THE COMPRESSED ANNIHILATOR GRAPH OF A COMMUTATIVE RING Sh. Payrovi and S. Babaei
Department of Mathematics, Imam Khomeini International University,
Postal Code: 34149-1-6818 Qazvin - Iran
e-mails: [email protected]; [email protected]
(Received 29 December 2014; after final revision 25 July 2017;
accepted 7 September 2017)
LetRbe a commutative ring. In this paper, we introduce and study the compressed annihilator graph ofR. The compressed annihilator graph ofR is the graphAGE(R), whose vertices are equivalence classes of zero-divisors ofRand two distinct vertices[x]and[y]are adjacent if and only ifann(x)∪ann(y)⊂ann(xy). For a reduced ringR, we show that compressed annihilator graph ofRis identical to the compressed zero-divisor graph ofRif and only if0is a 2-absorbing ideal ofR. As a consequence, we show that an Artinian ringRis either local or reduced whenever 0is a 2-absorbing ideal ofR.
Key words : Annihilator graph; compressed annihilator graph; zero divisor graph; 2-Absorbing ideal.
1. INTRODUCTION
LetRbe a commutative ring with nonzero identity, and letZ(R)denote the set of zero-divisors of R. The zero-divisor graph ofR, denoted byΓ(R), is the graph with verticesZ(R)∗ =Z(R)\ {0},
the set of nonzero zero-divisors ofR, and for distinctx, y∈Z(R)∗, the verticesxandyare adjacent
if and only if xy = 0, see [2, 6]. The compressed zero-divisor graph of R, that is constructed
from equivalence classes of zero-divisors by annihilator ideals, rather than individual zero-divisors
themselves was introduced by Mulay in [8] and studied in some literature, for examples [1, 7, 11].
It will be denoted byΓE(R). This graph has some advantages over the zero-divisor graph. In many
cases, the compressed zero-divisor graph ofR is finite when the zero-divisor graph is infinite and
another important aspect of the compressed zero-divisor graph is the connection to the associated
The annihilator graph AG(R) for a commutative ring R was introduced and investigated by
Badawi in [5]. Let a ∈ Z(R) and let ann(a) = {r ∈ R|ra = 0}. AG(R) is the graph with
verticesZ(R)∗, and two distinctxandyare adjacent if and only ifann(x)∪ann(y)⊂ann(xy). In
this paper, inspired by ideas from Mulay in [8], we introduce the compressed annihilator graph ofR,
which is constructed from classes of zero-divisors determined by annihilator ideals, and two distinct
classes[x]and[y]are adjacent if and only ifann(x)∪ann(y) ⊂ ann(xy). It will be denoted by AGE(R).
In section two, we determine the girth and the diameter ofAGE(R)and we show that ifAGE(R)
is complete, thenZ(R) is a prime ideal ofR and|Ass(R)| = 1. In section three, we show that AG(R) = Γ(R)if and only if0is a 2-absorbing ideal ofR. We investigate the compressed annihilator
graph of the direct product of two rings. As a consequence, we show that an Artinian ringRis either
local or reduced whenever0is a 2-absorbing ideal ofR.
LetGbe a graph. Recall thatGis connected if there is a path between any two distinct vertices
ofG. For verticesxandyofG, letd(x, y)be the length of a shortest path fromxtoy(d(x, x) = 0
andd(x, y) =∞if there is no such path). The diameter ofGisdiam(G)=sup{d(x, y)|xandyare vertices of G}. The girth ofG, denoted bygr(G), is the length of a shortest cycle inG(gr(G) =∞
ifGcontains no cycles). A cycle withnvertices will be denoted by Cn. The degree of a vertexv,
denoted bydegv, is the number of edges incident tov.
A graph G is complete if any two distinct vertices are adjacent. The complete graph with n
vertices will be denoted byKn. A complete bipartite graph is a graph Gwhich may be partitioned
into two disjoint nonempty vertex setsAandBsuch that two distinct vertices are adjacent if and only
if they are in distinct vertex sets. If one of the vertex set is a singleton, then we callGa star graph. Let G0 be an another graph. IfGis identical toG0, then we writeG=G0; otherwise, we writeG6=G0.
Throughout, R will denote a commutative ring with non-zero identity, Z(R) its set of
zero-divisors,Nil(R) its set of nilpotent elements,Ass(R) its set of associated primes and Min(R) its
set of minimal prime ideals. We sayRis reduced ifNil(R) = 0and thatRis local if it has a unique
maximal ideal. As usual,ZandZnwill denote the integers and integers modulon, respectively. For
notations and terminologies not given in this article, the reader is referred to [10].
2. BASICRESULTS
Forx, y∈R, we say thatx∼yif and only ifann(x) = ann(y). As noted in [8],∼is an equivalence
other equivalence classes form a partition ofZ(R)∗.
Definition 2.1 — The compressed annihilator graph ofR, denotedAGE(R), is the graph
associ-ated toZ(R)whose vertices are the classes of elements inZ(R)∗, and two distinct classes[x]and[y]
are adjacent if and only ifann(x)∪ann(y)⊂ann(xy).
Remark 2.2 : Adjacency is well-defined: let [x1] = [x2]and [y1] = [y2]; that is ann(x1) = ann(x2)andann(y1) = ann(y2). If z ∈ ann(x1y1)\(ann(x1)∪ann(y1)), thenzx1y1 = 0and
sozy1 ∈ ann(x1) = ann(x2). Thus zy1x2 = 0 and sozx2 ∈ ann(y1) = ann(y2). Therefore, z ∈ ann(x2y2)\(ann(x2)∪ann(y2)). Hence, ann(x1) ∪ann(y1) ⊂ ann(x1y1) if and only if ann(x2)∪ann(y2)⊂ann(x2y2).
The graphAGE(R)has same vertices asΓE(R). It is obvious that each edge ofΓE(R)is an edge
ofAGE(R). It is shown thatΓE(R)is connected anddiam(ΓE(R))≤3, see [11, Proposition 1.4].
Lemma 2.3 — Lety1, y2∈Rbe such thatann(y1)andann(y2)are distinct associated primes of R. Then[y1]and[y2]are adjacent inAGE(R). IfRis Noetherian and [x]is a vertex ofAGE(R),
then eitherann(x)∈Ass(R)or there is a vertex[x0]adjacent to[x]such thatann(x0)∈Ass(R).
PROOF: The proof is similar to that of Lemma 1.2 in [11]. 2
Theorem 2.4 — The graphAGE(R)is connected anddiam AGE(R)≤2.
PROOF: Let[x]and[y]be two non-adjacent vertices. Thenann(x)∪ann(y) = ann(xy). It is
clear thatxy 6= 0and alsoann(x)⊆ann(y)orann(y) ⊆ann(x). Assume thatann(y)⊆ann(x).
There is a non-zero element w ∈ ann(y). Thuswy = wx = 0. Furthermore, [w] 6= [x] and [w]6= [y]sincex ∈ann(w)\ann(y)andy ∈ann(w)\ann(x). Hence,[x]−[w]−[y]is a path in
AGE(R). 2
Theorem 2.5 — IfRis Noetherian, thengr(AGE(R))∈ {3,4,∞}.
PROOF: If|Ass(R)|≥3, then in view of Lemma 2.3, we have a cycle with length 3. Therefore,
assume that|Ass(R) |≤2. By Theorem 2.4diam AGE(R)≤2which implies thatgr(AGE(R))≤ 5. Let[x1]− · · · −[x5]−[x1]be a cycle. If|Ass(R) |= 1andp = ann(y)is a prime ideal ofR,
then any vertex ofAGE(R)is adjacent to[y]and so we have a cycle with length 3. Now, assume
that|Ass(R) |= 2andp1 = ann(y1)andp2 = ann(y2)are prime ideals ofR. If[y1] = [x1], then [x3],[x4]are adjacent to[y1]or[y2]and we have a cycle with length 3. By a similar argument one can
show that, in the other cases, we have a cycle with length 3. Therefore,gr(AGE(R))≤4, ifAGE(R)
Example 2.6 : LetR = Zp2q, wherep, qare two prime numbers. Then[p],[p2],[q]and[pq]are
vertices ofAGE(R)andAGE(R) =C4. Therefore,gr(AGE(R)) = 4.
Example 2.7 : LetR =Z4[X, Y]/(X2, Y X,2X)and letxandydenote the image ofXandY
inR, respectively. The proof of Corollary 1.6 in [11] shows thatΓE(R)is the path[2]−[x]−[y]. It
is clear thatann(2y) = ann(2)and so[2]−[y]is not an edge ofAGE(R). Therefore,AGE(R) = ΓE(R)andgr(AGE(R)) =∞.
It is shown that there are no complete graphΓE(R)with three or more vertices, see [11,
Propo-sition 1.5]. But the following example shows that for any integer n there is a ring for which the
compressed annihilator graph is complete withnvertices.
Example 2.8 : Suppose that p is a prime number and n ≥ 2. Then[p],[p2],· · ·,[pn]are all
distinct vertices ofAGE(Zpn+1). Assume that1≤i, j≤n. It is obvious that[pi][pj] = 0whenever i+j ≥n, so that[pi],[pj]are adjacent. Ifi+j < n, thenpn−(i+j)∈ann(pi+j)\(ann(pi)∪ann(pj))
so that[pi],[pj]are adjacent. Hence,AGE(Zpn+1)is a complete graph withnvertices.
Theorem 2.9 — LetAGE(R)be a complete graph with at least three vertices. ThenZ(R)is an
ideal ofR.
PROOF: Suppose thatx, y∈Z(R)∗. It is enough to show thatx+y∈Z(R). If[x] = [y], then
there is a non-zero elementz∈Rsuch thatz(x+y) = 0and in this case we are done. Now, assume
that[x]6= [y]. There isw∈Z(R)∗such that[w]6= [x],[y]and[w]−[y]is an edge ofΓE(R)since ΓE(R) is connected. Thuswy = 0. If wx = 0, thenw(x+y) = 0 and we are done. Otherwise wx6= 0and[w]−[x]is an edge ofAGE(R), so there isz∈ann(wx)\(ann(w)∪ann(x)). Hence, zw6= 0andzw(x+y) = 0. Therefore,Z(R)is an ideal ofR. 2
Lemma 2.10 — Letx∈Randp= ann(x)be a prime ideal ofRand lety∈Z(R)∗ be such that [x]6= [y]. If[x]−[y]is not an edge ofΓE(R), then[x]−[y]is not an edge ofAGE(R).
PROOF: By assumption[x]−[y]is not an edge ofΓE(R)so thatxy 6= 0. Assume that[x]−[y]
is an edge ofAGE(R). Thusann(x)∪ann(y)⊂ann(xy)and so there isz∈ann(xy)\(ann(x)∪ ann(y)). Hence,zxy= 0andzx6= 0. Therefore,zy ∈ann(x) =pbutz 6∈pandy6∈pwhich is a
contradiction. Thus[x]−[y]is not an edge ofAGE(R). 2
Corollary 2.11 — LetAGE(R) be a complete graph withnvertices. Ifann(x)is a prime ideal
ofRfor somex∈Z(R)∗, thendeg[x] =n−1inΓE(R).
Theorem 2.12 — LetRbe Noetherian and AGE(R) be a complete graph withn ≥ 3vertices.
ThenZ(R)is a prime ideal ofRand|Ass(R)|= 1.
PROOF : Assume that z ∈ Z(R)∗ is such that ann(z) is a prime ideal of R. We show that ann(z) = Z(R). Letx ∈Z(R)∗. Then eitherxz = 0orann(x) = ann(z)sincedeg[z] = n−1
inΓE(R), by Corollary 2.11. Ifxz = 0 we are done. Now, assume that ann(x) = ann(z). It
is enough to show that z2 = 0. Let [x],[x1],· · · ,[xn−1]be all vertices of AGE(R). In view of
[11, Proposition 1.5] ΓE(R) is not complete, so we can assume that [x1]is not adjacent to[x2]in ΓE(R). Thusx1x2 6= 0. By Theorem 2.9,Z(R) is an ideal ofRsoz+x1 ∈ Z(R)∗. Therefore, ann(z+x1) = ann(z)orann(z+x1) = ann(xi)for someiwith1≤i≤n−1. In the first case, the
assumptionzx2 = 0implies thatx1x2 = 0which is a contradiction. Thusann(z+x1) = ann(xi)
for someiwith1 ≤ i ≤ n−1. We havezxj = 0for all j with1 ≤ j ≤ n−1, in particular for
j= 1, i. Hence,z2= 0. 2
Theorem 2.13 — Let R be Noetherian andΓE(R)be a star graph with at least four vertices.
ThenAGE(R)is a complete graph.
PROOF: In view of [11, Proposition 2.4],Ass(R) ={p}andp3 = 0. Let[y]be the vertex with
maximal degree and let[x1],[x2],· · · be the other vertices ofΓE(R). Ifi 6= j, thenxixj 6= 0and
annihilated byxiandxj sincep3 = 0. Hence, the only choice for[xixj]is[y]. If[xi]−[xj]is not
an edge ofAGE(R), thenann(xi)∪ann(xj) = ann(xixj) = ann(y). Soann(xi) = ann(y)or ann(xj) = ann(y), which is a contradiction. Thus[xi]−[xj]is an edge ofAGE(R)and therefore,
AGE(R)is a complete graph. 2
Example 2.14 : LetR=Z2[X, Y, Z]/(X2, Y2). By [11, Example 2.6]ΓE(R)is an infinite star
graph. Therefore,AGE(R)is an infinite complete graph.
Corollary 2.15 — LetRbe Noetherian. Then there is no star graphAGE(R)with more than three
vertices.
Proposition 2.16 — LetRbe Noetherian. Thengr(AGE(R)) = 4if and only ifAGE(R) =C4.
PROOF: If the girth ofΓE(R)is finite, then it must be three by Theorem 6.6, in [7]. Thus the
girth ofAGE(R)is three contrary to assumption. Hence,ΓE(R)is acyclic and therefore,ΓE(R)is
a star graph or a path of length three, by Proposition 6.4, in [7]. Now, Theorem 2.13 implies that ΓE(R)is not a star graph. ThusΓE(R)is a path of length three and soAGE(R) =C4. 2
Theorem 2.17 — Let R be reduced. Then|AGE(R)| = 2if and only if Γ(R) is a complete
PROOF: Letx, y ∈Z(R)∗ and[x]6= [y]. Thenann(x) 6= ann(y). Suppose thatz ∈ann(x)\ ann(y). Thusann(z) = ann(x)orann(z) = ann(y)and sox2 = 0orxy = 0. SinceRis reduced,
we havexy= 0. Furthermore, for anyw∈Z(R)∗, we haveann(w) = ann(x)orann(w) = ann(y).
Ifann(w) = ann(x), thenwy = 0andwx6= 0and ifann(w) = ann(y), thenwx= 0andwy 6= 0.
It follows that Γ(R) is a complete bipartite graph. Conversely, assume that Γ(R) is a complete
bipartite graph. ThusΓE(R)is the graphK2sinceRis reduced. Hence,|AGE(R)|= 2. 2
Theorem 2.18 — IfRis Noetherian and|AGE(R)|= 3, thenRis not reduced.
PROOF: Assume that[x],[y]and[z]are distinct vertices ofAGE(R). It follows that ΓE(R) is
the graph[x]−[y]−[z]by Corollary 1.6, in [11]. So we haveAGE(R) = ΓE(R)orAGE(R) =K3.
IfAGE(R) = ΓE(R), thenann(x)∪ann(z) = ann(xz). Without loss of generality, assume that ann(x) = ann(xz). Thusann(z) ⊂ann(x). Selects∈ann(x)\ann(z). Thenann(s) = ann(x)
and sox2 = 0, soRis not reduced. Now, assume thatAG
E(R) = K3. Thus[x]−[z]is an edge of AGE(R). Henceann(x)∪ann(z) ⊂ann(xz)and thereforeann(xz) = ann(y). We havexy = 0
andx2z= 0. So that(xz)2= 0thusRis not reduced. 2
Corollary 2.19 — IfRis Noetherian andgr(AGE(R)) =∞, thenAGE(R)is a path of length≤
2.
PROOF: By hypothesis we haveΓE(R) = AGE(R). Hence,ΓE(R)is a star graph or a path of
length three by Proposition 6.4, in [7]. On the other hand, Theorem 2.13 implies thatAGE(R)is not a
star graph with more than three vertices. We have to show thatAGE(R)is not a path of length three.
Suppose thatAGE(R)is the path[z]−[x]−[y]−[w]. It follows thatann(z)∪ann(w) = ann(zw)
and therefore one can assume thatann(z) = ann(zw). Now, we haveyzw= 0sinceyw = 0and so yz= 0butyz 6= 0. HenceAGE(R)is not a path of length three. 2
Proposition 2.20 — LetR, S be commutative rings andϕ : R −→ S be a flat ring
monomor-phism. ThenAGE(R)is isomorphic to an induced subgraph ofAGE(S).
PROOF: By the proof of Proposition 4.1, in [7],Z(R)∗maps intoZ(S)∗ andann(x) = ann(y)
if and only ifann(ϕ(x)) = ann(ϕ(y))for everyx, y ∈ Z(R)∗. On the other hand,z ∈ann(xy)\ (ann(x) ∪ann(y)) if and only if ϕ(z) ∈ ann(ϕ(xy))\ (ann(ϕ(x))∪ann(ϕ(y))) since ϕ is a
monomorphism. This means thatAGE(R)is isomorphic to an induced subgraph ofAGE(S). 2
Corollary 2.21 — IfT =R\Z(R), thenAGE(R)∼= AGE(T−1R).
annT−1R(x/1), for everys ∈T. Also, we haveann(x) 6= ann(y)if and only ifannT−1R(x/1)6= annT−1R(y/1), for allx, y∈Z(R). It follows thatAGE(R)∼= AGE(T−1R)by Proposition 2.15.2
3. 2-ABSORBINGIDEAL ANDANNIHILATORGRAPH
The concept of 2-absorbing ideal is a generalization of prime ideals and introduced in [4]. A proper
ideal I of R is called 2-absorbing if whenever abc ∈ I for a, b, c ∈ R, thenab ∈ I or ac ∈ I
orbc ∈ I. In the following we investigate the connection between Γ(R) andAG(R), when0is a
2-absorbing ideal ofR.
Theorem 3.1 —Γ(R) =AG(R)if and only if0is a 2-absorbing ideal ofR.
PROOF: Suppose that 0is a 2-absorbing ideal ofR. Letx, y ∈ Z(R)∗ andx−yis an edge of AG(R). Thenann(x)∪ann(y)⊂ann(xy). Therefore, there isz∈ann(xy)\(ann(x)∪ann(y))and
so we havezxy = 0,zx6= 0andzy 6= 0. Thusxy = 0since0is a 2-absorbing ideal ofR. Hence, x−yis an edge ofΓ(R). Conversely, assume thatΓ(R) =AG(R). We show that0is a 2-absorbing
ideal ofR. Letx, y, z ∈R,xyz= 0withxy 6= 0andxz6= 0. Thenx∈ann(yz)\(ann(y)∪ann(z))
and thusy−zis an edge ofAG(R). Therefore,y−zis an edge ofΓ(R)and soyz= 0. 2
Corollary 3.2 — IfΓ(R) =AG(R), then{ann(x)|x∈R}is a totally ordered set or is the union
of two totally ordered sets.
PROOF: By Theorem 3.1,0is a 2-absorbing ideal ofR. Now the result follows by Theorem 2.1,
in [9]. 2
The following corollary is a generalization of Theorem 3.5, in [5].
Corollary 3.3 — If|M in(R)| ≥3, thenΓ(R)6=AG(R).
PROOF: Assume thatΓ(R) =AG(R). Then in view of Theorem 3.1,0is a 2-absorbing ideal of R. Thus Theorem 2.3, in [4], shows that|M in(R)| ≤2contrary with assumption. 2
Proposition 3.4 — Let0be a 2-absorbing ideal ofRandZ(R) = Nil(R). ThenΓ(R)is complete.
PROOF: Since0 is a 2-absorbing ideal, we havez2 = 0 for everyz ∈ Z(R). Letx andybe
distinct vertices. Thus(x+y)xy = 0. Hence,(x+y)x = 0or(x+y)y = 0orxy = 0. In each
case,xy = 0. That impliesΓ(R)is complete. 2
Proposition 3.5 — LetR be a reduced ring. ThenΓE(R) = AGE(R) if and only ifΓ(R) = AG(R).
2-absorbing ideal ofR. LetΓE(R) = AGE(R)and letx, y, z ∈Rbe such thatxyz= 0,xy 6= 0and xz6= 0. Theny, z∈Z(R)∗. SinceRis reduced, we have[y]6= [z]. Ifyz 6= 0, then[y]−[z]is not an
edge ofΓE(R). Therefore,[y]−[z]is not an edge ofAGE(R). Thenann(y)∪ann(z) = ann(yz),
and soann(y) = ann(yz) orann(z) = ann(yz). It follows that xy = 0 orxz = 0 and it is a
contradiction. Henceyz= 0and0is a 2-absorbing ideal ofR. The converse is clear. 2
In the following, we show that ifR1, R2 are commutative rings such that eitherZ(R1)∗ 6=∅or Z(R2)∗ 6=∅, then0is not a 2-absorbing ideal ofR1×R2.
Theorem 3.6 — LetR1, R2 be commutative rings such that eitherZ(R1)∗ 6= ∅orZ(R2)∗ 6=∅
andR=R1×R2. Then
(i) AGE(R)is not complete.
(ii) AGE(R)6= ΓE(R).
PROOF : Suppose Z(R1)∗ 6= ∅ and select a ∈ Z(R1)∗. Then there isb ∈ Z(R1)∗ such that ab= 0.
(i) It is clear that(a,0),(1,0)∈Z(R)∗and[(a,0)]6= [(1,0)]. We haveann(a,0)∪ann(1,0) = ann((a,0)(1,0)). Therefore,[(a,0)]−[(1,0)]is not an edge ofAGE(R) and thusAGE(R) is not
complete.
(ii) It is easily verified that(a,1),(1,0)∈Z(R)∗ and[(a,1)]6= [(1,0)]. On the other hand, we
have(1,0)(a,1)6= 0and(b,1)∈ann(a,0)\(ann(1,0)∪ann(a,1)). Thus[(a,1)]−[(1,0)]is an
edge ofAGE(R)but it is not an edge ofΓE(R). Hence,AGE(R)6= ΓE(R). 2
Theorem 3.7 — LetR=R1×R2and letR1, R2be commutative rings such that eitherAGE(R1)
orAGE(R2)has at least an edge. Thengr(AGE(R)) = 3anddiam AGE(R) = 2.
PROOF: Leta, b ∈Z(R)∗ and[a]−[b]be an edge ofAGE(R1). Then there isc ∈ ann(ab)\ (ann(a)∪ann(b)). Thus(c,0) ∈ ann((a,1)(b,1))but(c,0) 6∈ ann(a,1)∪ann(b,1). It follows
that[(a,1)]−[(b,1)]is an edge ofAGE(R). Now, the proof of Theorem 3.6(ii) shows that[(1,0)]− [(a,1)]−[(b,1)]−[(1,0)]is a cycle inAGE(R). Therefore,gr(AGE(R)) = 3. Furthermore,AGE(R)
is not complete by Theorem 3.6(i). Hence,diam AGE(R) = 2by Theorem 2.4. 2
Corollary 3.8 — LetR be an Artinian ring such that0 is a 2-absorbing ideal ofR. Then R is
either local or reduced.
PROOF : By assumption 0 is a 2-absorbing ideal, thus R has at most two maximal ideals by
k >0such thatmk1 ∩m2k= 0andR ∼=R/mk
1 ×R/mk2, by Proposition 8.4 and Theorem 8.7, in [3]. Ifk > 1, thenZ(R/mk1)∗ 6=∅. Hence,AGE(R) 6= ΓE(R) by Theorem 3.6(ii). It follows that0is
not 2-absorbing which is a contradiction. Therefore,k= 1and soNil(R) =m1∩m2 = 0. Hence,R
is reduced. 2
Corollary 3.9 — Let 0 be a 2-absorbing ideal of R and let 0 = q1 ∩ · · · ∩qn be a primary
decomposition of0, whereqi,qjare coprime wheneveri6=j. Then0is either primary or intersection
of two prime ideals.
PROOF: Suppose that0is not primary. Thus Proposition 1.10, in [3] shows thatR ∼=
Q iR/qi. Assume thatqiis not prime for somei. ThenZ(R/qi)∗ 6=∅. Hence,AGE(R)6= ΓE(R)by Theorem
3.8. It follows that0is not 2-absorbing, contrary to assumption. Hence,qiis prime ideal ofR, for all i. On the other hand,Rhas at most two prime ideals that are minimal, by [4, Theorem 2.3]. Hence, 0 =q1∩q2, whereq1,q2are prime ideals ofR. 2
ACKNOWLEDGEMENT
We would like to thank the referee for a careful reading of our article and insightful comments which
saved us from several errors.
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