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.$)$
