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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 refered 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 Kolmogroff 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:

Question1: Does there exist a maximal ideal in $Ba_1[0 , 1]$ other than maximal ideals of the form $(M_x^1$ for $x\in X)$

Question2: Is the ring $Ba_1[0 , 1]$ a 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

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