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.$)$