Abstract
Abstract. The aim of this paper is to study the dominant dimension of two important classes of finite dimensional algebras, namely, hereditary algebras and tree algebras. We derive an explicit formula for the dominant dimension of each class.
It is quite common to classify algebras by certain homological invariants. One such classification of finite dimensional algebras with respect to the length of an exact sequence of their projective-injective bimodules was proposed by Nakayama In applied sense, dominant dimension has been used not only to characterize the double centralizer property but also to classify certain algebras. In [10] the dominant dimension has been used to prove several Schur-Weyl-dualities. Though the theory of dominant dimension is growing rapidly in applied context, see [6, 9] , the precise value of dominant dimension for many well-known classes of algebras is still unknown. Algebras of infinite dominant dimension also have been of interest of many ( e.g. [3]) in connection with Nakayama's conjecture, but the above perspective also suggests to investigate the information itself about the dominant dimension (finite) of many important classes of finite dimensional algebras.
In this paper we study the dominant dimensions of two well-known classes of algebras, namely hereditary algebras and tree algebras. We use quiver-theoretic techniques and give explicit combinatorial proofs of the results.
In Section 3, we consider hereditary algebras (quiver), and establish that a branching vertex plays a key role to characterize such class of algebras in terms of dominant dimensions. We conclude this section by Theorem. 3.6. Let A = KQ be a path algebra of a finite, connected and acyclic quiver Q. Then
where A n is linearly oriented.
Indroduction
It is quite common to classify algebras by certain homological invariants. One such classification of finite dimensional algebras with respect to the length of an exact sequence of their projective-injective bimodules was proposed by Nakayama [15] . In [18] Tachikawa characterized QF -3 algebras by such length. Subsequently in 1964, Tachikawa [19] introduced the notion of dominant dimension, where he studied the dominant dimension of QF -3 algebras as well. Later on, the classical theory of dominant dimension has been developed by Mueller [14] , Tachikawa [20] , Morita [13] and few others (e.g. [17] ). The dominant dimension also provides one of the two conditions (the other one also is about a homological dimension: global dimension) in Auslander's [2] celebrated characterization of finite representation type, that is of the representation category being finite.
In applied sense, dominant dimension has been used not only to characterize the double centralizer property but also to classify certain algebras. In [10] the dominant dimension has been used to prove several Schur-Weyl-dualities. Though the theory of dominant dimension is growing rapidly in applied context, see [6, 9] , the precise value of dominant dimension for many well-known classes of algebras is still unknown. Algebras of infinite dominant dimension also have been of interest of many ( e.g. [3] ) in connection with Nakayama's conjecture, but the above perspective also suggests to investigate the information itself about the dominant dimension (finite) of many important classes of finite dimensional algebras.
In this paper we study the dominant dimensions of two well-known classes of algebras, namely hereditary algebras and tree algebras. We use quiver-theoretic techniques and give explicit combinatorial proofs of the results.
In Section 3, we consider hereditary algebras (quiver), and establish that a branching vertex plays a key role to characterize such class of algebras in terms of dominant dimensions. We conclude this section by Theorem. 3.6. Let A = KQ be a path algebra of a finite, connected and acyclic quiver Q. Then
where A n is linearly oriented.
This research was fully supported by Kohat University of Science and Technology (KUST), Kohat, Pakistan, through the Human Resource Development (HRD) program of KUST.
In Section 4, we pass through the quotients of A n , and show that the quotients having free vertices have dominant dimension not greater than two. We establish that Theorem. 4.3 . Let A be a bound quiver algebra of A n . Then for a fixed n ≥ 3 1 ≤ dom.dimA ≤ n − 1.
It is also shown that these bounds are attained by quotients of A n , and that every natural number occurs as dominant dimension. For the full set of fully overlapped zero relations of the same length m, we derive an explicit formula for dom.dimA. In Section 5, we study the dominant dimension of tree ( = A n ) algebras. We define arms of a tree, and split trees into two classes, namely, trees without arms and trees with arms. The following is an example of such two classes. Like hereditary algebras, it turns out that the dominant dimension of tree ( = A n ) algebras also can not exceed one. To deal with trees without arms, we define ( see Definition 5.4) conditions ( * ) as a set of relations satisfying:
(i) For each source a and sink c, both P (a) and I(c) are uniserial.
(ii) For each i ∈ Q In Subsection 5.2, we pay attention to the trees with arms. Of course, trees with arms having, as set of relations, the conditions ( * ) only act just like trees without arms, as Proposition 5.13 says. But in general, sets of relations on trees with arms are bigger than and might be containing the conditions ( * ). Hence this leads to the conditions ( * * ), an extension of the conditions ( * ). We define such conditions as (see Definition 5.17):
Let R and R ′ be sets of zero relations on Q and Q ′ , respectively, such that R ′ ⊆ R. Then R is said to satisfy the conditions ( * * ) if 
This paper is a part of a comprehensive project on dominant dimensions where finite dimensional algebras are to be characterized explicitly by precise values or by a range of values of their dominant dimensions.
Preliminaries
Here we recall some basic notions from quiver theory and make some useful conventions. We also give few elementary results.
Throughout, K is assumed to be a field, and Q = (Q 0 , Q 1 , s, t) a finite, connected and acyclic quiver, and A n a linearly oriented quiver having Q 0 = {1, 2, 3 · · · , n} as the set of vertices, where n ∈ N. We call Q a tree if there is a unique path between any two vertices in Q 0 . Let x be a path in Q. We denote by Q x 0 and Q x 1 respectively the set of all vertices in x and the set of all arrows in x. We say the path x contains a vertex a if a ∈ Q x 0 . If there exists in Q a path from a to b, then a is said to be a predecessor of b, and b is said to be a successor of a. In particular, if there exists an arrow a → b, then a, written b − , is said to be an immediate predecessor of b, and b, written a + , is said to be an immediate successor of a. We define a relation in Q with coefficients in K as a K-linear combination of paths of length at least two having the same source and target. A zero (or monomial) relation in Q is a relation comprising only one term of K-linear combination, see [1] for details. Any zero relation, by definition, is minimal. By length of a relation we mean the number of arrows in the relation. The source and the target of a zero relation are defined as the source (target) of the first (last) arrow in the zero relation. A path x of length at least one in Q is said to be maximal if it is not a subpath of any other path in KQ or KQ/I.
All the projective P (j) and the injective I(j) modules under consideration are the indecomposable left A-modules corresponding to some vertex j ∈ Q 0 , where A is either a path algebra or a bound quiver algebra. Any projective (injective) module which is injective (projective), up to isomorphism, will be called projectiveinjective. Any zero module, by definition, is projective-injective. Throughout, an injective envelope of a module M is denoted by EM .
The following definition has a fundamental role when dealing with hereditary and tree algebras. Definition 2.1. A vertex a in Q is said to be a branching vertex if there exist distinct arrows α, β ∈ Q 1 such that s(α) = a = s(β) or t(α) = a = t(β).
By definition, A n is a branching-free tree. Definition 2.2. A vertex a in Q is said to be a free vertex if it is neither the source nor the target of any zero relation. Definition 2.3. Let M be an A-module. M is said to have the dominant dimension at least n ∈ N, written dom.dimM ≥ n, if there exists a minimal injective resolution
of M such that all the modules I j with 1 ≤ j ≤ n are projective-injective.
If the injective envelope I 1 of M is not projective, we set dom.dimM = 0. In case dom.dimM ≥ n and dom.dimM n + 1, we say dom.dimA = n. If no such n exists, we write dom.dimM = ∞. The dominant dimension of an algebra A is defined as the dominant dimension of the left regular module A A, that is, dom.dimA = dom.dim A A. A self-injective algebra has infinite dominant dimension, since all of its projective modules are injective. An obvious consequence of the definition is that
where M and N are finite dimensional A-modules. Because the dominant dimension of a projective-injective module is infinite, this consequence implies to forget the trivial part ( where every projective is injective ) of the minimal injective resolution
• is minimal, then I n is not projective.
Proof. Let us assume, on the contrary, that I n is projective. Then the epimorphism I n−1 → I n → 0 splits. This implies that I n is a direct summand of I n−1 . Consequently, I
• is not minimal, but I • was minimal. Hence I n is not projective.
In general, an upper bound of dom.dimA for many algebras is not known yet. But, in particular, directed algebras are bounded above by the number of their projective-injective modules, as shown below. Proof. Because A is not self-injective, there are at most n − 1 projective-injective A-modules, and therefore d ≤ n − 1. Since Q has no oriented cycles, A is a directed algebra. Let
be a minimal injective resolution of A such that I j = ⊕I(a ij ) is non-zero projective for 1 ≤ j ≤ k, where I(a ij ∈ Q 0 ) is indecomposable projective-injective A-module, 1 ≤ i ≤ n j and n j is the number of direct summands of I j . Now for j = 1, 2, · · · , k− 1, each matrix φ j : I j −→ I j+1 has as entries the morphisms φ ij :
Since A is a directed algebra, so φ ij is non-zero, non-invertible, and for each j = 1, 2, · · · , k − 1, a ij b i ′ j+1 . This implies that the injective resolution I
• is finite, as |Q 0 | = n is fixed. Hence there exists some j = k + 2 (say) such that I j = 0 and k + 2 ≤ n + 1. Since I • is minimal, it follows from Lemma 2.4 that I k+1 is not projective. Consequently,
The following Lemma is used frequently to settle many results. 
, which is contrary to the supposition. Since Q is a tree, and both P (a) and I(b) have simple socle and simple top, therefore both P (a) and I(b) are uniserial.
Conversely, assume that there exists a maximal path x from a to b, and both P (a) and I(b) are uniserial. Because x is a maximal path from a to b, and P (a) is uniserial, therefore socP (a) = S(b). This shows that EP (a) = I(b). Thus P (a) ֒→ I(b). Now since I(b) is also uniserial, the maximality of the path x gives topI(b) = S(a). This implies that the projective cover of I(b) is P (a), and hence
Obviously, all the projectives and injectives are uniserial if Q = A n . Let socP (1) = S(j = n) and topI(n) = S(i = 1) where i = j ∈ Q 0 . Since 1 ∈ Q 0 being a source has no predecessors, the path from 1 to j is maximal. Hence P (1) ∼ = I(j). Similarly, the path from i to n is maximal because n being a sink has no successors. Thus I(n) ∼ = P (i).
Hereditary algebras
Throughout the section, except in Proposition 3.1, it is assumed that Q = A n and A = KQ is a path algebra of Q, where A n is a linearly oriented quiver with n vertices.
Proposition 3.1. The path algebra of A n has dominant dimension equal to one.
Proof. Let A be the path algebra of A n . We first show that I(n) is the only injective which, up to isomorphism, is projective. Since every projective P (i) has the simple socle S(n), therefore EP (i) = I(n). As P (1) and I(n) have the same dimension, P (1) ∼ = I(n). Next we show that for all j = 1, 2, ..., n − 1, I(j) is not projective. Suppose, on the contrary, that I(j) is projective. That is, I(j) ∼ = P (k) for a k such that dimI(j) = dimP (k). Now P (k) ∼ = I(j) implies EP (k) = I(j) where j = n, a contradiction to the fact that EP (k) = I(n). Thus I(j) is not projective for j = 1, 2, ..., n − 1. Now the minimal injective resolution of A becomes
where I(n) n is the direct sum of n copies of I(n). Hence dom.dimA = 1.
Lemma 3.2. Every longest path in Q contains at least one branching vertex.
Proof. Let x be a longest path in Q and Q x 0 = {a 1 , . . . , a n } such that s(x) = a 1 and t(x) = a n . Since x is longest, so there does not exist a path, say w such that s(w) = a n or t(w) = a 1 . Now there may or may not exist a vertex
Since Q is connected, there exist: an unoriented path, say y between b and some a i ∈ Q x 0 , and an arrow α ∈ Q y 1 such that
Now by definition, a i is a branching vertex.
Next suppose there does not exist any Proof. Since x is a longest path in Q, it follows from Lemma 3.2 that x contains at least one branching vertex. Thus we can assume that c or d in Q x 0 are the branching vertices.
Assume that x 1 and x 2 be two paths with s(
, where b 1 is the sink and a 1 is the source, and ℓ(x) is the length of the path x. Hence x 1 , x 2 are not the subpaths. Now since x 1 is not a subpath and b 1 is a sink, so x 1 is maximal and hence annihilated by the radical rad(A) of the path algebra A = KQ. i.e. rad(A).x 1 = 0. This implies that S(b 1 ) is a summand of socP (a). Also S(b) is a summand of socP (a). Note that if b 1 = b then we have two copies of the simple A-module S(b) in the socle. Hence socP (a) has at least two simple summands.
Similarly x 2 is not a subpath and hence is maximal, so that < x * 2 >⊂ top(I(b)). Hence the simple A-modules S(a) and S(a 1 ) are summands of topI(b). If a 1 = a then we have two copies of the simple A-module S(a) in the topI(b). Thus topI(b) has at least two simple summands. This proves the Lemma. Proof. Since x is a longest path in Q, so the proof of Lemma 3.3 gives
Next suppose, on the contrary, that I(b) is projective. Then I(b) ∼ = P (a) gives EP (a) = I(b). This implies that socP (a) and the topI(b) are simple, a contradiction to the fact that socP (a) or topI(b) has at least two simple summands. Hence I(b) is not projective. Proof. It is enough to prove that I 1 = ⊕EP (i) in the minimal injective resolution of A contains a non-projective summand. Since Q is finite, connected and acyclic, it contains a longest path, say, from a to b. Then by Lemma 3.4, I(b) is a nonprojective summand of EP (a) and hence of I 1 . Hence I 1 is not projective showing that dom.dimA = 0.
We summarize this chapter as Theorem 3.6. Let A = KQ be a path algebra of a finite, connected and acyclic quiver Q. Then
In the following two sections we concentrate on the bound quiver algebras of finite trees. Before we proceed further, we observe from the above section that a branching vertex has a central role in computing the dominant dimension of a path algebra of a finite, connected and acyclic quiver. Since every tree, except A n , has at least one branching vertex, it motivates us to consider first the bound quiver algebras of the branching free tree A n .
Bound quiver algebras of A n
We consider the quotient algebras of A n for n ≥ 3. Throughout this section, we assume that Q = A n = Q n and that A = KQ/I is a bound quiver algebra of Q, where I is an admissible of KQ generated by a certain set of zero relations. We go through different sets of zero relations to investigate how dom.dimA depends on the choice of zero relations. We find lower and upper bound of dom.dimA and show by examples that these bounds are optimal.
For convenience, we denote by dd(P (i), Q j ) the dominant dimension of the projective module P (i) when Q = Q j where 3 ≤ j < n. If Q = Q n , we write dd(P (i), Q n ) = ddP (i). Given a set R of zero relations on Q, we denote by R Proof. Let a ∈ Q 0 be an arbitrary vertex. Then a may or may not be the source of a maximal path. If a is the source of a maximal path, say x with target t(x), then by Lemma 2.6 P (a) ∼ = I(t(x)). Hence the injective envelope EP (a) = I(t(x)) of P (a) is projective. Now assume that a is not the source of any maximal path. Let y be the longest path with source a. Then EP (a) = E(socP (a)) = E(S(t(y))) = I(t(y)). The lemma follows if we show that t(y) is the target of some maximal path. Since the path y is not maximal, there exists the smallest predecessor c of a such that the path z from c to t(y) is maximal. Thus t(z) = t(y) and hence I(t(y)) = I(t(z)) is projective.
The following Proposition gives lower bound of dom.dimA. Proof. We have dom.dimA ≥ 1 from Proposition 4.2. We need to find a projective P such that ddP ≯ 1.
Proof.
First we assume that the source of A n is free, that is, 1 is not the source of any zero relation in R. Then the path from 1 to t − is maximal, where t ∈ R t 0 is smallest. Hence P (1) ∼ = I(t − ) and socP (2) = S(t − ). Now ddP (2) = 1, as obvious from the following resolution
where I (1) is not projective. Hence dom.dimA = 1. Next, suppose that the sink n of A n is free. Then I(n) ∼ = P (s + ) and topI(n−1) = S(s + ) where s ∈ R s 0 is largest. Because I(n − 1) is not projective, the resolution 0
shows that ddP (n) = 1, and ultimately dom.dimA = 1.
In view of Lemma 4.4, from now on we assume that 1 ∈ R s 0 and n ∈ R t 0 for every set R of zero relations on Q.
Proposition 4.5. Let R be a set of zero relations on Q.
(
Proof. (i) Assume that a / ∈ R s 0 and a + / ∈ R t 0 for some a ∈ Q 0 . Then P (a + ) is not injective and I(a) is not projective. Now it is easy to see that P (s + ) ∼ = I(t − ), where s ∈ R s 0 is largest but s ≤ a − and t ∈ R t 0 is the target of zero relation with smallest
with I(a) not projective. Hence dom.dimA = 1.
(ii) Let a be a free vertex for R. All we need is to find a projective P such that
where I(a − ) is not projective, and therefore ddP (a + ) = 2. Hence dom.dimA can not exceed two, or dom.dimA ≤ 2.
In the following we list some of those sets which satisfy the conditions of the Proposition 4.5.
Remark 4.6. We observe that there always exists a free vertex for the following sets R of relations on A n , where n is fixed.
(1) Every two relations in R are disjoint. Thus, in order to generate larger dominant dimensions, it is now essential to consider the sets of fully overlapped zero relations on Q. (
where n ∈ mN + j.
Proof. (i) If m = 2, then P (i) ∼ = I(i + 1) for all i = 1, 2, · · · , n − 1, and thus P (n) is the only projective which is not injective. The minimal injective resolution of P (n) becomes
where I(1) is not projective, and hence dom.dimA = ddP (n) = n − 1.
(ii) The projectives P (i = n − m + 2, n − m + 3, · · · , n − 1, n) are not injective. We first find ddP (n − m + 2). We note that the injectives I(i = 1, 2, · · · , m − 1) are not projective.
Consider the resolution
where the second cokernel n − 2m + 2 n − 2m + 3 . . .
n − m is the indecomposable projective KQ n−m /I m -module P ′ (n − 2m + 2), and Q n−m with (Q n−m ) 0 = {1, 2, · · · , n − m} is a subquiver of Q n . Because exactly two injectives are projective to obtain the subquiver Q n−m , and since all the projectiveinjective KQ n−m /I m -modules are the projective-injective A-modules, so we have
where I m is an admissible ideal of KQ n−m generated by the full set of zero relations of the same length m. By the similar arguments, we obtain
Hence proceeding in this way, we get where n ∈ mN + j, r = m + j with j = 1, 2, · · · , m − 2, m − 1, m, and x times m is subtracted from n to obtain such r. Thus n − mx = r ⇒ x = n − r m and 2x = 2n − 2r m
Substituting the values of x and r in (4.1) and get
where j = 1, 2, · · · , m − 1, m. Now the same process gives
Next we prove that
for all k = n − m + 3, n − m + 4, · · · , n − 1, n. To prove (4.2) we have to show for all j = 1, 2,
where i = 3, 4, · · · , m. Let us consider the resolution
Since I(j + 1) is not projective for each j = 1, 2, · · · , m − 2, it follows from the above resolution that
where (1) is not projective and therefore
with n ∈ mN + j.
Bound quiver algebras of general trees
Throughout the following we assume that Q = A n is a finite tree. is not projective, and again dom.dimB = 0. Definition 5.3. By a left arm of a tree Q we mean a subquiver of Q which is linearly oriented A n with n ≥ 1 from a source to an immediate predecessor of a branching vertex. Similarly, a right arm of Q is a subquiver of Q which is linearly oriented A n with n ≥ 1 from an immediate successor of a branching vertex to a sink.
A left (right) arm is said to be trivial if it is A n with n = 1, otherwise it is called non-trivial. A tree is said to be a tree without arms if it has no non-trivial arms. We always denote by Q ′ = (Q Proof. Assume, to the contrary, that the number of sources is not equal to the number of sinks. Let X and Y be the sets of sources and sinks of Q ′ , respectively. (i) and (ii) of the conditions ( * ) imply that for each source a, socP (a) = S(c) for some sink c. This defines a map, say, f : X → Y sending each source a to a unique sink c. If the number of sources is greater than the number of sinks, then there exist at least two sources, say a 1 and a 2 , and a sink c such that f (a 1 ) = c = f (a 2 ). This implies that f is not injective. Consequently I(c) is not uniserial, a contradiction to (i).
Dually, (i) and (iii) together define a map g : Y → X by g(c) = a such that topI(c) = S(a). Now if there are more sinks than sources, then there are at least two sinks, say c 1 and c 2 , and a source a such that g(c 1 ) = a = g(c 2 ). Hence g is not injective as well, and consequently P (a) is not uniserial, again a contradiction to (i). Hence Q ′ has equal number of sources and sinks.
An immediate consequence is the following Corollary 5.7. The conditions ( * ) imply: sources and sinks are in one-to-one correspondence: there is a unique maximal path from each source a to a unique sink c such that P (a) ∼ = I(c).
Proof. Suppose conditions ( * ) hold. It follows from Lemma 5.6, that Q ′ has equal number of sources and sinks. Let a be a source in Q ′ . Then we have from the conditions ( * ) that socP (a) = S(c) and topI(c) = S(a) where c is a unique sink. This implies that the path from a to c is maximal, and hence P (a) ∼ = I(c).
The reverse implication is not true in general, as shown in the following

DOMINANT DIMENSIONS OF TWO CLASSES OF FINITE DIMENSIONAL ALGEBRAS 13
Example 5.8. Let Q ′ be the following tree without arms:
Let {βα, γβ} be a set of zero relations on Q ′ . Then sources and sinks are in oneto-one correspondence, but it does not imply the conditions ( * ). For socP (2) = S(3) ⊕ S(5), while the vertex 5 is not a sink.
Proof. We assume that R ′ does not satisfy the conditions ( * ). First, let a be a source such that P (a) is not uniserial. Then P (a) is not injective. Hence EP (a) is not projective, because a is a source. This gives dom.dimB ′ = 0. Similarly, if I(c) is not uniserial for some sink c, then obviously I(c) is not projective and so is EP (c) = I(c). Consequently, dom.dimB ′ = 0. Now we assume that R ′ satisfies (i) but does not satisfy (ii). Then obviously (iii) is also not satisfied. Because (ii) does not hold, there exist two leftmost vertices, say, i and h such that socP (i) ⊇ S(h), where h is not a sink. Now if there exist a source j and a path x from j to h, then x is not zero in the algebra B ′ because otherwise it would contradict the fact that h is leftmost. This shows that topI(h) containing S(j) and S(i) is not simple. Hence EP (i) ⊇ I(h) is not projective, and dom.dimB ′ = 0. If no such j and x exist, then topI(h) = S(i) is simple. Since Q ′ has no arms,
, then i must be a source of Q ′ , since i is leftmost. Thus P (i) is uniserial by (i). Now there exists at least one successor h Proof. By Lemma 5.6, Q ′ has equal number of sources and sinks. Since R ′ satisfies the conditions ( * ), it follows immediately from Lemma 5.10 that the injective envelope EP (a) of P (a) is projective for each vertex a ∈ Q ′ 0 . This implies that dom.dimB ′ ≥ 1.
Next we show that dom.dimB ′ = 1. Let c ∈ Q ′ 0 be a sink. Then there exists a unique maximal path from the corresponding source a to c such that 
Trees with arms.
In this subsection, a bound quiver algebra KQ/I of a tree Q with arms will be denoted simply by B, where I is an admissible ideal of KQ generated by a set R of zero relations on Q.
In general, dom.dimB is not equal to dom.dimB ′ , but we have the following Proposition 5.13. Let R and R ′ be sets of zero relations on Q and Q ′ respectively. 
DOMINANT DIMENSIONS OF TWO CLASSES OF FINITE DIMENSIONAL ALGEBRAS 15
The reverse implication in part (b) of Proposition 5.13 is not true in general. For we have the following Example 5.14. Let Q be the following tree with arms:
Let R ′ = {θα, βδ} and R = {θα, βδ, γβα} be two sets of zero relations. Then dom.dimB = 0, since EP (3) contains a non-projective summand I(6), but R Proof. We assume that R ′ ⊆ R. Let i be arbitrary in Q 0 Q ′′ 0 . There are two cases: either i is contained in a left arm or it belongs to some right arm. First, we suppose that i is contained in a left arm of Q. Then it is trivial to see that I(i) is uniserial, because arms of Q, by definition, are linearly oriented. Also socP (i) is necessarily simple S(j) for some successor j ∈ Q 0 of i, because R ′ satisfies the conditions ( * ) and R ′ R. Hence P (i) is uniserial. Next, suppose that i belongs to some right arm of Q. Then trivially P (i) is uniserial, because arms are linearly oriented. Again by the same argument that R ′ satisfies the conditions ( * ) and R ′ R, it follows that topI(i) is simple S(h) for some predecessor h ∈ Q 0 of i, and thus I(i) is uniserial. Hence both projective and injective B-modules P (i) and I(i) are uniserial for each i ∈ Q 0 Q ′′ 0 . We note that the assumptions in Lemma 5.15 are not sufficient for the indecomposable projective (injective) B-module P (a) (I(c)) to be injective (projective) for each source a (sink c) in Q, as shown in the following Example 5.16. Let Q be the following tree with arms:
be two sets of zero relations. Then clearly R ′ satisfies the conditions ( * ) and R ′ ⊂ R. We see that the B-modules P (i) and I(i) are uniserial for each i ∈ Q 0 Q is not uniserial.
We tackle this problem by defining on trees with arms the following natural analogue of the conditions ( * ).
Definition 5.17. Let R and R ′ be sets of zero relations on Q and Q ′ , respectively, such that R ′ ⊆ R. Then R is said to satisfy the conditions ( * * ) if
It is important to mention that when R satisfies (i) the conditions ( * * ), then (ii) implies (iii), and vice versa. For if i in a left arm is such that socP (i) = S(j) where j belongs to some right arm. Then topI(j) is necessarily simple S(h) with h either i or some of its predecessors, since otherwise it would contradict the conditions ( * ). Similarly, (iii) implies (ii) can be justified. Proof. We assume that R does not satisfy the conditions ( * * ). Suppose R does not satisfy (i) of the conditions ( * * ), that is, R ′ does not satisfy the condition ( * ). Then it follows from Theorem 5.12 that dom.dimB ′ = 0. Since R ∩ S ′ = R ′ , it follows immediately from Proposition 5.13 that dom.dimB = 0. Now we assume that (i) holds but R does not satisfy (ii) of the conditions ( * * ). Then, by Remark 5.18, (iii) is also not satisfied by R. Since (i) holds, it follows from Lemma 5.19 that Q has equal number of sources and sinks. Let an i in a left arm be such that socP (i) = S(j) for some j ∈ Q ′′ 0 . Because j ∈ Q ′′ 0 and (i) holds, so I(j) is not uniserial, and thus it can not be projective. Hence EP (i) = I(j) is not projective, and ultimately we get dom.dimB = 0. Proof. We assume that R satisfies the conditions ( * * ). Let a be an arbitrary source in Q. Then from (ii) of the conditions ( * * ), socP (a) = S(j) for some successor j ∈ Q 0 Q ′′ 0 of a. Because j / ∈ Q ′′ 0 , I(j) is uniserial by Lemma 5.15. Now the path starting from the source a to j is necessarily maximal, and hence P (a) ∼ = I(j) by Lemma 2.6.
Next we show that I(c) is projective for each sink c in Q. Obviously, c belongs to some right arm. It follows from (iii) of the conditions ( * * ) that topI(c) = S(i) for some predecessor i ∈ Q 0 Q ′′ 0 of c. By Lemma 5.15, P (i) is uniserial. Since the path from i to the sink c is maximal, it follows that P (i) ∼ = I(c). Proof. We assume that R satisfies the conditions ( * * ). Let i ∈ Q 0 be an arbitrary vertex.
First, let i belong to some left arm of Q. From (ii) of the conditions ( * * ), socP (i) = S(j) for some successor j ∈ Q 0 Q ′′ 0 . It follows from Lemma 5.15, that I(j) is also uniserial. Now if the path from i to j is maximal, then Lemma 2.6 gives P (i) ∼ = I(j). Otherwise, there exists in the left arm a predecessor h of i such that the path from h to j is maximal. It follows again from Lemma 2.6, that P (h) ∼ = I(j). This implies that EP (i) = I(j) is projective. Hence EP (i) is projective for each i in a left arm. Now suppose that i does not belong to any left arm of Q. Then P (i) may or may not be uniserial. As the conditions ( * ) also hold, we can assume that socP (i) = ⊕S(j), where each j belongs to some right arm and is the target of some maximal path. From (iii) of the conditions ( * * ), we have topI(j) = S(h) for some predecessor h / ∈ Q ′′ 0 of j. Since each P (h) is uniserial and paths from h to j are maximal, we have P (h) ∼ = I(j) by Lemma 2.6. Hence EP (i) = ⊕I(j) is projective.
Proposition 5.23. Let R and R ′ be sets of zero relations on Q and Q ′ respectively, such that R ′ ⊆ R. If R satisfies the conditions ( * * ), then dom.dimB = 1.
Proof. First we assume that R satisfies the conditions ( * * ). Then it follows immediately from Lemma 5.22 that the injective envelope EP (a) of a B-module P (a) is projective for each vertex a in Q 0 . This implies that dom.dimB ≥ 1. Now to show that dom.dimB ≯ 1, let b ∈ Q ′′ 0 be a branching vertex such that P (b) is not uniserial. I(b) is also not uniserial, since b ∈ Q ′′ 0 and R ′ R. We assume that socP (b) = S(j 1 ) ⊕ S(j 2 ), where j 1 = j 2 , belonging to two distinct right arms, are the targets of some maximal paths. Therefore, there exist i 1 = i 2 in the respective two distinct left arms such that dom.dimB ′ = 0 if R ′ does not satisfy the conditions ( * ).

{'paper_id': '116303892', 'title': 'Dominant dimensions of two classes of finite dimensional algebras', 's2_url': 'https://api.semanticscholar.org/CorpusID:116303892', 'source': 'S2ORC'}