@ Appl. Gen. Topol. 20, no. 2 (2019), 379-393 doi:10.4995/agt.2019.11417 c© AGT, UPV, 2019 Ideals in B1(X) and residue class rings of B1(X) modulo an ideal A. Deb Ray a and Atanu Mondal b a Department of Pure Mathematics, University of Calcutta, 35, Ballygunge Circular Road, Kolkata-700019, India (debrayatasi@gmail.com) b Department of Mathematics and Statistics, Commerce Evening, St. Xavier’s College, Kolkata- 700016, India (atanu@sxccal.edu) Communicated by F. Lin Abstract This paper explores the duality between ideals of the ring B1(X) of all real valued Baire one functions on a topological space X and typical families of zero sets, called ZB-filters, on X. As a natural outcome of this study, it is observed that B1(X) is a Gelfand ring but non- Noetherian in general. Introducing fixed and free maximal ideals in the context of B1(X), complete descriptions of the fixed maximal ideals of both B1(X) and B ∗ 1(X) are obtained. Though free maximal ideals of B1(X) and those of B ∗ 1(X) do not show any relationship in general, their counterparts, i.e., the fixed maximal ideals obey natural relations. It is proved here that for a perfectly normal T1 space X, free maximal ideals of B1(X) are determined by a typical class of Baire one functions. In the concluding part of this paper, we study residue class ring of B1(X) modulo an ideal, with special emphasize on real and hyper real maximal ideals of B1(X). 2010 MSC: 26A21; 54C30; 13A15; 54C50. Keywords: ZB- filter; ZB-ultrafilter; ZB-ideal; fixed ideal; free ideal; residue class ring; real maximal ideal; hyper real maximal ideal. Received 20 February 2019 – Accepted 01 August 2019 http://dx.doi.org/10.4995/agt.2019.11417 A. Deb Ray and A. Mondal 1. Introduction In [1], we have introduced the ring of Baire one functions defined on any topo- logical space X and have denoted it by B1(X). It has been observed that B1(X) is a commutative lattice ordered ring with unity containing the ring C(X) of continuous functions as a subring. The collection of bounded Baire one functions, denoted by B∗1 (X), is a commutative subring and sublattice of B1(X). Certainly, B ∗ 1 (X) ∩C(X) = C∗(X). In this paper, we study the ideals, in particular, the maximal ideals of B1(X) (and also of B∗1 (X)). There is a nice interplay between the ideals of B1(X) and a typical family of zero sets (which we call a ZB-filter) of the underlying space X. As a natural consequence of this duality of ideals of B1(X) and ZB-filters on X, we obtain that B1(X) is Gelfand and in general, B1(X) is non-Noetherian. Introducing the idea of fixed and free ideals in our context, we have char- acterized the fixed maximal ideals of B1(X) and also those of B ∗ 1 (X). We have shown that although fixed maximal ideals of the rings B1(X) and B ∗ 1 (X) obey a natural relationship, the free maximal ideals fail to do so. However, for a perfectly normal T1 space X, free maximal ideals of B1(X) are determined by a typical class of Baire one functions. In the last section of this paper, we have discussed residue class ring of B1(X) modulo an ideal and introduced real and hyper-real maximal ideals in B1(X). 2. ZB-filters on X and Ideals in B1(X) Definition 2.1. A nonempty subcollection F of Z(B1(X)) ([1]) is said to be a ZB-filter on X, if it satisfies the following conditions: (1) ∅ /∈ F (2) if Z1,Z2 ∈ F , then Z1 ∩Z2 ∈ F (3) if Z ∈ F and Z′ ∈ Z(B1(X)) is such that Z ⊆ Z′, then Z′ ∈ F . Clearly, a ZB-filter F on X has finite intersection property. Conversely, if a subcollection B ⊆ Z(B1(X)) possesses finite intersection property, then B can be extended to a ZB-filter F (B) on X, given by F (B) = {Z ∈ Z(B1(X)): there exists a finite subfamily {B1,B2, ...,Bn} of B with Z ⊇ n⋂ i=1 Bi}. Indeed this is the smallest ZB-filter on X containing B. Definition 2.2. A ZB-filter U on X is called a ZB-ultrafilter on X, if there does not exist any ZB-filter F on X, such that U $ F . Example 2.3. Let A0 = {Z ∈ Z(B1(R)) : 0 ∈ Z}. Then A0 is a ZB-ultrafilter on R. Applying Zorn’s lemma one can show that, every ZB-filter on X can be ex- tended to a ZB-ultrafilter. Therefore, a family B of Z(B1(X)) with finite intersection property can be extended to a ZB-ultrafilter on X. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 380 Ideals in B1(X) and residue class rings of B1(X) modulo an ideal Remark 2.4. A ZB-ultrafilter U on X is a subfamily of Z(B1(X)) which is maximal with respect to having finite intersection property. Conversely, if a family B of Z(B1(X)) has finite intersection property and maximal with respect to having this property, then B is a ZB-ultrafilter on X. In what follow, by an ideal I of B1(X) we always mean a proper ideal. Theorem 2.5. If I is an ideal in B1(X), then ZB[I] = {Z(f) : f ∈ I} is a ZB-filter on X. Proof. Since I is a proper ideal in B1(X), we claim ∅ /∈ ZB[I]. If possible let ∅ ∈ ZB[I]. So, ∅ = Z(f), for some f ∈ I. As f ∈ I =⇒ f2 ∈ I and Z(f2) = Z(f) = ∅, hence 1 f2 ∈ B1(X) [1]. This is a contradiction to the fact that, I is a proper ideal and contains no unit. Let Z(f),Z(g) ∈ ZB[I], for some f,g ∈ I. Our claim is Z(f) ∩Z(g) ∈ ZB[I]. Z(f) ∩Z(g) = Z(f2 + g2) ∈ ZB[I], as I is an ideal and so, f2 + g2 ∈ I. Now assume that Z(f) ∈ ZB[I] and Z′ ∈ Z(B1(X)) is such that Z(f) ⊆ Z′. Then we can write Z′ = Z(h), for some h ∈ B1(X). Z(f) ⊆ Z′ =⇒ Z(h) = Z(h) ∪ Z(f). So, Z(h) = Z(hf) ∈ ZB[I], because hf ∈ I. Hence, ZB[I] is a ZB-filter on X. � Theorem 2.6. Let F be a ZB-filter on X. Then Z −1 B [F ] = {f ∈ B1(X) : Z(f) ∈ F} is an ideal in B1(X). Proof. We note that, ∅ /∈ F . So the constant function 1 /∈ Z−1B [F ]. Hence Z−1B [F ] is a proper subset of B1(X). Choose f,g ∈ Z−1B [F ]. Then Z(f),Z(g) ∈ F and F being a ZB-filter Z(f) ∩ Z(g) ∈ F . Now Z(f) ∩ Z(g) ⊆ Z(f − g). Hence Z(f − g) ∈ F , F being a ZB-filter on X. This implies f −g ∈ Z−1B [F ]. For f ∈ Z−1B [F ] and h ∈ B1(X), Z(f.h) = Z(f) ∪Z(h). As Z(f) ∈ F and F is a ZB-filter on X, it follows that Z(f.h) ∈ F . Hence f.h ∈ Z−1B [F ]. Thus Z−1B [F ] is an ideal of B1(X). � We may define a map Z : B1(X) → Z(B1(X)) given by f 7→ Z(f). Certainly, Z is a surjection. In view of the above results, such Z induces a map ZB between the collection of all ideals of B1(X), say IB and the collection of all ZB-filters on X, say FB(X), i.e., ZB : IB → FB(X) given by ZB(I) = ZB[I], ∀ I ∈ IB. The map ZB is also a surjective map because for any F ∈ FB(X), Z−1B [F ] is an ideal in B1(X). We also note that ZB[Z −1 B [F ]] = F . So each ZB-filter on X is the image of some ideal in B1(X) under the map ZB : IB → FB(X). Observation. The map ZB : IB → FB(X) is not injective in general. Be- cause, for any ideal I in B1(X), Z −1 B [ZB[I]] is an ideal in B1(X), such that I ⊆ Z−1B [ZB[I]] and by our previous result ZB[Z −1 B [ZB[I]]] = ZB[I]. If one gets an ideal J in B1(X) such that I ⊆ J ⊆ Z−1B [ZB[I]], then we must have ZB[I] = ZB[J]. The following example shows that such an ideal is indeed pos- sible to exist. In fact, in the following example, we get countably many ideals In in B1(R) such that the images of all the ideals are same under the map ZB. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 381 A. Deb Ray and A. Mondal Example 2.7. Let f0 : R → R be defined as, f0(x) =   1 q if x = p q , where p ∈ Z,q ∈ N and g.c.d. (p,q) = 1 1 if x = 0 0 otherwise It is well known that f0 ∈ B1(R) (see [2]). Consider the ideal I in B1(X) gener- ated by f0, i.e., I = 〈f0〉. We claim that f 1 3 0 /∈ I. If possible, let f 1 3 0 ∈ I. Then there exists g ∈ B1(R), such that f 1 3 0 = gf0. When x = p q , where p ∈ Z,q ∈ N and g.c.d (p,q) = 1, g(x) = q 2 3 . We show that such g does not exist in B1(R). Let α be any irrational number in R. We show that g is not continuous at α, no matter how we define g(α). Suppose g(α) = β. There exists a sequence of rational numbers {pm qm }, such that {pm qm } converges to α and pm ∈ Z,qm ∈ N with g.c.d (pm,qm) = 1, ∀ m ∈ N. If g is continuous at α then {g(pmqm )} con- verges to g(α), which implies that q 2 3 m converges to β. But qm ∈ N, so {q 2 3 m} must be eventually constant. Suppose there exists n0 ∈ N such that ∀ m ≥ n0, qm is either c or −c or qm oscillates between c and −c, for some natural num- ber c, i.e., {pm c } converges to α or −α or oscillates. In any case, {pm qm } cannot converges to α. Hence we get a contradiction. So, g is not continuous at any irrational point. It is well known that, if, f ∈ B1(X,Y ), where X is a Baire space, Y is a metric space and B1(X,Y ) stands for the collection of all Baire one functions from X to Y then the set of points where f is continuous is dense in X [4]. Therefore, the set of points of R where g is continuous is dense in R and is a subset of Q. Hence it is a countable dense subset of R (Since R is a Baire space). But using Baire’s category theorem it can be shown that, there exists no function f : R → R, which is continuous precisely on a countable dense subset of R. So, we arrive at a contradiction and no such g exists. Hence f 1 3 0 /∈ I. Observe that, Z(f0) = Z(f 1 3 0 ) and I ⊆ Z −1 B [ZB[I]]. Again, f 1 3 0 /∈ I but f 1 3 0 ∈ Z −1 B ZB[I], which implies I $ Z −1 B [ZB[I]]. By an earlier result ZB[I] = ZB[Z −1 B [ZB[I]]], proving that the map ZB : IB → FB(X) is not injective when X = R. Observation: 〈f0〉 ( 〈f 1 3 0 〉. Analogously, it can be shown that 〈f0〉 ( 〈f 1 3 0 〉 ( 〈f 1 5 0 〉 ( ... ( 〈f 1 2m+1 0 〉 ( ... is a strictly increasing chain of proper ideals in B1(R). Hence B1(R) is not a Noetherian ring. Theorem 2.8. If M is a maximal ideal in B1(X) then ZB[M] is a ZB- ultrafilter on X. Proof. By Theorem 2.5, ZB[M] is a ZB-filter on X. Let F be a ZB-filter on X such that, ZB[M] ⊆ F . Then M ⊆ Z−1B [ZB[M]] ⊆ Z −1 B [F ]. Z −1 B [F ] being a proper ideal and M being a maximal ideal, we have Z−1B [F ] = M =⇒ c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 382 Ideals in B1(X) and residue class rings of B1(X) modulo an ideal ZB[M] = ZB[Z −1 B [F ]] = F . Hence every ZB-filter that contains ZB[M] must be equal to ZB[M]. This shows ZB[M] is a ZB-ultrafilter on X. � Theorem 2.9. If U is a ZB-ultrafilter on X then Z −1 B [U ] is a maximal ideal in B1(X). Proof. By Theorem 2.6, we have Z−1B [U ] is a proper ideal in B1(X). Let I be a proper ideal in B1(X) such that Z −1 B [U ] ⊆ I. It is enough to show that Z−1B [U ] = I. Now Z −1 B [U ] ⊆ I =⇒ ZB[Z −1 B [U ]] ⊆ ZB[I] =⇒ U ⊆ ZB[I]. Since U is a ZB-ultrafilter on X, we have U = ZB[I] =⇒ Z−1B [U ] = Z−1B [ZB[I]] ⊇ I. Hence Z −1 B [U ] = I � Remark 2.10. Each ZB-ultrafilter on X is the image of a maximal ideal in B1(X) under the map ZB. Let M(B1(X)) be the collection of all maximal ideals in B1(X) and ΩB(X) be the collection of all ZB-ultrafilters on X. If we restrict the map ZB to the class M(B1(X)), then it is clear that the map ZB ∣∣∣∣ M(B1(X)) : M(B1(X)) → ΩB(X) is a surjective map. Further, this restriction map is a bijection, as seen below. Theorem 2.11. The map ZB ∣∣∣∣ M(B1(X)) : M(B1(X)) → ΩB(X) is a bijection. Proof. It is enough to check that ZB ∣∣∣∣ M(B1(X)) : M(B1(X)) → ΩB(X) is in- jective. Let M1 and M2 be two members in M(B1(X)) such that ZB[M1] = ZB[M2] =⇒ Z−1B [ZB[M1]] = Z −1 B [ZB[M1]]. But M1 ⊆ Z −1 B [ZB[M1]] and M2 ⊆ Z−1B [ZB[M2]]. By maximality of M1 and M2 we have, M1 = Z −1 B [ZB[M1]] = Z−1B [ZB[M2]] = M2. � Definition 2.12. An ideal I in B1(X) is called a ZB-ideal if Z −1 B [ZB[I]] = I, i.e., ∀ f ∈ B1(X), f ∈ I ⇐⇒ Z(f) ∈ ZB[I]. Since ZB[Z −1 B [FB]] = FB, Z −1 B [FB] is a ZB-ideal for any ZB-filter FB on X. If I is any ideal in B1(X), then, Z −1 B [ZB[I]] is the smallest ZB-ideal containing I. It is easy to observe (1) Every maximal ideal in B1(X) is a ZB ideal. (2) The intersection of arbitrary family of ZB-ideals in B1(X) is always a ZB-ideal. (3) The map ZB ∣∣∣∣ JB : JB → FB(X) is a bijection, where JB denotes the collection of all ZB-filters on X. Example 2.13. Let I = {f ∈ B1(R) : f(1) = f(2) = 0}. Then I is a ZB ideal in B1(R) which is not maximal, as I ( M̂1 = {f ∈ B1(R) : f(1) = 0}. The ideal I is not a prime ideal, as the functions x− 1 and x− 2 do not belong to I, but their product belongs to I. Also no proper ideal of I is prime. More c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 383 A. Deb Ray and A. Mondal generally, for any subset S of R,IS = {f ∈ B1(R) : f(S) = 0} is a ZB-ideal in B1(R). It is well known that in a commutative ring R with unity, the intersection of all prime ideals of R containing an ideal I is called the radical of I and it is denoted by √ I. For any ideal I, the radical of I is given by {a ∈ R : an ∈ I, for some n ∈ N} ([3]) and in general I ⊆ √ I. For if I = √ I, I is called a radical ideal. Theorem 2.14. A ZB-ideal I in B1(X) is a radical ideal. Proof. √ I = {f ∈ B1(X) : ∃ n ∈ N such that fn ∈ I} = {f ∈ B1(X) : such that Z(fn) ∈ ZB[I] for some n ∈ N} ( As I is a ZB-ideal in B1(X) ) = {f ∈ B1(X) : Z(f) ∈ ZB[I] } = {f ∈ B1(X) : f ∈ I} = I. So I is a radical ideal in B1(X). � Corollary 2.15. Every ZB-ideal I in B1(X) is the intersection of all prime ideals in B1(X) which contains I. Next theorem establishes some equivalent conditions on the relationship among ZB-ideals and prime ideals of B1(X). Theorem 2.16. For a ZB-ideal I in B1(X) the following conditions are equiv- alent: (1) I is a prime ideal of B1(X). (2) I contains a prime ideal of B1(X). (3) if fg = 0, then either f ∈ I or g ∈ I. (4) Given f ∈ B1(X) there exists Z ∈ ZB[I], such that f does not change its sign on Z. Proof. (1) =⇒ (2) and (2) =⇒ (3) are immediate. (3) =⇒ (4): Let (3) be true. Choose f ∈ B1(X). Then (f ∨0).(f ∧0) = 0. So by (3), f ∨ 0 ∈ I or f ∧ 0 ∈ I. Hence Z(f ∨ 0) ∈ ZB[I] or Z(f ∧ 0) ∈ ZB[I]. It is clear that f ≤ 0 on Z(f ∧ 0) and f ≥ 0 on Z(f ∨ 0). (4) =⇒ (1): Let (4) be true. To show that I is prime. Let g,h ∈ B1(X) be such that gh ∈ I. By (4) there exists Z ∈ ZB[I], such that |g| − |h| ≥ 0 on Z (say). It is clear that, Z ∩Z(g) ⊆ Z(h). Consequently Z ∩Z(gh) ⊆ Z(h). Since ZB[I] is a ZB-filter on X, it follows that Z(h) ∈ ZB[I]. So h ∈ I, since I is a ZB-ideal. Hence, I is prime. � Theorem 2.17. In B1(X), every prime ideal P can be extended to a unique maximal ideal. Proof. If possible let P be contained in two distinct maximal ideals M1 and M2. So, P ⊆ M1 ∩ M2. Since maximal ideals in B1(X) are ZB-ideals and intersection of any number of ZB-ideals is ZB-ideal, M1 ∩ M2 is a ZB-ideal containing the prime ideal P . By Theorem 2.16, M1∩M2 is a prime ideal. But in a commutative ring with unity, for two ideals I and J, if, I * J and J * I, c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 384 Ideals in B1(X) and residue class rings of B1(X) modulo an ideal then I ∩J is not a prime ideal. Thus M1 ∩M2 is not prime ideal and we get a contradiction. So, every prime ideal can be extended to a unique maximal ideal. � Corollary 2.18. B1(X) is a Gelfand ring for any topological space X. Definition 2.19. A ZB-filter FB on X is called a prime ZB-filter on X, if, for any Z1,Z2 ∈ Z(B1(X)) with Z1 ∪Z2 ∈ FB either Z1 ∈ FB or Z2 ∈ FB. The next two theorems are analogous to Theorem 2.12 in [3] and therefore, we state them without proof. Theorem 2.20. If I is a prime ideal in B1(X), then ZB[I] = {Z(f) : f ∈ I} is a prime ZB-filter on X. Theorem 2.21. If FB is a prime ZB-filter on X then Z −1 B [FB] = {f ∈ B1(X) : Z(f) ∈ FB} is a prime ideal in B1(X). Corollary 2.22. Every prime ZB-filter can be extended to a unique ZB-ultrafilter on X. Corollary 2.23. A ZB-ultrafilter U on X is a prime ZB-filter on X, as U = ZB[M], for some maximal ideal M in B1(X). 3. Fixed ideals and free ideals in B1(X) In this section, we introduce fixed and free ideals of B1(X) and B ∗ 1 (X) and completely characterize the fixed maximal ideals of B1(X) as well as those of B∗1 (X). It is observed here that a natural relationship exists between fixed maximal ideals of B∗1 (X) and the fixed maximal ideals of B1(X). However, free maximal ideals do not behave the same. In the last part of this section, we find a class of Baire one functions defined on a perfectly normal T1 space X which precisely determines the fixed and free maximal ideals of the corresponding ring. Definition 3.1. A proper ideal I of B1(X) (respectively, B ∗ 1 (X)) is called fixed if ⋂ Z[I] 6= ∅. If I is not fixed then it is called free. For any Tychonoff space X, the fixed maximal ideals of the ring B1(X) and those of B∗1 (X) are characterized. Theorem 3.2. {M̂p : p ∈ X} is a complete list of fixed maximal ideals in B1(X), where M̂p = {f ∈ B1(X) : f(p) = 0}. Moreover, p 6= q =⇒ M̂p 6= M̂q. Proof. Choose p ∈ X. The map Ψp : B1(X) → R, defined by Ψp(f) = f(p) is clearly a ring homomorphism. Since the constant functions are in B1(X), Ψp is surjective and ker Ψp = {f ∈ B1(X) : Ψp(f) = 0} = {f ∈ B1(X) : f(p) = 0} = M̂p (say). By First isomorphism theorem of rings we get B1(X)/M̂p is isomorphic to the field R. B1(X)/M̂p being a field we conclude that M̂p is a maximal ideal in B1(X). Since p ∈ ⋂ ZB[M], the ideal M̂p is a fixed ideal. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 385 A. Deb Ray and A. Mondal For any Tychonoff space X, we know that p 6= q =⇒ Mp 6= Mq, where Mp = {f ∈ C(X) : f(p) = 0} is the fixed maximal ideal in C(X). Since M̂p∩C(X) = Mp, it follows that for any Tychonoff space X, p 6= q =⇒ M̂p 6= M̂q. Let M be any fixed maximal ideal in B1(X). There exists p ∈ X such that for all f ∈ M, f(p) = 0. Therefore, M ⊆ M̂p. Since M is a maximal ideal and M̂p is a proper ideal, we get M = M̂p. � Theorem 3.3. {M̂∗p : p ∈ X} is a complete list of fixed maximal ideals in B∗1 (X), where M̂ ∗ p = {f ∈ B∗1 (X) : f(p) = 0}. Moreover, p 6= q =⇒ M̂∗p 6= M̂∗q . Proof. Similar to the proof of Theorem 3.2. � The following two theorems show the interrelations between fixed ideals of B1(X) and B ∗ 1 (X). Theorem 3.4. If I is any fixed ideal of B1(X) then I ∩B∗1 (X) is a fixed ideal of B∗1 (X). Proof. Straightforward. � Lemma 3.5. Given any f ∈ B1(X), there exists a positive unit u of B1(X) such that uf ∈ B∗1 (X). Proof. Consider u = 1|f|+1 . Clearly u is a positive unit in B1(X) [1] and uf ∈ B∗1 (X) as |uf| ≤ 1. � Theorem 3.6. Let an ideal I in B1(X) be such that I∩B∗1 (X) is a fixed ideal of B∗1 (X). Then I is a fixed ideal of B1(X). Proof. For each f ∈ I, there exists a positive unit uf of B1(X) such that uff ∈ I ∩ B∗1 (X). Therefore, ⋂ f∈I Z(f) = ⋂ f∈I Z(uff) ⊇ ⋂ g∈B∗1 (X)∩I Z(g) 6= ∅. Hence I is fixed in B1(X). � Since for any discrete space X, C(X) = B1(X) and C ∗(X) = B∗1 (X), consid- ering the example 4.7 of [3], we can conclude the following: (1) For any maximal ideal M of B1(X), M∩B∗1 (X) need not be a maximal ideal in B∗1 (X). (2) All free maximal ideals in B∗1 (X) need not be of the form M ∩B∗1 (X), where M is a maximal ideal in B1(X). Theorem 3.7. If X is a perfectly normal T1 space then for each p ∈ X, χp : X → R given by χp(x) = { 1 if x = p 0 otherwise. is a Baire one function. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 386 Ideals in B1(X) and residue class rings of B1(X) modulo an ideal Proof. For any open set U of R, χ−1p (U) =   X if 0, 1 ∈ U X \{p} if 0 ∈ U but 1 /∈ U {p} if 0 /∈ U but 1 ∈ U ∅ if 0 /∈ U but 1 /∈ U. Since X is a perfectly normal space, the open set X\{p} is a Fσ set. Hence in any case χp pulls back an open set to a Fσ set. So χp is a Baire one function [5]. � In view of Theorem 3.7 we obtain the following facts about any perfectly normal T1 space. Observation 3.8. If M is a maximal ideal of B1(X) where X is a perfectly normal T1 space then (1) For each p ∈ X either χp ∈ M or χp − 1 ∈ M. This follows from χp(χp − 1) = 0 ∈ M and M is prime. (2) If χp − 1 ∈ M then χq ∈ M for all q 6= p. For if χq − 1 ∈ M for some q 6= p then Z(χp − 1),Z(χq − 1) ∈ ZB[M]. This implies ∅ = Z(χp − 1) ∩ Z(χq − 1) ∈ ZB[M] which contradicts that ZB[M] is a ZB-ultrafilter. (3) M is fixed if and only if χp − 1 ∈ M for some p ∈ X. If M is fixed then M = M̂p for some p ∈ X and therefore, χp−1 ∈ M. Conversely let χp − 1 ∈ M for some p ∈ X. Then {p} = Z(χp − 1) ∈ ZB[M] shows that M is fixed. (4) M is free if and only if M contains {χp : p ∈ X}. Follows from Observation (3). The following theorem ensures the existence of free maximal ideals in B1(X) where X is any infinite perfectly normal T1 space. Theorem 3.9. For a perfectly normal T1 space X, the following statements are equivalent: (1) X is finite. (2) Every maximal ideal in B1(X) is fixed. (3) Every ideal in B1(X) is fixed. Proof. (1) =⇒ (2): Since a finite T1 space is discrete, C(X) = B1(X) = XR. X being finite, it is compact and therefore all the maximal ideals of C(X)( = B1(X) ) are fixed. (2) =⇒ (3): Proof obvious. (3) =⇒ (1): Suppose X is infinite. We shall show that there exists a free (proper) ideal in B1(X). Consider I = {f ∈ B1(X) : X \Z(f) is finite} (Here finite includes ∅). Of course I 6= ∅, as 0 ∈ I. Since X is infinite, 1 /∈ I and so, I is proper. We show that, I is an ideal in B1(X). Let f,g ∈ I. Then X \Z(f) and c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 387 A. Deb Ray and A. Mondal X \Z(g) are both finite. Now X \Z(f −g) ⊆ X \Z(f) ∪ X \Z(g) implies that X \Z(f −g) is finite. Hence f −g ∈ I. Similarly, X \Z(f.g) ⊆ X \Z(f) for any f ∈ I and g ∈ B1(X). So, X \Z(f.g) is finite and hence f.g ∈ I. Therefore, I is an ideal in B1(X). We claim that I is free. For any p ∈ X, consider χp : X → R given by χp(x) = { 1 if x = p 0 otherwise. Using Theorem 3.7, χp is a Baire one function. Also, X \Z(χp) = X \ (X \{p}) = {p} = {p} = finite and χp(p) 6= 0. Hence, I is free. � 4. Residue class ring of B1(X) modulo ideals An ideal I in a partially ordered ring A is called convex if for all a,b,c ∈ A with a ≤ b ≤ c and c,a ∈ I =⇒ b ∈ I. Equivalently, for all a,b ∈ A, 0 ≤ a ≤ b and b ∈ I =⇒ a ∈ I. If A is a lattice ordered ring then an ideal I of A is called absolutely convex if for all a,b ∈ A, |a| ≤ |b| and b ∈ I =⇒ a ∈ I. Example 4.1. If t : B1(X) → B1(Y ) is a ring homomorphism, then ker t is an absolutely convex ideal. Proof. Let g ∈ ker t and |f| ≤ |g|, where f ∈ B1(X). g ∈ ker t =⇒ t(g) = 0 =⇒ t(|g|) = |t(g)| = 0. Since any ring homomorphism t : B1(X) → B1(Y ) preserves the order, t(|f|) = 0 =⇒ |t(f)| = 0 =⇒ t(f) = 0 =⇒ f ∈ ker t. � Let I be an ideal in B1(X). In what follows we shall denote any member of the quotient ring B1(X)/I by I(f) for f ∈ B1(X). i.e., I(f) = f + I. Now we begin with two well known theorems. Theorem 4.2 ([3]). Let I be an ideal in a partially ordered ring A. The corresponding quotient ring A/I is a partially ordered ring if and only if I is convex, where the partial order is given by I(a) ≥ 0 iff ∃ x ∈ A such that x ≥ 0 and a ≡ x( mod I). Theorem 4.3 ([3]). On a convex ideal I in a lattice-ordered ring A the follow- ing conditions are equivalent. (1) I is absolutely convex. (2) x ∈ I implies |x| ∈ I. (3) x,y ∈ I implies x∨y ∈ I. (4) I(a∨ b) = I(a) ∨ I(b), whence A/I is a lattice ordered ring. (5) ∀ a ∈ A, I(a) ≥ 0 iff I(a) = I(|a|). Remark 4.4. For an absolutely convex ideal I of A, I(|a|) = I(a ∨ −a) = I(a) ∨ I(−a) = |I(a)|, ∀ a ∈ A. Theorem 4.5. Every ZB-ideal in B1(X) is absolutely convex. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 388 Ideals in B1(X) and residue class rings of B1(X) modulo an ideal Proof. Suppose I is any ZB-ideal and |f| ≤ |g|, where g ∈ I and f ∈ B1(X). Then Z(g) ⊆ Z(f). Since g ∈ I, it follows that Z(g) ∈ ZB[I], hence Z(f) ∈ ZB[I]. Now I being a ZB-ideal, f ∈ I. � Corollary 4.6. In particular every maximal ideal in B1(X) is absolutely con- vex. Theorem 4.7. For every maximal ideal M in B1(X), the quotient ring B1(X)/M is a lattice ordered field. Proof. Proof is immediate. � The following theorem is a characterization of the non-negative elements in the lattice ordered ring B1(X)/I, where I is a ZB-ideal. Theorem 4.8. Let I be a ZB-ideal in B1(X). For f ∈ B1(X), I(f) ≥ 0 if and only if there exists Z ∈ ZB[I] such that f ≥ 0 on Z. Proof. Let I(f) ≥ 0. By condition (5) of Theorem 4.3, we write I(f) = I(|f|). So, f −|f| ∈ I =⇒ Z(f −|f|) ∈ ZB[I] and f ≥ 0 on Z(f −|f|). Conversely, let f ≥ 0 on some Z ∈ ZB[I]. Then f = |f| on Z =⇒ Z ⊆ Z(f −|f|) =⇒ Z(f −|f|) ∈ ZB[I]. I being a ZB-ideal we get f −|f| ∈ I, which means I(f) = I(|f|). But |f| ≥ 0 on Z gives I(|f|) ≥ 0. Hence, I(f) ≥ 0. � Theorem 4.9. Let I be any ZB-ideal and f ∈ B1(X). If there exists Z ∈ ZB[I] such that f(x) > 0, for all x ∈ Z, then I(f) > 0. Proof. By Theorem 4.8, I(f) ≥ 0. But Z ∩ Z(f) = ∅ and Z ∈ ZB[I] =⇒ Z(f) /∈ ZB[I] =⇒ f /∈ I =⇒ I(f) 6= 0 =⇒ I(f) > 0. � The next theorem shows that the converse of the above theorem holds if the ideal is a maximal ideal in B1(X). Theorem 4.10. Let M be any maximal ideal in B1(X) and M(f) > 0 for some f ∈ B1(X) then there exists Z ∈ ZB[M] such that f > 0 on Z. Proof. By Theorem 4.8, there exists Z1 ∈ ZB[M] such that f ≥ 0 on Z1. Now M(f) > 0 =⇒ f /∈ M which implies that there exists g ∈ M, such that Z(f) ∩ Z(g) = ∅. Choosing Z = Z1 ∩ Z(g), we observe Z ∈ ZB[M] and f(x) > 0, for all x ∈ Z. � Corollary 4.11. For a maximal ideal M of B1(X) and for some f ∈ B1(X), M(f) > 0 if and only if there exists Z ∈ ZB[M] such that f(x) > 0 on Z. Now we show Theorem 4.10 doesn’t hold for every non-maximal ideal I. Theorem 4.12. Suppose I is any non-maximal ZB-ideal in B1(X). There exists f ∈ B1(X) such that I(f) > 0 but f is not strictly positive on any Z ∈ ZB[I]. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 389 A. Deb Ray and A. Mondal Proof. Since I is non-maximal, there exists a proper ideal J of B1(X) such that I $ J. Choose f ∈ J r I. f2 /∈ I =⇒ I(f2) > 0. Choose any Z ∈ ZB[I]. Certainly, Z ∈ ZB[J] and so, Z ∩Z(f2) ∈ ZB[J] =⇒ Z ∩Z(f2) 6= ∅. So f is not strictly positive on the whole of Z. � In what follows, we characterize the ideals I in B1(X) for which B1(X)/I is a totally ordered ring. Theorem 4.13. Let I be a ZB-ideal in B1(X), then the lattice ordered ring B1(X)/I is totally ordered ring if and only if I is a prime ideal. Proof. B1(X)/I is a totally ordered ring if and only if for any f ∈ B1(X), I(f) ≥ 0 or I(−f) ≤ 0 if and only if for all f ∈ B1(X), there exists Z ∈ ZB[I] such that f does not change its sign on Z if and only if I is a prime ideal (by Theorem 2.16). � Corollary 4.14. For every maximal ideal M in B1(X), B1(X)/M is a totally ordered field. Theorem 4.15. Let M be a maximal ideal in B1(X). The function Φ : R → B1(X)/M (respectively, Φ : R → B∗1 (X)/M) defined by Φ(r) = M(r), for all r ∈ R, where r denotes the constant function with value r, is an order preserving monomorphism. Proof. It is clear from the definitions of addition and multiplication of the residue class ring B1(X)/M that the function is a homomorphism. To show φ is injective. Let M(r) = M(s) for some r,s ∈ R with r 6= s. Then r − s ∈ M. This contradicts to the fact that M is a proper ideal. Hence M(r) 6= M(s), when r 6= s. Let r,s ∈ R with r > s. Then r−s > 0. The function r − s is strictly positive on X. Since X ∈ Z(B1(X)), by Theorem 4.9, M(r − s) > 0 =⇒ M(r) > M(s) =⇒ Φ(r) > Φ(s). Thus Φ is an order preserving monomorphism. � For a maximal ideal M in B1(X), the residue class field B1(X)/M (respectively B∗1 (X)/M) can be considered as an extension of the field R. Definition 4.16. The maximal ideal M of B1(X) (respectively, B ∗ 1 (X)) is called real if Φ(R) = B1(X)/M (respectively, Φ(R) = B∗1 (X)/M) and in such case B1(X)/M is called real residue class field. If M is not real then it is called hyper-real and B1(X)/M is called hyper-real residue class field. Definition 4.17 ([3]). A totally ordered field F is called archimedean if given α ∈ F, there exists n ∈ N such that n > α. If F is not archimedean then it is called non-archimedean. If F is a non-archimedean ordered field then there exists some α ∈ F such that α > n, for all n ∈ N. Such an α is called an infinitely large element of F and 1 α is called infinitely small element of F which is characterized by the relation 0 < 1 α < 1 n , ∀ n ∈ N. The existence of an infinitely large (equivalently, infinitely small) element in F assures that F is non-archimedean. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 390 Ideals in B1(X) and residue class rings of B1(X) modulo an ideal In the context of archimedean field, the following is an important theorem available in the literature. Theorem 4.18 ([3]). A totally ordered field is archimedean iff it is isomorphic to a subfield of the ordered field R . We thus get that the real residue class field B1(X)/M is archimedean if M is a real maximal ideal of B1(X). Theorem 4.19. Every hyper-real residue class field B1(X)/M is non-archimedean. Proof. Proof follows from the fact that the identity is the only non-zero homo- morphism on the ring R into itself. � Corollary 4.20. A maximal ideal M of B1(X) is hyper-real if and only if there exists f ∈ B1(X) such that M(f) is an infinitely large member of B1(X)/M. Theorem 4.21. Each maximal ideal M in B∗1 (X) is real. Proof. It is equivalent to show that B∗1 (X)/M is archimedean. Choose f ∈ B∗1 (X). Then |f(x)| ≤ n, for all x ∈ X and for some n ∈ N. i.e., |M(f)| = M(|f|) ≤ M(n). So there does not exist any infinitely large member in B∗1 (X)/M and hence B ∗ 1 (X)/M is archimedean. � Corollary 4.22. If X is a topological space such that B1(X) = B ∗ 1 (X) then each maximal ideal in B1(X) is real. The following theorem shows how an unbounded Baire one function f on X is related to an infinitely large member of the residue class field B1(X)/M. Theorem 4.23. Given a maximal ideal M of B1(X) and f ∈ B1(X), the following statements are equivalent: (1) |M(f)| is infinitely large member in B1(X)/M. (2) f is unbounded on each zero set in ZB[M]. (3) for all n ∈ N, Zn = {x ∈ X : |f(x)| ≥ n}∈ ZB[M]. Proof. (1) ⇐⇒ (2): |M(f)| is not infinitely large in B1(X)/M if and only if ∃ n ∈ N such that, |M(f)| = M(|f|) ≤ M(n) if and only if |f| ≤ n on some Z ∈ ZB[M] if and only if f is bounded on some Z ∈ ZB[M]. (2) =⇒ (3): Choose n ∈ N, we shall show that Zn ∈ ZB[M]. By (2), Zn intersects each member in ZB[M]. Now ZB[M] being a ZB-ultrafilter, Zn ∈ ZB[M]. (3) =⇒ (2): Let each Zn ∈ ZB[M], for all n ∈ N. Then for each n ∈ N, |f| ≥ n on some zero set in ZB[M]. Hence |M(f)| = M(|f|) ≥ M(n), for all n ∈ N. That means |M(f)| is infinitely large member in B1(X)/M. � Theorem 4.24. f ∈ B1(X) is unbounded on X if and only if there exists a maximal ideal M in B1(X) such that M(f) is infinitely large in B1(X)/M. Proof. Let f be unbounded on X. So, each Zn in Theorem 4.23 is non-empty. We observe that {Zn : n ∈ N} is a subcollection of Z(B1(X)) having finite c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 391 A. Deb Ray and A. Mondal intersection property. So there exists a ZB-ultrafilter U on X such that {Zn : n ∈ N} ⊆ U . Therefore, there is a maximal ideal M in B1(X) for which U = ZB[M] and so, Zn ∈ ZB[M], for all n ∈ N. By Theorem 4.23 M(f) is infinitely large. Converse part is a consequence of (1) =⇒ (2) of Theorem 4.23. � Corollary 4.25. If a completely Hausdorff space X is not totally disconnected then there exists a hyper-real maximal ideal M in B1(X). Proof. It is enough to prove that there exists an unbounded Baire one func- tion in B1(X). We know that if a completely Hausdorff space is not totally disconnected, then there always exists an unbounded Baire one function [1]. � In the next theorem we characterize the real maximal ideals of B1(X). Theorem 4.26. For the maximal ideal M of B1(X) the following statements are equivalent: (1) M is a real maximal ideal. (2) ZB[M] is closed under countable intersection. (3) ZB[M] has countable intersection property. Proof. (1) =⇒ (2): Assume that (2) is false, i.e., there exists a sequence of functions {fn} in M for which ∞⋂ n=1 Z(fn) /∈ ZB[M]. Set f(x) = ∞∑ n=1 ( |fn(x)|∧ 1 4n ) , ∀x ∈ X. It is clear that, the function f defined on X is actually a Baire one function ([1]) and Z(f) = ∞⋂ n=1 Z(fn). Thus, Z(f) /∈ ZB[M]. Hence f /∈ M =⇒ M(f) > 0 in B1(X)/M. Fix a natural number m. Then Z(f1) ⋂ Z(f2) ⋂ Z(f3)... ⋂ Z(fm) = Z(say) ∈ ZB[M]. Now for any point x ∈ Z, f(x) = ∞∑ n=m+1 ( |fn(x)|∧ 14n ) ≤ ∞∑ n=m+1 1 4n = 3−14−m. This shows that, 0 < M(f) ≤ M(3−14−m), ∀ m ∈ N. Hence M(f) is an infinitely small member in B1(X)/M. So, M becomes a hyper-real maximal ideal and then (1) is false. (2) =⇒ (3): Trivial, as ∅ /∈ ZB[M]. (3) =⇒ (1): Assume that (1) is false, i.e. M is hyper-real. So, there exists f ∈ B1(X) so that |M(f)| is infinitely large in B1(X)/M. Therefore for each n ∈ N, Zn defined in Theorem 4.23, belongs to ZB[M]. Since R is archimedean, we have ∞⋂ n=1 Zn = ∅. Thus (3) is false. � So far we have seen that for any topological space X, all fixed maximal ideals of B1(X) are real. Though the converse is not assured in general, we show in the next example that in B1(R) a maximal ideal is real if and only if it is fixed. Example 4.27. Suppose M is any real maximal ideal in B1(R). We claim that M is fixed. The identity i : R → R belongs to B1(R). Since M is a real maximal ideal, there exists a real number r such that M(i) = M(r). This c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 392 Ideals in B1(X) and residue class rings of B1(X) modulo an ideal implies i− r ∈ M. Hence Z(i− r) ∈ ZB[M]. But Z(i− r) is a singleton. So, ZB[M] is fixed, i.e., M is fixed. In view of Observation 3.8(3), we conclude that a maximal ideal M in B1(R) is real if and only if there exists a unique p ∈ R such that χp − 1 ∈ M. If X is a P-space then C(X) possesses real free maximal ideals. In such case however, B1(X) = C(X). Consequently, B1(X) possesses real free maximal ideals, when X is a P-space. It is still a natural question, what are the topo- logical spaces X for which B1(X) (⊇ C(X)) contains a free real maximal ideal? References [1] A. Deb Ray and A. Mondal, On rings of Baire one functions, Applied Gen. Topol. 20, no. 1 (2019), 237–249. [2] J. P. Fenecios and E. A. Cabral, On some properties of Baire-1 functions, Int. Journal of Math. Analysis 7, no. 8 (2013), 393–402. [3] L. Gillman and M. Jerison, Rings of Continuous Functions, New York: Van Nostrand Reinhold Co., 1960. [4] J. R. Munkres, Topology, Second edition, Pearson Education, Delhi, 2003. [5] L. Vesely, Characterization of Baire-one functions between topological spaces, Acta Uni- versitatis Carolinae. Mathematica et Physica 33, no. 2 (1992), 143–156. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 393