LATTICE STRUCTURE OF TORSION CLASSES FOR HEREDITARY ARTIN ALGEBRAS

Abstract

Let $\unicode[STIX]{x1D6EC}$ be a connected hereditary artin algebra. We show that the set of functorially finite torsion classes of $\unicode[STIX]{x1D6EC}$-modules is a lattice if and only if $\unicode[STIX]{x1D6EC}$ is either representation-finite (thus a Dynkin algebra) or $\unicode[STIX]{x1D6EC}$ has only two simple modules. For the case of $\unicode[STIX]{x1D6EC}$ being the path algebra of a quiver, this result has recently been established by Iyama–Reiten–Thomas–Todorov and our proof follows closely some of their considerations.

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 [IRTT], 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 [IRTT] 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 [Ro], 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 [Ro], and later by Auslander–Smalø [AS]. It is also a consequence of the following Lemma which will be needed for our further considerations.

Lemma 1.

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 $Y$ .

2. (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 [ONFR] or also [R3].

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:

1. (i) The class ${\mathcal{T}}_{U}$ is a torsion class.

2. (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 [DR], 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 [R2, 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 [DR]. 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 [DR], 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 [DR]), 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 [IRTT, 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 $\operatorname{f - tors}\unicode[STIX]{x1D6EC}$ .◻