Ring of continuous functions is a Jacobson ring

Question

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

Answer 1

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, 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.