International Journal of Algebra, Vol. 3, 2009, no. 20, 981-991 Gelfand Semirings, m-semirings and the Zariski Topology L. M. Ruza Departamento de Matemáticas, Facultad de Ciencias Universidad de los Andes, Merida, Venezuela ruza@ula.ve J. Vielma Departamento de Matemáticas, Facultad de Ciencias Universidad de los Andes, Merida, Venezuela vielma@ula.ve Abstract In this work by a semiring R we understand a commutative semiring with identity and we consider in its prime spectrum the Zariski topology t z.we denote by t z the smallest Alexandroff topology containing t z. Also t z denotes its corresponding cotopology. We say that R is a Gelfand semiring if every prime ideal is contained in a unique maximal ideal. We say that R is an m-semiring if each prime ideal contains only one minimal prime ideal.we give a characterization of such semirings in terms of the clopen subsets of the smallest Alexandroff topology t z in Spec(R) containing the Zariski topology t z. Also, characterizations of the compactness and connectedness of the spectrum of such semirings, in terms of the topologies t z and t z, are given. Mathematics Subject Classification: 54E18, 54F65, 13C05 Keywords: Alexandroff topology, Zariski topology, Gelfand semirings, m- semirings, prime spectrum 1 Introduction Let R be a commutative semiring with non-zero identity. Spec(R) denotes the set of all prime ideals of R, equipped with the Zariski topology t z ([7]). For every proper set I of R, we denote by (I) 0 the set of all prime ideals of R containing I, and D 0 (I) =Spec(R) (I) 0. Max(R) and Min(R) denote
982 L. M. Ruza and J. Vielma the set of all maximal and minimal prime ideals of R, respectively. If we view t z as a subset of 2 Spec(R) with the product topology, then its closure t z is also a topology and it is the smallest Alexandroff topology containing t z ([8]). A topology is said to be Alexandroff if it is closed under arbitrary intersections. For any point x in a topological space (X, τ), Ker(x) will denote the intersection of all τ-open sets containing x. Also A τ is a τ-closed set and is just the union of the τ-closure of each of its points. We say that R is a Gelfand semiring if every prime ideal is contained in a unique maximal ideal. Following Avila in [3], we say that R is an m-semiring if each prime ideal contains only one minimal prime ideal. The Gelfand rings were characterized in [6], as those rings in which the maximal prime spectrum is a retract of (Spec(R),t z ). We extend this result to the more general case when R is a Gelfand semiring. In addition, we show that R is a Gelfand semiring if and only if for every M in Max(R), Ker(M) is clopen in t z. We show that R is an m-semiring if and only if for every prime ideal M Min(R), (M) 0 is clopen in t z if and only if Min(R) is a retract of (Spec(R), t z ). We give a characterization of the compactness and connectedness of the spectrun of such semirings in terms of the topologies t z and t z. A semiring R is said to be semilocal if it has a finite number of maximal ideals. R is called local if it has only one maximal ideal. A topological space (X, τ) is said to be nearly compact (almost compact) if every τ-open cover Θ of X contains a finite subfamily {U i : i = 1...n} such that X = n i=1 intu i (X = n i=1 U i). In fact, we prove that for Gelfand semirings, t z -compactness, t z -nearly compactness and t z -almost compactness are equivalent to the fact that R is a semilocal semiring. Also for m-semirings, t z -compactness, t z -nearly compactness and t z -almost compactness are also equivalent to the condition that R has a finite number of minimal prime ideals. 2 Terminology On the following, N := {0, 1, 2,...} denotes the set of natural numbers. Also, by a space we understand a topological space, (X, τ) always denotes a space and τ the family of τ-closed subsets of X ([8]). If A 2 X, we denote by A the intersection of all the elements of A. Recall that a semiring (commutative with non-zero identity) is an algebra (R, +,, 0, 1), where R is a set with 0, 1 R, and + and are binary operations on R called sum and multiplication, respectively, which satisfy the following: (S 1 )(R, +, 0) and (R,, 1) are commutative monoids with 1 0. (S 2 ) a (b + c) =a b + a c for every a, b, c R.
Gelfand semirings, m-semirings and the Zariski topology 983 (S 3 ) a 0 = 0 for every a R. As is usual, we denote a semiring (R, +,, 0, 1) by R. The notions of (proper) ideal, prime ideal and maximal ideal of a semiring R are defined as in commutative rings ([7]). Example 2.1 The set N with the usual addition and multiplication is a semiring, which it has an unique maximal ideal M = N {1} ([7]). Example 2.2 The set Q + of all nonnegative rational numbers, with the usual addition and multiplication is a semiring; the same is true for the set R + of all nonnegative real numbers ([7]). Example 2.3 If (X, τ) is a topological space, then τ is a semiring with operations of addition and multiplication given by A + B = A B and AB = A B. The additive identity is and multiplicative identity X. (Easy verification). Example 2.4 Let R = R { }. Then (R, min, +) is a commutative semiring in which addition is the operation of taking minimum and multiplication is ordinary addition. This semiring is important in solving the shortest-path problem in optimization (example 1.22 in [7]). Example 2.5 If X is a Hausdorff topological space then the set R of all continuous bounded functions from X to R + is a commutative semiring ([5]). The following examples are discussed with great detail in [1]. 3 The Semirings B(n, i) Let R be a semiring with multiplicative identity 1 R and additive identity 0 R. The set N1 ={n1 R : n N} is a commutative subsemiring of R, where 0 R = 0.1 R N1 1 R = 1.1 R N1 n1 R + m1 R = (n + m)1 R N1. n1 R.m1 R = (nm)1 R N1. Let us consider the case when k1 R 0 R for k>1 and a1 R = b1 R for some a b. Let n be the least positive integer with n1 R = i1 R where 1 i n 1. Here we write j for j1 R so N1 ={0, 1, 2,..., (n 1)}. It such case, if m = n i the followings holds:
984 L. M. Ruza and J. Vielma a + b = { a + b, if 0 a + b n 1 l, with l a + b mod m, if a + b n. where, l is the unique number such that l a + b modm with i l n 1. And the multiplication { ab, if 0 ab n 1 ab = l, with l ab mod m, if ab n. (l is the unique number such that l ab modm with i l n 1). Conversely, given n 2 and 0 i n 1 there is a unique (up to isomorphism) semiring of the kind described above. Let B(n, i) ={0, 1,.., n 1} and put m = n i. Make B(n, i) into a semiring by defining the operation of addition as following a + b = { a + b, si 0 a + b n 1 l, con l a + b mod m, si a + b n. then l is the unique number such that l a + b modm with i l n 1. The product is defined similarly. The following theorem, which appears in [1] as Theorem 24, is included so that the reader can appreciate some of the properties of such semirings. Theorem 3.1 (1) dimb(n, i) =0if i =0or i =1and n =2. (2) dimb(n, i) =1if i =1and n>2. In this case the prime ideals of B(n, i) are 0 and pb(n, i) where p is a prime with p n 1. (3) dimb(n, i) =1if n>2 and n = i +1. In this case the prime ideals of B(n, i) are 0 and {0, 2, 3,..., n 1}. (4) dimb(n, i) =2si n 1 >i 2. In this case the prime ideals are 0, M = {0, 2, 3,..., n 1} and pb(n, i) where p is prime and p n i. 4 Gelfand Semirings Remember that a semiring R is a Gelfand semiring if each prime ideal is contained in a unique maximal prime ideal. Example 4.1 The nonnegative integers N, with the usual addition and multiplication, is an example of a Gelfand semiring. More than that, N is a local semiring whose only maximal ideal is M = N {1}. For more details see ([1]). A good list of Gelfand semirings is the one that comes out of (4) in Theorem 3.1.
Gelfand semirings, m-semirings and the Zariski topology 985 Lemma 4.1 If A Spec(R), then A = {P Spec(R) : A P } Proof. First observe that B = {P Spec(R) : A P } is a closed subset of Spec(R), since B =( A) 0. Also, A B. Then, A B. Now, since A is closed in Spec(R) we have that A =(I) 0 for some ideal I. Take P Spec(R) such that A P. Then since A A =(I) 0, it follows that I A P. So, P (I) 0. Therefore, P A. Lemma 4.2 Let μ : Spec(R) Max(R), be the function defined by μ(p )= M P, where M P is the unique maximal ideal that contains P, when R be Gelfand semiring. If D be a closed subset of Max(R), F = D, Q Spec(R) and Q B = {M : M D}, then μ(q) D. Proof. Since D is closed in Max(R), there exist a closed set A in Spec(R) such that D = A Max(R). Also, A D. On the other hand, Q + F B, then there exist a maximal ideal M such that Q + F M. Also, since F M, then A F,soM A Max(R) which implies that M D. Then μ(q) D.. What follows, is the version for semirings of the well known Krull s Lemma [2], which will be stated without proof. Lemma 4.3 (Krull) Let S be a multiplicative closed subset of a semiring R and I an ideal of R such that I S = φ. Then, there exists an ideal P of R maximal with respect to the property P S = φ and I P. Further, every such an ideal is prime. The following theorem is well known for commutative rings [6]. Theorem 4.1 A semiring R is Gelfand if and only if Max(R) is a retract of (Spec(R),t z ). Proof. Let μ : Spec(R) Max(R), be the function defined by μ(p )=M P, where M P is the unique maximal ideal that contains P. We show that μ is continuous. Let D be a closed subset of Max(R) and consider F = D and I = {P Spec(R) :μ(p ) D}. Let P Spec(R) and I P. We show that P contains a prime ideal Q B = {M : M D}, which implies that μ(p )=μ(q) D by Lemma 4.2. Let S = R B and T = R P, choose s S, t T. Since I P, there exist P μ 1 (D) such that t/ P. Also, since s/ P, it follows that st / P. Therefore, st / I. That means that the closed multiplicative system ST does not intersect I. By Krull s Lemma, there exists a prime ideal Q containing I, and disjoint from ST, Since Q B and Q P, it follows that μ(p )=μ(q) D. So, μ is continuous.
986 L. M. Ruza and J. Vielma Conversely, if φ is a retract from Spec(R)ontoMax(R). Take P Spec(R) with φ(p )=M. Then, P φ 1 ({M}). Since every maximal prime ideal is a closed point, we have that φ 1 ({M}) is closed in Spec(R). Then, {P } φ 1 ({M}). Therefore, if M 1 Max(R) and P M 1 we have that M 1 {P } =(P ) 0. So, M 1 = φ(m 1 )=M, R is Gelfand. Lemma 4.4 Let R be a semiring. If M is a prime ideal of R and Ker(M) is t z -clopen, then M Max(R). Proof. Suppose P is a prime ideal of R with Ker(P ) t z -clopen. Then (P ) 0 Ker(P ). Since t z is a T 0 topology, we have that {P } =(P ) 0 Ker(P ), and then {P } =(P ) 0 which implies that P is maximal. Theorem 4.2 A semiring R es Gelfand if and only if for every M in Max(R), Ker(M) is clopen in t z. Proof. Let M be a maximal prime. Since Ker(M) ist z -open, it remains only to prove that Ker(M) ist z -closed. Let P Ker(M) and we see that (P ) 0 Ker(M). Let Q (P ) 0, then P Q. If M Q is the only maximal prime containing Q, it follows that M Q = M. So, Q Ker(M). Conversely, if P is a prime ideal and M 1, M 2 are maximal ideals containing P. Then, P Ker(M 1 ) and P Ker(M 2 ). By hypothesis, we have that (P ) 0 Ker(M 1 ) and (P ) 0 Ker(M 2 ). So, M 1 Ker(M 2 ) and therefore M 1 = M 2. Then R is a Gelfand semiring. Remember that a semiring R is said to be semilocal if it has a finite number of maximal ideals. R is called local if it has only one maximal ideal. Theorem 4.3 Let R be a Gelfand semiring. The following are equivalent. (a) (Spec(R), t z ) is compact (b) (Spec(R), t z ) is nearly-compact (c) (Spec(R), t z ) is almost-compact (d) R is a semilocal semiring (e) (Spec(R), t z t z ) is compact Proof. Is obvious that (a) (b), (b) (c). By Lemma 4.1, (c) (d). Let us prove that (d) (a). Let {U α } a covering by t z -open subsets of Spec(R). By (d), there exits a finite number of maximal ideals, say M 1,.., M k. Then for each M i there exits an U αi such that M i U αi. Then, Ker(M i ) U αi and {U αi } is a finite subcover of Spec(R). Clearly (a) implies (e). Now, If {U α } is a covering by t z -open subsets of Spec(R), for each M Max(R) there exits
Gelfand semirings, m-semirings and the Zariski topology 987 an U αm such that M U αm. Then, Ker(M) U αm. Now, since R is Gelfand, it follows that {Ker(M),M Max(R)} is an t z t z open covering of Spec(R), so there is a finite subcovering {Ker(M i ):i =1,..., n, M i Max(R)} and then {U αi : i =1,..., n} is a finite subcover of Spec(R). Proposition 4.1 If R is a Gelfand semiring and (Spec(R), t z ) is connected, then R is a local semiring. Proof. Let M be a maximal ideal in R, then Ker(M) ist z -clopen. Since Ker(M), it follows that Spec(R) =Ker(M). Therefore, M is the unique maximal ideal of R. Lemma 4.5 If R is a local semiring, then (Spec(R), t z ) is connected. Proof. Let M be the unique maximal ideal of R. and Ω a nonempty τ z -clopen subset of Spec(R). If P U, then (P ) 0 U, and then M U y Ker(M) U. Now since Ker(M) = Spec(R), it follows that U = Spec(R). Let us remember that two topologies on a space X are said to be complementary if their supremum is the discrete topology and their intersection is the indiscrete one. Theorem 4.4 Let R be a Gelfand semiring. The following are equivalent: (a) (Spec(R), t z ) is connected. (b) R is a local semiring. (c) t z y t z are complementary topologies. Proof. Proposition 4.1 and Lemma 4.5 implies that (a) and (b) are equivalent. Clearly (c) implies (a). Now, since t 0 is a T 0 topology [9], then the supremum of t z and t z is the discrete topology. Also, since t z is connected then the intersection of t z and t z is the indiscrete topology. 5 m-semirings Remember that a semiring R is an m-semiring if each prime ideal contains only one minimal prime ideal. Example 5.1 The nonnegative integers N, with the usual addition and multiplication, is also a good example of an m-semiring which is not an m-ring. More than that, N is a local semiring with only one minimal ideal. For more details see [1].An extense list of m-semirings can be found if if we look at the examples that come out from (2) in Theorem 3.1.
988 L. M. Ruza and J. Vielma Remember that a topological space X is supercompact if X belongs to every open cover of X. The supercompact elements in complete lattices are introduced in [4]. Theorem 5.1 (Spec(R), t z) is supercompact if and only if R has only one minimal prime ideal. Proof. Let Π = {(M) 0 : M Min(R)}. Then Π is an t z-open covering of Spec(R). It follows that for some (M) 0 Π, Spec(R) =(M) 0. Therefore Min(R) (M) 0, which implies that M is the unique minimal prime ideal of R. Conversely, if Π is an t z-open covering of Spec(R), then the unique minimal prime ideal M of R belongs to one of the elements in Π, say A M. But A M = N A M (N) 0. That implies that M (M ) 0 for some M A M. Therefore A M = Spec(R), which proves that (Spec(R), t z ) is supercompact. Lemma 5.1 Let R be a semiring. If M is a prime ideal of R and (M) 0 is t z -clopen, then M Min(R). Proof. Let P be a prime ideal and (P ) 0 t z -clopen. Then Ker(P ) (P ) 0. Since Since t z is a T 0 topology, {P } =(P ) 0 Ker(P ), then {P } = Ker(P ) which implies that P is minimal. Theorem 5.2 A semiring R is an m-semiring if and only if for each minimal prime ideal M, (M) 0 is clopen in t z. Proof. Let M be a minimal prime. Since (M) 0 is t z -closed we only need to prove that (M) 0 es kernelled. Let P (M) 0 and take Q Ker(P ), then Q P. If M Q is the unique minimal ideal contained in Q it follows that M Q P. Therefore M Q is the unique minimal ideal contained in P. Then M Q = M and Q (M) 0. Conversely, take P prime and M 1, M 2 minimal prime ideals contained in P. Since (M 1 ) 0 and (M 2 ) 0 are clopen in t z and P (M 1 ) 0, it follows that Ker(P ) (M 1 ) 0. Also P (M 2 ) 0, therefore M 2 (M 1 ) 0. Then M 1 = M 2, which completes the proof. Lemma 5.2 Let R be a semiring. A prime ideal P of R is minimal if and only if Ker(P )={P }. Proof. It is trivial Theorem 5.3 Let R be a semiring. R is an m-semiring if and only if Min(R) is a retract of (Spec(R), t z ).
Gelfand semirings, m-semirings and the Zariski topology 989 Proof. Suppose φ be a retract from Spec(R) ontomin(r). Let P be a prime ideal, M 1 and M 2 minimal prime ideals contained in P. Since {M 1 } is kernelled in Spec(R), then {M 1 } is kernelled in Min(R) and therefore φ 1 ({M 1 })isa kernelled subset of Spec(R) which contains P. Then it follows that Ker(P )isa subset of φ 1 ({M 1 }) and that M 2 belongs to φ 1 ({M 1 }). Therefore M 1 = M 2 and R is an m-semiring. Conversely, If R is an m-semiring, the map φ : Spec(R) Min(R) defined by φ(p )=m P, where m P is the unique minimal prime ideal contained in P, is well defined and from Theorem 5.2 and Lemma 5.2 it is easy to show that φ is continuous. Lemma 5.3 Let R be an m-semiring. If (Spec(R), t z ) is almost compact then there exists a finite number of minimal prime ideals in the semiring. Proof. Let us consider the covering {(m) 0 : m Min(R)} by t z -open subsets of Spec(R). Then there exits a finite subcollection {(m i ) 0 : i =1,..., n} such that Spec(R) = (m i ) 0 = (m i ) 0. Therefore, the only minimal prime ideals of the semiring are {m i }. Theorem 5.4 Let R be an m-semiring. The following are equivalent. (a) (Spec(R), t z ) is compact (b) (Spec(R), t z ) is nearly-compact (c) (Spec(R), t z ) is almost-compact (d) R has a finite number of minimal prime ideals (e) (Spec(R), t z t z ) is compact Proof. The following implications are trivial: (a) (b), (b) (c), and by Lemma??, (c) (d). Let see that (d) (a). Let {U α } be an t z -open covering of Spec(R). For each minimal ideal M there exists an U αm with M U αm. Since (M) o is a subset of {U αm }, and the fact that {(M) o : M Min(R)} is an (t z t z )-open covering of Spec(R), then by (d), there exist a finite number of minimal prime ideals, say M 1,...,M k such that (M 1 ) o,...,(m k ) 0 covers Spec(R). So, {U αmi } is a finite subcovering of Spec(R). Proposition 5.1 If R is an m-semiring and (Spec(R), t z ) is connected, then there exits a unique minimal prime ideal. Proof. It follows form the fact that (m) 0 is a t z -clopen if m is a minimal prime ideal. Lemma 5.4 If R is a semiring with a unique minimal prime ideal, then (Spec(R), t z ) is connected.
990 L. M. Ruza and J. Vielma Proof. Let m be the unique minimal prime ideal of R and take U Spec(R) a nontrivial t z -clopen subset. If m U, then (m) 0 U and since (m) 0 = Spec(R) we get a contradiction. Similarly if m/ U. Remember that a space is irreducible if every open subset is dense. Lemma 5.5 Let R be a semiring. (Spec(R),t z ) is irreducible if and only if η(0) is a prime ideal. In such case, η(0) is the only minimal prime ideal. Proof. Let ab η(0), then ab P for every P Spec(R). Take P and Q Spec(R) such that a/ P y b/ Q. Then, P D 0 (a) yq D 0 (b), by hypothesis there exist W D 0 (a) D 0 (b). Then, a/ W y b/ W, contradicting the fact that ab η(0) W. Conversely, if η(0) is a prime ideal, let P D 0 (a) yq D 0 (b). Since, η(0) P D 0 (a) and η(0) Q D 0 (b). We get that D 0 (a) D 0 (b). The last part is evident. Theorem 5.5 Let R be a semiring. The following are equivalent: (a) (Spec(R), t z ) is connected. (b) R has a unique minimal prime ideal. (c) (Spec(R),t z ) is an irreducible space. (d) η(0) is the unique minimal prime ideal. (e) t z y t z are complementary topologies. Proof. Proposition 5.1 and lemma 5.4 implies that (a) and (b) are equivalent. Lemma 5.5 implies the equivalence of (c) and (d). Since t z is a T 0 topology and t z is connected, then (a) and (d) are equivalent. The equivalence of (b) and (d) is trivial. ACKNOWLEDGEMENTS: We are grateful to Dr Daniel Anderson for his comments and suggestions. References [1] F. Alarcón, D.D. Anderson, Commutative semirings and their lattices of ideals, Houston J. Math. 20 (4) (1994), 571-590. [2] M. Atiyah, I. Macdonald, Introduction to commutative algebra, Addison- Wesley, P.C., 1969.
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