname{Hom}(Y/Y_{3},X)=0.$
Remark.
Concerning the cases 3 and 4, there is an alternative proof which uses dimension vectors and the Euler form on the Grothendieck group
$K_{0}(\unicode[STIX]{x1D6EC})$
. But for this approach, one needs to deal with the valuation of
$Q(\unicode[STIX]{x1D6EC})$
as in [Reference Dlab and RingelDR], attaching to any arrow
$i\rightarrow j$
a pair
$(a,b)$
of positive numbers instead of the single number
$v(i,j)=ab$
.
Proof of Proposition 5.
Let
$\unicode[STIX]{x1D6EC}$
be connected, hereditary, representation-infinite, with at least 3 simple modules. Case 2 shows that we can assume that
$Q(\unicode[STIX]{x1D6EC})$
is a tree.
Assume that there is a subquiver
$Q^{\prime }$
such that at least two of the arrows have valuation at least
$2$
, choose such a
$Q^{\prime }$
of minimal length. We are to construct an
$\operatorname{Ext}$
-pair for the restriction of
$\unicode[STIX]{x1D6EC}$
to
$Q^{\prime }$
. Using reflection functors (see [Reference Dlab and RingelDR]), we can assume that
$Q^{\prime }$
has orientation
$1\rightarrow 2\rightarrow \cdots \rightarrow n\!-\!1\rightarrow n$
. If
$n=3,$
then this is case 3. Thus assume
$n\geqslant 4.$
The minimality of
$Q^{\prime }$
asserts that
$\unicode[STIX]{x1D708}(i,i+1)=1$
for
$2\leqslant i\leqslant n-2.$
If we denote by
$\unicode[STIX]{x1D6EC}^{\prime }$
the restriction of
$\unicode[STIX]{x1D6EC}$
to
$Q^{\prime }$
, then
$\unicode[STIX]{x1D6EC}^{\prime }$
has a full exact abelian subcategory
${\mathcal{U}}$
which is equivalent to the module category of an algebra as discussed in case 3 (namely the subcategory of all
$\unicode[STIX]{x1D6EC}^{\prime }$
-modules which do not have submodules of the form
$S(i)$
with
$2\leqslant i\leqslant n-2$
and no factor modules of the form
$S(i)$
with
$3\leqslant i\leqslant n-1$
). Since
${\mathcal{U}}$
has
$\operatorname{Ext}$
-pairs, also
$\operatorname{mod}\unicode[STIX]{x1D6EC}$
has
$\operatorname{Ext}$
-pairs.
Thus, we can assume that at most one arrow
$i\rightarrow j$
has valuation greater than
$2$
. If
$v(i,j)\geqslant 3$
, then we take a connected subquiver
$Q^{\prime }$
with 3 vertices containing this arrow
$i\rightarrow j$
. If necessary, we use again reflection functors in order to change the orientation so that we are in case 4.
Thus we are left with the representation-infinite algebras
$\unicode[STIX]{x1D6EC}$
with the following properties:
$Q(\unicode[STIX]{x1D6EC})$
is a tree, there is no arrow with valuation greater than
$2$
and at most one arrow with valuation equal to
$2$
. It is easy to see that
$Q(\unicode[STIX]{x1D6EC})$
contains a subquiver
$Q^{\prime }$
such that the restriction of
$\unicode[STIX]{x1D6EC}$
to
$Q^{\prime }$
is tame, thus we can use case 1.◻
Proof of Theorem.
Let
$\unicode[STIX]{x1D6EC}$
be connected and hereditary. If
$\unicode[STIX]{x1D6EC}$
is representation-finite, then
$\operatorname{tors}\unicode[STIX]{x1D6EC}=\operatorname{f - tors}\unicode[STIX]{x1D6EC}$
, thus
$\operatorname{f - tors}\unicode[STIX]{x1D6EC}$
is a lattice. If
$\unicode[STIX]{x1D6EC}$
has precisely two simple modules, then
$\operatorname{f - tors}\unicode[STIX]{x1D6EC}$
can be described easily (see [Reference Iyama, Reiten, Thomas and TodorovIRTT, proof of Proposition 2.2] which works in general), thus
$\operatorname{f - tors}\unicode[STIX]{x1D6EC}$
obviously is a lattice.
On the other hand, if
$\unicode[STIX]{x1D6EC}$
is representation-infinite and has at least three simple modules, then Proposition 5 asserts that
$\unicode[STIX]{x1D6EC}$
has an
$\operatorname{Ext}$
-pair, say
$X,Y$
. Since
$X,Y$
are exceptional modules, Proposition 1 shows that
${\mathcal{T}}(X)={\mathcal{G}}(X)$
and
${\mathcal{T}}(Y)={\mathcal{G}}(Y)$
both belong to
$\operatorname{f - tors}\unicode[STIX]{x1D6EC}$
. The join of
${\mathcal{T}}(X)$
and
${\mathcal{T}}(Y)$
in
$\operatorname{tors}\unicode[STIX]{x1D6EC}$
is
${\mathcal{T}}(X,Y)$
. According to Proposition 4,
${\mathcal{T}}(X,Y)$
cannot belong to
$\operatorname{f - tors}\unicode[STIX]{x1D6EC}$
.◻
