On the ring of all real valued functions (continuous)

Question

Let $X$ be a metric space , let $C(X)$ be the ring of all real values continuous functions on $X$ and $F(X)$ be the ring of all real valued functions on $X$ (the ring operations are pointwise addition and multiplication of functions ). If $C(X) \cong F(X)$ as rings , then is it true that $C(X)=F(X)$ i.e. is it true that the topology of $X$ then becomes discrete ?

Answer 1

$F(X)$ is a commutative von Neumann regular ring. That is, for every element $f$, there is an element $g$ such that $f^2g=f$. Namely, $g(x)=f(x)^{-1}$ whenever $f(x)\neq0$, and $g(x)$ can be chosen arbitrarily when $f(x)=0$.

But suppose $X$ is not discrete, and let $x_0\in X$ be a non-isolated point. Let $f(x)=d(x,x_0)$ be the "distance from $x_0$" function. Suppose there were $g\in C(X)$ such that $f^2g=f$. Then $g(x)=d(x,x_0)^{-1}$ for $x\neq x_0$. But then $g(x)\to\infty$ as $x\to x_0$, and so there is no way to choose $g(x_0)$ so that $g$ is continuous. So $C(X)$ is not a von Neumann regular ring, and so $C(X)\not\cong F(X)$ unless $X$ is discrete.