Gelfand Semirings, m-semirings and the Zariski Topology

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1 International Journal of Algebra, Vol. 3, 2009, no. 20, Gelfand Semirings, m-semirings and the Zariski Topology L. M. Ruza Departamento de Matemáticas, Facultad de Ciencias Universidad de los Andes, Merida, Venezuela J. Vielma Departamento de Matemáticas, Facultad de Ciencias Universidad de los Andes, Merida, Venezuela 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

2 982 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.

3 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:

4 984 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.

5 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.

6 986 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

7 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.

8 988 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 ).

9 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.

10 990 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), [2] M. Atiyah, I. Macdonald, Introduction to commutative algebra, Addison- Wesley, P.C., 1969.

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