Let $X$ be an infinite discrete topological space. Is $$C\_b(X)=\{ f \colon X \to \mathbb{R} \text{ bounded }\}$$ a Jacobson ring ?



Ring of continuous functions is a Jacobson ring
First, let us consider the question of when the ring $C(X)$ of *all* continuous real-valued functions on a topological space $X$ (not necessarily discrete) is Jacobson, keeping in mind that $C\_b(X) = C(\beta X)$ where $\beta X$ is the Stone-Čech compactification.


In fact, every prime ideal of $C(X)$ is contained in a *unique* maximal ideal: see Gillman & Jerison, *Rings of Continuous Functions* (1960) theorem 7.15 on page 107. So $C(X)$ is a Jacobson ring iff every prime ideal is maximal, meaning that $X$ is a P-space, a rather strong condition which, [see the reference in this other answer](https://mathoverflow.net/questions/266808/when-c-x-is-zero-dimensional/266820#266820), is equivalent to the condition “every $f\in C(X)$ which vanishes at $p$ vanishes in some neighborhood of $p$“.


(Note for example that $C([0,1]) = C\_b([0,1])$ is *not* a Jacobson ring: the ring of functions which vanish in some neighborhood of $0$, while not prime itself, is contained in a prime ideal, which itself is contained in a a unique maximal ideal, namely that of functions which vanish at $0$.)


Now if $X$ is infinite discrete, it is indeed a P-space, so $C(X)$ is a Jacobson ring. However, $\beta X$ is *not* a P-space, as every compact P-space is finite (Gillman & Jerison, problem 4K.1 on page 63, and remarks following theorem 14.29 on page 212). So $C\_b(X) = C(\beta X)$ is *not* a Jacobson ring.


