SIMPLICITY OF LEAVITT PATH ALGEBRAS VIA GRADED RING THEORY

Abstract Suppose that R is an associative unital ring and that 
$E=(E^0,E^1,r,s)$
 is a directed graph. Using results from graded ring theory, we show that the associated Leavitt path algebra 
$L_R(E)$
 is simple if and only if R is simple, 
$E^0$
 has no nontrivial hereditary and saturated subset, and every cycle in E has an exit. We also give a complete description of the centre of a simple Leavitt path algebra.


Introduction
The Leavitt path algebra of a row-finite graph, over a field, was introduced in [2,5] and has since then been successively generalised (see, for example, [3,20]).The Leavitt path algebra of an arbitrary directed graph, over a unital ring, was introduced in [12].For an account of the development of the field of Leavitt path algebras, we refer the reader to [1].Here is our first main result.THEOREM 1.1.Suppose that R is an associative unital ring and that E = (E 0 , E 1 , r, s) is a directed graph.The Leavitt path algebra L R (E) is simple if and only if R is simple, E 0 has no nontrivial hereditary and saturated subset, and every cycle in E has an exit.
Characterisations of simple Leavitt path algebras over fields have previously been established in [19,Theorem 6.18], [3,Theorem 3.1] and [11,Theorem 3.5].Theorem 1.1 generalises all of those results, and also partially generalises [20,Theorem 7.20].Our second main result, stated below, completely describes the centre of a simple Leavitt path algebra.It generalises [6,Theorem 4.2] from the case where R is a field and E is a row-finite graph.THEOREM 1.2.Suppose that R is an associative unital ring and that E = (E 0 , E 1 , r, s) is a directed graph.Furthermore, suppose that L R (E) is a simple Leavitt path algebra.The following assertions hold.
Whereas earlier proofs of Theorems 1.1 and 1.2 (when R is a field) use ad hoc arguments, specifically designed for graph algebras, we use the general theory of graded rings to obtain our results.This makes our proofs shorter and, we believe, clearer.Indeed, we show that L R (E) is graded simple if and only if R is simple and E 0 has no nontrivial hereditary and saturated subset (see Proposition 3.6).We also show that every cycle in E has an exit if and only if the centre of each corner subring of L R (E) at a vertex has degree zero (see Proposition 3.14).
We point out that there are various generalisations of Leavitt path algebras in the literature (see, for example, [1, Section 5] and [9,18]).A simplicity result for Steinberg algebras was obtained in [8], and when translated to Leavitt path algebras, one recovers Theorem 1.1 in the special case where R is a commutative unital ring.Note that [6, Theorem 4.2] was generalised to Kumjian-Pask algebras in [7], and in [10], Steinberg algebra techniques were used to give a complete description of the centre of a general (not necessarily simple) Leavitt path algebra L R (E), where R is a commutative unital ring.

Simple Z-graded rings
Let Z denote the rational integers and write N := {1, 2, 3, . ..}. Suppose that S is a ring.By this, we mean that S is associative but not necessarily unital.If S is unital, then we let 1 S denote the multiplicative identity of S. Furthermore, we let Z(S) denote the centre of S, that is, the set of all s ∈ S satisfying st = ts for every t ∈ S. Recall that S is said to be Z-graded if, for each n ∈ Z, there is an additive subgroup S n of S such that S = ⊕ n∈Z S n and S n S m ⊆ S n+m , for all n, m ∈ Z.In that case, each element s ∈ S may be written as s = n∈Z s n , where s n ∈ S n is zero for all but finitely many n ∈ Z.The support of s is defined as the finite set Supp(s) := {n ∈ Z | s n 0}.An ideal I of a Z-graded ring S is said to be graded if I = ⊕ n∈Z (I ∩ S n ).If {0} and S are the only graded ideals of S, then S is said to be graded simple.
We recall some properties of graded rings.
LEMMA 2.1.Suppose that S is a unital Z-graded ring.Next, we state a special case of [17,Theorem 1.2] and [13,Theorem 5].For the convenience of the reader, we include a shortened version of the proof from these sources adapted to the situation at hand.PROPOSITION 2.2.Suppose that S is a unital Z-graded ring.Then, the following assertions are equivalent: (i) S is simple; (ii) S is graded simple and Z(S) is a field; (iii) S is graded simple and Z(S) ⊆ S 0 .
PROOF.(i)⇒(ii) is clear and (ii)⇒(iii) follows from Lemma 2.1.Now we show that (iii)⇒(i).Suppose that S is graded simple and that Z(S) ⊆ S 0 .Let I be a nonzero ideal of S. We wish to show that 1 S ∈ I. Amongst all nonzero elements of I, choose s such that |Supp(s)| is minimal.Take m ∈ Supp(s).Since S is graded simple, there are n ∈ N and homogeneous elements p 1 , . . ., p n , q 1 , . . ., q n ∈ S, such that n i=1 p i s m q i = 1 S and p i s m q i ∈ S 0 \ {0} for every i ∈ {1, . . ., n}.Write t := n i=1 p i sq i .Note that t ∈ I, t 0 = Let S be a ring.Recall from [4] (see also [16]) that a set U of idempotents in S is called a set of local units for S if for every n ∈ N and all s 1 , . . ., s n ∈ S, there is some e ∈ U such that es i = s i e = s i for every i ∈ {1, . . ., n}.REMARK 2.3.Suppose that S is a Z-graded ring.If e ∈ S 0 is an idempotent, then the corner subring eSe inherits a natural Z-grading defined by (eSe) n := eS n e for n ∈ Z.
For future reference, we recall the following two results.PROPOSITION 2.4.Suppose that S is a Z-graded ring equipped with a set of local units U ⊆ S 0 .Then, S is (graded) simple if and only if, for every f ∈ U, the ring f S f is (graded) simple.
PROOF.First we show the 'only if' statement.Suppose that S is (graded) simple and that f ∈ U. Let J be a nonzero (graded) ideal of f S f .By (graded) simplicity of S, it follows that SJS = S. Thus, PROPOSITION 2.5.Suppose that S is a Z-graded ring equipped with a set of local units and that f ∈ S 0 is a nonzero idempotent.If S is graded simple and f S f is simple, then S is simple.
PROOF.Suppose that S is graded simple and that f S f is simple.Let I be a nonzero ideal of S. Take a nonzero s ∈ I and write s = n∈Supp(s) s n .Fix m ∈ Supp(s) and define J := Ss m S.Then, J is a nonzero graded ideal of S. By graded simplicity of S, it follows that J = S and, in particular, that f ∈ J.Note that f ∈ f J f .Since f 0, it follows that there exist nonzero homogeneous y, z ∈ S such that f ys m z f is nonzero and deg(y) + deg(z) = −m.Now, define s := f ysz f .By the construction of s , it follows that s ∈ I ∩ f S f and that s is nonzero.In particular, I ∩ f S f {0}.Hence, by simplicity of f S f , we see that I ∩ f S f = f S f .Thus, f ∈ I.Note that S f S is a nonzero graded ideal of S. Hence, by graded simplicity of S, we have I ⊇ S f S = S.This shows that I = S.

Simple Leavitt path algebras
Let R be an associative unital ring and let E = (E 0 , E 1 , r, s) be a directed graph.Recall that r (range) and s (source) are maps E 1 → E 0 .The elements of E 0 are called vertices and the elements of E 1 are called edges.The elements of E 1 are called real edges, while for 1 denote the corresponding ghost path.For any vertex v ∈ E 0 , we put s(v) := v and r(v) := v.We let r( f * ) denote s( f ) and we let s( f * ) denote r( f ).For n ≥ 2, we define E n to be the set of paths of length n and E * := n≥0 E n the set of all finite paths.
Following Hazrat [12], we make the following definition.
DEFINITION 3.1.The Leavitt path algebra of E with coefficients in R, denoted by L R (E), is the algebra generated by the sets with the coefficients in R, subject to the relations: Moreover, the ideal coming from relations (1)-(4) in Definition 3.1 is graded.Using this, it is easy to see that the natural Z-grading on F R (E) carries over to a Z-grading on the quotient algebra (c) Motivated by Definition 3.1(2), for u ∈ E 0 , we write u * := u.
REMARK 3.4.Note that ∅ and E 0 are always hereditary and saturated subsets of E 0 .They are referred to as trivial.
We wish to show that H I is hereditary and saturated.To this end, take v ∈ H I .Suppose that e ∈ E 1 with s(e) = v.Then, r(e) = e * e = e * ve ∈ I. Thus, H I is hereditary.Now, take v ∈ E 0 such that 0 < |s −1 (v)| < ∞, and suppose that r(s −1 (v)) ⊆ H I .For each e ∈ s −1 (v), we have r(e) ∈ H I and hence ee * = er(e)e * ∈ I. Thus, v = e∈s −1 (v) ee * ∈ I and v ∈ H I .Therefore, H I is saturated.By our assumption, H I = E 0 .This shows that I must contain all the local units of L R (E) and thus I = L R (E).Hence, L R (E) is graded simple.Now, we show the 'only if' statement.Suppose that L R (E) is graded simple.Let J be a proper ideal of R. We wish to show that J = {0}.To this end, let L J (E) denote the graded ideal of L R (E) consisting of all elements of L R (E) with coefficients coming from J. Consider the natural ring homomorphism ϕ : L R (E) → L R/J (E).Clearly, ϕ is well defined.Note that L J (E) ⊆ ker(ϕ).Choose some u ∈ E 0 .By Lemma 3.5, applied to L R/J (E), it follows that u ker(ϕ) and hence u L J (E).Thus, L J (E) is a proper graded ideal of L R (E).By graded simplicity of L R (E), it follows that L J (E) = {0}.Thus, in particular, Ju = {0}.By Lemma 3.5 applied to L R (E), we see that J = {0}.
REMARK 3.12.The definition of a cycle in a directed graph varies in the literature on Leavitt path algebras.In contrast to the most common definition of a cycle (see [2, page 320], following [20], we allow a cycle to 'intersect' itself.In Theorem 1.1, the condition that 'every cycle in E has an exit' appears.That condition is commonly known as Condition (L).It is easy to see that Condition (L) is satisfied with the first definition of a cycle [2] if and only if it is satisfied with the second definition of a cycle [20].REMARK 3.13.Let x be a nonzero element of L R (E).It is clear from the definition of L R (E) that x can be represented as a finite sum x = n i=1 r i α i β * i , where r i ∈ R \ {0} and α i , β i ∈ E * .Following [20,Definition 4.8], we define the real degree (respectively ghost degree) of this representation as max{deg(α Note that, in general, the real degree and ghost degree of x depend on the particular choice of representation.If, however, x has a representation in only real (respectively ghost) edges, that is, if x = n i=1 r i α i (respectively x = n i=1 r i β * i ), then, by Lemma 3.5, the real (respectively ghost) degree is independent of the choice of representation of x in real (respectively ghost) edges.
Now we show the 'only if' statement.Suppose that every cycle in E has an exit.Take u ∈ E 0 .We wish to show that Z(uL R (E)u) ⊆ (uL R (E)u) 0 .By Lemma 2.1(a) and Lemma 3.9, it is enough to show that (Z(uL R (E)u)) N = {0} for every negative integer N.
We now adapt parts of the proof of [3, Theorem 3.1] to our situation.Take N < 0. Seeking a contradiction, suppose that the set is nonempty.If (u, x), (v, y) ∈ M, then we write (u, x) ≤ (v, y) if x has a representation in L R (E) of real degree less than or equal to all real degrees of representations of y in L R (E).We write (u, x) = (v, y) whenever (u, x) ≤ (v, y) and (v, y) ≤ (u, x).Clearly, ≤ is a total order on M which therefore has a minimal element (u, x).Choose a minimising representation x = n i=1 e i a i + b, where e 1 , . . ., e n ∈ E 1 are all distinct, each a i ∈ L R (E) is either zero or nonzero and representable as an element of smaller real degree than that of x, and b is a polynomial (possibly zero) in only ghost paths whose source and range equals u.Take i ∈ {1, . . ., n}.Write v i := r(e i ).By Lemma 3.10, we get f * x f 0. Note that the real degree of f * x f is less than or equal to the real degree of x.Hence, by the assumption made on (u, x), and possibly after replacing (u, x) by (v, f * x f ), we may assume that a i = 0 for every i ∈ {1, . . ., n}.Therefore, suppose that x = m j=1 r j β * j for some nonzero r j ∈ R and some distinct paths β j ∈ E −N with s(β j ) = r(β j ) = u.Take k ∈ {1, . . ., m}.By Lemma 3.10, it follows that By assumption, the cycle β k has an exit at some w ∈ E 0 .Thus, there are γ, δ ∈ E * and ∈ E 1 such that β k = γδ, r(γ) = s( ) = w and * δ = 0.By Lemma 3.10, it follows that r k (δγ We now reach a contradiction, because 0 * r k (δγ) * = r k (δγ) * * = 0. Now, we prove our main result.
PROOF OF THEOREM 1.1.First, we show the 'only if' statement.Suppose that L R (E) is simple.Then L R (E) is graded simple and hence, by Proposition 3.6, it follows that R is simple and that E 0 has no nontrivial hereditary and saturated subset.Furthermore, Proposition 2.4 implies that uL R (E)u is simple for every u ∈ E 0 , and hence, by Proposition 2.2, Z(uL R (E)u) ⊆ (uL R (E)u) 0 for every u ∈ E 0 .Thus, by Proposition 3.14, every cycle in E has an exit.Now we show the 'if' statement.Suppose that R is simple, E 0 has no nontrivial hereditary and saturated subset, and every cycle in E has an exit.By Proposition 3.6, L R (E) is graded simple.Take u ∈ E 0 .It follows from Proposition 3.14 that Z(uL R (E)u) ⊆ (uL R (E)u) 0 .Furthermore, by Proposition 2.4, uL R (E)u is graded simple.Thus, by Proposition 2.2, uL R (E)u is simple.Hence, by Proposition 2.5, L R (E) is simple.

The centre of a simple Leavitt path algebra
In this section, we prove Theorem 1.2 using results from the previous sections together with some auxiliary observations.REMARK 4.1.Let E = (E 0 , E 1 , r, s) be a directed graph.
(a) Take v ∈ E 0 .We write w ≤ v, for w ∈ E 0 , if there is μ ∈ E * with s(μ) = v and r(μ) = w.The set T(v) := {w ∈ E 0 | w ≤ v} is the smallest hereditary subset of E 0 containing v.
The following result can be proved by induction (see [14,Proposition 14.11] and [20,Lemma 5.2]).PROPOSITION 4.2.Suppose that R is an associative unital ring and that E = (E 0 , E 1 , r, s) is a directed graph.If a ∈ (L R (E)) 0 is nonzero, then there exist α, β ∈ E * , v ∈ E 0 and a nonzero k ∈ R such that α * aβ = kv.Now, we prove our second main result.

( a )
The ring Z(S) is Z-graded with respect to the grading Z(S) n which is defined by Z(S) n := Z(S) ∩ S n for n ∈ Z.(b) If S is a field, then S = S 0 .PROOF.Item (a) is [15, page 15, Exercise 8] and item (b) is [15, Remark 1.3.10].
nonempty and finite.Here, elements of the ring R commute with the generators.REMARK 3.2.(a) The Leavitt path algebra L R (E) carries a natural Z-grading.Indeed, put deg(v) := 0 for each v ∈ E 0 .For each f ∈ E 1 , we put deg( f ) := 1 and deg( f * ) := −1.By assigning degrees to the generators in this way, we obtain a Z-grading on the free algebra

DEFINITION 3 . 7 .
Define an additive map L : L R (E) → L R (E) by requiring that L(λαβ * ) = λβα * for all λ ∈ R and α, β ∈ E * .REMARK 3.8.The map L is an isomorphism of additive groups such that L

PROPOSITION 3 . 14 .
Every cycle in E has an exit if and only if for every u∈ E 0 , the inclusion Z(uL R (E)u) ⊆ (uL R (E)u) 0 holds.PROOF.First, we show the 'if' statement by contrapositivity.Suppose that there is a cycle p ∈ E * \ E 0 without any exit.Set u := s(p) and write p 0 := u.Take r ∈ R and α, β ∈ E * with s(α) = s(β) = u and r(α) = r(β).Since p has no exit, there are m, n ∈ N ∪ {0} and γ ∈ E * such that α = p m γ and β = p n γ.Note that γγ * = u = pp * .This yields prαβ * = prp m γγ * (p * ) n = rp m+1 (p * ) n and rαβ * p = rp m γγ * (p * ) n p = rp m (p * ) n p.If n = 0, then p m+1 (p * ) n = p m+1 = p m (p * ) n p, and if n > 0, then p m+1 (p * ) n = p m pp * (p * ) n−1 = p m (p * ) n−1 p * p = p m (p * ) n p.In either case, we get prαβ i is of smaller real degree than x, it follows that e * i xe i = 0. Further, since x ∈ (Z(uL R (E)u)) N , it follows that e * i x = e * i e i e * i x = e * i xe i e * i = 0. Thus, 0 = e * i x = a i + e * i b and hence a i = −e * i b.Now, 0 x = (u − n i=1 e i e * i )b.Thus, u n i=1 e i e * i and b 0. This implies that there is some f ∈ E 1 \ {e 1 , . . ., e n } with s( f ) = u.Furthermore, f * x = f * b, and, by Lemma 3.5, f * b 0 since it is a sum of distinct ghost paths.Write v := r( f ).By Lemma 3.10, it follows that f [20,the set of real (respectively ghost) paths is linearly independent in the left R-module L R (E) and in the right R-module L R (E).PROOF.The proof of [20, Proposition 4.9] immediately carries over to the case where R is a noncommutative unital ring.The same holds for the proof of[20,  Proposition 3.4]in case E 0 and E 1 are countable sets.Otherwise, the proof may be adapted by taking ℵ to be an infinite cardinal at least as large as card(E [20,E 1 ) and defining Z := ⊕ ℵ R (with the notation of[20, Proposition 3.4]).PROPOSITION 3.6.Te Leavitt path algebra L R (E) is graded simple if and only if R is simple and E 0 has no nontrivial hereditary and saturated subset.PROOF. Frst we show the 'if' statement.Suppose that R is simple and that E 0 has no nontrivial hereditary and saturated subset.Let I be a nonzero graded ideal of L R (E).Consider the set H I := {v ∈ E 0 | kv ∈ I for some nonzero k ∈ R}.By the same argument as in [20, Lemma 5.1], H I is nonempty.Furthermore, since R is simple, it follows that