Irreducible Divisor Graphs

Irreducible Divisor Graphs Jim Coykendall Department of Mathematics North Dakota State University Fargo, ND 58105-5075 Jack Maney Department of Mathematical Sciences The University of South Dakota 414 E. Clark St. Vermillion, SD 57069 October 21, 2005 Abstract In [6], I. Beck introduces the idea of a zero-divisor graph of a commutative ring. We generalize this idea to study factorization in integral domains and define irreducible divisor graphs. We use these irreducible divisor graphs to characterize certain classes of domains, including UFDs. 1 Introduction In this paper, R will denote an integral domain (commutative, with nonzero identity). Often, but not always, we will also assume that R is an atomic domain (that is, every nonzero nonunit element of R can be written as a product of irreducibles). If S is a subset of R, we denote S \ {0} by S. Also, N and F q denote the natural numbers and the field of q elements, respectively. If G = (V, E) is a graph and if v is a verte of G, then by v G we will mean v V. Also, V and E may also be respectively denoted by V (G) and E(G) if there is any danger of confusion as to which graph is being referred. A recent subject of study linking commutative ring theory with graph theory has been the concept of the zero-divisor graph of a commutative ring. Let R be a commutative ring with identity and denote the set of (nonzero) zero-divisors of R as Z(R). The zero-divisor graph of R, denoted by Γ(R), is the graph with verte set Z(R) and if a, b Z(R), ab E(Γ(R)) if and only if ab = 0. Zero-divisor graphs were introduced by I. Beck in [6], and they have been studied and generalized even more recently (cf. [5], [3], [11], [8], [10], [14], [4], 1

[15], [1], [2], [12], [16], [9], and [17]). In this paper, we stretch this idea in a different direction, applying it to factorization in integral domains. If R is atomic, it is well-known that nonzero nonunit elements of R need not factor uniquely into irreducibles. If a nonzero nonunit R (with R atomic) does not factor uniquely into irreducibles, then it follows that there eist two nonassociate irreducible divisors, π and ξ of such that = πα 1 α 2 α m = ξβ 1 β 2 β m, with α i, β j irreducible, and with π nonassociate to each β j and likewise with ξ and each α i. When this happens, we will say that π and ξ show up in two different factorizations of. Equivalently, if it is not the case that any two nonassociate irreducible divisors of show up in two different factorizations of, then factors uniquely into irreducibles. Thus, if is a nonzero nonunit element of an (atomic) domain R, then one approach to studying the factorization(s) of this (fied) lies in looking at how many pairs of nonassociate irreducible divisors of are both in a factorization of (or, in other words, looking at how many pairs of nonassociate irreducible divisors of have their product dividing ). This leads to the development of the irreducible divisor graph. If R is a domain, we denote the set of irreducibles of R by Irr(R). However, as is the case when studying factorization in integral domains, we will generally not distinguish between an irreducible and its associates. So, we define Irr(R) to be a (pre-chosen) set of coset representatives, one representative from each coset in the collection {πu(r) π Irr(R)}. We also recall that a graph G (with no loops or multiple edges) is a clique if G is a complete graph, that is, if there is an edge between every pair of vertices. A clique on n vertices (n N) will be denoted by K n. If G is a graph with a subgraph H, we recall that H is an induced subgraph of G if for all u, v H, uv E(G) implies that uv E(H). For v G, we denote the neighborhood of v (in G) by N(v) = N G (v) := {u G uv E(G)}. We will need a new graph-theoretic definition that we will find helpful. Definition 1.1 Let G be a graph. We say that G is a pseudo-clique if G is a complete graph having some number (possibly zero) of loops. It immediately follows that a clique is a pseudo-clique. Definition 1.2 Let R be a domain, and let R be a nonzero nonunit that can be factored into irreducibles. 1. We define the irreducible divisor graph of to be the graph G() = (V, E) where V = {y Irr(R) y }, and given y 1, y 2 Irr(R), y 1 y 2 E if and only if y 1 y 2. Further, we attach n 1 loops to the verte y if y n divides. 2

2. We define the reduced divisor graph of to be the subgraph of G() containing no loops, and we denote the reduced divisor graph of as G(). 3. If R is an antimatter domain, that is, a domain having no irreducibles (and this includes the class of fields), then we define G() = G() to be the empty graph for each R. 4. If A V (G()), then by G A (), we mean the induced subgraph of G() on N(A). If A = {π 1, π 2,, π n }, then G A () will be denoted by G (π1,π 2,,π n )() and if A = {π}, then G A () will be denoted by G π (). Note that in G() it is possible for π to be in N(π). In fact, this happens precisely when π 2. We will mainly be interested in the case of R being atomic. Throughout the rest of this paper, we will assume that R is not an antimatter domain. 2 Eamples Eample 2.1 Let R = Z[ 5]. Using norms, it is easy to see that the only nonassociate irreducible factorizations of 6 are: 6 = 2 3 = (1 5)(1 + 5). So, G(6) is as follows: 2 1+ 5 3 1 5 Also, the only irreducible factorizations of 18 in R are: 18 = 2 3 2 = 3(1 5)(1 + 5) = 2(2 + 5)(2 5). So, G(18) is as follows: 2 3 2+ 5 2 5 1 5 1+ 5 Eample 2.2 Let R = Z[ 14]. Then, again using norms, we see that the only irreducible factorizations of 81 are: 81 = 3 4 = (5 2 14)(5 + 2 14). 3

Thus G(81) is as follows: 5+2 14 3 5 2 14 Eample 2.3 Let R be a UFD, and let be any nonzero nonunit. Then we may factor as p a 1 1 pa 2 2 pa n n where a i N and the p i are distinct nonassociate primes. Since this is the only way to factor into irreducibles, we see that G() is isomorphic to K n with a i 1 loops on each verte p i. It also follows that G() = K n. Eample 2.4 Let R = Q[ 2, 3 ] and let f() = 9 10. The only irreducible factorizations of f in R are: f() = 2 2 3 ( 2 3 ) = 2 2 2 ( 3 4 ) = 3 3 ( 3 4 ). And G(f) is as follows: 3 4 2 3 2 3 Eample 2.5 Let F = F 2, and let y be a root of the (irreducible) polynomial X 3 + X + 1 F []. Let K = F [y] and let V be the F -subspace of K with basis {1, y}. Finally, let R = F + V + 2 K[[]]. After some calculations, it is apparent that V = {0, 1, y, y 3 } and that Irr(R) = {, y, y 3, y 5 2 }. Furthermore, the only nonassociate irreducible factorizations of 5 are: () 5 = (y 5 2 )(y 3 ) 3 = (y 5 2 )(y) 2 = 2 (y 3 ) 2 (y) = (y) 4 (y 3 ). 4

Therefore G( 5 ) is as follows: y 3 y y 5 2 and G( 5 ) has four loops on, two loops on y 3, and three loops on y. We now notice a contrast between these eamples. In Eamples 2.1, 2.2, and 2.3 every (maimal) pseudo-clique corresponds to a factorization. For eample, in the second graph in 2.1, we have a triangle formed by 2 and 2 ± 5 and we also have 2(2 + 5)(2 5) = 18. Similarly, in Eample 2.2, the verte 3 (along with its 3 loops) makes a factorization of 81, as does the K 2 formed by 5 ± 2 14. These eamples are nice in the sense that it is easy to pick out the irreducible factorizations just by looking at the irreducible divisor graph. In contrast, in Eample 2.4, we have a triangle formed by 2, 3, and 3 4. This triangle also forms a maimal pseudo-clique (since G(f) is not a psuedoclique). However, 2 3 ( 3 4 ) does not divide f(). Also, in Eample 2.5, G( 5 ) is a pseudo-clique. However, y 9 5 = y 2 5 = (y)(y 3 )(y 5 2 ) does not divide 5 in R. The phenomenon behind this contrast will be studied in the sequel. 3 Preliminary Results Recall that R is a Finite Factorization Domain (FFD) if R is atomic and every nonzero nonunit element of R has finitely many nonassociate irreducible divisors. Proposition 3.1 Let R be an atomic domain. Then R is an FFD if and only if G() is finite for each nonzero nonunit R. Proof: If R is an FFD, then every nonzero nonunit R has but finitely many nonassociate irreducible factors, whence the verte set of G() is finite. On the other hand, if G() is finite for each, then each has only finitely many nonassociate irreducible divisors, and R is an FFD. Lemma 3.2 Let R be a domain and let R be a nonzero nonunit that can be factored into irreducibles, and let π be an irreducible divisor of. Then the verte sets of G π () and G( π ) coincide. In other words, the vertices of G( π ) are precisely those in the neighborhood of π. What is more, the edge set of G( π ) is contained in the edge set of G π (). Proof: Note that the result is vacuously true if Irr(R). 5

Assume that there eists another irreducible divisor of that is adjacent to π, call it ξ. Since ξ and π are adjacent in G(), we must have that ξπ whence ξ G( π ). Conversely, if ξ is a verte of G( π ), then ξ is an irreducible divisor of π whence ξπ and ξ is a verte in G π(). Now, let α, β G( π ) have an edge between them. Then αβ π, yielding that αβπ Thus αβ combined with the fact that α, β N(π) yields that αβ E(G π ()). If no other irreducible divisors of are adjacent to π, then (up to a unit of R), we have = π k for some k 2. It follows that π is adjacent to itself in G(), whence it is easy to see that the assertions follow. We recall that an atomic domain R is a half factorial domain, or HFD, if given any collection of irreducibles {α 1, α 2,, α m, β 1, β 2,, β n } with α 1 α m = β 1 β n, then m = n. Theorem 3.3 Let R be an atomic domain. Then the following are equivalent: 1. R is an HFD. 2. For any nonzero nonunit of R, and for any irreducible factorization π a1 1 πa2 2 πan n with the π i s pairwise nonassociate, the sum of the number of vertices and the number of loops in G (π1,π 2,,π n )() is constant. Proof: (1 2): Clear from the definition of HFD. (2 1): Let π a 1 1 πa 2 2 πan n = ξ b 1 1 ξb 2 2 ξbm m be two irreducible factorizations of. In G (π1,π 2,,π n)() (resp. G (ξ1,ξ 2,,ξ n)()), π i (ξ j ) has a i 1 (b j 1) loops. Thus, by hypothesis, n m n + (a i 1) = m + (b j 1) j=1 n m a i = b j, j=1 and R is an HFD. 4 Boundary Valuation Domains We will now characterize Boundary Valuation Domains (which were defined and studied by the second author in [13]) by their irreducible divisor graphs. We first require some definitions. Definition 4.1 Let R be an HFD with quotient field K. If R K, we define the boundary map R : K Z by R (α) = t s where α = π 1π 2 π t where δ 1 δ 2 δ s π i, δ j Irr(R) for each i and j. If R = K, then we declare R (α) = 0 for all α K. The boundary map was first introduced in [7]. It is clear that R is a homomorphism of abelian groups. 6

Definition 4.2 ([13]) Let R be an HFD with quotient field K. We say that R is a boundary valuation domain (BVD) if given any α K with R (α) 0, either α R or α 1 R. It is easy to see that an HFD R with quotient field K is a BVD if for all α K with R (α) > 0, α R (cf. Theorem 2.3 in [13]). Clearly, any rank-1 DVR is a BVD. A non-trivial class of eamples includes domains of the form F + K[[]] where F K is an etension of fields. Eample 4.3 Let R be a BVD with complete integral closure R. Then R is a rank 1 DVR (cf. [13]). Denote the maimal ideal of R by zr. Then every irreducible of R is of the form uz for u U(R ). Of course, if Irr(R), then G() is merely isomorphic to K 1. If R () = 2, then we may write = (u 1 z)(u 2 z) for u i U(R ). Let uz be any other irreducible of R nonassociate (in R) to u 1 z and u 2 z. Then u 1 u 1 u 2 z is an irreducible of R and clearly (uz)(u 1 u 1 u 2 z) =. Therefore V (G()) = Irr(R). Now, suppose that (uz)(u 1 z) E(G()). Then (up to a unit in R), (uz)(u 1 z) = = (u 1 z)(u 2 z) whence u = u 2, a contradiction. Therefore uz is not connected to either u 1 z or u 2 z in G(). In fact, the only irreducible in R that is connected to uz is u 1 u 1 u 2 for if (uz)(vz) E(G()), then (again, up to a unit in R), (uz)(vz) = = (u 1 z)(u 2 z) and v = u 1 u 1 u 2. Therefore, in the case of R () = 2, G() consists of the disjoint union of K 2 s and single vertices with one loop (in the case where u 1 u 2 U(R)). Finally, suppose that R () = n 3. Then we may write = (u 1 z)(u 2 z) (u n z). By the same argument used above, V (G()) = Irr(R). So, let uz, vz Irr(R) be arbitrary. Then = (u 1 z)(u 2 z) (u n z) = (uz)(vz)(u 1 v 1 u 1 u 2 u 3 z) (u n z), and it follows that uz and vz are connected in E(G()), whence G() is a pseudo-clique on V vertices. Proposition 4.4 Let R be an atomic domain. Then the following are equivalent: 1. R is a BVD. 2. For all nonzero nonunit in R, the following hold: a). Either Irr(R) or V (G()) = Irr(R). b). If can be written as a product of two irreducibles, then G() is a disjoint union of graphs, each of which is (graph) isomorphic to K 2 or to a verte with a single loop. c). If can be written as the product of three or more irreducibles, then G() is a pseudo-clique on V (G()). 7

Proof: Eample 4.3 shows that 1 implies 2. For the other implication, suppose that condition 2 holds. We will first show that R is an HFD. If Irr(R), then we are done. Otherwise, suppose that = π a1 1 πa2 2 πa n n = ξ b1 1 ξb2 2 ξb m m, with a i, b j N and π i, ξ j Irr(R). We may assume that each of the π s and ξ s are pairwise nonassociate and that n a i is the minimal length of any irreducible factorization of. If Suppose net that However, m b j = 3. j=1 π 1 n a i = 2, then by property 2b, it is clear that m b j = 2. j=1 n a i = 3. Since G() is a pseudo-clique, ξ 1 G( π 1 ). can be written as a product of two irreducibles. It follows that So we may now assume that 4 n a i < m b j. Without loss of generality, we will also assume that amongst all elements of R with different length factorizations, that n a i is minimal. Again, ξ 1 G( π 1 ) whence we have for α l irreducible. Thus and by the minimality of However, we also have = π 1 ξ 1 α 1 α t j=1 ξ 1 α 1 α t = π a 1 1 1 π a 2 2 πan n n a i, we have t + 2 = n a i. π 1 α 1 α t = ξ b 1 1 1 ξ b m m. ( n ) Counting irreducibles, we have t + 1 = a i 1 irreducibles on the left and ( ) m n b j 1 irreducibles on the right. By the minimality of a i, we must j=1 have equal lengths, implying that n a i = m b j, a contradiction. Therefore R is an HFD. Now we show that R is a BVD. Let α = π 1π 2 π n+t be an element of the δ 1 δ 2 δ n quotient field of R with π i, δ j Irr(R) and R (α) = t > 0. Our aim is to show that α R. Let = π 1 π n+t and let y = δ 1 δ n. If n + t = 1 then n = 0 = R (y), whence y U(R) and α R. j=1 8

If n + t = 2 then either n = 0 (and y U(R)) or n = t = 1 and α = π 1π 2. δ 1 Since V (G()) = Irr(R), it follows that δ 1 and α R. Suppose now that n + t 3. Again, since V (G()) = Irr(R), we have that δ 1. Thus there eists r R such that δ 1 r = and R (r) = R () 1. Also, r we have α =. Continuing inductively, we conclude that y and that δ 2 δ n α R. 5 A Characterization of Unique Factorization We close with a characterization of unique factorization domains (or UFDs) via divisor graphs. Theorem 5.1 Let R be atomic. The following are equivalent: 1. R is a UFD. 2. For each nonzero nonunit R, G() is a pseudo-clique. 3. For each nonzero nonunit R, G() is a clique. 4. For each nonzero nonunit R, G() is connected. Proof: It is clear that 2 3 and 2 4 both hold. (1 2,3): This assertion follows from Eample 2.3. (2 1): Assume that R is not a UFD. Let A denote the set of all nonzero nonunit elements of R that do not factor uniquely into irreducibles. Assume that A is nonempty. We set n := min Choose A and an {m z = δ 1δ 2 δ m ; δ i Irr(R)}. z A irreducible factorization of, say = π a 1 1 πa 2 2 πam n with π i and π j nonassociate for each i j and a 1 + a 2 + + a m = n. Since A, we have another (distinct) irreducible factorization of : = π a 1 1 πa 2 2 πam n = ξ b 1 1 ξb 2 2 ξbt m, with ξ i Irr(R), each ξ i and ξ j nonassociate for each i j, and b 1 + b 2 + + b t n. Note that no π i can be associate to any ξ j, for if this were to occur, then we could cancel, violating the minimality of n. Now, we can factor π 1 n. Since G() is a pseudo-clique, ξ 1 is adjacent to π 1 in G(), and hence ξ 1 is in the verte set of G( π 1 ). This means that ξ 1 is an irreducible divisor of π 1. So we have the factorizations as π a 1 1 1 π a 2 2 πan π 1 = π a 1 1 1 π a 2 2 πan n = rξ 1 for some r R. Factoring r into irreducibles, we obtain π 1 = π a1 1 1 π a2 2 πa n n = α 1 α 2 α t ξ 1. 9

Note that since ξ 1 is pairwise nonassociate to any π i, the factorizations above are necessarily distinct. But this violates the minimality of a 1 + a 2 + + a m as the minimal length of factorization of any element of R admitting non-unique irreducible factorizations. Therefore A = {}, no such non-unique factorizations eist, and R is a UFD. (4 1): It will suffice to show that G() is a pseudo-clique for each. Again, let A denote the set of all nonzero nonunit elements of R that do not factor uniquely into irreducibles. Assume that A is nonempty. We again set n := min {m z = δ 1δ 2 δ m ; δ i Irr(R)}, and we take note that n 2. To z A finish off the proof, we induct on n. First, suppose that n = 2. Let A with = yz, y, z Irr(R) nonassociate. Since A, we have another factorization: = yz = π 1 π 2 π t, with π i Irr(R). Now G() is connected, so we may declare, without loss of generality that π 1 is adjacent to y in G() and no π i is associate to z. To see why this is true, we first note that there is a path in G() between π 1 and y, so we may replace π 1 with the verte on this path that is adjacent to y, implying that π 1 is nonassociate to y. We may then take π 1 π 2 π t to be an irreducible factorization of involving π 1. If some π i, for 1 i t, were to be associate to some z, then we may cancel and conclude that t = 2 and that factors uniquely, contradicting the fact that A. Thus our claim is established. So, we have that π 1 y, whence there eists r R such that π 1 yr = = yz, implying that π 1 is associate to z, a contradiction. If n = 2 and y is an associate to z (without loss of generality, y = z), then = y 2. The eact same argument above yields the same contradiction. Thus if has a factorization consisting of eactly two irreducibles, then G() is a pseudo-clique. Net, we assume that n 3 and that any nonzero nonunit of R with minimal length factorization of length less than n factors uniquely. We now suppose that there eists A with y 1 y 2 y n a minimal length irreducible factorization of. We will show that G() is a pseudo-clique. Since A, we have another irreducible factorization = y 1 y 2 y n = π 1 π 2 π t, where t n. Although the y i s may not be distinct, we claim that each y i is nonassociate to each π j. For if not, say without loss of generality, uy 1 = π 1, where u U(R), then we would have u 1 y 2 y 3 y n = π 2 π 3 π t. However, (u 1 y 2 )y 3 y n and π 2 π 3 π t are nonassociate irreducible factorizations of y 1 (since y 1 y n and π 1 π t are nonassociate irreducible factorizations of ). We conclude that y 1 A, violating the minimality of n. Therefore no y i is associate to any π j. 10

The fact that G() is connected again implies that, without loss of generality, π 1 and y 1 are adjacent in G(). Consider G( y 1 ). Since y 1 π 1, π 1 is a verte in G( y 1 ), which is to say that π 1 divides y 1 = y 2 y 3 y n. By induction, G( y 1 ) is a pseudo-clique. Now, π 1, y 2, y 3,, y n are all vertices in G( y 1 ). Since π 1 is nonassociate to each of y 2, y 3,, y n and since G( y 1 ) is a pseudo-clique, we must have π 1 y 2 y 1. Thus π 1 G( y 1 y 2 ). Continuing via use of induction, we see that π 1 G( y 1y 2 y n 1 ). However, = y n, implying that π 1 and y n are associate, a contradiction. Therefore A = {} and R is a UFD. y 1y 2 y n 1 Corollary 5.2 Let R be atomic and suppose that for all nonzero nonunit R, G() is connected. Then each G() is a finite pseudo-clique. Proof: If G() is connected for all nonzero nonunit R, then R is a UFD and hence an FFD. The result clearly follows. Corollary 5.3 Let R be atomic. Then R is a UFD if and only if for each nonzero nonunit R, diam(g()) = 1. Proof: This follows from the fact that a graph has a diameter of 1 if and only if it is complete. References [1] S. Akbari, H. R. Maimani, and S. Yassemi, When a zero-divisor graph is planar or a complete r-partite graph, J. Algebra 270 (2003), no. 1, 169 180. MR MR2016655 (2004h:13025) [2] S. Akbari and A. Mohammadian, On the zero-divisor graph of a commutative ring, J. Algebra 274 (2004), no. 2, 847 855. MR MR2043378 (2004k:13009) [3] David F. Anderson, Andrea Frazier, Aaron Lauve, and Philip S. Livingston, The zero-divisor graph of a commutative ring. II, Ideal theoretic methods in commutative algebra (Columbia, MO, 1999), Lecture Notes in Pure and Appl. Math., vol. 220, Dekker, New York, 2001, pp. 61 72. MR MR1836591 (2002e:13016) [4] David F. Anderson, Ron Levy, and Jay Shapiro, Zero-divisor graphs, von Neumann regular rings, and Boolean algebras, J. Pure Appl. Algebra 180 (2003), no. 3, 221 241. MR MR1966657 (2003m:13007) [5] David F. Anderson and Philip S. Livingston, The zero-divisor graph of a commutative ring, J. Algebra 217 (1999), no. 2, 434 447. MR MR1700509 (2000e:13007) 11

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