Ideals in $B_1(X)$ and residue class rings of $B_1(X)$ modulo an ideal

TL;DR

This paper investigates the structure of ideals in the ring of Baire one functions, exploring duality with zero sets, characterizing maximal ideals, and analyzing residue class rings, revealing algebraic properties and relationships in topological spaces.

Contribution

It provides a detailed duality framework between ideals and zero sets in $B_1(X)$, characterizes fixed and free maximal ideals, and studies residue class rings, advancing understanding of $B_1(X)$'s algebraic structure.

Findings

$B_1(X)$ is a Gelfand ring but not Noetherian.

Fixed maximal ideals are fully described for certain spaces.

Residue class rings are analyzed with respect to real and hyper real maximal ideals.

Abstract

This paper explores the duality between ideals of the ring $B_1(X)$ of all real valued Baire one functions on a topological space $X$ and typical families of zero sets, called $Z_B$-filters, on $X$. As a natural outcome of this study, it is observed that $B_1(X)$ is a Gelfand ring but non-Noetherian in general. Introducing fixed and free maximal ideals in the context of $B_1(X)$, complete descriptions of the fixed maximal ideals of both $B_1(X)$ and $B_1^*(X)$ are obtained. Though free maximal ideals of $B_1(X)$ 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 $T_1$ space $X$, free maximal ideals of $B_1(X)$ are determined by a typical class of Baire one functions. In the concluding part of this paper, we study residue class ring of $B_1(X)$ modulo an ideal, with special emphasize on real and hyper real maximal ideals of $B_1(X)$.

Equations

$$f_{0}(x)=\begin{cases}\frac{1}{q}&$ if $x=\frac{p}{q}$, where $p\in\mathbb{Z},q\in\mathbb{N}$ and g.c.d. $(p,q)=1\\ 1&$ if $x=0$ $\\ 0&$ otherwise $\end{cases}$$

$$\chi_{p}(x)=\begin{cases}1&$ if $x=p$$

\\ 0&$otherwise.$\end{cases}$$

$$\chi_{p}^{-1}(U)=\begin{cases}X&$ if $0,1\in U$$

\\ X\setminus\{p\}&$if$0\in U$but$1\notin U

$$\\ \{p\}&$ if $0\notin U$ but $1\in U$$

\\ \emptyset&$if$0\notin U$but$1\notin U

$$.\par\par\end{cases}$$

$$\chi_{p}(x)=\begin{cases}1&$ if $x=p$$

\\ 0&$otherwise.$\end{cases}$$

Full text

The second author is supported by Council of Scientific and Industrial Research, HRDG, India. Sanction No.- 09/028(0998)/2017-EMR-1

1. Introduction

In [1], we have introduced the ring of Baire one functions defined on any topological space $X$ and have denoted it by $B_{1}(X)$. It has been observed that $B_{1}(X)$ 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 $B_{1}(X)$. Certainly, $B_{1}^{*}(X)\cap C(X)=C^{*}(X)$.

In this paper, we study the ideals, in particular, the maximal ideals of $B_{1}(X)$ (and also of $B_{1}^{*}(X)$). There is a nice interplay between the ideals of $B_{1}(X)$ and a typical family of zero sets (which we call a $Z_{B}$-filter) of the underlying space $X$. As a natural consequence of this duality of ideals of $B_{1}(X)$ and $Z_{B}$-filters on $X$, we obtain that $B_{1}(X)$ is Gelfand and in general, $B_{1}(X)$ is non-Noetherian.

Introducing the idea of fixed and free ideals in our context, we have characterized the fixed maximal ideals of $B_{1}(X)$ and also those of $B_{1}^{*}(X)$. We have shown that although fixed maximal ideals of the rings $B_{1}(X)$ and $B_{1}^{*}(X)$ obey a natural relationship, the free maximal ideals fail to do so. However, for a perfectly normal $T_{1}$ space $X$, free maximal ideals of $B_{1}(X)$ are determined by a typical class of Baire one functions.

In the last section of this paper, we discuss residue class ring of $B_{1}(X)$ modulo an ideal and introduce real and hyper-real maximal ideals in $B_{1}(X)$.

2. $Z_{B}$-filters on $X$ and Ideals in $B_{1}(X)$

Definition 2.1**.**

A nonempty subcollection $\mathscr{F}$ of $Z(B_{1}(X))$ is said to be a $Z_{B}$-filter on $X$, if it satisfies the following conditions:

1. (1)

$\emptyset\notin\mathscr{F}$ 2. (2)

if $Z_{1},Z_{2}\in\mathscr{F}$, then $Z_{1}\cap Z_{2}\in\mathscr{F}$ 3. (3)

if $Z\in\mathscr{F}$ and $Z^{\prime}\in Z(B_{1}(X))$ is such that $Z\subseteq Z^{\prime}$, then $Z^{\prime}\in\mathscr{F}$.

Clearly, a $Z_{B}$-filter $\mathscr{F}$ on $X$ has finite intersection property. Conversely, if a subcollection $\mathscr{B}\subseteq Z(B_{1}(X))$ possesses finite intersection property, then $\mathscr{B}$ can be extended to a $Z_{B}$-filter $\mathscr{F}(\mathscr{B})$ on $X$, given by $\mathscr{F}(\mathscr{B})=\{Z\in Z(B_{1}(X))$: there exists a finite subfamily $\{B_{1},B_{2},...,B_{n}\}$ of $\mathscr{B}$ with $Z\supseteq\bigcap\limits_{i=1}^{n}B_{i}$}. Indeed this is the smallest $Z_{B}$-filter on $X$ containing $\mathscr{B}$.

Definition 2.2**.**

A $Z_{B}$-filter $\mathscr{U}$ on $X$ is called a $Z_{B}$-ultrafilter on $X$, if there does not exist any $Z_{B}$-filter $\mathscr{F}$ on $X$, such that $\mathscr{U}\subsetneqq\mathscr{F}$.

Example 2.3**.**

Let, $A_{0}=\{Z\in Z(B_{1}(\mathbb{R})):0\in Z\}$. Then $A_{0}$ is a $Z_{B}$-ultrafilter on $\mathbb{R}$.

Applying Zorn’s lemma one can show that, every $Z_{B}$-filter on $X$ can be extended to a $Z_{B}$-ultrafilter. Therefore, a family $\mathscr{B}$ of $Z(B_{1}(X))$ with finite intersection property can be extended to a $Z_{B}$-ultrafilter on $X$.

Remark 2.4*.*

A $Z_{B}$-ultrafilter $\mathscr{U}$ on $X$ is a subfamily of $Z(B_{1}(X))$ which is maximal with respect to having finite intersection property. Conversely, if a family $\mathscr{B}$ of $Z(B_{1}(X))$ has finite intersection property and maximal with respect to having this property, then $\mathscr{B}$ is a $Z_{B}$-ultrafilter on $X$.

In what follow, by an ideal $I$ of $B_{1}(X)$ we always mean a proper ideal.

Theorem 2.5**.**

If $I$ is an ideal in $B_{1}(X)$, then $Z_{B}[I]=\{Z(f):f\in I\}$ is a $Z_{B}$-filter on $X$.

Proof.

Since $I$ is a proper ideal in $B_{1}(X)$, we claim $\emptyset\notin Z_{B}[I]$. If possible let $\emptyset\in Z_{B}[I]$. So, $\emptyset=Z(f)$, for some $f\in I$. As $f\in I\implies f^{2}\in I$ and $Z(f^{2})=Z(f)=\emptyset$, hence $\frac{1}{f^{2}}\in B_{1}(X)$ [1]. This is a contradiction to the fact that, $I$ is a proper ideal and contains no unit.

Let $Z(f),Z(g)\in Z_{B}[I]$, for some $f,g\in I.$ Our claim is $Z(f)\cap Z(g)\in Z_{B}[I]$. $Z(f)\cap Z(g)=Z(f^{2}+g^{2})\in Z_{B}[I]$, as $I$ is an ideal and so, $f^{2}+g^{2}\in I$.

Now assume that $Z(f)\in Z_{B}[I]$ and $Z^{\prime}\in Z(B_{1}(X))$ is such that $Z(f)\subseteq Z^{\prime}$. Then we can write $Z^{\prime}=Z(h)$, for some $h\in B_{1}(X)$. $Z(f)\subseteq Z^{\prime}\implies Z(h)=Z(h)\cup Z(f)$. So, $Z(h)=Z(hf)\in Z_{B}[I]$, because $hf\in I.$ Hence, $Z_{B}[I]$ is a $Z_{B}$-filter on $X$. ∎

Theorem 2.6**.**

Let $\mathscr{F}$ be a $Z_{B}$-filter on $X$. Then $Z_{B}^{-1}[\mathscr{F}]=\{f\in B_{1}(X):Z(f)\in\mathscr{F}\}$ is an ideal in $B_{1}(X)$.

Proof.

We note that, $\emptyset\notin\mathscr{F}$. So the constant function 1 $\notin Z_{B}^{-1}[\mathscr{F}]$. Hence $Z_{B}^{-1}[\mathscr{F}]$ is a proper subset of $B_{1}(X)$.

Choose $f,g\in Z_{B}^{-1}[\mathscr{F}]$. Then $Z(f),Z(g)\in\mathscr{F}$ and $\mathscr{F}$ being a $Z_{B}$-filter $Z(f)\cap Z(g)\in\mathscr{F}$. Now $Z(f)\cap Z(g)\subseteq Z(f-g)$. Hence $Z(f-g)\in\mathscr{F}$, $\mathscr{F}$ being a $Z_{B}$-filter on $X$. This implies $f-g\in Z_{B}^{-1}[\mathscr{F}]$.

For $f\in Z_{B}^{-1}[\mathscr{F}]$ and $h\in B_{1}(X)$, $Z(f.h)=Z(f)\cup Z(h)$. As $Z(f)\in\mathscr{F}$ and $\mathscr{F}$ is a $Z_{B}$-filter on $X$, it follows that $Z(f.h)$ $\in$ $\mathscr{F}$. Hence $f.h\in Z_{B}^{-1}[\mathscr{F}]$.

Thus $Z_{B}^{-1}[\mathscr{F}]$ is an ideal of $B_{1}(X)$. ∎

We may define a map $Z:B_{1}(X)\rightarrow Z(B_{1}(X))$ given by $f\mapsto Z(f)$. Certainly, $Z$ is a surjection. In view of the above results, such $Z$ induces a map $Z_{B}$ between the collection of all ideals of $B_{1}(X)$, say $\mathscr{I}_{B}$ and the collection of all $Z_{B}$-filters on $X$, say $\mathscr{F}_{B}(X)$, i.e., $Z_{B}:\mathscr{I}_{B}\rightarrow\mathscr{F}_{B}(X)$ given by $Z_{B}(I)=Z_{B}[I]$, $\forall$ $I\in\mathscr{I}_{B}$. The map $Z_{B}$ is also a surjective map because for any $\mathscr{F}\in\mathscr{F}_{B}(X)$, $Z_{B}^{-1}[\mathscr{F}]$ is an ideal in $B_{1}(X)$. We also note that $Z_{B}[Z_{B}^{-1}[\mathscr{F}]]=\mathscr{F}$. So each $Z_{B}$-filter on $X$ is the image of some ideal in $B_{1}(X)$ under the map $Z_{B}:\mathscr{I}_{B}\rightarrow\mathscr{F}_{B}(X)$.

Observation. The map $Z_{B}:\mathscr{I}_{B}\rightarrow\mathscr{F}_{B}(X)$ is not injective in general. Because, for any ideal $I$ in $B_{1}(X)$, $Z_{B}^{-1}[Z_{B}[I]]$ is an ideal in $B_{1}(X)$, such that $I\subseteq Z_{B}^{-1}[Z_{B}[I]]$ and by our previous result $Z_{B}[Z_{B}^{-1}[Z_{B}[I]]]=Z_{B}[I]$. If one gets an ideal $J$ in $B_{1}(X)$ such that $I\subseteq J\subseteq Z_{B}^{-1}[Z_{B}[I]]$, then we must have $Z_{B}[I]=Z_{B}[J]$. The following example shows that such an ideal is indeed possible to exist. In fact, in the following example, we get countably many ideals $I_{n}$ in $B_{1}(\mathbb{R})$ such that the images of all the ideals are same under the map $Z_{B}$.

Example 2.7**.**

Let $f_{0}:\mathbb{R}\rightarrow\mathbb{R}$ be defined as,

[TABLE]

It is well known that $f_{0}\in B_{1}(\mathbb{R})$ (see [2]). Consider the ideal $I$ generated by $f_{0}$, i.e., $I=\langle f_{0}\rangle$. We claim that $f_{0}^{\frac{1}{3}}\notin I$. If possible, let $f_{0}^{\frac{1}{3}}\in I$. Then there exists $g\in B_{1}(\mathbb{R})$, such that $f_{0}^{\frac{1}{3}}=gf_{0}$. When $x=\frac{p}{q}$, where $p\in\mathbb{Z},q\in\mathbb{N}$ and g.c.d $(p,q)=1$, $g(x)=q^{\frac{2}{3}}$. We show that such $g$ does not exist in $B_{1}(\mathbb{R})$.

Let $\alpha$ be any irrational number in $\mathbb{R}$. We show that $g$ is not continuous at $\alpha$, no matter how we define $g(\alpha)$. Suppose $g(\alpha)=\beta$. There exists a sequence of rational numbers $\{\frac{p_{m}}{q_{m}}\}$, such that $\{\frac{p_{m}}{q_{m}}\}$ converges to $\alpha$ and $p_{m}\in\mathbb{Z},q_{m}\in\mathbb{N}$ with g.c.d $(p_{m},q_{m})=1$, $\forall$ $m\in\mathbb{N}$. If $g$ is continuous at $\alpha$ then $\{g(\frac{p_{m}}{q_{m}})\}$ converges to $g(\alpha)$, which implies that $q_{m}^{\frac{2}{3}}$ converges to $\beta$. But $q_{m}\in\mathbb{N}$, so $\{q_{m}^{\frac{2}{3}}\}$ must be eventually constant. Suppose there exists $n_{0}\in\mathbb{N}$ such that $\forall$ $m\geq n_{0}$, $q_{m}$ is either $c$ or $-c$ or $q_{m}$ oscillates between $c$ and $-c$, for some natural number $c$, i.e., $\{\frac{p_{m}}{c}\}$ converges to $\alpha$ or $-\alpha$ or oscillates. In any case, $\{\frac{p_{m}}{q_{m}}\}$ cannot converges to $\alpha$. Hence we get a contradiction. So, $g$ is not continuous at any irrational point. It is well known that, if, $f\in B_{1}(X,Y)$, where $X$ is a Baire space, $Y$ is a metric space and $B_{1}(X,Y)$ 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 $\mathbb{R}$ where $g$ is continuous is dense in $\mathbb{R}$ and is a subset of $\mathbb{Q}$. Hence it is a countable dense subset of $\mathbb{R}$ (Since $\mathbb{R}$ is a Baire space). But using Baire’s category theorem it can be shown that, there exists no function $f:\mathbb{R}\rightarrow\mathbb{R}$, which is continuous precisely on a countable dense subset of $\mathbb{R}$. So, we arrive at a contradiction and no such $g$ exists. Hence $f_{0}^{\frac{1}{3}}\notin I$.

Observe that, $Z(f_{0})=Z(f_{0}^{\frac{1}{3}})$ and $I\subseteq Z_{B}^{-1}[Z_{B}[I]]$. Again, $f_{0}^{\frac{1}{3}}\notin I$ but $f_{0}^{\frac{1}{3}}\in Z_{B}^{-1}Z_{B}[I]$, which implies $I\subsetneqq Z_{B}^{-1}[Z_{B}[I]]$. Hence by an earlier result $Z_{B}[I]=Z_{B}[Z_{B}^{-1}[Z_{B}[I]]]$, proving that the map $Z_{B}:\mathscr{I}_{B}\rightarrow\mathscr{F}_{B}(X)$ is not injective when $X=\mathbb{R}$.

Observation: $\langle f_{0}\rangle\subsetneq\langle f_{0}^{\frac{1}{3}}\rangle$. Analogously, it can be shown that $\langle f_{0}\rangle\subsetneq\langle f_{0}^{\frac{1}{3}}\rangle\subsetneq\langle f_{0}^{\frac{1}{5}}\rangle\subsetneq...\subsetneq\langle f_{0}^{\frac{1}{2m+1}}\rangle\subsetneq...$ is a strictly increasing chain of proper ideals in $B_{1}(\mathbb{R})$. Hence $B_{1}(\mathbb{R})$ is not a Noetherian ring.

Theorem 2.8**.**

If $M$ is a maximal ideal in $B_{1}(X)$ then $Z_{B}[M]$ is a $Z_{B}$-ultrafilter on $X$.

Proof.

By Theorem 2.5, $Z_{B}[M]$ is a $Z_{B}$-filter on $X$. Let $\mathscr{F}$ be a $Z_{B}$-filter on $X$ such that, $Z_{B}[M]\subseteq\mathscr{F}$. Then $M\subseteq Z_{B}^{-1}[Z_{B}[M]]\subseteq Z_{B}^{-1}[\mathscr{F}]$. $Z_{B}^{-1}[\mathscr{F}]$ being a proper ideal and $M$ being a maximal ideal, we have $Z_{B}^{-1}[\mathscr{F}]=M\implies Z_{B}[M]=Z_{B}[Z_{B}^{-1}[\mathscr{F}]]=\mathscr{F}$. Hence every $Z_{B}$-filter that contains $Z_{B}[M]$ must be equal to $Z_{B}[M]$. This shows $Z_{B}[M]$ is a $Z_{B}$-ultrafilter on $X$. ∎

Theorem 2.9**.**

If $\mathscr{U}$ is a $Z_{B}$-ultrafilter on $X$ then $Z_{B}^{-1}[\mathscr{U}]$ is a maximal ideal in $B_{1}(X)$.

Proof.

By Theorem 2.6, we have $Z_{B}^{-1}[\mathscr{U}]$ is a proper ideal in $B_{1}(X)$. Let $I$ be a proper ideal in $B_{1}(X)$ such that $Z_{B}^{-1}[\mathscr{U}]\subseteq I$. It is enough to show that $Z_{B}^{-1}[\mathscr{U}]=I$. Now $Z_{B}^{-1}[\mathscr{U}]\subseteq I\implies Z_{B}[Z_{B}^{-1}[\mathscr{U}]]\subseteq Z_{B}[I]\implies\mathscr{U}\subseteq Z_{B}[I]$. Since $\mathscr{U}$ is a $Z_{B}$-ultrafilter on $X$, we have $\mathscr{U}=Z_{B}[I]\implies Z_{B}^{-1}[\mathscr{U}]=Z_{B}^{-1}[Z_{B}[I]]\supseteq I$. Hence $Z_{B}^{-1}[\mathscr{U}]=I$ ∎

Remark 2.10*.*

Each $Z_{B}$-ultrafilter on $X$ is the image of a maximal ideal in $B_{1}(X$) under the map $Z_{B}$.

Let $\mathcal{M}(B_{1}(X))$ be the collection of all maximal ideals in $B_{1}(X$) and $\Omega_{B}(X)$ be the collection of all $Z_{B}$-ultrafilters on $X$. If we restrict the map $Z_{B}$ to the class $\mathcal{M}(B_{1}(X))$, then it is clear that the map Z_{B}\bigg{|}_{\mathcal{M}(B_{1}(X))}:\mathcal{M}(B_{1}(X))\rightarrow\Omega_{B}(X) is a surjective map. Further, this restriction map is a bijection, as seen below.

Theorem 2.11**.**

The map Z_{B}\bigg{|}_{\mathcal{M}(B_{1}(X))}:\mathcal{M}(B_{1}(X))\rightarrow\Omega_{B}(X) is a bijection.

Proof.

It is enough to check that Z_{B}\bigg{|}_{\mathcal{M}(B_{1}(X))}:\mathcal{M}(B_{1}(X))\rightarrow\Omega_{B}(X) is injective. Let $M_{1}$ and $M_{2}$ be two members in $\mathcal{M}(B_{1}(X))$ such that $Z_{B}[M_{1}]=Z_{B}[M_{2}]\implies Z_{B}^{-1}[Z_{B}[M_{1}]]=Z_{B}^{-1}[Z_{B}[M_{1}]]$. But $M_{1}\subseteq Z_{B}^{-1}[Z_{B}[M_{1}]]$ and $M_{2}\subseteq Z_{B}^{-1}[Z_{B}[M_{2}]]$. By maximality of $M_{1}$ and $M_{2}$ we have, $M_{1}=Z_{B}^{-1}[Z_{B}[M_{1}]]=Z_{B}^{-1}[Z_{B}[M_{2}]]=M_{2}$. ∎

Definition 2.12**.**

An ideal $I$ in $B_{1}(X)$ is called a $Z_{B}$-ideal if $Z_{B}^{-1}[Z_{B}[I]]=I$, i.e., $\forall$ $f\in B_{1}(X)$, $f\in I\iff Z(f)\in Z_{B}[I]$.

Since $Z_{B}[Z_{B}^{-1}[\mathscr{F}_{B}]]=\mathscr{F}_{B}$, $Z_{B}^{-1}[\mathscr{F}_{B}]$ is a $Z_{B}$-ideal for any $Z_{B}$-filter $\mathscr{F}_{B}$ on $X$.

If $I$ is any ideal in $B_{1}(X)$, then, $Z_{B}^{-1}[Z_{B}[I]]$ is the smallest $Z_{B}$-ideal containing $I$. It is easy to observe

1. (1)

Every maximal ideal in $B_{1}(X)$ is a $Z_{B}$ ideal. 2. (2)

The intersection of arbitrary family of $Z_{B}$-ideals in $B_{1}(X)$ is always a $Z_{B}$-ideal. 3. (3)

The map Z_{B}\bigg{|}_{\mathscr{J}_{B}}:\mathscr{J}_{B}\rightarrow\mathscr{F}_{B}(X) is a bijection, where $\mathscr{J}_{B}$ denotes the collection of all $Z_{B}$-filters on $X$.

Example 2.13**.**

Let $I=\{f\in B_{1}(\mathbb{R}):f(1)=f(2)=0\}$. Then $I$ is a $Z_{B}$ ideal in $B_{1}(\mathbb{R})$ which is not maximal, as $I\subsetneq\widehat{M}_{1}=\{f\in B_{1}(\mathbb{R}):f(1)=0\}$, which is in fact a maximal ideal. 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 generally, for any subset $S$ of $\mathbb{R},I_{S}=\{f\in B_{1}(\mathbb{R}):f(S)=0\}$ is a $Z_{B}$-ideal in $B_{1}(\mathbb{R})$.

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 $\sqrt{I}$. For any ideal $I$, the radical of $I$ is given by $\{a\in R:a^{n}\in I$, for some $n\in\mathbb{N}\}$ ([3]) and in general $I\subseteq\sqrt{I}$. For if $I=\sqrt{I}$, $I$ is called a radical ideal.

Theorem 2.14**.**

A $Z_{B}$-ideal $I$ in $B_{1}(X)$ is a radical ideal.

Proof.

$\sqrt{I}=\{f\in B_{1}(X):\exists$ $n\in\mathbb{N}$ such that $f^{n}\in I\}=\{f\in B_{1}(X):$ such that $Z(f^{n})\in Z_{B}[I]$ for some $n\in\mathbb{N}\}$ (As $I$ is a $Z_{B}$-ideal in $B_{1}(X)$ ) $=\{f\in B_{1}(X):$ such that $Z(f)\in Z_{B}[I]$ $\}=\{f\in B_{1}(X):f\in I\}=I$. So $I$ is a radical ideal in $B_{1}(X)$. ∎

Corollary 2.15**.**

Every $Z_{B}$-ideal $I$ in $B_{1}(X)$ is the intersection of all prime ideals in $B_{1}(X)$ which contains $I$.

Next theorem establishes some equivalent conditions on the relationship among $Z_{B}$-ideals and prime ideals of $B_{1}(X)$.

Theorem 2.16**.**

For a $Z_{B}$-ideal $I$ in $B_{1}(X)$ the following conditions are equivalent:

1. (1)

$I$* is a prime ideal of $B_{1}(X)$.* 2. (2)

$I$* contains a prime ideal of $B_{1}(X)$.* 3. (3)

if $fg=0$, then either $f\in I$ or $g\in I$. 4. (4)

Given $f\in B_{1}(X)$ there exists $Z\in Z_{B}[I]$, such that $f$ does not change its sign on $Z$.

Proof.

$(1)\implies(2)$ and $(2)\implies(3)$ are immediate.

$(3)\implies(4)$: Let $(3)$ be true. Choose $f\in B_{1}(X).$ Then $(f\vee 0).(f\wedge 0)=0$. So by $(3)$, $f\vee 0\in I$ or $f\wedge 0\in I$. Hence $Z(f\vee 0)\in Z_{B}[I]$ or $Z(f\wedge 0)\in Z_{B}[I]$. It is clear that $f\leq 0$ on $Z(f\wedge 0)$ and $f\geq 0$ on $Z(f\vee 0)$.

$(4)\implies(1)$: Let $(4)$ be true. To show that $I$ is prime. Let $g,h\in B_{1}(X)$ be such that $gh\in I$. By $(4)$ there exists $Z\in Z_{B}[I]$, such that $|g|-|h|\geq 0$ on $Z$ (say). It is clear that, $Z\cap Z(g)\subseteq Z(h)$. Consequently $Z\cap Z(gh)\subseteq Z(h)$. Since $Z_{B}[I]$ is a $Z_{B}$-filter on $X$, it follows that $Z(h)\in Z_{B}[I]$. So $h\in I$, since $I$ is a $Z_{B}$-ideal. Hence, $I$ is prime. ∎

Theorem 2.17**.**

In $B_{1}(X)$, every prime ideal $P$ can be extended to a unique maximal ideal.

Proof.

If possible let $P$ be contained in two distinct maximal ideals $M_{1}$ and $M_{2}$. So, $P\subseteq M_{1}\cap M_{2}$. Since maximal ideals in $B_{1}(X)$ are $Z_{B}$-ideals and intersection of any number of $Z_{B}$-ideals is $Z_{B}$-ideal, $M_{1}\cap M_{2}$ is a $Z_{B}$-ideal containing the prime ideal $P$. By Theorem 2.16, $M_{1}\cap M_{2}$ is a prime ideal. But in a commutative ring with unity, for two ideals $I$ and $J$, if, $I\nsubseteq J$ and $J\nsubseteq I$, then $I\cap J$ is not a prime ideal. Thus $M_{1}\cap M_{2}$ is not prime ideal and we get a contradiction. So, every prime ideal can be extended to a unique maximal ideal. ∎

Corollary 2.18**.**

$B_{1}(X)$ is a Gelfand ring for any topological space $X$.

Definition 2.19**.**

A $Z_{B}$-filter $\mathscr{F}_{B}$ on $X$ is called a prime $Z_{B}$-filter on $X$, if, for any $Z_{1},Z_{2}\in Z(B_{1}(X))$ with $Z_{1}\cup Z_{2}\in\mathscr{F}_{B}$ either $Z_{1}\in\mathscr{F}_{B}$ or $Z_{2}\in\mathscr{F}_{B}$.

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 $B_{1}(X)$, then $Z_{B}[I]=\{Z(f):f\in I\}$ is a prime $Z_{B}$-filter on $X$.

Theorem 2.21**.**

Let $\mathscr{F}_{B}$ be a prime $Z_{B}$-filter on $X$ then $Z_{B}^{-1}[\mathscr{F}_{B}]=\{f\in B_{1}(X):Z(f)\in\mathscr{F}_{B}\}$ is a prime ideal in $B_{1}(X)$.

Corollary 2.22**.**

Every prime $Z_{B}$-filter can be extended to a unique $Z_{B}$-ultrafilter on $X$.

Corollary 2.23**.**

A $Z_{B}$-ultrafilter $\mathscr{U}$ on $X$ is a prime $Z_{B}$-filter on $X$, as $\mathscr{U}=Z_{B}[M]$, for some maximal ideal $M$ in $B_{1}(X)$.

3. Fixed ideals and free ideals in $B_{1}(X)$

In this section, we introduce fixed and free ideals of $B_{1}(X)$ and $B_{1}^{*}(X)$ and completely characterize the fixed maximal ideals of $B_{1}(X)$ 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 $B_{1}(X)$. 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 $T_{1}$ space $X$ which precisely determine the fixed and free maximal ideals of the corresponding ring.

Definition 3.1**.**

A proper ideal $I$ of $B_{1}(X)$ (respectively, $B_{1}^{*}(X)$) is called fixed if $\bigcap Z[I]\neq\emptyset$. If $I$ is not fixed then it is called free.

For any Tychonoff space $X$, the fixed maximal ideals of the ring $B_{1}(X)$ and those of $B_{1}^{*}(X)$ are characterized.

Theorem 3.2**.**

$\{\widehat{M}_{p}:p\in X\}$* is a complete list of fixed maximal ideals in $B_{1}(X)$, where $\widehat{M}_{p}=\{f\in B_{1}(X):f(p)=0\}$. Moreover, $p\neq q\implies\widehat{M}_{p}\neq\widehat{M}_{q}$.*

Proof.

Choose $p\in X$. The map $\Psi_{p}:B_{1}(X)\rightarrow\mathbb{R}$, defined by $\Psi(f)=f(p)$ is clearly a ring homomorphism. Since the constant functions are in $B_{1}(X)$, $\Psi_{p}$ is surjective and $\ker\Psi_{p}=\{f\in B_{1}(X):\Psi_{p}(f)=0\}=\{f\in B_{1}(X):f(p)=0\}=\widehat{M}_{p}$ (say).

By First isomorphism theorem of rings we get $B_{1}(X)/\widehat{M}_{p}$ is isomorphic to the field $\mathbb{R}$. $B_{1}(X)/\widehat{M}_{p}$ being a field we conclude that $\widehat{M}_{p}$ is a maximal ideal in $B_{1}(X)$. Since $p\in\bigcap Z_{B}[M]$, the ideal $\widehat{M}_{p}$ is a fixed ideal.

For any Tychonoff space $X$, we know that $p\neq q\implies M_{p}\neq M_{q}$, where $M_{p}=\{f\in C(X):f(p)=0\}$ is the fixed maximal ideal in $C(X)$. Since $\widehat{M_{p}}\cap C(X)=M_{p}$, it follows that for any Tychonoff space $X$, $p\neq q\implies\widehat{M_{p}}\neq\widehat{M_{q}}$.

Let $M$ be any fixed maximal ideal in $B_{1}(X)$. There exists $p\in X$ such that for all $f\in B_{1}(X),$ $f(p)=0$. Therefore, $M\subseteq\widehat{M_{p}}$. Since $M$ is a maximal ideal and $\widehat{M_{p}}$ is a proper ideal, we get $M=\widehat{M_{p}}$. ∎

Theorem 3.3**.**

$\{\widehat{M}^{*}_{p}:p\in X\}$* is a complete list of fixed maximal ideals in $B_{1}(X)$, where $\widehat{M}^{*}_{p}=\{f\in B_{1}^{*}(X):f(p)=0\}$. Moreover, $p\neq q\implies\widehat{M}^{*}_{p}\neq\widehat{M}^{*}_{q}$.*

Proof.

Similar to the proof of Theorem 3.2. ∎

The following two theorems show the interrelations between fixed ideals of $B_{1}(X)$ and $B_{1}^{*}(X)$.

Theorem 3.4**.**

If $I$ is any fixed ideal of $B_{1}(X)$ then $I\cap B_{1}^{*}(X)$ is a fixed ideal of $B_{1}^{*}(X)$.

Proof.

Straightforward. ∎

Lemma 3.5**.**

Given any $f\in B_{1}(X)$, there exists a positive unit $u$ of $B_{1}(X)$ such that $uf\in B_{1}^{*}(X)$.

Proof.

Consider $u=\frac{1}{|f|+1}$. Clearly $u$ is a positive unit in $B_{1}(X)$ [1] and $uf\in B_{1}^{*}(X)$ as $|uf|\leq 1$. ∎

Theorem 3.6**.**

Let an ideal $I$ in $B_{1}(X)$ be such that $I\cap B_{1}^{*}(X)$ is a fixed ideal of $B_{1}^{*}(X)$. Then $I$ is a fixed ideal of $B_{1}(X)$.

Proof.

For each $f\in I$, there exists a positive unit $u_{f}$ of $B_{1}(X)$ such that $u_{f}f\in I\cap B_{1}^{*}(X)$. Therefore, $\bigcap\limits_{f\in I}Z(f)=\bigcap\limits_{f\in I}Z(u_{f}f)\supseteq\bigcap\limits_{g\in B_{1}^{*}(X)\cap I}Z(g)\neq\emptyset$. Hence $I$ is fixed in $B_{1}(X)$. ∎

Since for any discrete space $X$, $C(X)=B_{1}(X)$ and $C^{*}(X)=B_{1}^{*}(X)$, then considering the example 4.7 of [3], we can conclude the following:

1. (1)

For any maximal ideal $M$ of $B_{1}(X)$, $M\cap B_{1}^{*}(X)$ need not be a maximal ideal in $B_{1}^{*}(X)$. 2. (2)

All free maximal ideals in $B_{1}^{*}(X)$ need not be of the form $M\cap B_{1}^{*}(X)$, where $M$ is a maximal ideal in $B_{1}(X)$.

Theorem 3.7**.**

If $X$ is a perfectly normal $T_{1}$ space then for each $p\in X,$ $\chi_{p}:X\rightarrow\mathbb{R}$ given by

[TABLE]

is a Baire one function.

Proof.

For any open set $U$ of $\mathbb{R}$,

[TABLE]

Since $X$ is a perfectly normal space, the open set $X\setminus\{p\}$ is a $F_{\sigma}$ set. Hence in any case $\chi_{p}$ pulls back an open set to a $F_{\sigma}$ set. So $\chi_{p}$ is a Baire one function. ∎

In view of Theorem 3.7 we obtain the following facts about any perfectly normal $T_{1}$ space.

Observation 3.8**.**

If $M$ is a maximal ideal of $B_{1}(X)$ where $X$ is a perfectly normal $T_{1}$ space then

1. (1)

For each $p\in X$ either $\chi_{p}\in M$ or $\chi_{p}-1\in M$.

This follows from $\chi_{p}(\chi_{p}-1)=0\in M$ and $M$ is prime. 2. (2)

If $\chi_{p}-1\in M$ then $\chi_{q}\in M$ for all $q\neq p$.

For if $\chi_{q}-1\in M$ for some $q\neq p$ then $Z(\chi_{p}-1),Z(\chi_{q}-1)\in Z_{B}[M]$. This implies $\emptyset=Z(\chi_{p}-1)\cap Z(\chi_{q}-1)\in Z_{B}[M]$ which contradicts that $Z_{B}[M]$ is a $Z_{B}$-ultrafilter. 3. (3)

$M$ is fixed if and only if $\chi_{p}-1\in M$ for some $p\in X$.

If $M$ is fixed then $M=\widehat{M_{p}}$ for some $p\in X$ and therefore, $\chi_{p}-1\in M$. Conversely let $\chi_{p}-1\in M$ for some $p\in X$. Then $\{p\}=Z(\chi_{p}-1)\in Z_{B}[M]$ shows that $M$ is fixed. 4. (4)

$M$ is free if and only if $M$ contains $\{\chi_{p}:p\in X\}$.

Follows from Observation (3).

The following theorem ensures the existence of free maximal ideals in $B_{1}(X)$ where $X$ is any infinite perfectly normal $T_{1}$ space.

Theorem 3.9**.**

For a perfectly normal space $T_{1}$ space $X$, the following statements are equivalent:

1. (1)

$X$* is finite.* 2. (2)

Every maximal ideal in $B_{1}(X)$ is fixed. 3. (3)

Every ideal in $B_{1}(X)$ is fixed.

Proof.

$(1)\implies(2)$: Since a finite $T_{1}$ space is discrete, $C(X)=B_{1}(X)=X^{\mathbb{R}}$. $X$ being finite, it is compact and therefore all the maximal ideals of $C(X)$ \big{(}=B_{1}(X)\big{)} are fixed.

$(2)\implies(3)$: Proof obvious.

$(3)\implies(1)$: Suppose $X$ is infinite. We shall show that there exists a free (proper) ideal in $B_{1}(X)$.

Consider $I=\{f\in B_{1}(X):\overline{X\setminus Z(f)}$ is finite$\}$ (Here finite includes $\emptyset$).

Of course $I\neq\emptyset$, as $\mathbf{0}\in I$. Since $X$ is infinite, $\mathbf{1}\notin I$ and so, $I$ is proper. We show that, $I$ is an ideal in $B_{1}(X)$. Let $f,g\in I$. Then $\overline{X\setminus Z(f)}$ and $\overline{X\setminus Z(g)}$ are both finite. Now $\overline{X\setminus Z(f-g)}\subseteq\overline{X\setminus Z(f)}\cup\overline{X\setminus Z(g)}$ implies that $\overline{X\setminus Z(f-g)}$ is finite. Hence $f-g\in I$. Similarly, $\overline{X\setminus Z(f.g)}\subseteq\overline{X\setminus Z(f)}$ for any $f\in I$ and $g\in B_{1}(X)$. So, $\overline{X\setminus Z(f.g)}$ is finite and hence $f.g\in I$. Therefore, $I$ is an ideal in $B_{1}(X)$. We claim that $I$ is free.

For any $p\in X$, consider $\chi_{p}:X\rightarrow\mathbb{R}$ given by

[TABLE]

Using Theorem 3.7, $\chi_{p}$ is a Baire one function. Also, $\overline{X\setminus Z(\chi_{p})}=\overline{X\setminus(X\setminus\{p\})}=\overline{\{p\}}=\{p\}=$ finite and $\chi_{p}(p)\neq 0$. Hence, $I$ is free. ∎

4. Residue class ring of $B_{1}(X)$ modulo ideals

An ideal $I$ in a partially ordered ring $A$ is called convex if for all $a,b,c\in A$ with $a\leq b\leq c$ and $c,a\in I\implies b\in I$. Equivalently, for all $a,b\in A,0\leq a\leq b$ and $b\in I\implies a\in I$.

If $A$ is a lattice ordered ring then an ideal $I$ of $A$ is called absolutely convex if for all $a,b\in A$, $|a|\leq|b|$ and $b\in I\implies a\in I$.

Example 4.1**.**

If $t:B_{1}(X)\rightarrow B_{1}(Y)$ is a ring homomorphism, then $\ker t$ is an absolutely convex ideal.

Proof.

Let $g\in\ker t$ and $|f|\leq|g|$, where $f\in B_{1}(X)$. $g\in\ker t\implies t(g)=0\implies t(|g|)=|t(g)|=0$. Since any ring homomorphism $t:B_{1}(X)\rightarrow B_{1}(Y)$ preserves the order, $t(|f|)=0\implies|t(f)|=0\implies t(f)=0\implies f\in\ker t$. ∎

Let $I$ be an ideal in $B_{1}(X)$. In what follows we shall denote any member of the quotient ring $B_{1}(X)/I$ by $I(f)$ for $f\in B_{1}(X)$. 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 $I(a)\geq 0\ \textnormal{iff}\ \exists$ $x\in A\ \textnormal{ such that }x\geq 0\textnormal{ and }a\equiv x(\mod I).$

Theorem 4.3**.**

[3]** On a convex ideal $I$ in a lattice-ordered ring $A$ the following conditions are equivalent.

1. (1)

$I$* is absolutely convex.* 2. (2)

$x\in I$* implies $|x|\in I.$* 3. (3)

$x,y\in I$* implies $x\vee y\in I.$* 4. (4)

$I(a\vee b)=I(a)\vee I(b)$, whence $A/I$ is a lattice ordered ring. 5. (5)

$\forall$* $a\in A$, $I(a)\geq 0$ iff $I(a)=I(|a|)$.*

Remark 4.4*.*

For an absolutely convex ideal $I$ of $A$, $I(|a|)=I(a\vee-a)=I(a)\vee I(-a)=|I(a)|$, $\forall$ $a\in A$.

Theorem 4.5**.**

Every $Z_{B}$-ideal in $B_{1}(X)$ is absolutely convex.

Proof.

Suppose $I$ is any $Z_{B}$-ideal and $|f|\leq|g|$, where $g\in I$ and $f\in B_{1}(X)$. Then $Z(g)\subseteq Z(f)$. Since $g\in I$, it follows that $Z(g)\in Z_{B}[I]$, hence $Z(f)\in Z_{B}[I]$. Now $I$ being a $Z_{B}$-ideal, $f\in I$. ∎

Corollary 4.6**.**

In particular every maximal ideal in $B_{1}(X)$ is absolutely convex.

Theorem 4.7**.**

For every maximal ideal $M$ in $B_{1}(X)$, the quotient ring $B_{1}(X)/M$ 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 $B_{1}(X)/I$, where $I$ is a $Z_{B}$-ideal.

Theorem 4.8**.**

Let $I$ be a $Z_{B}$-ideal in $B_{1}(X)$. For $f\in B_{1}(X)$, $I(f)\geq 0$ if and only if there exists $Z\in Z_{B}[I]$ such that $f\geq 0$ on $Z$.

Proof.

Let $I(f)\geq 0$. By condition $(5)$ of Theorem 4.3, we write $I(f)=I(|f|)$. So, $f-|f|\in I\implies Z(f-|f|)\in Z_{B}[I]$ and $f\geq 0$ on $Z(f-|f|)$.

Conversely, let $f\geq 0$ on some $Z\in Z_{B}[I]$. Then $f=|f|$ on $Z\implies Z\subseteq Z(f-|f|)\implies Z(f-|f|)\in Z_{B}[I]$. $I$ being a $Z_{B}$-ideal we get $f-|f|\in I$, which means $I(f)=I(|f|)$. But $|f|\geq 0$ on $Z$ gives $I(|f|)\geq 0$. Hence, $I(f)\geq 0$. ∎

Theorem 4.9**.**

Let $I$ be any $Z_{B}$-ideal and $f\in B_{1}(X)$. If there exists $Z\in Z_{B}[I]$ such that $f(x)>0$, for all $x$ $\in Z$, then $I(f)>0$.

Proof.

By Theorem 4.8, $I(f)\geq 0$. But $Z\cap Z(f)=\emptyset$ and $Z\in Z_{B}[I]\implies Z(f)\notin Z_{B}[I]\implies f\notin I\implies I(f)\neq 0\implies I(f)>0$. ∎

The next theorem shows that the converse of the above theorem holds if the ideal is a maximal ideal in $B_{1}(X)$.

Theorem 4.10**.**

Let $M$ be any maximal ideal in $B_{1}(X)$ and $M(f)>0$ for some $f\in B_{1}(X)$ then there exists $Z\in Z_{B}[M]$ such that $f>0$ on $Z$.

Proof.

By Theorem 4.8, there exists $Z_{1}\in Z_{B}[M]$ such that $f\geq 0$ on $Z_{1}$. Now $M(f)>0\implies f\notin M$ which implies that there exists $g$ $\in M$, such that $Z(f)\cap Z(g)=\emptyset$. Choosing $Z=Z_{1}\cap Z(g)$, we observe $Z\in Z_{B}[M]$ and $f(x)>0$, for all $x\in Z$. ∎

Corollary 4.11**.**

For a maximal ideal $M$ of $B_{1}(X)$ and for some $f\in B_{1}(X)$, $M(f)>0$ if and only if there exists $Z\in Z_{B}[M]$ 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 $Z_{B}$-ideal in $B_{1}(X)$. There exists $f\in B_{1}(X)$ such that $I(f)>0$ but $f$ is not strictly positive on any $Z\in Z_{B}[I]$.

Proof.

Since $I$ is non-maximal, there exists a proper ideal $J$ of $B_{1}(X)$ such that $I\subsetneqq J$. Choose $f\in J\smallsetminus I$. $f^{2}\notin I\implies I(f^{2})>0$. Choose any $Z\in Z_{B}[I]$. Certainly, $Z\in Z_{B}[J]$ and so, $Z\cap Z(f^{2})\in Z_{B}[J]\implies Z\cap Z(f^{2})\neq\emptyset$. So $f$ is not strictly positive on the whole of $Z$. ∎

In what follows, we characterize the ideals $I$ in $B_{1}(X)$ for which $B_{1}(X)/I$ is a totally ordered ring.

Theorem 4.13**.**

Let $I$ be a $Z_{B}$-ideal in $B_{1}(X)$, then the lattice ordered ring $B_{1}(X)/I$ is totally ordered ring if and only if $I$ is a prime ideal.

Proof.

$B_{1}(X)/I$ is a totally ordered ring if and only if for any $f\in B_{1}(X)$, $I(f)\geq 0$ or $I(-f)\leq 0$ if and only if for all $f\in B_{1}(X)$, there exists $Z\in Z_{B}[I]$ 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 $B_{1}(X)$, $B_{1}(X)/M$ is a totally ordered field.

Theorem 4.15**.**

Let $M$ be a maximal ideal in $B_{1}(X)$. The function $\Phi:\mathbb{R}\rightarrow B_{1}(X)/M$ (respectively, $\Phi:\mathbb{R}\rightarrow B_{1}^{*}(X)/M$) defined by $\Phi(r)=M(\mathbf{r})$, for all $r\in\mathbb{R}$, where $\mathbf{r}$ 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 $B_{1}(X)/M$ that the function is a homomorphism.

To show $\phi$ is injective. Let $M(\mathbf{r})=M(\mathbf{s})$ for some $r,s\in\mathbb{R}$ with $r\neq s$. Then $\mathbf{r}-\mathbf{s}\in M$. This contradicts to the fact that $M$ is a proper ideal. Hence $M(\mathbf{r})\neq M(\mathbf{s})$, when $r\neq s$.

Let $r,s\in\mathbb{R}$ with $r>s$. Then $r-s>0$. The function $\mathbf{r}-\mathbf{s}$ is strictly positive on $X$. Since $X\in Z(B_{1}(X))$, by Theorem 4.9, $M(\mathbf{r}-\mathbf{s})>0\implies M(\mathbf{r})>M(\mathbf{s})\implies\Phi(r)>\Phi(s)$. Thus $\Phi$ is an order preserving monomorphism. ∎

For a maximal ideal $M$ in $B_{1}(X)$, the residue class field $B_{1}(X)/M$ (respectively $B_{1}^{*}(X)/M$) can be considered as an extension of the field $\mathbb{R}$.

Definition 4.16**.**

The maximal ideal $M$ of $B_{1}(X)$ (respectively, $B_{1}^{*}(X)$) is called real if $\Phi(\mathbb{R})=B_{1}(X)/M$ (respectively, $\Phi(\mathbb{R})=B_{1}^{*}(X)/M$) and in such case $B_{1}(X)/M$ is called real residue class field. If $M$ is not real then it is called hyper-real and $B_{1}(X)/M$ is called hyper-real residue class field.

Definition 4.17**.**

[3] A totally ordered field $F$ is called archimedean if given $\alpha\in F$, there exists $n\in\mathbb{N}$ such that $n>\alpha$. If $F$ is not archimedean then it is called non-archimedean.

If $F$ is a non-archimedean ordered field then there exists some $\alpha\in F$ such that $\alpha>n$, for all $n\in\mathbb{N}$. Such an $\alpha$ is called an infinitely large element of $F$ and $\frac{1}{\alpha}$ is called infinitely small element of $F$ which is characterized by the relation $0<\frac{1}{\alpha}<\frac{1}{n}$, $\forall$ $n\in\mathbb{N}$. The existence of an infinitely large (equivalently, infinitely small) element in $F$ assures that $F$ is non-archimedean.

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 $\mathbb{R}$ .

We thus get that the real residue class field $B_{1}(X)/M$ is archimedean if $M$ is a real maximal ideal of $B_{1}(X)$.

Theorem 4.19**.**

Every hyper-real residue class field $B_{1}(X)/M$ is non-archimedean.

Proof.

Proof follows from the fact that the identity is the only non-zero homomorphism on the ring $\mathbb{R}$ into itself. ∎

Corollary 4.20**.**

A maximal ideal $M$ of $B_{1}(X)$ is hyper-real if and only if there exists $f\in B_{1}(X)$ such that $M(f)$ is an infinitely large member of $B_{1}(X)/M$.

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 $f\in B_{1}^{*}(X)$. Then $|f(x)|\leq n$, for all $x\in X$ and for some $n\in\mathbb{N}$. i.e., $|M(f)|=M(|f|)\leq M(\mathbf{n})$. 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 $B_{1}(X)=B_{1}^{*}(X)$ then each maximal ideal in $B_{1}(X)$ 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 $B_{1}(X)/M$.

Theorem 4.23**.**

Given a maximal ideal $M$ of $B_{1}(X)$ and $f\in B_{1}(X)$, the following statements are equivalent:

1. (1)

$|M(f)|$* is infinitely large member in $B_{1}(X)/M$.* 2. (2)

$f$* is unbounded on each zero set in $Z_{B}[M]$.* 3. (3)

for all $n\in\mathbb{N}$, $Z_{n}=\{x\in X:|f(x)|\geq n\}\in Z_{B}[M]$.

Proof.

$(1)\iff(2)$: $|M(f)|$ is not infinitely large in $B_{1}(X)/M$ iff $\exists$ $n\in\mathbb{N}$ such that, $|M(f)|=M(|f|)\leq M(\mathbf{n})$ iff $|f|\leq\mathbf{n}$ on some $Z\in Z_{B}[M]$ if and only if $f$ is bounded on some $Z\in Z_{B}[M]$.

$(2)\implies(3)$: Choose $n\in\mathbb{N}$, we shall show that $Z_{n}\in Z_{B}[M]$. By $(2)$, $Z_{n}$ intersects each member in $Z_{B}[M]$. Now $Z_{B}[M]$ being a $Z_{B}$-ultrafilter, $Z_{n}\in Z_{B}[M]$.

$(3)\implies(2)$: Let each $Z_{n}\in Z_{B}[M]$, for all $n\in\mathbb{N}$. Then for each $n\in\mathbb{N}$, $|f|\geq n$ on some zero set in $Z_{B}[M]$. Hence $|M(f)|=M(|f|)\geq M(\mathbf{n})$, for all $n\in\mathbb{N}$. That means $|M(f)|$ is infinitely large member in $B_{1}(X)/M$. ∎

Theorem 4.24**.**

$f\in B_{1}(X)$* is unbounded on $X$ if and only if there exists a maximal ideal $M$ in $B_{1}(X)$ such that $M(f)$ is infinitely large in $B_{1}(X)/M$.*

Proof.

Let $f$ be unbounded on $X$. So, each $Z_{n}$ in Theorem 4.23 is non-empty. We observe that $\{Z_{n}:n\in\mathbb{N}\}$ is a subcollection of $Z(B_{1}(X))$ having finite intersection property. So there exists a $Z_{B}$-ultrafilter $\mathscr{U}$ on $X$ such that $\{Z_{n}:n\in\mathbb{N}\}\subseteq\mathscr{U}$. Therefore, there is a maximal ideal $M$ in $B_{1}(X)$ for which $\mathscr{=}Z_{B}[M]$ and so, $Z_{n}\in Z_{B}[M]$, for all $n\in\mathbb{N}$. By Theorem 4.23 $M(f)$ is infinitely large.

Converse part is a consequence of $(1)\implies(2)$ 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 $B_{1}(X)$.

Proof.

It is enough to prove that there exists an unbounded Baire one function in $B_{1}(X)$. 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 $B_{1}(X)$.

Theorem 4.26**.**

For the maximal ideal $M$ of $B_{1}(X)$ the following statements are equivalent:

1. (1)

$M$* is a real maximal ideal.* 2. (2)

$Z_{B}[M]$* is closed under countable intersection.* 3. (3)

$Z_{B}[M]$* has countable intersection property.*

Proof.

$(1)\implies(2)$: Assume that $(2)$ is false, i.e. there exists a sequence of functions $\{f_{n}\}$ in $M$ for which $\bigcap\limits_{n=1}^{\infty}Z(f_{n})\notin Z_{B}[M]$. Set f(x)=\sum\limits_{n=1}^{\infty}\big{(}|f_{n}(x)|\wedge\frac{1}{4^{n}}\big{)}, $\forall x\in X$. It is clear that, the function $f$ defined on $X$ is actually a Baire one function ([1]) and $Z(f)=\bigcap\limits_{n=1}^{\infty}Z(f_{n})$. Thus, $Z(f)\notin Z_{B}[M]$. Hence $f\notin M\implies M(f)>0$ in $B_{1}(X)/M$.

Fix a natural number $m$. Then $Z(f_{1})\bigcap Z(f_{2})\bigcap Z(f_{3})...\bigcap Z(f_{m})=Z$(say) $\in Z_{B}[M]$. Now for any point $x\in Z$, f(x)=\sum\limits_{n=m+1}^{\infty}\big{(}|f_{n}(x)|\wedge\frac{1}{4^{n}}\big{)}\leq\sum\limits_{n=m+1}^{\infty}\frac{1}{4^{n}}=3^{-1}4^{-m}. This shows that, $0<M(f)\leq M(3^{-1}4^{-m})$, $\forall$ $m\in\mathbb{N}$. Hence $M(f)$ is an infinitely small member in $B_{1}(X)/M$. So, $M$ becomes a hyper-real maximal ideal and then $(1)$ is false.

$(2)\implies(3)$: Trivial, as $\emptyset\notin Z_{B}[M]$.

$(3)\implies(1)$: Assume that $(1)$ is false, i.e. $M$ is hyper-real. So, there exists $f\in B_{1}(X)$ so that $|M(f)|$ is infinitely large in $B_{1}(X)/M$. Therefore for each $n\in\mathbb{N}$, $Z_{n}$ defined in Theorem 4.23, belongs to $Z_{B}[M]$. Since $\mathbb{R}$ is archimedean, we have $\bigcap\limits_{n=1}^{\infty}Z_{n}=\emptyset$. Thus $(3)$ is false. ∎

So far we have seen that for any topological space $X$, all fixed maximal ideals of $B_{1}(X)$ are real. Though the converse is not assured in general, we show in the next example that in $B_{1}(\mathbb{R})$ a maximal ideal is real if and only if it is fixed.

Example 4.27**.**

Suppose $M$ is any real maximal ideal in $B_{1}(\mathbb{R})$. We claim that $M$ is fixed. The identity $i:\mathbb{R}\rightarrow\mathbb{R}$ belongs to $B_{1}(\mathbb{R})$. Since $M$ is a real maximal ideal, there exists a real number $r$ such that $M(i)=M(\mathbf{r})$. This implies $i-\mathbf{r}\in M$. Hence $Z(i-\mathbf{r})\in Z_{B}[M]$. But $Z(i-\mathbf{r})$ is a singleton. So, $Z_{B}[M]$ is fixed, i.e., $M$ is fixed.

In view of Observation 3.8(3), we conclude that a maximal ideal $M$ in $B_{1}(\mathbb{R})$ is real if and only if there exists a unique $p\in\mathbb{R}$ such that $\chi_{p}-1\in M$.

If $X$ is a P-space then $C(X)$ possesses real free maximal ideals. In such case however, $B_{1}(X)=C(X)$. Consequently, $B_{1}(X)$ possesses real free maximal ideals, when $X$ is a P-space. It is still a natural question, what are the topological spaces $X$ for which $B_{1}(X)$ ($\supseteq C(X)$) contains a free real maximal ideal?