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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2020; 60(1): 21-43

Published online March 31, 2020

### The Leavitt Path Algebras of Ultragraphs

Mostafa Imanfar and Abdolrasoul Pourabbas∗, Hossein Larki

Faculty of Mathematics and Computer Science, Amirkabir University of Technology, 424 Hafez Avenue, 15914 Tehran, Iran e-mail : m.imanfar@aut.ac.ir and arpabbas@aut.ac.ir
Department of Mathematics, Faculty of Mathematical Sciences and Computer, Shahid Chamran University of Ahvaz, Iran e-mail : h.larki@scu.ac.ir

We introduce the Leavitt path algebras of ultragraphs and we characterize their ideal structures. We then use this notion to introduce and study the algebraic analogy of Exel-Laca algebras.

Keywords: ultragraph C∗,-algebra, Leavitt path algebra, Exel-Laca algebra.

The Cuntz-Krieger algebras were introduced and studied in [6] for binary-valued matrices with finite index. Two immediate and important extensions of the Cuntz-Krieger algebras are: (1) the class of C*-algebras associated to (directed) graphs [5, 8, 10, 11] and (2) the Exel-Laca algebras of infinite matrices with {0, 1}-entries [7]. It is shown in [8] that if E is a graph with no sinks and sources, then the C*-algebra C*(E) is canonically isomorphic to the Exel-Laca algebra , where AE is the edge matrix of E. However, the class of graph C*-algebras and Exel-Laca algebras are differ from each other.

To study both graph C*-algebras and Exel-Laca algebras under one theory, Tomforde [15] introduced the notion of an ultragraph and its associated C*-algebra. Briefly, an ultragraph is a directed graph which allows the range of each edge to be a nonempty set of vertices rather than a singleton vertex. We see in [15] that for each binary-valued matrix A there exists an ultragraph so that the C*-algebra of is isomorphic to the Exel-Laca algebra of A. Furthermore, every graph C*-algebra can be considered as an ultragraph C*-algebra, whereas there is an ultragraph C*-algebra which is not a graph C*-algebra nor an Exel-Laca algebra.

Recently many authors have discussed the algebraic versions of matrix and graph C*-algebras. For example, in [3] the algebraic analogue of the Cuntz-Krieger algebra , denoted by , was studied for finite matrix A, where K is a field. Also the Leavitt path algebra LK(E) of directed graph E was introduced in [1, 2] as the algebraic version of graph C*-algebra C*(E). The class of Leavitt path algebras includes naturally the algebras of [3] as well as the well-known Leavitt algebras L(1, n) of [14]. More recently, Tomforde defined a new version of Leavitt path algebras with coefficients in a unital commutative ring [18]. In the case R = ℂ, the Leavitt path algebra L(E) is isomorphic to a dense *-subalgebra of the graph C*-algebra C*(E) [17].

The purpose of this paper is to define the algebraic versions of ultragraphs C*-algebras and Exel-Laca algebras. For an ultragraph and unital commutative ring R, we define the Leavitt path algebra . To study the ideal structure of , we use the notion of quotient ultragraphs from [13]. Given an admissible pair (H, S) in , we define the Leavitt path algebra associated to the quotient ultragraph and we prove two kinds of uniqueness theorems, namely the Cuntz-Krieger and the graded-uniqueness theorems, for . Next we apply these uniqueness theorems to analyze the ideal structure of . Although the construction of Leavitt path algebra of ultragraph will be similar to that of ordinary graph, we see in Sections 3 and 4 that the analysis of its structure is more complicated. The aim of the definition of ultragraph Leavitt path algebras can be summarized as follows:

• Every Leavitt path algebra of a directed graph can be embedded as an subalgebra in a unital ultragraph Leavitt path algebra. Also, the ultragraph Leavitt path algebra is isomorphic to a dense *-subalgebra of .

• By using the definition of ultragraph Leavitt path algebras, we give an algebraic version of Exel-Laca algebras.

• The class of ultragraph Leavitt path algebras is strictly larger than the class of Leavitt path algebras of directed graphs.

The article is organized as follows. We define in Section 2 the Leavitt path algebra of an ultragraph over a unital commutative ring R. We continue by considering the definition of quotient ultragraphs of [13]. For any admissible pair (H, S) in an ultragraph , we associate the Leavitt path algebra to the quotient ultragraph and we see that the Leavitt path algebras and have a similar behavior in their structure. Next, we prove versions of the graded and Cuntz-Krieger uniqueness theorems for by approximating with R-algebras of finite graphs.

By applying the graded-uniqueness theorem in Section 3, we give a complete description of basic graded ideals of in terms of admissible pairs in . In Section 4, we use the Cuntz-Krieger uniqueness theorem to show that an ultragraph satisfies Condition (K) if and only if every basic ideal in is graded.

In Section 5, we generalize the algebraic Cuntz-Krieger algebra of [3], denoted by ℰℒA(R), for every infinite matrix A with entries in {0, 1} and every unital commutative ring R. In the case R = ℂ, the Exel-Laca ℂ-algebra ℰℒA(ℂ) is isomorphic to dense *-subalgebra of . We prove that the class of Leavitt path algebras of ultragraphs contains the Leavitt path algebras as well as the algebraic Exel-Laca algebras. Furthermore, we give an ultragraph such that the Leavitt path algebra is neither a Leavitt path algebra of graph nor an Exel-Laca R-algebra.

In this section, we follow the standard constructions of [1] and [15] to define the Leavitt path algebra of an ultragraph. Since the quotient of ultragraph is not an ultragraph, we will have to define the Leavitt path algebras of quotient ultragraphs and prove the uniqueness theorems for them to characterize the ideal structure in Section 3.

### 2.1. Ultragraphs

Recall from [15] that an ultragraph consists of a set of vertices G0, a set of edges , the source map and the range map , where ℘(G0) is the collection of all subsets of G0. Throughout this work, ultragraph will be assumed to be countable in the sense that both G0 and are countable.

For a set X, a subcollection of ℘(X) is said to be lattice if and it is closed under the set operations ∩ and ∪. An algebra is a lattice such that for all . If is an ultragraph, we write for the algebra in ℘(G0) generated by .

A path in ultragraph is a sequence α = e1e2 · · · en of edges with for 1 ≤ in − 1 and we say that the path α has length |α| := n. We write for the set of all paths of length n and $Path(G):=∪n=0∞Gn$ for the set of finite paths. We may extend the maps and on by setting and for |α| ≥ 1 and for . For every edge e, We say that e* is the ghost edge associated to e. The following definition is the algebraic version of [15, Definition 2.7].

Definition 2.1

Let be an ultragraph and let R be a unital commutative ring. A Leavitt-family in an R-algebra A is a set {pA, se, se* : and } of elements in A such that

• p∅︀ = 0, pApB = pA∩B and pA∪B = pA + pBpA∩B for all ;

• and for all ;

• for all e, ;

• for every vertex v with $0<∣sG-1(v)∣<∞$,

where pv denotes p{v}. The R-algebra generated by the Leavitt -family {s, p} is denoted by LR(s, p).

We say that the Leavitt -family {s, p} is universal, if B is an R-algebra and {S, P} is a Leavitt -family in B, then there exists an algebra homomorphism φ : LR(s, p) → B such that φ(pA) = PA, φ(se) = Se and φ(se*) = Se* for every and every . The Leavitt path algebra ofwith coefficients in R, denoted by , is the (unique up to isomorphism) R-algebra generated by a universal Leavitt -family.

### 2.2. Quotient Ultragraphs

We will use the notion of quotient ultragraphs and we generalize the definition of Leavitt path algebras for quotient ultragraphs.

Definition 2.2. ([16, Definition 3.1])

Let be an ultragraph. A subcollection is called hereditary if satisfying the following conditions:

• implies for all e.

• ABH for all A, BH.

• AH, and BA, imply BH.

Also, is called saturated if for any vG0 with $0<∣sG-1(v)∣<∞$, we have

${rG(e):e∈G1 and sG(e)=v}⊆H implies {v}∈H.$

For a saturated hereditary subcollection , the breaking vertices of H is denoted by

$BH:={v∈G0:∣sG-1(v)∣=∞ but 0<∣SG-1(v)∩{e:rG(e)∉H}∣<∞}.$

An admissible pair in is a pair (H, S) of a saturated hereditary set and some SBH.

In order to define the quotient of ultragraphs we need to recall and introduce some notations from [13, Section 3]. Let (H, S) be an admissible pair in . For each , we denote Ā := A ∪ {w′ : wA ∩ (BH S)}, where w′ denotes another copy of w. Consider the ultragraph , where , 0 := G0 ∪ {w′ : wBH S} and the maps , are defined by $r¯(e):=rG(e)¯$ and

$s¯(e):={sG(e)′if sG(e)∈BHS and rG(e)∈H,sG(e)otherwise,$

for every , respectively. We write for the algebra generated by the sets {v}, {w′} and r(e).

Lemma 2.3. ([13, Lemma 3.5])

Let (H, S) be an admissible pair in an ultragraphand let ~ be a relation ondefined by A ~ B if and only if there exists VH such that AV = BV. Then ~ is an equivalence relation onand the operations

$[A]∪[B]:=[A∪B],[A]∩[B]:=[A∩B] and [A][B]:=[AB]$

are well-defined on the equivalence classes.

We usually denote [v] instead of [{v}] for every v0. For , we write [A] ⊆ [B] whenever [A] ∩ [B] = [A]. The set ∪AH A is denoted by ∪H.

Definition 2.4. ([13, Definition 3.6])

Let (H, S) be an admissible pair in . The quotient ultragraph ofby (H, S) is the quadruple , where

$Φ(G0):={[v]:v∈G0⋃H}∪{[w′]:w∈BHS},Φ(G1):={e∈G1:rG(e)∉H},$

and and are the maps defined by and $r(e):=[rG(e)¯]$ for every , respectively.

Lemma 2.5

Ifis a quotient ultragraph, then

$Φ(G0)={⋃j=1k⋂i=1njAi,jBi,j:Ai,j,Bi,j∈Φ(G0)∪{r(e):e∈Φ(G1)}},$

whereis the smallest algebra incontaining

${[v],[w′]:v∈G0⋃H,w∈BHS}∪{r(e):e∈Φ(G1)}.$
Proof

We denote by X the right hand side of the above equality. It is clear that , because is an algebra generated by the elements [v], [w′] and r(e). For the reverse inclusion, we note that X is a lattice. Furthermore, one can show that X is closed under the operation . Thus X is an algebra contains [v], [w′] and r(e) and consequently .

### Remark 2.6

If , then $A∪B¯=A¯∪B¯,A∩B¯=A¯∩B¯$ and $AB¯=A¯B¯$. Thus, by applying an analogous lemma of Lemma 2.5 for and , we deduce that for all . One can see that

$G¯0={A¯∪F1∪F2:A∈G0,F1 and F2 are finite subsets of G0 and {w′:w∈BH},respectively}.$

For example we have

$A¯{v}=A{v}¯∪({v}¯{v}∩A¯A),$

and

$A¯{w′}=A{w}¯∪(A∩{w}).$

Furthermore, it follows from Lemma 2.3 that .

### Remark 2.7

The hereditary property of H and Remark 2.6 imply that A ~ B if and only if both A B and B A belong to H.

Similar to ultragraphs, a path in is a sequence α := e1e2 · · · en of edges in such that s(ei+1) ⊆ r(ei) for 1 ≤ in − 1. We say the path α has length |α| := n and we consider the elements in to be paths of length zero. We denote by , the union of paths with finite length. The maps r and s can be naturally extended on . Let be the set of ghost edges. We also define the ghost path$α*:=en*en-1*⋯e1*$ for every and [A]* := [A] for every .

Using Theorem 3.4, we define the Leavitt path algebra of a quotient ultragraph which is corresponding to the quotient R-algebra . We use this concept to characterize the ideal structure of in Section 3. The following definition is the algebraic version of [13, Definition 3.8].

### Definition 2.8

Let be a quotient ultragraph and let R be a unital commutative ring. A Leavitt-family in an R-algebra A is a set and of elements in A such that

• q[∅︀] = 0, q[A]q[B] = q[A][B] and q[A][B] = q[A] + q[B]q[A][B];

• qs(e)te = teqr(e) = te and qr(e)te* = te*qs(e) = te*;

• te*tf = δe, f qr(e);

• q[v] = ∑s(e)=[v]tete* for every [v] ∈ Φ(G0) with 0 < |s−1([v])| < ∞.

The R-algebra generated by the Leavitt -family {t, q} is denoted by LR(t, q). The Leavitt path algebra ofwith coefficients in R, denoted by , is the (unique up to isomorphism) R-algebra generated by a universal Leavitt -family (the definition of universal Leavitt -family is similar to ultragraph case).

If (H, S) = (∅︀, ∅︀), then we have [A] = {A} for each . In this case, every Leavitt -family is a Leavitt -family and vice versa. So, we can consider the ultragraph Leavitt path algebra as .

Let R be a unital commutative ring. For a nonempty set X, we write w(X) for the set of words w := w1w2 · · ·wn from the alphabet X. The free R-algebra generated by X is denoted by . For definition of the free R-algebra we refer the reader to [4, Section 2.3].

Now, we show that for every quotient ultragraph , there exists a universal Leavitt -family. Suppose that and I is the ideal of the free R-algebra generated by the union of the following sets:

• {[∅︀], [A][B] − [AB], },

• {es(e)e, eer(e), e*e*s(e), },

• {e*fδe, f r(e) : e, },

• {v – ∑s(e)=[v]ee* : 0 < |s−1([v])| < ∞}.

If is the quotient map, then it is easy to check that the collection { } is a Leavitt -family. For our convenience, we denote q[A] := π(A), te := π(e) and te* := π(e*) for every and , and we show that the Leavitt -family {t, q} has the universal property.

Assume that {T, Q} is a Leavitt -family in an R-algebra A. If we define f : XA by f([A]) = P[A], f(e) = Te and f(e*) = Te*, then by [4, Proposition 2.6], there is an R-algebra homomorphism such that φf | X = f. Since {T, Q} is a Leavitt -family, φf vanishes on the generator of I. Hence, we can define an R-algebra homomorphism by φ(π(x)) = φf (x) for every . Furthermore, we have φ(q[A]) = Q[A], φ(te) = Te and φ(te*) = Te*.

From now on we denote the universal Leavitt -family and -family by {s, p} and {t, q}, respectively. So, we suppose that {s, p} and {t, q} are the canonical generators of and , respectively.

### Theorem 2.9

Letbe a quotient ultragraph and let R be a unital commutative ring. Thenis of the form

$spanR{tαq[A]tβ*∈LR(G/H,S)):α,β∈Path(G/(H,S)),r(α)∩[A]∩r(β)≠[∅]}.$

$LR(G/(H,S))n=spanR{tαq[A]tβ*∈LR(G/(H,S)):∣α∣-∣β∣=n} (n∈ℤ).$
Proof

If tαq[A]tβ*, , then

$(tαq[A]tβ*) (tμq[B]tν*)={tαμ′q[B]tν*if μ=βμ′ and s(μ′)⊆A∩r(α),tαq[A]∩r(β)∩[B]tν*if μ=β,tαq[A]t(νβ′)*if β=μβ′ and s(β′)⊆B∩r(ν),0otherwise,$

which proves the first statement. Let and consider as defined before. If we define a degree map d : X → ℤ by d([A]) = 0, d(e) = 1 and d(e*) = −1 for every and , then by [4, Proposition 2.7], is a ℤ-graded ring with the grading

$LR(G/(H,S))n=spanR{tαq[A]tβ*∈LR(G/(H,S)):∣α∣-∣β∣=n}.$

### Theorem 2.10

Letbe an ultragraph and let R be a unital commutative ring. Then for all nonempty setsand all rR {0}, the elements rpA ofare nonzero. In particular, rse ≠ 0 and rse*≠ 0 for everyand every rR {0}.

Proof

By the universality, it suffices to generate a Leavitt -family {s̃, p̃} in which rp̃A ≠ 0 for every nonempty set and every rR {0}. For each , we define a disjoint copy Ze := ⊕ R, where the direct sum is taken over countably many copies of R and for each vG0, let

$Zv:={⊕s(e)=vZeif ∣sG-1(v)∣≠0,⊕Rif ∣sG-1(v)∣=0,$

where ⊕ R is another disjoint copy for each v. For every , define PA : ⊕vA Zv → ⊕vA Zv to be the identity map. Also, for each choose an isomorphism and let $Se*:=Se-1:Ze→⊕v∈rG(e)Zv$. Now if Z := ⊕vG0Zv, then we naturally extend all PA, Se, Se* to homomorphisms A, s̃e, s̃e* ∈ HomR(Z, Z), respectively by setting the yet undefined components to zero. It is straightforward to verify that {s̃, p̃} is a Leavitt -family in HomR(Z, Z) such that rp̃A ≠ 0 for every and every rR {0}.

Note that we cannot follow the argument of Theorem 2.10 to show that rq[A] ≠ 0. For example, suppose that is the ultragraph

and let H be the collection of all finite subsets of {v1, v2, . . .}, which is a hereditary and saturated subcollection of . If we consider the quotient ultragraph , then {[∅︀] ≠ [v] : [v] ⊆ r(e)} = ∅︀. So we can not define the idempotent qr(e) : ⊕ [v]⊆r(e)Z[v] → ⊕ [v]⊆r(e)Z[v] as in the proof of Theorem 2.10. In Section 3 we will solve this problem.

### Remark 2.11

Every directed graph E = (E0, E1, r, s) can be considered as an ultragraph , where G0 := E0, and the map is defined by for every . In this case, the algebra is the collection of all finite subsets of G0. The Leavitt path algebra LR(E) is naturally isomorphic to (see [1, 18] for more details about the Leavitt path algebras of directed graphs). So the class of ultragraph Leavitt path algebras contains the class of Leavitt path algebras of directed graphs.

### Lemma 2.12

Letbe an ultragraph and let R be a unital commutative ring. Thenis unital if and only ifand in this case.

Proof

If , then the Relations of Definition 2.1 imply that pG0 is a unit for .

Conversely, suppose that is unital and write

$1LR(G)=∑k=1nrksαkpAksβk*,$

where rkR, and . Let $A:=∪k=1ns(αk)∈G0$. If , then we can choose an element vG0 A and derive

$pv=pv·1LR(G)=∑k=1nrkpvsαkpAksβk*=∑k=1nrkpvpsG(αk)sαkpAksβk*=0,$

this contradicts Theorem 2.10 and it follows , as desired.

We note that, for directed graph E, the algebra LR(E) is unital if and only if E0 is finite. If E0 is infinite, then we can define an ultragraph associated to E such that is unital and LR(E) is an embedding subalgebra of . More precisely, Consider the ultragraph where G0 = E0 ∪ {v}, and . Since , by Lemma 2.12, is unital. Define Se = se and Pv = pv for eE1 and vE0, respectively. It is straightforward to see that {S, P} is a Leavitt E-family for directed graph E. The result now follows by [18, Theorem 5.3].

Let be a quotient ultragraph. We prove the graded and Cuntz-Krieger uniqueness theorems for and . We do this by approximating the Leavitt path algebras of quotient ultragraphs with the Leavitt path algebras of finite graphs. Our proof in this section is standard (see [13, Section 4]), and we give the details for simplicity of further results of the paper.

A vertex [v] ∈ Φ(G0) is called a sink if s−1([v]) = ∅︀ and [v] is called an infinite emitter if |s−1([v])| = ∞. A singular vertex is a vertex that is either a sink or an infinite emitter. The set of all singular vertices is denoted by Φsg(G0).

Let be a finite subset and write F0 := F ∩ Φsg(G0) and . Following [13], we construct a finite graph EF as follows. First, for every ω = (ω1, . . . , ωn) ∈ {0, 1}n {0n}, we define r(ω) := ∩ωi=1r(ei) ωj=0r(ej) and R(ω) := r(ω) [v]∈F0 [v] which belong to . We also set

$Γ0:={ω∈{0,1}n{0n}:there are vertices [v1],…,[vm]} such that R(ω)=∪i=1m[vi] and ∅≠s-1([vi])⊆F1 for 1≤i≤m},$

and

$ΓF:={ω∈{0,1}n{0n}:R(ω)≠[∅] and ω∉Γ0}.$

Now we define the finite graph $EF=(EF0,EF1,rF,sF)$, where

$EF0:=F0∪F1∪ΓF,EF1:={(e,f)∈F1×F1:s(f)⊆r(e)}∪{(e,[v])∈F1×F0:[v]⊆r(e)}∪{(e,ω)∈F1×ΓF:ωi=1 whenever e=ei},$

with sF (e, f) = sF (e, [v]) = sF (e, ω) = e and rF (e, f) = f, rF (e, [v]) = [v], rF (e, ω) = ω.

### Lemma 2.13

Letbe a quotient ultragraph and let R be a unital commutative ring. Then we have the following assertion:

• For every finite set, the elements

$Pe:tete*,P[v]:=q[v](1-∑e∈F1tete*),Pω:=qR(ω)(1-∑e∈F1tete*),S(e,f):=tePf,S(e,[v]):=teP[v],S(e,ω):=tePω,S(e,f)*:=Pfte*,S(e,[v])*:=P[v]te*,S(e,ω)*:=Pωte*,$

form a Leavitt EF -family which generates the subalgebra ofgenerated by {q[v], te, te* : [v] ∈ F0, eF1}.

• For rR {0}, if rq[A] ≠ 0 for all [A] ≠ [∅︀] in, then rPz ≠ 0 for all$z∈EF0$. In this case, we have

$LR(EF)≅LR(S,P)=Alg{q[v],te,te*∈LR(G/(H,S)):[v]∈F0,e∈F1}.$

Proof

The statement (i) follows from the fact that {t, q} is a Leavitt -family, or see the similar [13, Proposition 4.2].

For (ii), fix rR {0}. If rq[A] ≠ 0 for every , then rte ≠ 0 and rte* ≠ 0 for every edge e. Thus rPe ≠ 0 for every edge eF1. Let [v] ∈ F0. If [v] is a sink, then rP[v] = rq[v] ≠ 0. If [v] is an infinite emitter, then there is such that s(f) = [v]. In this case, we have rP[v]tf = rq[v]tf = tf ≠ 0. Therefore rP[v] ≠ 0 for all [v] ∈ F0. Moreover, for each ω ∈ ΓF, there is a vertex [v] ⊆ R(ω) such that either [v] is a sink or there is an edge with s(f) = [v]. In the former case, we have q[v](rPω) = rq[v] ≠ 0 and in the later, tf* (rPω) = rtf* ≠ 0. Thus all rPω are nonzero. Consequently, rPz ≠ 0 for every vertex $z∈EF0$.

Now we show that LR(EF) ~= LR(S, P). Note that for $z∈EF0$ and $g∈EF1$, the degree of Pz, Sg and Sg* as elements in are 0, 1 and −1, respectively. So LR(S, P) is a graded subalgebra of with the grading

$LR(S,P)n:=LR(S,P)∩LR(G/(H,S))n.$

Let {s̃, p̃} be the canonical generators of LR(EF). By the universal property, there is an R-algebra homomorphism π : LR(EF) → LR(S, P) such that π(rp̃z) = rPz ≠ 0, π(g) = Sg and π(g*) = Sg* for $z∈EF0,g∈EF1$ and rR{0}. Since π preserves the degree of generators, the graded-uniqueness theorem for graphs [18, Theorem 5.3] implies that π is injective. As π is also surjective, we conclude that LR(EF) is isomorphic to LR(S, P) as R-algebras.

### Theorem 2.14. (The graded-uniqueness theorem)

Letbe a quotient ultragraph and let R be a unital commutative ring. If T is a-graded ring andis a graded ring homomorphism with π(rq[A]) ≠ 0 for all [A] ≠ [∅︀] inand rR {0}, then π is injective.

Proof

Let {Fn} be an increasing sequence of finite subsets of such that $∪n=1∞Fn=Φsg(G0)∪Φ(G1)$. For each n, the graded subalgebra of generated by {$q[v],te,te*:[v]∈Fn0$ and $e∈Fn1$ }is denoted by Xn. Since π(rq[A]) ≠ 0 for all and rR {0}, by Lemma 2.13, there is an graded isomorphism πn : LR(EFn) → Xn. Thus ππn : LR(EFn) → T is a graded homomorphism.

Fix rR {0}. We show that ππn(rp̃z) = π(rPz) ≠ 0 for all $z∈EFn0$. Since π(rq[A]) ≠ 0 for all , the elements π(rte) and π(rte*) are nonzero for every edge e. So π(rPete) = π(rte) ≠ 0 and thus π(rPe) ≠ 0. Let [v] ∈ F0. If [v] is a sink, then π(rP[v]) = π(rq[v]) ≠ 0. If [v] is an infinite emitter, then there is such that s(f) = [v]. In this case, we have π(rP[v]tf) = π(rq[v]tf) = π(tf) ≠ 0, hence π(rP[v]) ≠ 0. For every ω ∈ ΓF, there is a vertex [v] ⊆ R(ω) such that either [v] is a sink or there is an edge with s(f) = [v]. In the former case, we have π(q[v](rPω)) = π(rq[v]) ≠ 0 and in the later, π(tf* (rPω)) = π(rtf*) ≠ 0. Thus ππn(rp̃z) ≠ 0 for every $z∈EFn0$ and rR {0}. Hence, we may apply the graded-uniqueness theorem for graphs [18, Theorem 5.3] to obtain the injectivity of ππn.

If [v] is a non-singular vertex, then we have q[v] = ∑s(e)=[v]tete*. Furthermore, q[A][B] = q[A]q[A]q[B] for every . Thus, by Lemma 2.5, is an R-algebra generated by

${q[v],te,te*:[v]∈Φsg(G0) and e∈Φ(G1)},$

and so $∪n=1∞Xn-LR(G/(H,S))$. It follows that π is injective on , as desired.

### Corollary 2.15

Letbe an ultragraph, R a unital commutative ring and T a ℤ-graded ring. Ifis a graded ring homomorphism such that π(rpA) ≠ 0 for alland rR {0}, then π is injective.

### Definition 2.16

A loop in a quotient ultragraph is a path α with |α| ≥ 1 and s(α) ⊆ r(α). An exit for a loop α1 · · · αn is an edge with the property that s(f) ⊆ r(αi) but fαi+1 for some 1 ≤ in, where αn+1 := α1. We say that satisfies Condition (L) if every loop α := α1 · · · αn in has an exit, or r(αi) ≠ s(αi+1) for some 1 ≤ in.

### Theorem 2.17. (The Cuntz-Krieger uniqueness theorem)

Letbe a quotient ultragraph satisfying Condition (L) and let R be a unital commutative ring. If T is a ring andis a ring homomorphism such that π(rq[A]) ≠ 0 for everyand rR {0}, then π is injective.

Proof

Choose an increasing sequence {Fn} of finite subsets of such that $∪n=1∞Fn=Φsg(G0)∪Φ(G1)$. Let Xn be the subalgebras of as in Theorem 2.14. Since π(rq[A]) ≠ 0 for all and rR {0}, by Lemma 2.13, there exists an isomorphism πn : LR(EFn) → Xn for each n ∈ ℕ. Furthermore, ππn(rp̃z) ≠ 0 for every $z∈EFn0$ and rR {0}. Since satisfies Condition (L), By [13, Lemma 4.8], all finite graphs EFn satisfy Condition (L). So, the Cuntz-Krieger uniqueness theorem for graphs [18, Theorem 6.5] implies that ππn is injective for n ≥ 1. Now by the fact $∪n=1∞Xn=LR(G/(H,S))$, we conclude that π is an injective homomorphism.

### Corollary 2.18

Letbe an ultragraph satisfying Condition (L), R a unital commutative ring and T a ring. Ifis a ring homomorphism such that π(rpA) ≠ 0 for alland rR {0}, then π is injective.

In this section, we apply the graded-uniqueness theorem for quotient ultragraphs to investigate the ideal structure of . We would like to consider the ideals of that are reflected in the structure of the ultragraph . For this, we give the following definition of basic ideals.

Let (H, S) be an admissible pair in an ultragraph . For any wBH, we define the gap idempotent

$pwH:=pw-∑s(e)=w, r(e)∉Hsese*.$

Let I be an ideal in . We write , which is a saturated hereditary subcollection of . Also, we set $SI:={w∈BHI:pwHI∈I}$. We say that the ideal I is basic if the following conditions hold:

• rpAI implies pAI for and rR {0}.

• $rpwHI∈I$ implies $pwHI∈I$ for wBHI and rR {0}.

For an admissible pair (H, S) in , the (two-sided) ideal of generated by the idempotents ${pA:A∈H}∪{pwH:w∈S}$ is denoted by I(H, S).

### Lemma 3.1. (cf. [12, Lemma 3.9])

If (H, S) is an admissible pair in ultragraph, then

$I(H,S):=spanR{sαpAsβ*,sμpwHsν*∈LR(G):A∈H and w∈S}$

and I(H, S)is a graded basic ideal of.

Proof

We denote the right-hand side of the above equality by J. The hereditary property of H implies that J is an ideal of being contained in I(H, S). On the other hand, all generators of I(H, S) belong to J and so we have I(H, S) = J.

Note that the elements sαpAsβ* and $sμpwHsν*$ are homogeneous elements of degrees |α| – |β| and |μ| – |ν|, respectively. Thus I(H, S) is a graded ideal.

To show that I(H, S) is a basic ideal, suppose rpAI(H, S) for some and rR {0} and write

$rpA=x:=∑i=1nrisαipAisβi*+∑j=1msjsμjpwjHsνj*,$

where AiH, wjS and ri, sjR for all i, j. We first show assertion (1) of the definition of basic ideal in several steps.

Step I

If A = {v} ∉ H and vS, then $0<∣sG-1(v)∣<∞$.

Note that pvx = x, so the assumption v ∉ ∪ HS and the hereditarity of H imply that we may assume that |αi|, |μj| ≥ 1 and for every i, j (because a sum like ∑riAi could be zero). Hence v is not a sink. Set αi = αi,1αi,2 · · · αi,|α|. If $∣sG-1(v)∣=∞$, then there is an edge eα1,1, . . . , αn,1, μ1,1, . . . , μm,1 with . So rse* = se* (rpv) = se*x = 0, contradicting Theorem 2.3. Thus $0<∣sG-1(v)∣<∞$.

Step II

If AH, then there exists vA such that {v} ∉ H.

Let rpA = x. Assume first that |μj | = 0 for some j. Then, since rpA = pArpA = pAx = x, wjA. As wjBH we deduce that {wj} ∉ H. So let |μj| ≥ 1 for all j. If |αi| = 0 for some i then we let . In this case, we have , then rpv = rpvpA = pvx = 0, a contradiction). Thus . Suppose that {{v} : vA} ⊆ H. Since H is hereditary, AH, which is impossible. Hence there is a vertex vA such that {v} ∉ H.

Step III

If A = {v}, then {v} ∈ H.

We go toward a contradiction and assume {v} ∉ H. Set v1 := v. If v1BH, then there is an edge such that and . If v1BH, we have $0<∣sG-1(v)∣<∞$ by Step I. The saturation of H gives an edge e1 with and . By Step II, there is a vertex such that {v2} ∉ H. We may repeat the argument to choose a path γ = γ1. . . γL for L = maxi, j {|βi|, |νj |} + 1, such that and for all 1 ≤ kL. Note that and so for all i, k. Moreover, since r(γk) ∉ H we have $pwjHsγk=0$ for all j, k. It follows that rsγ = (rpv)sγ = xsγ = 0, a contradiction. Therefore {v} ∈ H.

Step IV

If rpAI(H, S), then AH.

If AH, then by Step II there is a vertex vA such that {v} ∉ H, which contradicts the Step III. Hence AH and consequently pAI(H, S), as desired.

Now, we show that I(H, S) satisfies assertion (2) of the definition of basic ideal. Note that, by Step IV, we have HI(H, S) = H. Let wBH, rR {0} and $rpwH∈I(H,S)$. Using the first part of the lemma, write

$rpwH=x:=∑i=1nrisαipAisβi*+∑j=1msjsμjpwjHsνj*,$

where AiH, wjS and ri, sjR for all i, j. Since $rpwH=pwHrpwH=pwHx=x$ and $pwHAi=0$, We may assume that |αi| ≥ 1 for all i. As wBH, we can choose an edge eα1,1, . . . , αn,1, μ1,1, . . . , μm,1 such that and . If wS, then $rse*=se*(rpwH)=se*x=0$ which is a contradiction. Therefore wS and $pwH∈I(H,S)$

Let (H, S) be an admissible pair in ultragraph and let rR{0}. The argument of Lemma 3.1 implies that

$H={A∈G0:rpA∈I(H,S)} and S={w∈BH:rpwH∈I(H,S)}.$

Letbe an ultragraph and let R be a unital commutative ring. If (H, S) is an admissible pair in, then.

### Proof

Suppose that {S̃, P̃} is a universal Leavitt -family. If we define

$PA:=P˜A¯for A∈G0,Se:=S˜efor e∈G1,Se*:=S˜e*for e∈G1,$

then it is straightforward to see that {S, P} is a Leavitt -family in . Note that LR(S, P) inherits the graded structure of . Since rPA ≠ 0 for all and rR {0}, by Corollary 2.15, .

We show that . For , A = PĀLR(S, P). If vBH S, then $P˜{v}¯=P˜v=Pv∈LR(S,P)$. We note that

$s¯(e)={sG(e)′if sG(e)∈BHS and rG(e)∈H,sG(e)otherwise,$

for every . Thus for wBH S, we have 0 < |−1(w)| < ∞ and for e−1(w). Hence

$P˜w=∑s¯(e)=wS˜eS˜e*=∑sG(e)=w, rG(e)∉HSeSe*∈LR(S,P).$

Also,

$P˜w′=P˜{w,w′}-P˜w=Pw-∑sG(e)=w,rG(e)∉HSeSe*=PwH∈LR(S,P).$

Thus, by Remark 2.6, ALR(S, P) for all . Since e, S̃e*LR(S, P), we deduce that . Consequently, .

Letbe an ultragraph and let R be a unital commutative ring. Then

• For any admissible pair (H, S) in, we have.

• The map (H, S) ↦ I(H, S)is a bijection from the set of all admissible pairs ofto the set of all graded basic ideals of.

### Proof

(1) Let {S̃, P̃} be a universal Leavitt -family and let , where {S, P} is the Leavitt -family as defined in 3.1. Define

$Q[A]:=P˜A+I(H,S)for A∈G¯0Te:=S˜e+I(H,S)for e∈Φ(G1),Te*:=S˜e*+I(H,S)for e∈Φ(G1).$

It can be shown that the family is a Leavitt -family that generates . Furthermore, by Remark 3.2, rQ[A] ≠ 0 for all and rR {0}.

Now we use the universal property of to get an R-homomorphism such that π(te) = Te, π(te*) = Te* and π(rq[A]) = rQ[A] ≠ 0 for and rR {0}. Since I(H, S) is a graded ideal, the quotient is graded. Moreover, the elements Q[A], Te and Te* have degrees 0, 1 and −1 in , respectively and thus π is a graded homomorphism. It follows from the graded-uniqueness Theorem 2.14 that π is injective. Since is generated by , we deduce that π is also surjective. Hence .

(2) The injectivity of the map (H, S) ↦ I(H, S) is a consequence of Remark 3.2. To see that it is onto, let I be a graded basic ideal in . Then I(HI, SI)I. Consider the ultragraph with respect to admissible pair (HI, SI). Since I is a graded ideal, the quotient ring is graded. Let be the quotient map. For (HI, SI), consider {S̃, P̃} and {T, Q} as defined in Equations 3.1 and 3.2, respectively. Since I is basic, we have rpAI and $rpwHI∉I$ for , wBHI SI and rR {0}.

We show that π(rq[A]) = π(rQ[A]) = r P̃A +I ≠ 0 for all and rR {0}. Fix rR {0}. We know that . Let A = B for some . If [A] ≠ [∅︀], then BHI and thus r P̃A = rpBI. Therefore π(rq[A]) ≠ 0. If wBHI SI, then $rP˜w′=rpwHI∉I$. Hence π(rq[w′]) ≠ 0. In view of Remark 2.10, we deduce that π(rq[A]) ≠ 0 for every . Furthermore, π is a graded homomorphism. It follows from Theorem 2.14 that the quotient map π is injective. Hence I = I(HI, SI).

As we have seen in the proof of Theorem 3.4, if (H, S) is an admissible pair in , then there is a Leavitt -family {T, Q} in such that rQ[A] ≠ 0 for all and rR {0}. So we have the following proposition.

Let (H, S) be an admissible pair in ultragraphand let R be a unital commutative ring. If {t, q} is the universal Leavitt-family, then rq[A] ≠ 0 for everyand every rR {0}.

In this section we recall Condition (K) for ultragraphs and we consider the ultragraph that satisfy Condition (K) to describe that all basic ideals of are graded.

Let be an ultragraph and let vG0. A first-return path based at v in is a loop α = e1e2 · · · en such that and for i = 2, 3, . . . , n.

### Definition 4.1. ([9, Section 7])

An ultragraph satisfies Condition (K) if every vertex in G0 is either the base of no first-return path or it is the base of at least two first-return paths.

Let be a quotient ultragraph. By rewriting Definition 2.2 for , one can define the hereditary property for the subcollections of . More precisely, a subcollection is called hereditary if satisfying the following conditions:

• s(e) ∈ K implies r(e) ∈ K for all .

• [A] ∪ [B] ∈ K for all [A], [B] ∈ K.

• [A] ∈ K, and [B] ⊆ [A], imply [B] ∈ K.

For any hereditary subcollection , the ideal IK in generated by {q[A] : [A] ∈ K} is equal to

$spanR{tαq[A]tβ*∈LR(G/(H,S)):α,β∈Path(G/(H,S)) and [A]∈K}.$
Lemma 4.2

Letbe a quotient ultragraph and let R be a unital commutative ring. Ifdoes not satisfy Condition (L), thencontains a non-graded ideal I such that rq[A]I for alland rR{0}.

Proof

Suppose that contains a closed path γ := e1e2 · · · en without exits and r(ei) = s(ei+1) for 1 ≤ in where en+1 := e1. Thus we can assume that s(ei) ≠ s(ej) for all i, j. Let s(ei) = [vi] for 1 ≤ in and let X be the subalgebra of generated by {tei, $tei*$, q[vi] : 1 ≤ in}.

Claim 1

The subalgebra X is isomorphic to the Leavitt path algebra LR(E), where E is the graph containing a single simple closed path of length n, that is,

$E0={w1,…,wn}, E1={f1,…,fn},r(fi)=s(fi+1)=wi+1 for 1≤i≤n where wn+1:=w1.$
Proof of Claim 1

Set Pwi := q[vi], Sfi := tei and $Sfi*:=tei*$ for all i. Then {S, P} is a Leavitt E-family in X such that LR(S, P) = X. So there is a homomorphism π : LR(E) → X such that π(sfi) = tei, $π(sfi*)=tei*$ and π(pwi) = q[vi] for all i. Since π preserves the degree of generators, π is a graded homomorphism. By Proposition 3.5, π(rpwi) = rq[vi] ≠ 0 for all rR {0} and all i. Now apply [18, Theorem 5.3] to obtain the injectivity of π. Hence, X ~= LR(E) which proves the claim.

By [18, Lemma 7.16], LR(E) contains an ideal J that is basic but not graded. Thus π(J) is a non-graded ideal of X.

### Claim 2

rq[A]π(J) for every rR {0} and .

Proof of Claim 2

We show that rpwiJ for all rR{0} and all i. Let rpwjJ for some rR {0} and some 1 ≤ jn. Since J is a basic ideal, pwjJ. For every 1 ≤ in, there is a path α ∈ Path(E) such that s(α) = wj and r(α) = wi. Hence $pwi=sαi*pwjsαi∈J$ for all i. Thus J = LR(E), contradicting the hypothesis that J is non-graded.

Suppose that there exist rR{0} and such that rq[A]π(J). If we identify X with the subalgebra

$spanR{tαtβ*∈LR(G/(H,S)):s(α),s(β)∈{[v1],…,[vn]}}$

of , then $rq[A]=r1tα1tβ1*+…+rmtαmtβm*$, where rkR {0} and s(αk), s(βk) ∈ {[v1], . . . , [vn]}. Thus

$rq[A]=rq[A]q[A]=r1q[A]tα1tβ1*+…+rmq[A]tαmtβm*≠0,$

which implies that $[A]∩(∪i=1n[vi])≠[∅]$. So there is j such that [vj ] ⊆ [A]. In this case we have rq[vj] = q[vj]rq[A]π(J). Hence rpwj = π−1(rq[vj]) ∈ J, a contradiction.

Let

$K:={⋃k=1m[vnk]:1≤m≤n and nk∈{1,2,…,n}}$

and set $[A]:=∪i=1n[vi]$. Since γ is a closed path without exits and r(ei) = s(ei+1), K is a hereditary subset of .

### Claim 3

If rq[v]IK for some and some rR{0}, then there exists such that s(α) = [v] and r(α) ∩ A ≠ [∅︀].

Proof of Claim 3

Suppose that

$rq[v]=x:=r1tα1q[vn1]tβ1*+…+rmtαmq[vnm]tβm*,$

where riR {0} and . If there is no such path, then we can choose an edge e such that s(e) = [v] and r(e) ∩ [A] = [∅︀]. Since rqr(e)IK, there is a vertex [w] ⊆ r(e) such that rq[w]IK. By a continuing process, one can choose a path η = η1 · · · ηL for L = maxi {|βi|} + 1, such that s(η) = [v] and r(ηk) ∩ [A] = [∅︀] for all 1 ≤ kL. Thus we have rtη = (rq[v])tη = xtη = 0, a contradiction. Therefore, there is a path α such that s(α) = [v] and r(α)∩[A] ≠ [∅︀].

### Claim 4

I := IKπ(J)IK is a non-graded ideal of .

Proof of Claim 4

Suppose that I is graded. Thus I = ⊕ In, where . For xπ(J) and we have q[A]xq[A] = x and q[A]y, yq[A]X. Thus π(J) ⊆ I. Let xπ(J). Since I = ⊕ In we have x = rn1yn1xn1zn1 + . . . + rnmynmxnmznm, where xnkπ(J), ynk, znkIK and ynkxnkznkInk. Thus

$x=q[A]xq[A]=rn1q[A]yn1xn1zn1q[A]+…+rnmq[A]ynmxnmznmq[A].$

Since q[A]ynk, znkq[A]X, we have rnkq[A]ynkxnkznkq[A]π(J) ∩Xn, where . Hence π(J) is a graded ideal of X, a contradiction.

Finally, we show that rq[B]I for all and rR{0}. Assume rq[B]I for some rR{0} and some . Since rq[B]IK, there is a vertex [v] ⊆ [B] such that rq[v]IIK. By Claim 2, there is a path α such that s(α) = [v] and [vi] ⊆ r(α)∩[A] for some i ∈ {1, . . . , n}. Sorq[vi] = tα*rq[v]tαq[vi]I. If we consider rq[vi] in terms of its representation in IKπ(J)IK, then one can show that rq[vi](J)X = π(J) which is a contradiction. Therefore, we have rq[B]I for every rR {0} and every .

Letbe an ultragraph and let R be a unital commutative ring. Thensatisfies Condition (K) if and only if every basic ideal inis graded.

### Proof

Suppose that satisfies Condition (K). If I is a basic ideal of , then by Theorem 3.4, we have the quotient map

$π:LR(G/(HI,SI))≅LR(G)/I(HI,SI)→LR(G)/I,$

such that π(rq[A]) ≠ 0 for all and all rR {0}. By Proposition [13, Proposition 6.2] and the Cuntz-Krieger uniqueness Theorem 2.17, π is injective and I = I(HI, SI). It follows from Lemma 3.1 that I is a graded ideal.

Conversely, if does not satisfy Condition (K), then [13, Proposition 6.2] and Lemma 4.2 imply that there exist H and S such that contains a non-graded ideal J such that rq[A]J for all and all rR {0}. If is the quotient map, then I := π−1(J) is a non-graded ideal of . Similar to the last paragraph of the proof of Theorem 3.4, we have rpAI for ∅︀ ≠ AH and rR {0}, that is, . Also, $rpwH∉I$

for wBH S and rR {0}. Consequently I is a basic ideal. Therefore satisfies Condition (K).

Letbe an ultragraph and let R be a unital commutative ring. Ifsatisfies Condition (K), then the map (H, S) ↦ I(H, S)is a bijection from the set of all admissible pairs ofinto the set of all basic ideals of.

The Exel-Laca algebras are generated by partial isometries whose relations are determined by a countable {0, 1}-valued matrix A with no identically zero rows [7, Definition 8.1]. In this section, we define an algebraic version of the Exel-Laca algebras.

### Definition 5.1

Let I be a countable set and let A be an I ×I matrix with entries in {0, 1}. The ultragraph $GA:=(GA0,GA1,r,s)$ is defined by $GA0:={vi:i∈I},GA1:=I$, s(i) := vi and r(i) := {vj : A(i, j) = 1} for all iI.

By [15, Theorem 4.5], the Exel-Laca algebra is canonically isomorphic to . Motivated by this fact we give the following definition.

### Definition 5.2

Let R be a unital commutative ring, let I be a countable set and let A be a {0, 1}-valued matrix over I with no identically zero rows. The Exel-Laca R-algebra associated to A, denoted by ℰℒA(R), is defined by .

### Example 5.3

Let R be a unital commutative ring and let A = (A(i, j)) be an n × n matrix, with with no identically zero rows and A(i, j) ∈ {0R, 1R} for all i, j. The Cuntz-Krieger R-algebra associated to A, as defined in [3, Example 2.5], is the free R-algebra generated by a set {x1, y1, . . . , xn, yn}, modulo the ideal generated by the following relations:

• xiyixi = xi and yixiyi = yi for all i,

• yixj = 0 for all ij,

• $yixi=∑j=1nA(i,j)xjyj$ for all i,

• $∑j=1nxjyj=1$.

The Cuntz-Krieger R-algebra associated to A is denoted by . We will show that . Note that is a finite ultragraph. If we define PB := ∑viB xiyi, Si := xi and Si* := yi for all i and all $B∈GA0$, then it is easy to verify that {$PB,Si,Si*:B∈GA0,1≤i≤n$ }is a Leavitt -family.

On the other hand, if {s, p} is the universal Leavitt -family for , then the elements Xi := si and Yi := si*, 1 ≤ in, satisfy relations (1)–(4). Now the universality of and conclude that .

Note that for any {0, 1}-valued matrix A with no identically zero rows, the ultragraph contains no singular vertices. So, each Exel-Laca algebra is the Leavitt path algebra of a non-singular ultragraph. In Theorem 5.7 below, we will prove the converse. For this, given any ultragraph with no singular vertices, we first reform to an ultragraph as follows. Associated to each vG0 and es−1(v), we introduce a vertex ve and set Ã := {ve : vA and es−1(v)} for every . Next we define the ultragraph , where 0 := {ve : vG0 and es−1(v)}, , the source map (e) := s(e)e and the range map $r˜(e):=r(e)˜$, where $r(e)˜={ve:v∈r(e) and e∈s-1(v)}$.

### Remark 5.4

We notice that each vertex ve in emits exactly one edge e. Moreover, Lemma 2.5 implies that

$G˜0={A˜∪F:A∈G0 and F is a finite subset of G˜0}.$

### Lemma 5.5

Letbe a row-finite ultragraph with no singular vertices and let R be a unital commutative ring. Thenis isomorphic toas algebras.

Proof

Let {t, q} be the universal Leavitt -family for . Since is row-finite, for every vG0 we have . Define

$PA:=qA˜for A∈G0,Se:=tefor e∈G1,Se*:=te*for e∈G1.$

By Remark 5.4, for each , the idempotent PA satisfies the condition (1) of Definition 2.1. It is straightforward to check that {S, P} is a Leavitt -family in . For example, to verify condition (2) of Definition 2.1 suppose that . Then SePr(e) = teq(e) = te = Se and

$Ps(e)Se=(∑f∈G˜1qs(e)f)qs˜(e)te=(∑f∈G˜1qs(e)f)qs(e)ete=qs(e)ete=te=Se.$

Thus there is an R-algebra homomorphism such that φ(se) = te, φ(se*) = te* and for vG0, and rR {0}. So (pA) ≠ 0 for all . As φ is a graded homomorphism, Corollary 2.15 implies that φ is injective. Moreover, for any ve0, we have ()−1(ve) = {e} and hence SeSe* = tete* = qve. Therefore, φ is an isomorphism and we get the result.

### Definition 5.6. ([15, Definition 2.4])

Let be an ultragraph. The edge matrix of an ultragraph is the matrix given by

$AG(e,f):={1if s(f)∈r(e),0otherwise.$

We can check that, if is an ultragraph with no singular vertices, then . So, by Lemma 5.5 we have the following.

### Theorem 5.7

Let R be a unital commutative ring and letbe an ultragraph with no singular vertices. Thenis isomorphic to.

### Example 5.8

Suppose that E is a graph with one vertex E0 = {v} and n edges E1 = {e1, . . . .en}. We know that LR(E) is isomorphic to the Leavitt algebra LR(1, n). If all entries of matrix An×n is 1, then A = AE. Hence, Theorem 5.7 shows that ℰℒA(R) ~= LR(1, n).

### Remark 5.9

Theorem 5.7 shows that the Leavitt path algebras of ultragraphs with no singular vertices are precisely the Exel-Laca R-algebras.

### Remark 5.10

We know that for every directed graph E, L(E) is isomorphic to a dense *-subalgebra of the C*-algebra C*(E) (see [17, Section 7]). Using a similar argument to [17, Theorem 7.3], one can show that is isomorphic to a dense *-subalgebra of the C*-algebra introduced in [15]. In particular, for any countable {0, 1}-valued matrix A, the Exel-Laca algebra ℰℒA(ℂ) is isomorphic to a dense *-subalgebra of [7].

Last part of this section is an example to emphasize that the class of the Leavitt path algebras of ultragraphs is strictly larger than the class of the Leavitt path algebras of directed graphs as well as the class of algebraic Exel-Laca algebras.

### Example 5.11

Let R = ℤ2 and let be the ultragraph

with one ultraedge e such that s(e) = v0 and r(e) = {v1, v2, . . .}. Note that satisfies Condition (K) and so all ideals of are basic and graded by Theorem 4.3. Moreover, is unital, because by Lemma 2.12 . We claim that the algebra is not isomorphic to the Leavitt path algebra of a graph. Suppose on the contrary that there is a such graph E. Let {t, q} be the canonical generators for L2 (E). If E has a loop α, then for tαL2 (E) we have $tαn≠tαm$ for all mn, but one can show that does not include such member. Thus L2 (E) does not have any loop and consequently E satisfies Condition (K) for directed graphs. Hence by [12, Theorem 3.18] any ideal of L2 (E) is graded. Since L2 (E) is unital, E0 must be finite and hence L2 (E) contains finitely many (graded) ideals by Theorem 3.4(2), which contradicts the fact that has infinitely many pairwise orthogonal ideals I{vn} for n ≥ 1.

Now assume for some matrix A and let be an isomorphism. We may consider the ideal I{v1} in . Since v1 is a sink, we have

$I{v1}=spanℤ2{pv1,sepv1,pv1se*,sepv1se*},$

and so |I{v1}| < ∞. On the other hand, we know that has no sinks. Hence the graded ideal φ(I{v1}) in includes infinitely many elements, which is a contradiction.

The authors are grateful to the referee for carefully reading the paper, pointing out a number of misprints and for some helpful comments.

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