UNIQUE PRIME CARTESIAN FACTORIZATION OF GRAPHS OVER FINITE FIELDS

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1 UNIQUE PRIME CARTESIAN FACTORIZATION OF GRAPHS OVER FINITE FIELDS RICHARD H. HAMMACK Abstract. A fundamental result, due to Sabidussi and Vizing, states that every connected graph has a unique pre factorization relative to the Cartesian product; but disconnected graphs are not uniquely pre factorable. This paper describes a system of modular arithmetic on graphs under which both connected and disconnected graphs have unique pre Cartesian factorizations. 1. Introduction The Cartesian product of two sple graphs G = (V (G), E(G)) and H = (V (H), E(H)) is the graph G H with V (G H) = V (G) V (H), and (u, x)(v, y) E(G H) if either u = v and xy E(H), or uv E(G) and x = y. This product is commutative and associative: G H = H G and G (H K) = (G H) K (up to isomorphism) for all graphs G, H and K. Also G H is connected if and only if both G and H are connected. For a full treatment of this product, see Chapter 4 of Imrich and Klažar [2]. We denote the empty graph (i.e. the graph with no vertices) as O, and the complete graph on n vertices as K n. Notice that G O = O and G K 1 = G for all graphs G. If n N, then ng denotes the graph that is the disjoint union of n copies of G (or O if n = 0). Note n(g H) = ng H = G nh. For a positive integer n, we define G n = G G G (n factors) and we adopt the convention G 0 = K 1. A graph G is pre if it is nontrivial and G = G 1 G 2 plies G 1 = K 1 or G 2 = K 1. Every graph G has a pre factorization G = G 1 G 2 G p, where each factor G i is pre. A fundamental theorem, proved independently by Sabidussi [3] and Vizing [4] states that the pre factorization of a connected graph is unique, that is if a connected graph G has pre factorizations G 1 G 2 G p and H 1 H 2 H q, then p = q and G i = H i for 1 i p (after reindexing, if necessary). But disconnected graphs are not uniquely pre factorable, in general. One standard example is the graph G = K 1 + K 2 + K K3 2 + K4 2 + K5 2, where the sum represents disjoint union. It is proved in [2] (Theorem 4.2) that G has two distinct pre factorizations OCTOBER

2 MISSOURI JOURNAL OF MATHEMATICAL SCIENCES (K 1 + K 2 + K 2 2 ) (K 1 + K 3 2 ) and (K 1 + K 2 ) (K 1 + K K4 2 ). In this example we may think of G as having been obtained by substituting K 2 for x in the polynomial f = 1 + x + x 2 + x 3 + x 4 + x 5. This polynomial has two distinct factorizations into irreducibles over N, namely (1 + x + x 2 )(1 + x 3 ) and (1 + x)(1 + x 2 + x 4 ), which yield the two factorizations of G. Of course, f can be uniquely pre factored over Z as f = (1 + x)(1 + x + x 2 )(1 x + x 2 ), but this does not translate into a factoring of G because the negative has no mediate meaning when applied to graphs. But what if the factoring is done over Z 2? Then f factors uniquely as (1 + x)(1 + x + x 2 )(1 + x + x 2 ). Substituting K 2 gives (K 1 + K 2 ) (K 1 + K 2 + K2 2) (K 1 + K 2 + K2 2) = K 1 + 3K 2 + 5K K K4 2 + K5 2. This is not G, but rather G + 2K 2 + 4K K K2. 4 However, if the coefficients are regarded as elements in Z 2, it seems reasonable to define 2K 2 = O, 4K2 2 = O, etc., so (K 1 + K 2 ) (K 1 + K 2 + K2 2) (K 1 + K 2 + K2 2) is a factorization of G over Z 2. The next section makes this idea precise. For each pre number k, we construct a ring G k of graphs that are added modulo k and multiplied with the Cartesian product. These rings are shown to be unique factorization domains, so every graph connected or disconnected has a unique pre factorization in G k. 2. Graphs Modulo k In this section, k denotes a pre number and Z k is the field Z/kZ. We regard Z k as the subset {0, 1, 2,..., k 1} Z with addition and multiplication done modulo k. So if n Z k and G is a graph, then ng denotes the graph that is the disjoint union of n copies of G. Let Γ be the set of all sple graphs, including O, and let Γ c Γ denote the set of all connected graphs, excluding O. Denote by G k the infinite densional vector space over Z k with basis Γ c. An element in G k is thus a sum A Γ c a A A with each a A in Z k and a A = 0 for all but finitely many A Γ c. Such a sum can be visualized as the graph that has a A components isomorphic to A, for each connected graph A. (If all a A are 0, the sum is identified with the empty graph.) Thus we will think of G k as a collection of graphs, and a nonzero G = A Γ c a A A in G k is connected provided exactly one coefficient a A is nonzero, and it equals 1. In words, G k consists of all graphs G having the property that G has no more than k 1 components that are isomorphic to any other graph A, so for large k, G k can be thought of as an approxation of Γ. But unlike 150 VOLUME 21, NUMBER 3

3 UNIQUE FACTORIZATION OF GRAPHS OVER FIELDS Γ, there is an operation + on G k. For G, H G k, graph G + H has the following property. If exactly m of G s components and exactly n if H s components are isomorphic to a connected graph A, then exactly m + n (mod k) components of G + H are isomorphic to A. Define a product k on G k as ( ) ( ) a A A k b A A = a A b B (A B). A Γ c A Γ c A,B Γ c Notice that G k H = G H if G and H are connected. If G and H are not both connected, then, intuitively, G k H can be regarded as the graph G H with all sets of k isomorphic components deleted. For example, in G 3, we have 2K 2 3 2K 3 = K 2 K 3, while 2K 2 2K 3 = 4(K 2 K 3 ). Deleting three of the four isomorphic components of 4(K 2 K 3 ) leaves K 2 K 3. Next, we verify that k is distributive and associative. For this, let G = A Γ c a A A, H = A Γ c b A A, and K = A Γ c c A A. For distributivity, observe the following. ( ) [( ) ( )] G k (H + K) = a A A k b A A + c A A A Γ c A Γ c A Γ c ( ) ( ) = a A A k (b A + c A )A A Γ c A Γ c = a A (b B + c B )(A B) = (a A b B + a A c B )(A B) A,B Γ c A,B Γ c = a A b B (A B) + a A c B (A B) = G k H + G k K. A,B Γ c A,B Γ c Next, associativity is verified. [( ) ( )] ( ) (G k H) k K = a A A k b A A k c A A A Γ c A Γ c A Γ c = ( ) a A b B (A B) k c C C A,B Γ c C Γ c = ( a A b B (A B) k ) c C C (distributivity from right) A,B Γ c C Γ c = a A b B c C (A B) C A,B Γ c C Γ c = a A b B c C A (B C) A Γ c B,C Γ c OCTOBER

4 MISSOURI JOURNAL OF MATHEMATICAL SCIENCES = [( ) ] a A A k b B c C (B C) B,C Γ c A Γ c ( ) = a A A k b B c C B C (distributivity from left) A Γ c B,C Γ c ( ) [( ) ( )] = a A A k b A A k c A A = G k (H k K). A Γ c A Γ c A Γ c From this it follows that G k is a commutative ring with zero element O. It is mediate from the definition of k that K 1 k G = G for all ring elements G, so G k has identity K 1. Notice that there is an injective homomorphism φ: Z k G k defined as φ(n) = nk 1. Additionally, observe that if G is connected then ng = (nk 1 ) k G. Thus, A Γ c a A A = A Γ c (a A K 1 ) k A, and this sum is O if and only if each a A is zero. The remainder of this paper hinges on the following construction. Let P 1, P 2, P 3,... be an enumeration of all connected pre graphs indexed so that V (P 1 ) V (P 2 ) V (P 3 ). (Thus P 1 = K 2, P 2 is the path on three vertices, P 3 = K 3, etc.) For each positive integer m, construct a map φ m : Z k [x 1, x 2,..., x m ] G k (ai1i2 i m K 1 ) k P i1 i2 1 k P2 ( defined as φ m ai1i2 i m x i1 1 xi2 2 ) x m =, where the sums are taken over all m- k k P m tuples (i 1, i 2,..., i m ) N m. This is easily seen to be a ring homomorphism. (Apply Theorem 4.3 of [1] with φ as defined in the previous paragraph.) Observe that the homomorphism φ m is injective. Suppose ) = (ai1i 2 i m K 1 ) k P i1 1 ( φ m ai1i2 i m x i1 1 xi2 2 x i2 m k P2 k k P O. Recall that k = for connected graphs, so by unique factorization of m = connected graphs P i1 i2 1 k P2 k k Pm P j1 j2 jm 1 k P2 k k Pm for distinct m-tuples (i 1, i 2,..., i m ) and (j 1, j 2,..., j m ), and therefore all coefficients a i1i 2 i m are zero. Lemma 1. For any pre number k, the ring G k is an integral domain. Proof. Suppose G k H = O for two elements G, H G k. Choose m large enough so that every component of both G and H has a pre factorization of form P i1 i2 1 P2 Pm. (The powers, of course, are allowed to be zero.) By letting a i1i 2 i m be the number of components of G that are isomorphic to P i1 i2 1 P2 Pm, it follows that G = φ ( m ai1i2 i m x i1 1 xi2 2 ) x m. Silarly, H is also in the age of φ m, so G = φ m (g) and H = φ m (h) for appropriate polynomials g and h. From G k H = O, it follows that φ m (gh) = φ m (g) k φ m (h) = O. Then gh = 0 since φ m is injective, hence, g = 0 or h = 0 since Z[x 1, x 2,..., x m ] is an integral domain. Consequently, G = O or H = O, hence G k is an integral domain. 152 VOLUME 21, NUMBER 3

5 UNIQUE FACTORIZATION OF GRAPHS OVER FIELDS It will be useful to examine the units in G k. If G k H = K 1, then, as in the above proof, we may take m large enough so that G = φ m (g) and H = φ m (h). Then φ m (gh) = φ m (g) k φ m (h) = K 1, so gh = 1 by injectivity of φ m. Consequently, g and h are constant polynomials, so G and H are of the form φ m (n) = nk 1 for some nonzero n Z k. Thus the units of G k are K 1, 2K 1, 3K 1,..., (k 1)K 1. Recall that an element a of a ring is irreducible if a = bc plies either b or c is a unit (i.e. invertible). Element a is pre if a bc plies a b or a c for all b, c in the ring. Every pre is irreducible, but in general the converse is not true. We take the approach of Grove [1] in defining a unique factorization domain (UFD) to be an integral domain in which every nonunit is a product of irreducible elements, and every irreducible is pre. By Theorem 5.11 of [1], every nonzero non-unit element of a UFD has a unique pre factorization, that is if a = b 1 b 2 b p = c 1 c 2 c q where each b i and c i is pre, then p = q and (after relabeling if necessary) b i = u i c i for units u i, 1 i p. Proposition 2. For any pre number k, the ring G k is a UFD. Proof. By Lemma 1, G k is an integral domain. By the above remarks, showing it is a UFD entails showing any G G k is a product of irreducibles, and if G is irreducible then G (H k K) plies G H or G K for all H, K G k. Suppose G G k. Observe G is a product of irreducibles: If G is irreducible, we are done. Otherwise suppose G = H k K for non-units H and K. As before, take m large enough so G = φ m (g), H = φ m (h) and K = φ m (κ), and argue g = hκ. Since H and K are non-units, h and κ are nonconstant polynomials and their degrees must be strictly lower than the degree of g. This process may be continued to decompose H and K into products of non-units, and in turn the factors of H and K may be silarly decomposed. But since each iteration yields factors that are ages of polynomials of lower degree than those of the previous iteration, the process must eventually terminate. Consequently G is a product of irreducibles. Now suppose G is irreducible and G (H k K), that is G k F = H k K for some graph F. Take m large enough so G = φ m (g), F = φ m (f), H = φ m (h) and K = φ m (κ). Then φ m (gf) = φ m (hκ), so gf = hκ because φ m is injective, and hence g hκ. Now, g is irreducible in Z k [x 1, x 2,..., x m ], for any factorization g = g 1 g 2 into non-units would produce a factorization G = φ m (g 1 ) k φ m (g 2 ) into non-units. Then, since Z k [x 1, x 2,..., x m ] is a UFD ([1], Theorem 5.16), the relation g hκ means g h or g κ, that is gh 1 = h or gκ 1 = κ. Applying φ m, either G k φ m (h 1 ) = H or G k φ m (κ 1 ) = K, so G H or G K. OCTOBER

6 MISSOURI JOURNAL OF MATHEMATICAL SCIENCES The proposition plies that if a graph in G k factors into irreducibles as B 1 k B 2 k k B p and C 1 k C 2 k k C q, then p = q and (after relabeling) B i = (u i K 1 ) k C i for nonzero elements u i Z k. Because k and agree as operators on connected graphs, the usual pre factorization of a connected graph G will be a pre factorization over k. However, a pre factorization of G in G k may differ from the usual one by unit multiples of the factors. For example K 2 5 K 3 5 K 3 and 3K 2 5 3K 3 5 4K 3 are two factorizations of the same graph in G 5. Observe that 3K 2 5 3K 3 5 4K 3 = ((3K 1 ) 5 K 2 ) 5 ((3K 1 ) 5 K 3 ) 5 ((4K 1 ) 5 K 3 ), and it is evident that these two factorizations differ only by unit multiples of the factors. References [1] L. Grove, Algebra, Academic Press Series in Pure and Applied Mathematics, New York, [2] W. Imrich and S. Klavžar, Product Graphs; Structure and Recognition, Wiley Interscience Series in Discrete Mathematics and Optization, New York, [3] G. Sabidussi, Graph Multiplication, Math. Z., 72 (1960), [4] G. V. Vizing, The Cartesian Product of Graphs (Russian), Vyčisl Sistemy, 9 (1963), AMS Classification Numbers: 05C99 Department of Mathematics and Applied Mathematics, Virginia Commonwealth University, Richmond, VA , USA address: rhammack@vcu.edu 154 VOLUME 21, NUMBER 3