Investigating the Algebraic Structure Embedded in Zero-Divisor Graphs B. KELLY AND E. WILSON Abstract: Zero-divisor graphs have key algebraic information embedded in them. This paper investigates what algebraic information can be extracted from a zero-divisor graph.

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Introduction and De…nitions

In this paper, R will denote a …nite commutative ring with unity. Let r 2 R; if there exists a non-zero element x such that rx = 0 then r is called a zero-divisor. Let A denote An f0g for any set A. The zero-divisor graph of R, denoted (R) ; is given by the vertex set V = Z(R) and the edge set E = f(a; b) j ab = 0g. In this paper, if x2 = 0 then x is said to be looped, which means that there is an edge from x to x in (R). This is a variation of the de…nition given in most of the literature. An example of a graph with loops is (Z18 ) ; which is shown below. The illustrations in this paper were generated using a program which we created using Mathematica, the …les for which can be found online [3].

Figure 1:

(Z18 )

An element x 2 R is said to be nilpotent if there exists n 2 N such that xn = 0: A ring R is reduced if it has no non-zero nilpotent elements. A ring 1

R is local if it has a unique maximal ideal. The annihilator of x 2 R is the set ann(x) = fr 2 R j rx = 0g. The degree of x 2 R, deg(x); is the number of elements in ann (x) and corresponds to the number of edges incident with x in (R). A complete graph is a graph such that given any two vertices v1 ; v2 there exists an edge between v1 ; v2 :A k-clique of a graph is a complete subgraph with k vertices. A maximal k-clique is a complete subgraph with k vertices which is not a proper subgraph of any clique. A graph G is said to be star-shaped reducible if there exists a vertex v such that v is adjacent to all other vertices and looped to itself. It has been shown in [1] that if R1 ; R2 are …nite reduced rings that are not …elds, then (R1 ) = (R2 ) implies R1 = R2 : However, if R1 ; R2 are not reduced rings, there are examples where (R1 ) = (R2 ) but R1 R2 . Although the ring structure for R1 cannot be rebuilt by examining (R1 ), many algebraic properties are hidden in (R1 ). Progress has been made by Benson-Lender et al. in [2] ; where they show that Z(R) forms an ideal if and only if (R) is starshaped reducible. They also show that Z(R) forms an ideal if and only if R is local. Thus, for all (R) which are star-shaped reducible, R is a local ring. In this paper we look for more algebraic information that can be determined from (R). We will …rst discuss commutative rings with unity, and then speci…cally address local commutative rings with unity.

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Does

(R) Completely Determine the Multi-

plicative Structure of Z (R)? The following is an example of two rings where (R1 ) = (R2 ) but R1 Example 1 (Z16 ) = (Z2 = x4 ); but Z16 2 6= 16 = char (Z16 ) :

2

R2 :

Z2 = x4 since char Z2 = x4

=

Figure 2:

Figure 3:

(Z16 )

Z2 [x] = x4

Since (R) is constructed using the multiplicative structure of Z (R) it is reasonable to believe that (R) contains all the information necessary to reconstruct the multiplicative table of Z (R) : Examining Z (Z16 ) and Z Z2 = x4 we realize that Z (Z16 ) is a cyclic group under addition while Z Z2 = x4 is clearly not. Therefore there cannot be a ring isomorphism between Z (Z16 ) and Z Z2 = x4 : However we are able to construct a multiplicative isomorphism between Z (Z16 ) and Z Z2 = x4 : Zp [x] = (xn ) and Z (Zp ) Example 2 (Zpn ) = (Zp [x] = (xn )) but Zpn n n Z (Zp [x] = (x )) : However, : Z (Zp [x] = (x )) ! Z (Zpn ) where (x) = p and for r 2 Zp ; (r) = r is a multiplicative isomorphism. Example 3 (Z3 Z9 ) = Z3 Z3 [x] = x2 : 2 Let : Z Z3 Z3 [x] = x ! Z (Z3 Z9 ) be de…ned by the following: ((r; r0 )) = (r; (r0 )) where (1) = 1; (2) = 8; (x) = 3; (x + 1) = 4; (x + 2) = 2; (2x) = 6; (2x + 1) = 7; and (2x + 2) = 5: In this example, is a graph isomorphism that is also a multiplicative isomorphism. Mathematica found 69,120 graph isomorphisms, using the Combinatorica package, which counts graph isomorphisms, combined with our own programs [3], which construct zero-divisor graphs, many of which are not multiplicative isomorphisms.

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Figure 3:

(Z3

Z9 ) and

Z3

Z3 [x]= x2

In [4] Redmond fully classi…es (up to isomorphism) all rings that produce zero-divisor graphs with fewer than 14 vertices. It has been veri…ed that a multiplicative isomorphism between Z (R1 ) and Z (R2 ) exists whenever (R1 ) = (R2 ) in the catalogue of zero-divisor graphs with fewer than 14 vertices. Based on this information we make the following conjecture: Conjecture 4 (R1 ) = (R2 ) if and only if there exists a multiplicative isomorphism between Z (R1 ) and Z (R2 ).

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Algebraic Information from Degree and Cliques

The investigation of Conjecture 4 has led to the discovery of other algebraic information that is determined by (R) : The following propositions support Conjecture 4 by illuminating algebraic information which (R) captures. A potential proof of Conjecture 4 could be constructed by piecing propositions like 4

these together and concluding that (R) uniquely determines the multiplicative structure on Z (R) : The …rst two propositions exploit information gained from the degree of the vertices. Lemma 5 Let R be a commutative ring with unity. Let z 2 Z (R) and r 2 R. If r 2 = Z (R) then ann (zr) = ann (z). Proof. Assume r 2 = Z (R). Clearly ann (zr) ann (z) : Let x 2 ann (zr) : Then xzr = 0; but since r 2 = Z (R) this implies that xz = 0. Therefore x 2 ann (z). Corollary 6 Let R be a …nite commutative ring with unity. Let z 2 Z (R) and r 2 R. If r 2 = Z (R) then deg (zr) = deg (z). Proposition 7 Let R be a …nite commutative ring with unity. Let x; y 2 (R) such that x and y are of maximum degree. Then xy = 0 or ann (x) = ann (y) : Proof. Let x; y be vertices with maximum degree such that xy 6= 0. Clearly, ann (xy) ann (x) [ ann (y) : Since x; y are vertices with maximum degree, deg (x) deg (xy). Therefore ann (x) = ann (y) : Proposition 8 Let L = fx 2 (R) j deg (x) is minimalg. If a; b 2 Z (R) and ab 2 L then a; b 2 L and ann (a) = ann (b) : Proof. Let ab 2 L: Clearly, ann (ab) ann (a) [ ann (b) : Since deg (a) deg (ab) ; this implies that deg (a) = deg (ab) : Therefore, a 2 L and ann (ab) = ann (a). Similarly, b 2 L and ann (ab) = ann (b), and therefore ann (a) = ann (b) : Example 9 Consider the zero-divisor graph (Z21 ) : Consider the product 6 2 = 12 as Corollary 6 predicts deg (6) = deg (12) : Notice that 3; 6 2 Z (R) and that 3; 6 and 3 6 are all elements of minimal degree of (Z21 ) : Therefore by Proposition 8 we can conclude that ann (3) = ann (6)

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(Z21 ) The following propositions use properties of the cliques to gather algebraic information. ( ) [ Proposition 10 Let Kc = K j K is a k clique [ f0g. Then for any x 2 Kc ; xR

Kc :

Proof. Let x 2 K for some ; and let r 2 R. Then rx is either 0 or is connected to every element of K n fx g. Therefore, rx belongs to some k clique or rx = 0. In either case rx 2 Kc : Lemma 11 Let R be a …nite commutative ring with unity. Let x 2 Q where Q is a maximal k clique of (R) ; and let x2 = 0. Then xR Q [ f0g : Proof. Let x 2 Q where Q is a maximal k clique such that x2 = 0; and let r 2 R. The product xr is either 0 or is connected to every element in Q. Since Q is maximal xr 2 Q. In either case xr 2 Q [ f0g : Lemma 12 Let R be a …nite commutative ring with unity. Let M = fx j x is of maximum degreeg[f0g: Then for any r 2 R; rM

M:

Proof. Let m 2 M and let r 2 R: If rm = 0 then rm 2 M: Assume that rm 6= 0: Clearly ann (m) ann (rm) : But since deg (m) is maximum, ann (m) = ann (rm) and deg (rm) = deg (m) : Therefore rm 2 M:

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Theorem 13 Let Q be a maximal k clique, and let x2 = 0 for all x 2 Q. Then Q [ f0g is an ideal. Proof. Let w 2 Q and r 2 R. From Lemma 11 wr 2 Q: Let x; y 2 Q. For all q 2 Q, qx = qy = 0. Therefore q (x + y) = qx + qy = 0 + 0 = 0. Since Q is maximal, x + y 2 Q: Thus Q is an ideal. If x2 6= 0 for some x 2 Q; then Q [ f0g may not be an ideal. Example 14 Looking at the graph Theorem 13.

(Z81 ) we can easily identify an ideal using

(Z81 )

An ideal of Z81 is identi…ed by looking at

Example 15 A maximal 3 clique of (Z2 Z2 Z2 ) is given by f(1; 0; 0) ; (0; 1; 0) ; (0; 0; 1)g : However, this clique is not an ideal since it is not closed under addition. This demonstrates the necessary condition for x2 = 0 for all x 2 Q: It also demonstrates that including loops in zero-divisor graphs should become standard. Another way to exploit the graphical information given by (R) is to look at the 1 separations: A 1 seperation of a graph G is an ordered pair (A; B) of subgraphs, such that both A and B have at least one vertex, A [ B = G; and A \ B contains a single vertex called the cut vertex. Benson-Lender et al. proved the following theorem in [2]. Theorem 16 Let (R) be a zero-divisor graph with a 1 separation: If one of the subgraphs that forms the 1 separation forms a complete subgraph of order 3, then every vertex of that subgraph is looped. 7

(Z81 )

A corollary to this theorem is useful in identifying ideals in R by looking at subgraphs in (R) : Corollary 17 Let (R) be a zero-divisor graph with a 1 separation (A; B). If A is a complete subgraph then A is an ideal. Proof. This follows from Theorem 13 combined with the result from BensonLender et al., since the complete subgraph A will be a maximal k clique and from the above theorem x2 = 0 for all x 2 Q:Below Example 18 Three di¤ erent 1-seperations and their corresponding ideals are illustrated below in the zero-divisor graph of Z4 [x] = x2 :

Z4 [x] = x2

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Local Commutative Rings With Unity

If R is a local ring, then (R) reveals more algebraic information. It is possible that restricting Conjecture 4 to local rings may provide insight that leads to a more general proof.

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Theorem 19 Let R be a …nite local ring with unity. The set M = fx 2 R j x is of maximum degree in (R)g [ f0g forms an ideal in R. Proof. Let x 2 M and r 2 R. By Lemma 12, xr 2 M: Let w; v 2 M . Since R is a …nite local ring, (R) is star-shaped reducible by Theorem 2 in [2]. Because w; v 2 M it follows for all z 2 Z (R) : wz = vz = 0 Therefore for any z 2 Z (R) ; z (v + w) = 0 and v + w 2 M: The set M in the above theorem forms the looped centers of the star shaped reducible graph. Lemma 20 Let R be a …nite local ring with unity, let M = fx 2 R j x is of maximum degree in (R)g [ f0g ; and let ' : R ! R=M be the canonical map. Then 1) Z (R=M ) = ' (Z (R)) 2)Z (R=M ) forms an ideal in R=M: Proof. Since ' is onto we will write z + M as ' (z) : 1) ( ) Let ' (z) 2 Z (R=M ) such that z 2 = M . Assume for contradiction’s sake that z 2 = Z (R) : There exists r 2 RnM such that ' (r) ' (z) = 0 + M . Therefore rz 2 M . We divide the proof into two cases: rz = 0 and rz 6= 0 If rz = 0; then z 2 Z (R) which is a contradiction: If rz 6= 0; then rzrz = 0; because rz 2 M by proof of Theorem 19. However, (rz) z 6= 0 because z 2 = Z (R) : Thus r 2 Z (R) nM and deg (rz) > deg (r) which contradicts the contrapositive of Corollary 6. Therefore z 2 Z (R) : ( ) Let z 2 Z (R). If z 2 M then clearly ' (z) 2 Z (R=M ). So assume that z 2 = M . Since R is a …nite local ring Z (R) = nil (R). Let n 2 N such that n zn 2 = M but z n+1 2 M . Therefore ' (z) ' (z) = 0 + M and ' (z) 2 Z (R=M ) : 2) R Z (R) M . Since R is a …nite local ring Z (R) is an ideal by Theorems 1 and 2 in [2]. By the Correspondence Theorem, ' (Z (R)) = Z (R=M ) forms an ideal in R=M: Theorem 21 Let R be a …nite local ring with unity. Let M = fx 2 R j x is of maximum degree in (R)g [ f0g. Then the elements of second highest degree in (R) are of highest degree in (R=M ). Proof. Let Y = fy1 ; :::; yn g Z(R) be the set of all elements of second highest degree in (R). Let ' be the canonical map from R to R=M: By Lemma 20, Z(R=M ) is an ideal and thus (R=M ) is star-shaped reducible. If ' (yi ) is not of highest degree in (R=M ) ; then there exists an ' (a) 2 Z(R=M ) such that ' (a) ' (yi ) 6= M and thus ayi 2 = M . Note that deg(ayi ) = deg(yi ), since deg(ayi ) deg(yi ) and ayi does not have maximum degree in (R). Therefore ayi = yj for some yj 2 Y; 9

and ann(yi ) ann(yj ). Because deg(yj ) = deg(yi ); it follows that ann(yj ) = ann(yi ). Since R is local and …nite, we have that a is nilpotent from Lemmas 2 and 4 of [2]. So there exists some m such that am = 0: Clearly, yi am = 0: Since yj = ayi ; it follows that yj am 1 = 0. Because ann(yi ) = ann(yj ); yi am 1 = 0. If, for some n m; yi an = 0; then yj an 1 = 0; and hence yi an 1 = 0. By induction, yi a = 0 2 M . This is a contradiction. Theorem 21 suggets that there is a strong corespondence between the elements of highest degree in (R=M ) and the elements of second highest degree in (R). An unanswered question is if indeed the elements of second highest degree in (R) are the only constituents of elements of highest degree in (R=M ). It is clear that (R) provides much algebraic information about R; however, there are many questions open as to the extent of the knowledge gained. Though the results obtained so far are interesting, their usefulness in proving 4 remains uncertain.

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References [1] Anderson, D.F., Frazier, A., Lauve, A., and Livingston P., The zero-divisor graph of a commutative ring, II, Lecture Notes in Pure and Appl. Math., 220 (2001), 61-72 [2] Benson-Lender, L., Martinez, M., Roberts, J., Skalak, M., and Taylor, M., The Structure of Rings Based on Their Zero-Divisor Graphs [3] Kelly B., Wilson E., Mathematica Files and Zero-Divisor Graphs. http://kelly1618.googlepages.com/zerodivisors [4] Redmond, S., On Zero-Divisor Graphs of Small Finite Commutative Rings, Discrete Math. 307(9-10) (2007), 1155-1166.

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