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A real function $f:X\rightarrow \mathbb{R}$ is called Baire-one function, if there is a sequence $(f_n)_{n=1} ^\infty$ of continuous functions $f_n:X\rightarrow \mathbb{R}$ on $X$ so that for all $x\in X$ $$\lim_{n\rightarrow \infty}f_n(x)=f(x).$$

When $X$ is a Banach space, we have the following theorem referred to as Baire factorization theorem.

Theorem: The real function $f:X\rightarrow \mathbb{R}$ is in the class of Baire-one if and only if for all closed subset $K\subset X$, the restricted function $f|_K$ has a point of continuity with respect to $K$.

Definition: We denote the set of all Baire-one real functions on the space $X$ by $Ba_1(X)$.

As you could easily see, $Ba_1(X)$ forms a ring with pointwise addition and multiplication. For simplicity let me consider $X=[0 , 1]$.

Suppose $C[0 , 1]$ denotes the ring of all continuous real valued functions on the interval $[0 , 1]$. By the theorem of Gelfond and Kolmogrov we know that the set of all maximal ideals of the ring $C[0 , 1]$ is of the form $\{M_x: x\in X\}$, where $M_x=\{f\in C[0, 1]: f(x)=0\}$.

Compared with the ring $C[0 , 1]$ we could easily find that the sets of the form $M_x^1=\{f\in Ba_1[0 , 1]: f(x)=0\}$ are maximal ideals of the ring $Ba_1[0 , 1]$. From this property some questions came in my mind as follows:

Question 1: Does there exist a maximal ideal in $Ba_1[0 , 1]$ other than maximal ideals of the form $M_x^1$ for $x\in X$?

Question 2: Is the ring $Ba_1[0 , 1]$ a $\mathbf{PM}$-ring? $($i.e. a ring in which each prime ideal is contained in a unique maximal ideal.$)$

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It looks to me like the collection of functions with finite support (that is, those which are zero on a cofinite set) forms an ideal not contained in any of the maximal ideals you've listed. If I understand your question correctly, that means the answer to question 1 is "yes." –  Clinton Conley Aug 6 '12 at 23:00
    
Yes Dear Clinton. Very Good. Also we could consider the ideal generated by all $\chi_{(a,1]}$ for $a \in (0,1]$, where $\chi_{(a,1]}$ is the characteristic function of the interval $(a , 1]$. then we could find a maximal ideal which contains all of these functions.I think this over ring of $C[0,1]$ has a complex behavior, because as you consider, it is full of idempotents. –  Ali Reza Aug 6 '12 at 23:20
    
Isn't $Ba_1(X)$ a commutative C*-algebra (in which case it is certainly a PM-ring)? –  Douglas Somerset Mar 22 '13 at 22:43

1 Answer 1

The question may be answered in the comments. I thus collect the relevant comments here as an answer.

C. Conley: The collection of functions with finite support (that is, those which are zero on a cofinite set) forms an ideal not contained in any of the maximal ideals listed in the question.

D. Somerset: Perhaps $Ba_1(X)$ is a commutative C*-algebra, and thus a PM-ring.

Update: Y. Choi comments below that Somerset's approach does not work, since one is considering all the Baire-1 functions, not just the bounded ones.

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Douglas Somerset's comment (which also occurs in a deleted answer by someone else which is not visible unless one has sufficiently high reputation) is incorrect, because one is considering all the Baire-1 functions, not just the bounded ones. –  Yemon Choi Aug 17 at 3:05

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